Solutions to Exercises from Kenneth Brown's Cohomology of Groups

Home , Acyclic model, Direct sum, Direct sum of modules, Free abelian group, Split exact sequence, Tor functor

Solutions to Exercises from Kenneth Brown’s Cohomology of Groups

Christopher A. Gerig, Cornell University (College of Engineering) August 2008 - May 2009

I appreciate emails concerning any errors/corrections: [email protected]. Any errors would be due to solely myself, or at least the undergraduate-version of myself when I last looked over this. Remark made on 1/28/14.

1 major: Applied & Engineering Physics year: Sophomore Undergraduate

Hey Ken, thanks for taking me under your wing.

2 Contents

0 Errata to Cohomology of Groups 4

1 Chapter I: Some Homological Algebra 5

2 Chapter II: The Homology of a Group 12

3 Chapter III: Homology and Cohomology with Coeﬃcients 21

4 Chapter IV: Low-Dimensional Cohomology and Group Extensions 31

5 Chapter V: Products 41

6 Chapter VI: Cohomology Theory of Finite Groups 47

7 Chapter VII: Equivariant Homology and Spectral Sequences 55

8 Chapter VIII: Finiteness Conditions 58

9 Chapter IX: Euler Characteristics 59

10 Additional Exercises 61

11 References 76

3 0 Errata to Cohomology of Groups

pg62, line 11 missing a paranthesis ) at the end. pg67, line 15 from bottom missing word, should say “as an abelian group”. pg71, last line of Exercise 4 hint should be on a new line (for whole exercise). P −1 pg85, line 9 from bottom incorrect function, should be g∈C/H g ⊗ gm. pg114, ﬁrst line of Exercise 4 misspelled endomorphism with an extra r. pg115, line 5 from bottom missing word, should say “4.4 is a chain map”. ∗ pg141, line 4 from bottom missing a hat b on the last H . pg149, line 13 the ideal I should be italicized. pg158, line 20 there should be a space between SL2(Fp) and (p odd). ∗ pg160, line 5 from bottom missing the coeﬃcient in cohomology, Hb (H,M). pg165, line 26 the comma in the homology should be lowered, Hq(C∗,p).

4 1 Chapter I: Some Homological Algebra P P 2.1(a): The set A = {g − 1 | g ∈ G, g 6= 1} is linearly independent because αg(g − 1) = 0 ⇒ αgg = P P P P αg = αg1 + 0g, and since ZG is a free Z-module, αg has a unique expression, yielding P P αg = 0 ∀g ∈ A. To show that I = span(A), let αgg ∈ I and hence αg = 0. Thus we can write P P P P P αgg = αgg − 0 = αgg − αg = αg(g − 1). Since A is a linearly independent set which generates I, it is a basis for I.

2.1(b): Consider the left ideal A = ({s − 1 | s ∈ S}) over ZG. P P P P A ⊆ I since ε(A) = ε ( rgg)(s − 1) = ε( rgg)ε(s − 1) = y · 0 = 0. P s Pg Ps Pg If x ∈ I then x = αigi − βjgj such that αi = βj. P P P P P P x = ( αigi − αi) − ( βjgj − βj) + ( αi − βj) P P Thus x = αi(gi − 1) − βj(gj − 1) and it suﬃces to show g − 1 ∈ A for any g ∈ G so that x ∈ A and ±1 ±1 I ⊆ A. Since G = hSi, we have a representation x = s1 ··· sn . By using ab − 1 = a(b − 1) + (a − 1) and c−1 − 1 = −c−1(c − 1), the result follows immediately.

2.1(c): Suppose S ⊂ G | I = ({s − 1 | s ∈ S}). Then every element of I is a sum of elements of 0 0 ±1 Pn−1 0 the form g − g | g = g s . For g ∈ G, g − 1 ∈ I, so we have a ﬁnite sum g − 1 = i=1 (gi − gi).

Since this is a sum of elements in G where G is written multiplicatively, ∃ i0 | gi0 = g, say i0 = 1. 0 Pn−1 0 0 Pn−1 0 0 Thus g − 1 = g − g1 + i=2 (gi − gi) ⇒ g1 − 1 = i=2 (gi − gi). Another iteration yields g2 = g1 and 0 ±1 ±1 0 ±1 ±1 0 0 0 0 g = g1 = g1s1 = g2s1 = g2s2 s1 . At the last iteration, gn−2 − 1 = gn−1 − gn−1 ⇒ gn−1 = 1, gn−2 = 0 ±1 ±1 ±1 gn−1 = gn−1sn−1 = 1sn−1. Through this method we obtain a sequence g1, g2, ..., gn−1, gn | gi = gi+1si and g1 = g, gn = 1. ∴ g has a representation in terms of elements of S and so G = hSi since g was arbitrary.

2.1(d): If G is finitely generated, then by part(b) above, I is finitely generated. For the converse, suppose I = (a1, ..., an) is a left ideal over ZG. Noting from part(a) that I = hg − 1ig∈G as a Z-module, P each ai can be represented as a finite sum ai = j zj(gj −1). Since each ai is generated by finitely many elements, and there are finitely many ai, I is finitely generated as a left ideal by elements s − 1 where s ∈ G. Therefore, we apply part(c) to have G = hs1, . . . , ski and thus G is a finitely generated group.

2.2: Let G = hti with |G| = n and let t be the image of T in R = Z[T ]/(T n − 1) =∼ ZG. The element T − 1 is prime in Z[T ] because Z[T ]/(T − 1) =∼ Z which is an integral domain (An ideal P of a commutative ring R is prime iff the quotient ring R/P is an integral domain; Proposition 7.4.13[2]). By Proposition 8.3.10[2] (In an integral domain a prime element is always irreducible), T − 1 is irreducible in Z[T ]. We also could have obtained this result by applying Eisenstein’s Criterion with the substitution T = x + 3 and using the prime 2. Since Z[T ] is a Unique Factorization Domain, the specific factorization T n − 1 = (T − 1)(T n−1 + T n−2 + ··· + T + 1) with the irreducible T − 1 factor is unique, Pn−1 i considering the latter factor i=0 T as the expansion of its irreducible factors [note: if n is prime then Pn−1 i i=0 T = Φn(T ), a cyclotomic polynomial which is irreducible in Z[T ] by Theorem 13.6.41[2]]. Thus Pn−1 i every f ∈ R is annihilated by t − 1 iff it is divisible by N = i=0 t (and vice versa), and so the desired free resolution of M = Z = ZG/(t − 1) is: · · · → R −→t−1 R −→N R −→t−1 R → M → 0 .

F 3.1: The right cosets Hgi are H-orbits of G with the H-action as group multiplication. Since G = Hgi L ∼ L −1 where gi ranges over E, ZG = Z[Hgi] = Z[H/Hgi ]. G is a free H-set because hg = g ⇒ hgg = −1 gg ⇒ h = 1, i.e. the isotropy groups Hg are trivial. Therefore, ZG is a free ZH-module with basis E. P 3.2: Let hSi = H ⊆ G and consider Z[G/H]. Now x ∈ Z[G/H] has the expression x = zi(giH), and there exists an element ﬁxed by H, namely, x0 = g0H = H where g0 ∈ H. H is annihilated by I = Kerε = ({s − 1}) since (s − 1)H = sH − H = H − H = 0 ∀s ∈ S, and so Ix0 = 0. We have g − 1 ∈ I since ε(g − 1) = ε(g) − ε(1) = 1 − 1 = 0. Hence (g − 1)x0 = 0 ⇒ gx0 = x0 ∀g ∈ G ⇒ Gx0 = x0. Finally, GH = H ⇒ G ⊆ H. ∴ G = H = hSi.

n 4.1: Orienting each n-cell e gives a basis for Cn(X). If X is an arbitrary G-complex, then with

5 P n n ηiei ∈ Cn(X), g ∈ G can reverse the orientation of e by inversion (ﬁxing the cell). Thus G need not permute the basis, and hence Cn(X) is not necessarily a permutation module.

4.2: Since X is a free G-complex, it is necessarily a Hausdorff space with no fixed points under the G-action. First, assume G is finite and take the set of elements in Gx0, which are distinct points gix0 with 1x0 = x0 for an arbitrary point x0 ∈ X. Applying the Hausdorff condition, we have open sets Ugi containing gix0 where each such set is disjoint from U1 containing x0. Form the intersection T −1 W = ( i gi Ugi) ∩ U1 which contains x0. Since gkW ⊆ Ugk, we have gkW ∩ W = ∅ for all nonidentity gk ∈ G, and so W is the desired open neighborhood of x0 [This result does not follow for arbitrary G since an infinite intersection of open sets need not be open]. Assume the result has been proved for G infinite. Let ϕ : X → X/G be the quotient map, which sends the disjoint collection of giW ’s to ϕ(W ). Since −1 ` ϕ ϕ(W ) = i giW , giW → ϕ(W ) is a bijective map (restriction of ϕ) and thus it is a homeomorphism (ϕ continuous and open). This covering space is regular because G acts transitively on ϕ−1(Gx) by definition. Elements of G are obviously deck transformations since Ggx = Gx, hence G ⊆ Aut(X). Given Γ ∈ Aut(X) with Γ(a) = b, those two points are mapped to the same orbit in X/G (since ϕ ◦ Γ = ϕ), and so ∃ g ∈ G sending a to b. By the Lifting Lemma (uniqueness) we have Γ = g, hence Aut(X) ⊆ G ⇒ G is the group of covering transformations. ∼ ∼ If X is contractible then G = π1(X/G)/π1(X) = π1(X/G)/0 = π1(X/G) and ϕ is the universal cover of X/G, so X/G is a K(G, 1) with universal cover X. It suffices to show that the G-action is a “properly discontinuous” action on X when G is infinite: n n−1 n n Every CW-complex with given characteristic maps fj,n :(Bj ,Sj ) → (¯σj , ∂σ¯j ) admits a canonical open cover {Uσ} indexed by the cells, where Ua and Ub are disjoint open sets for distinct cells of equal dimension (for instance, if X is a simplicial complex we can take Uσ = St(ˆσ) which is the open star n n of the barycenter of σ in the barycentric subdivision of X). More precisely, for one cell σk ⊂ X in n n each G-orbit of cells define its “barycenter” asσ ˜k ≡ fk,n(bk), where bk = 0 ∈ B is the origin of the n n n n-disk; for the rest of the cells {σi = giσk } in each G-orbit define their “barycenters” asσ ˜i ≡ fi,n(bi), n n n 0 0 where bi ∈ Bi is chosen so thatσ ˜i = giσ˜k . Considering the 0-skeleton X , its cells σj are by defini- 0 0 tion open and so the canonical open cover of X is the collection {U 0 = σ }. Proceeding inductively σj j n−1 n n−1 S n (with Uσ open in X ), consider the n-skeleton X = X j σj and note that the preimage under n−1 n−1 −1 fj,n of the open cover of X is an open cover of the unit circle Sj . Take an open set fj,n (Uσ) n−1 n in Sj and form the “open sector” Wj,σ ⊂ Bj which is the union of all line segments emanating −1 −1 from bj and ending in fj,n (Uσ), minus bj and fj,n (Uσ); each Uσ determines such a Wj,σ. As fj,n is a n n homeomorphism of Int(Bj ) with σj , we have such open sectors fj,n(Wj,σ) in the n-cell. Noting the 0 S n n weak topology on X, the set Uσ = Uσ j fj,n(Wj,σ) is open in X iff its complement in X is closed n 0 i i 0 i i n 0 i i iff (X − Uσ) ∩ σ¯k ≡ σ¯k − (Uσ ∩ σ¯k) is closed inσ ¯k for all cells in X . For i < n, Uσ ∩ σ¯ = Uσ ∩ σ¯ n because fj,n(Wj,σ) ⊂ σj which is disjoint from the closure of all other cells, and the complement of this i n−1 intersection inσ ¯ ⊂ X is closed because Uσ is open by inductive hypothesis. Therefore, it suffices to n 0 n n 0 n show thatσ ¯k −(Uσ ∩σ¯k ) is closed inσ ¯k ∀ k, which is equivalent under topology of cells for Uσ ∩σ¯k to be n 0 n n S n n S open inσ ¯k . For arbitrary k we have Uσ ∩ σ¯k = (Uσ ∩ σ¯k ) j(fj,n(Wj,σ) ∩ σ¯k ) = (Uσ ∩ σ¯k ) fk,n(Wk,σ) −1 0 n −1 and taking the preimage we have Y = fk,n(Uσ ∩ σ¯k ) = fk,n(Uσ) ∪ Wk,σ. The n-disk is compact, the CW-complex Xn is Hausdorff, a closed subset of a compact space is compact (Theorem 26.2[6]), the image of a compact set under a continuous map is compact (Theorem 26.5[6]), and every compact subset of a Hausdorff space is closed (Theorem 26.3[6]); thus fk,n is a closed map and hence a quotient map for n n 0 n σ¯k (by Theorem 22.1[6]). It suffices to check that Y ⊂ Bk is open, for then fk,n(Y ) = Uσ ∩ σ¯k is open n n −1 inσ ¯k by definition of a quotient map. By construction, Y = {x ∈ Bk − bk | r(x) ∈ fk,n(Uσ)} where n n n−1 x−bk −1 r : B − bk → ∂B ≡ S is the radial projection r(x) = . As r is continuous and f (Uσ) is k k k ||x−bk|| k,n n −1 −1 n n open in ∂Bk , we have Y = r (fk,n(Uσ)) open in Bk − bk and hence in Bk (by Lemma 16.2[6]). Our n 0 n−1 0 n new collection for X is the open sets U (where σ ⊂ X ) plus the open n-cells U n = σ ; this is the σ σj j canonical open cover of Xn and hence completes the induction. 0 Any point x ∈ X will lie in an i-cell σ which lies in the open set Uσ, and we take this as our desired neighborhood of x: since any open set of our constructed cover is bounded by barycenters, and g ∈ G 0 0 0 maps barycenters to barycenters by construction, we have gUσ = Ugσ which is disjoint from Uσ by construction for all g 6= 1.

6 ∼ 2 4.3: Given G = Z ⊕ Z we have the torus T with π1T = G and its universal cover ρ : R → T . After drawing the lattice Z2 ⊂ R2, pick a square and label that surface L, with the bottom left cor- ner as the basepoint x0 and the bottom side as the edge es and the left side as the edge et (so the corners are x0, sx0, stx0, tx0 going counterclockwise around L, and the top and right sides of L are re- 2 spectively tes and set). Following Brown’s notation [in this section], x0 generates C0(R ) and es, et 2 2 generate C1(R ) with ∂1(es) = (s − 1)x0 and ∂1(et) = (t − 1)x0. Lastly, L generates C2(R ) with ∂2(L) = es + set − tes − et = (1 − t)es − (1 − s)et. Thus the desired free resolution of Z over ZG is: ∂2 ∂1 ε 0 → ZG −→ ZG ⊕ ZG −→ ZG −→ Z → 0 .

4.4:

5.1: The homotopy operator h in terms of the Z-basis g[g1| · · · |gn] for Fn is h(g[g1| · · · |gn]) = [g|g1| · · · |gn]. ¯ 5.2: Using G = Z2 = {1, s}, the elements of the normalized bar resolution F∗ = F∗/D∗ are [s|s| · · · |s], ¯ and each element forms a basis for the corresponding dimension, giving the identiﬁcation Fn = ZG. Denoting si = s ∀ i,  s[s| · · · |s] , i = 0  di[s1|s2| · · · |sn] = [s1| · · · |si−1|sisi+1| · · · |sn] = 0 , 0 < i < n [s| · · · |s] , i = n

2 The middle equation resulted from sisi+1 = s = 1 (so the element lies in D∗). The boundary operator ∂n then becomes s − 1 for n odd and s + 1 for n even. the normalized bar resolution is: ∴ s−1 s+1 s−1 ε · · · → ZG −→ ZG −→ ZG −→ ZG → Z → 0 .

5.3(a): [[Geometric Realization of a Semi-Simplicial Complex]] For each (n+1)-tuple σ = (g0, . . . , gn), let ∆σ be a copy of the standard n-simplex with vertices v0, . . . , vn. Let diσ = (g0,..., gî, . . . , gn) and let δi : ∆d σ → ∆σ be the linear embedding which sends v0, . . . , vn−1 i ` to v0,..., vî, . . . , vn. Consider the disjoint union X0 = σ ∆σ (topologize it as a topological sum) and def define the quotient space X = X0/∼ using the equivalence relation generated by (σ, δix) ∼ (diσ, x), where we rewrite ∆σ as σ × ∆σ for clarity of the relation properties. We assert that the geometric realization X is a CW-complex with n-skeleton Xn = (` ∆ )/∼. dim∆σ ≤n σ X0 is the collection of vertices and hence a 0-skeleton, and so we proceed by induction on n [sketch]: n (n) n−1 In X the equivalence relation ∼ identifies a point on a boundary ∂∆σ with a point in X , and (n) (n) it doesn’t touch the interior points of ∆σ . This means that the n-cells are {∆σ } with the attaching maps induced by di ∀ i. Refer to Theorem 38.2[4] for the analogous construction with adjunction spaces, n i providing Hausdorffness of X and weak topology w.r.t. {X }iisomorphism between the cellular chain complex C(X) and the standard resolution F∗, it suffices to determine the boundary operator on C(X) and see that it provides commutativity of ∼ the diagram F∗ → C(X), noting that Ci(X) = Fi by the correspondence ∆σ ↔ σ. Now C(X0) has the P i boundary ∂[v0, . . . , vn] = i(−1) [v0,..., vî, . . . , vn] and each resulting (n-1)-simplex is the image under P i a δi. Thus C(X) has boundary maps ∂(∆σ) = i(−1) ∆diσ, and these parallel those of F∗, giving commutativity of the diagram.

Note that this provides a solution to Exercise I.4.4: For any group G we can form the contractible

7 G-complex X as above, and since X → X/G is a regular covering map by the result of exercise I.4.2 above, the orbit space X/G is a K(G, 1), called the “classifying space.”

5.3(b): For the normalized standard resolution F¯∗, we simply follow part(a) while making these further identiﬁcations in X0 to collapse degenerate simplices. For each σ = (g0, . . . , gn) let siσ = (g0, . . . , gi, gi, . . . , gn),

and when forming the quotient X0 → X we also collapse ∆siσ to ∆σ via the linear map Li which sends v0, . . . , vn+1 to v0, . . . , vi, vi, . . . , vn (so the only simplices of X are those whose associated tuples have pairwise distinct coordinates). Thus the equivalence relation in part(a) is also generated by n n−1 (σ, Lix) ∼ (siσ, x). During the inductive process for X , if ∆σ is a degenerate simplex then X will already contain it and so those simplices need not be considered as n-cells. No problems arise when using the homotopy because for tuples of the form τ = (1, g0, . . . , gn) | g0 6= 1, the cone ∆τ ∗ v0 = ∆hτ = ∆τ has the identity map δ0 still being nulhomotopic [note: we actually have a deformation retraction since ∆(1) remains ﬁxed instead of looping around ∆(1,1) as in part(a)].

6.1: Given a finite CW-complex X with a map f : X → X such that every open cell satisfies S f(σ) ⊆ τ6=σ τ where dimτ ≤ dimσ, we have the condition f(σ) ∩ σ = ∅ and there are no fixed (n) (n−1) points. Viewing the open n-cell σ on the chain level in Hn(X ,X ) = Cn(X), f](σ) does not consist of σ and so the respective matrix has the value 0 at the row-column intersection for σ. Therefore, tr(f],Cn(X)) = 0 ∀ n since the diagonal of the matrix for the basis elements is zero. By the Hopf Trace P i P i Theorem, (−1) tr(fi,Hi(X)/torsion) = (−1) tr(f],Ci(X)) and so the Lefschetz number Λ(f) = 0. If X is a homology (2n − 1)-sphere, then Hi(X) is nontrivial only in dimensions 0 and 2n − 1 (in 0 which case it is isomorphic to Z). Thus, by the Lefschetz Fixed Point Theorem, (−1) tr(f0, Z) + 2n−1 (−1) tr(f2n−1, Z) = 1 − d = 0 ⇒ d = 1 and f∗ : H2n−1(X) → H2n−1(X) is the identity.

6.2: The group action G → Homeo(S2n) yields a degree map 2n ∼ φ : G → Aut(H2n(S )) = Aut(Z) = {±1} = Z/2Z which sends g ∈ G to the degree d = deg(g) of its associated homeomorphism g : S2n → S2n [note: deg(g) · deg(g−1) = deg(g · g−1) = deg(id) = 1 ⇒ |d| = 1]. Consider nontrivial G 6= Z/2Z and assert that this φ is not injective: Kerφ = 0 ⇒ 3 ≤ |G| = |Kerφ| · |Imφ| = 1 · |Imφ|. Since Imφ ⊆ Z/2Z, |Imφ| = 1 or 2 . In either case we arrive at a contradiction (since 1, 2 < 3). ∴ ∃ g ∈ Kerφ | g 6= id ⇒ degg = 1. Now assume this 2n 2n action is free, and use the notation fi : Hi(S ) → Hi(S ). By the Lefschetz Fixed Point Theorem, 0 2n (−1) tr(f0) + (−1) tr(f2n) = 1 + d = 0 ⇒ degf = d = −1 ∀ f 6= id. Our contradiction has now been reached (taking f = g from above).

7.1: Given the ﬁnite cyclic group G = hti, the free resolution F of Z over ZG with period two (chain complex with rotations of S1), and the bar resolution F 0, we obtain a commutative diagram where f : F → F 0 is the desired augmentation-preserving chain map: t−1 N t−1 ε ··· / ZG / ZG / ZG / ZG / Z / 0

f f f f id 3 2 1 0 Z 0 0 0 0 ··· / F3 / F2 / F1 / F0 / Z / 0 We deﬁne f inductively as fn+1 = knfn∂, where k is a contracting homotopy for the augmented complex associated to F 0, and each map is determined by where it sends the basis element:

f0(1) = k−1idZε(1) = k−1idZ(1) = k−1(1) = (1) ≡ [] f1(1) = k0f0∂(1) = k0f0(t − 1) = k0{tf0(1) − f0(1)} = k0{t[] − []} = [t] − [1] Pn−1 i Pn−1 i i f2(1) = k1f1∂(1) = k1f1(N) = k1{ i t f1(1)} = i ([t , t] − [t , 1]) Pn−1 i i i i f3(1) = k2f2∂(1) = k2f2(t − 1) = i ([t, t , t] + [1, t , 1] − [t, t , 1] − [1, t , t])

7.2: Here is the axiomatized version, Lemma 7.4: Under the additive category A, let (C, ∂) and (C0, ∂0) be chain complexes, let r be an integer, and let 0 0 (fi : Ci → Ci)i≤r be a class of morphisms such that ∂ifi = fi−1∂i for i ≤ r. If Ci is projective relative 0 0 0 to the class E of exact sequences for i > r, and Ci+1 → Ci → Ci−1 is in E for i ≥ r, then (fi)i≤r extends to a chain map f : C → C0 and f is unique up to homotopy. (Theorem 7.5 follows immediately)

8 7.3(a): Given an arbitrary category C, let A ∈ Ob(C) be an object and let hA = HomC(A, −): C → (Sets) be the covariant functor represented by A, with uA ∈ hA(A) as the identity map A → A. Let T : C → (Sets) be an arbitrary covariant functor. Any natural transformation ϕ : hA → T yields the commutative diagram (with f : A → B in hA(B) arbitrary): hA(f) hA(A) −−−−→ hA(B)   ϕ ϕ y y T (f) T (A) −−−−→ T (B) For any v ∈ T (A) suppose we have the natural transformation with ϕ(uA) = v. By commutativity, T (f)(v) = (ϕ ◦ hA(f))(uA) = ϕ(f ◦ uA) = ϕ(f), and hence the transformation is unique [determined by where it sends the identity]. For existence of the natural transformation ϕ(f) = T (f)(v), we assert that it satisﬁes the commutative diagram, using arbitrary g : B → C (and f as above): T (g)[ϕ(f)] = T (g)[T (f)(v)] = T (g ◦ f)(v) = ϕ(g ◦ f) ϕ[hA(g)(f)] = ϕ(g ◦ f) = T (g)[ϕ(f)] ∼ Thus, HomF(hA,T ) = T (A) where F is the category of functors C → (Sets), and we have ﬁnished proving Yoneda’s Lemma.

L 7.3(b): An M-free functor F : C → Ab is isomorphic to α ZhAα , where Aα ∈ M [M is a sub- class of Ob(C)] and ZhA : C → Ab is the composite of hA and the functor (Sets) → Ab which associates to a set the free abelian group it generates. Given the additive category A whose objects are covariant functors C → Ab and whose maps are natural transformations of functors, let E be the class of M-exact sequences in A. Consider the mapping problem (for all rows in E): F ψ 0 ϕ ~ T 0 / T / T 00 i j

By Yoneda’s Lemma (part(a) above), each component ZhAα of F with any natural transformation in the

above diagram is completely determined by the identity uAα , and so these identities “form a basis” for 0 00 F . In particular, for the identity uA we obtain the exact sequence T (A) → T (A) → T (A) of abelian groups from the above mapping problem since the associated row lies in E with A ∈ M. Thus, for each 0 identity we have ϕ(uAα ) ∈ Kerj = Imi, which implies ∃ xα ∈ T (Aα) | i(xα) = ϕ(uAα ), and so we

form ψ by ψ(uAα ) = xα. This means that F (an M-free functor) is projective relative to the class E of M-exact sequences.

7.3(c): There is a natural chain map in A from M-free complexes to M-acyclic complexes, and it is unique up to homotopy [using the categorial deﬁnitions from parts (a) and (b)]. This statement is a result of the combination of part(b) and Exercise 7.2 above, and is precisely the Acyclic Model Theorem in a rephrased form (M is the set of models).

7.4: Under the category of R-modules, let (C, δ) and (C,¯ δ¯) be cochain complexes, let r be an inte- ◦ ¯i i ◦ ¯ ◦ i ger, and let (fi : C → C )i≤r be a class of morphisms such that fi−1δi−1 = δi−1fi for i ≤ r. If C is injective relative to the class E◦ of exact sequences for i > r [yielding a cochain complex of injectives], ¯i−1 ¯i ¯i+1 ◦ ◦ and C → C → C is in E for i ≥ r [an acyclic cochain complex], then (fi )i≤r extends to a cochain map f ◦ : C¯ → C and f ◦ is unique up to homotopy. (The analogous “Theorem 7.5” follows immediately)

8.1: Obviously the trivial group is one, since Z[{1}] = Z and Z is a projective Z-module; so assume G is nontrivial. Give Z the trivial module structure (so that for r ∈ ZG, r · a = ε(r)a ∀ a ∈ Z). Considering ε the short exact sequence of modules 0 → I → ZG → Z → 0, we must ﬁnd a splitting µ : Z → ZG for Z to possibly be ZG-projective. Any such map is determined by where 1 ∈ Z is sent; say µ(1) = x. Then, P P P P for nontrivial α = rigi, µ(α · 1) = α · µ(1) = α · x = αx. But µ(α · 1) = µ( ri) = µ(1) ri = ( ri)x, P P P P P so αx = ( ri)x. Thus (α − ri)x = 0 ⇒ rigi = ri ⇒ ri(gi − 1) = 0 [this sum can be viewed

9 as having no gi = 1, and hence it lies in I]. Restricting our choice of nontrivial α to one which is not an integer, there is some nontrivial ri0 associated to gi0 6= 1 and hence we must have gi = 1 ∀ i (by freeness of I). But then G is the trivial group, and we are done.

8.2: Assume P is a projective ZG-module and consider the subgroup H ⊆ G. Then F = P ⊕ K where F is a free ZG-module, by Proposition I.8.2[1]. By restriction of scalars from ZG to its subring ZH [r · n = f(r)n with homomorphism f : ZG → ZH preserving identities], F has an inherent ZH- module structure. Since P is a direct summand of such a free module, it is a projective ZH-module, by L Proposition I.8.2[1]. Alternatively, we can note from Exercise 3.1 above that ZG = ZH and so F is a direct sum of H-modules, hence ZH-free.

8.3(a): Given F as a non-negative acyclic chain complex of projective modules Pn (over an arbitrary ¯ ∂n ¯ ring R), it will be contractible if each short exact sequence 0 → Zn → Pn → Zn−1 → 0 splits (where ∂ is induced by the boundary ∂), by Proposition I.0.3[1]. The case n = 0 is trivial using the zero map, and ¯ ¯ so arguing inductively we assume that ∂n−1 splits. Since Ker∂n−1 = Zn−1 is a direct summand of the projective module Pn−1 and a projective module is a direct summand of a free module by Proposition I.8.2[1], Zn−1 is necessarily a direct summand of a free module, and hence is projective by Lemma I.7.2[1]. ¯ Therefore, ∂n must have a splitting in the aforementioned short exact sequence by Proposition I.8.2[1].

8.3(b): If R is a principal ideal domain, then submodules of a free R-module are free by Theorem II.7.1[5]. Since a projective module is a direct summand of a free module by Proposition I.8.2[1], it is in particular a submodule and hence is free (over R as a PID). Therefore, submodules of a projective module over a PID are free and hence projective (free modules are projective by Lemma I.7.2[1]). The non-negativity hypothesis of Corollary I.7.7 can be dropped if we then restrict ourselves to PIDs, because we can follow part(a) above but not use induction since Zn−1 is already projective, being a submodule of the projective chain module Pn−1 (i.e. we don’t need any “starting point” in the resolution to obtain the desired splitting).

∼ L 8.4: Every permutation module admits the decomposition QX = Q[G/Gx] and a direct sum of projective modules is projective iff each summand is projective (by Lemma XVI.3.6[5]). Thus it suffices to show that Q[G/Gx] is a projective QG-module, where G is an arbitrary group and Gx is finite. Note Gx that HomQG(Q[G/Gx], −) is a left-exact functor (Corollary 10.5.32[2]); it is given by M → M because any homomorphism ϕ is determined by ϕ(Gx), and ϕ(Gx) = ϕ(g · Gx) = g · ϕ(Gx) for any g ∈ Gx. ¯ Thus, we must show that HomQG(Q[G/Gx], −) takes surjective homomorphisms M → M to surjective Gx ¯ Gx homomorphisms M → M [because then the functor is exact and then Q[G/Gx] is projective by definition]. Ifm ¯ ∈ M¯ Gx , liftm ¯ to m ∈ M. Then 1 P gm is also a lifting ofm ¯ which lies in M Gx |Gx| g∈Gx 1 P 1 P 1 Gx Gx (because gm 7→ gm¯ = · |Gx|m¯ =m ¯ ), and so M → M¯ is surjective. Thus, X is |Gx| |Gx| |Gx| Q a projective QG-module, where X is a G-set and Gx is finite for all x ∈ X.

8.5: If G is finite and k is a field of characteristic zero, consider any short exact sequence of kG- ϕ modules of the form 0 → M 0 → M → P → 0. Since k-vector spaces are free modules over k, and free modules are projective (by Lemma I.7.2[1]), P is k-projective and hence the sequence [as k-modules] splits by Proposition I.8.2[1]. Choosing a splitting f : P → M for the underlying sequence of k-vector 1 P −1 spaces, we form the homomorphism φ : P → M by x 7→ |G| g∈G gf(g x). Since f is a k-module homomorphism, it suffices to show that φ is equivariant (compatible with G-action) for it to thus be a kG-module homomorphism: 1 P −1 1 P −1 φ(g0x) = |G| gf(g g0x) = |G| g0hf(h x) −1 [where h = g0 g] 1 P −1 1 P −1 g0φ(x) = g0( |G| gf(g g0x)) = g0( |G| hf(h x)) = φ(g0x) P P −1 [because g∈G h = g∈G g, as g0 permutes the elements of G]

By Proposition 10.5.25[2] it suﬃces to show that ϕφ = idP for φ to thus be a splitting (noting that ϕf = idP ): 1 P −1 1 P −1 1 P −1 ϕ(φ(x)) = ϕ( |G| g∈G gf(g x)) = |G| g∈G ϕ(gf(g x)) = |G| g∈G gϕ(f(g x)) =

10 1 P −1 1 P 1 |G| g∈G gg x = |G| g∈G x = |G| |G|x = x Since φ is a kG-splitting, P is kG-projective by Proposition I.8.2[1].

∗ 8.6: Suppose P is an R-module such that ϕ : P ⊗R P → HomR(P,P ) is surjective, where this P map is given by ϕ(u ⊗ m)(x) = u(x) · m. Then in particular we have idP = ϕ( fi ⊗ ei), so that P P x = idP (x) = ϕ(fi ⊗ ei)(x) = fi(x)ei. Thus, P is projective by Proposition I.8.2[1], and it is ﬁnitely generated by the ei elements (noting that the summation is ﬁnite for tensor products).

8.7: Assume P is a finitely generated projective R-module and M is any (left) R-module, and take ∗ the canonical isomorphism ϕ : P ⊗R M → HomR(P,M) from Proposition I.8.3[1] which is given by ∗ ∗ ϕ(u ⊗ m)(x) = u(x) · m. For any z ∈ P ⊗R P we define the map ψz : HomR(P,M) → P ⊗R M as ∗ ψz(f) = (P ⊗ f)(z), which is a homomorphism since tensors are distributive over sums. −1 ∗ Method 1 : View ϕ as a natural transformation HomR(P, −) → P ⊗R −, where HomR(P, −) ∗ and P ⊗R − are exact covariant functors from the category R-mod to the category Ab by Corollary 10.5.41[2] and Corollary 10.5.32[2], noting that P ∗ is projective by Proposition I.8.3[1] and hence is a −1 −1 flat module by Corollary 10.5.42[2]. By Yoneda’s Lemma, ϕ is uniquely determined by ϕ (idP ) = z, −1 ∗ and the proof of the lemma (Exercise I.7.3(a) above) states that ϕ (f) = (P ⊗ f)(z) = ψz(f). Thus, −1 the inverse homomorphism ϕ is a map of the form ψz (independent of M). ∗ Method 2 : By Proposition I.8.2[1] we can choose elements ei ∈ P and fi ∈ P such that for every P P x ∈ P , x = fi(x)ei and fi(x) = 0 for cofinitely many x. Set z = fi ⊗ ei. We have ψzϕ = id ∗ P P because ψz(ϕ(u ⊗ m)) = (P ⊗ u · m)(z) = fi ⊗ u(ei) · m = fi · u(ei) ⊗ m = u ⊗ m [note: P P u(x) = u( fi(x)ei) = fi(x) · u(ei) as u is an R-module homomorphism]; we also have ϕψz = id ∗ P P P because ϕ(ψz(f)) = ϕ[(P ⊗ f)(z)] = ϕ( fi ⊗ f(ei)) = fi · f(ei) = f [note: f(x) = f( fi(x)ei) = P −1 fi(x) · f(ei) as f is also an R-module homomorphism]. Thus, the inverse ϕ is a map of the form ψz (independent of M).

8.8: Since P is finitely presented, we can form the obvious exact sequence F1 → F0 → P → 0 with F0 and F1 free of finite rank (the generators and relators, respectively). By Theorem 10.5.33[2], HomR(−,D) ∗ ∗ ∗ is a left exact contravariant functor, and so we obtain an exact sequence 0 → P → F0 → F1 of right ∗ R-modules [notation: M = HomR(M,R)]. Since P is a flat module, we can tensor this exact sequence ∗ ∗ ∗ with P to obtain the exact sequence 0 → P ⊗R P → F0 ⊗R P → F1 ⊗R P . Since Fi [i = 0, 1] is ∗ ∼ free, it is necessarily projective (by Lemma I.7.2[1]) and hence Fi ⊗R P = HomR(Fi,P ) by Proposition I.8.3[1]. Since F1 is contained in the quotient of F0 which gives P , HomR(F1,P ) = 0 and thus we have ∗ ∼ the isomorphism P ⊗R P = HomR(F0,P ). Now HomR(−,P ) as a functor on the original presentation sequence gives rise to the exact sequence 0 → ∼ HomR(P,P ) → HomR(F0,P ) → HomR(F1,P ) = 0, and thus we have the isomorphism HomR(P,P ) = HomR(F0,P ). ∼ ∗ = Since the discovered isomorphism P ⊗R P → HomR(P,P ) is surjective, P is a projective R-module by Exercise I.8.6 above.

11 2 Chapter II: The Homology of a Group ∼ L 2.1: For S an arbitrary G-set, ZS = Z[G/Gs]. By the Orbit-Stabilizer Theorem we have a bijection between Gs and G/Gs. When passing to the quotient for the group of co-invariants, the subset Gs ⊆ S is sent to the element Gs ⊆ S/G since s is identiﬁed with gs in S/G (giving trivial G-action). Therefore ∼ L ∼ L ∼ L ∼ (ZS)G = ( Z[G/Gs])G = (Z[G/Gs])G = Z[Gs] = Z[S/G]. ∼ 2.2: A weaker hypothesis “Let X be an arbitrary G-complex without inversions” suﬃces. Then C∗(X) = L Z[Xi] is a direct sum of permutation modules where Xi is the basis set of i-cells, so by the previous ∼ L ∼ L ∼ exercise, C∗(X)G = (Z[Xi])G = Z[Xi/G] = C∗(X/G) which has a Z-basis with one basis element for each G-orbit of cells of X.

2.3(a): With the G-module M and normal subgroup H/G, MH = M/IM where I is the augmentation ideal of ZH. The induced G/H-action is given by gH(m+IM) = gm+IM. The properties of an action are obviously satisﬁed, and it is well deﬁned because if g2 is another coset representative of g1H, then g2 = g1h and g2m+IM = g1hm+IM = g1hm−g1m+g1m+IM = (h−1)g1m+g1m+IM = g1m+IM.

2.3(b): We can form the group homomorphism ϕ : MG → (MH )G/H using part(a) by m 7→ m + IM, which is well-deﬁned because m = gm 7→ gm + IM = gH(m + IM) = m + IM; it is a homomorphism since ϕ(m1 m2) = ϕ(m1m2) = m1m2 + IM = (m1 + IM)(m2 + IM) = (m1 + IM)(m2 + IM) = ϕ(m1)ϕ(m2). For the inverse φ we use m + IM 7→ m which is well-deﬁned since given the equivalent 0 elements m + IM and gm + (h − 1)m + IM in (MH )G/H , φ(gm + (h − 1)m0 + IM) = gm+hm0 −1m0 = m+m0 −m0 = m = φ(m + IM); it is a homomorphism because φ(m1 + IM m2 + IM) = φ(m1m2 + IM) = m1m2 = m1 m2 = φ(m1 + IM)φ(m2 + IM). Thus, ∼ we have the isomorphism MG = (MH )G/H .

2.3(c): Let Z[G/H] ⊗ZG M be a G/H-module, where Z[G/H] is the obvious (G/H, G)-bimodule which forms the tensor product and gives it the desired module structure. The map Z[G/H] × M → MH given by (a, m) 7→ am¯ is clearly G-balanced, and so by the universal property of tensor products (Theorem

10.4.10[2]) there exists the group homomorpism ϕ : Z[G/H] ⊗ZG M → MH given by a ⊗ m 7→ am¯ , and it is clearly a G/H-module homomorphism. There is a well-deﬁned map φ : MH → Z[G/H] ⊗ZG M deﬁned asm ¯ 7→ 1 ⊗ m because of the identity 1H ⊗ hm = 1Hh ⊗ m = 1H ⊗ m, and it is a G/H-module homomorphism because φ(gH · m¯ ) = φ(gm¯ ) = 1H ⊗ gm = Hg ⊗ m = gH ⊗ m = gH1H ⊗ m = gH · φ(m ¯ ) [noting that gH = Hg since H is normal, and G/H acts on MH by part(a) above]. Since ϕ and φ are ∼ inverses of each other, they are isomorphisms and we obtain MH = Z[G/H] ⊗ZG M.

P sgnσ 3.1: Let g1, . . . gn ∈ G be pairwise-commutative elements and consider z = (−1) [gσ(1)| · · · |gσ(n)] ∈ Cn(G), where σ ranges over all permutations of {1, . . . , n}. The sign of a permutation is defined here to be the number of swaps between adjacent integers to bring the permuted set back to the identity. Looking P i at the boundary ∂z where ∂ = (−1) di, a particular dj with j 6= 0, n will provide elements in Cn−1(G) of the form [· · · |gkgk0 | · · · ]. Each of these appears twice because gkgk0 = gk0 gk, but the paired elements will have opposite signs and hence will cancel each other. For j = 0, n we have elements of the form [gi| · · · ] with one gk missing from each, and each of these elements also appears twice because d0 will take off gk from the beginning of some element while dn will take off gk from the end of some other element. The paired elements differ by the sign (−1)n due to the boundary map, and they also differ by the sign (−1)n−1 due to the permutation which takes the first slot and sends it to the last slot; since (−1)n(−1)n−1 = (−1)2n−1 = −1, these paired elements will also cancel each other. Thus, ∂z = 0 and z is a cycle in Cn(G).

3.2: Suppose Z admits a projective resolution of ﬁnite length over ZG where G = Zn. Then ∃ i0 | HiG = 0 ∀ i > i0, and we make note that HiG is independent of the choice of resolution (see Section II.1[1]). ∼ Yet by an earlier calculation II.3.1[1] (using an inﬁnite resolution), HiG = Zn for all positive odd integers i. Thus we have arrived at a contradiction.

3.3: If G has torsion, say Zn ⊆ G, and Z admits a projective resolution of ﬁnite length over ZG,

12 then since a projective ZG-module is projective as a ZH-module for any subgroup H ⊆ G (by Exercise I.8.2), we would obtain a corresponding ﬁnite projective resolution over Z[Zn]. But this cannot occur due to the previous exercise, and hence we arrive at a contradiction.

4.1: If Y is a path-connected space and has a contractible regular covering space X with covering ∼ group G, then X is its universal cover with free G-action as translation, and π1Y = G (so Y = X/G is sing a K(G, 1)-space). The singular chain module Cn (X) is a free Z-module with basis the set of singular simplices σn : ∆n → X (continuous maps of the standard simplex into the space). A G-action on this n n sing basis is given by the composition gσ : ∆ → X → X, and thus Cn (X) is a free ZG-module with th n n one basis element for every G-orbit of singular simplices. Denoting the i face of σ as σ Fi where n−1 n Fi = [v0,..., vˆi, . . . , vn] : ∆ → ∆ is the inclusion map, the induced G-action on the simplices maps faces of σn to faces of gσn. Thus, the boundary operator is equivariant and the singular [augmented] sing chain complex C∗ (X) is a free G-module chain complex. As X is contractible, the complex is exact and so it is a free resolution of Z over ZG. sing sing Consider the projection ϕ : C∗ (X) → C∗ (Y ) given by σ 7→ qσ, where q : X → X/G =∼ Y is the regular covering map. Noting that ϕ(gσ) = qgσ = qσ = ϕ(σ), the projection induces the map sing sing sing ϕ¯ : C∗ (X)G → C∗ (Y ) which sends basis elements to basis elements. Since C∗ (Y ) has a Z-basis sing with one basis element for each G-orbit of singular cells of X as does C∗ (X)G [by a property (II.2.3) ∼ of the coinvariants functor],ϕ ¯ is an isomorphism. Therefore, H∗G = H∗Y as the homologies of a group are independent of the choice of resolution up to canonical isomorphism.

4.2:

4.3:

5.1: Let Y be an n-dimensional connected CW-complex such that πiY = 0 for i < n (n ≥ 2), let π = π1Y , ∼ and let X be the universal cover of Y (so that π1X = 0). Since πiX = πiY for i > 1, πiX is trivial for i < n and so by the Hurewicz Theorem HiX = 0 for 0 < i < n and the Hurewicz map h : πnX → HnX is an isomorphism. In addition, we have a partial free resolution Cn(X) → · · · → C0(X) → Z → 0 whose th n homology group is ZnX = HnX (noting that X is n-dimensional). Lemma II.5.1[1] now gives us an ∼ exact sequence 0 → Hn+1π → (HnX)π → ZnY = HnY → Hnπ → 0, where Y = X/π because every universal cover is regular with covering transformation group π1Y (by Corollary 81.4[6]). Finally, noting that the coinvariants functor takes isomorphisms to isomorphisms (by right-exactness), the commutative ∼= diagram (πnX)π / (HnX)π

∼= (πnY )π / HnY yields the desired exact sequence 0 → Hn+1π → (πnY )π → HnY → Hnπ → 0.

5.2(a): Let G = hS ; r1, r2,...i = F (S)/R where R is the normal closure in F (S) of the words ri. Consider the abelianization map rirj 7→ [ri]+[rj] from R to the relation module Rab with the denotation Qn ±1 −1 [ri] = rimod[R,R]. Any element x ∈ R has a representation x = i=1(firi fi ) where fi ∈ F = F (S). Now F acts by conjugation on R and so induces an F -action on Rab, and R acts trivially on Rab due to the deﬁnition of abelianization; it is immediate that we obtain the (G = F/R)-action on Rab: −1 Pn g · [ri] = [frif ]. Subsequently, [x] = i=1(gi · [ri]) and hence Rab is generated as a G-module by the images of the presentation words.

W 1 2 2 5.2(b): Let Y = ( s S ) ∪r1 e ∪r2 e ∪ · · · be the 2-complex associated to the given presentation of G in part(a), and let Y˜ be its universal cover. Consider the augmented cellular chain complex

˜ ∂2 ˜ ∂1 ˜ (∗) 0 → C2(Y ) → C1(Y ) → C0(Y ) → Z → 0 ˜ ˜ L ˜ L Note that C0(Y ) = ZG and C1(Y ) = |S| ZG and C2(Y ) = |R| ZG. By Proposition II.5.4[1] we have the exact sequence ˜ ∂1 ˜ 0 / Rab / C1(Y ) / C0(Y ) / Z / 0

13 and hence Rab = Ker∂1 is always in the chain complex for Y˜ . If Y is a K(G, 1) then Y˜ is acyclic and ˜ (*) is exact by Proposition I.4.2[1], so C2(Y ) = Rab and hence Rab is a free ZG-module. Now suppose Y˜ is not acyclic, so that Y is not a K(G, 1). Then H2Y˜ is nontrivial (since HiY˜ = 0 for i > 2 by (*) and H1Y˜ = 0 by the simply-connected property of Y˜ ) which implies that the boundary map ∂2 is not injective, and by exactness we can refer to this non-injective map as C2(Y˜ ) → Ker∂1 = Rab. Therefore, there exists a nontrivial ZG-relation amongst the words ri in Ker∂2, so {[ri]} is not ZG- independent in Rab and hence does not generate Rab freely.

5.2(c): Let G = hS; ri be an arbitrary one-relator group and write r = un ∈ F = F (S), where n ≥ 1 is maximal. By a result of Lyndon-Schupp, the image t of u in G has order exactly n, and we let C = hti = Zn. If n > 1 then the relation module Rab is not freely generated by r mod[R,R] since this generator is ﬁxed by C, but a result of Lyndon shows that no other relations hold (i.e. the projection Z[G/C] → Rab is an isomorphism). Let Y be a bouquet of circles indexed by S, and let Y˜ be the connected regular covering space of Y corresponding to the normal subgroup R of F = π1Y . Choosing a basepointv ˜ ∈ Y˜ lying over the vertex of Y , we identify G with the group of covering transformations of Y˜ ; as explained in [1] on pg15, Y˜ is a (1-dimensional) free G-complex. Sinceu ˜ ends at tv˜, the liftingr ˜ is the composite path o u˜ tv˜ v˜O

tu˜ tn−1u˜ t2v˜ / tn−1v˜ Thus the map S1 → Y˜ corresponding tor ˜ is compatible with the action of C, where C acts on S1 as a group of rotations (i.e. tk is multiplication by e2πik/n). Consider the 2-complex X obtained by attaching 2-cells to Y˜ along the loops gr˜, where g ranges over a set of representatives for the cosets G/C. Given a 2-cell σ, each g ∈ G sends ∂σ homeomorphically to ∂σ0 for some 2-cell σ0, and so we just pick our favorite extension g : σ → σ0. This G-action makes X a G-complex since the permutations of 2-cells are determined by the permutations of their boundary loops. Thus, if σ is the 2-cell attached alongr ˜ then i W 1 2 only C ⊂ G will ﬁx σ, since t simply rotates the loop ∂σ (i.e. Gσ = C). Let Γ = ( s S ) ∪r e be the ˜ ˜ S standard 2-complex associated to the presentation of G; its universal cover is Γ = Y g∈G σg where σg ˜ ˜ is attached along gr˜, and C2(Γ) = ZG. Thus X is the quotient of Γ by identifying σg with σgti for all i ∼ ˜ (for each g), and C2(X) = Z[G/C] = Rab. If n = 1 then C = {1} and X = Γ. Lyndon’s theorem about one-relator groups says that Rab is freely generated by the image of r, provided r is not a power, which is equivalent to Γ being a K(G, 1) by part(b) above, and hence equivalent to X being contractible (X is the Cayley complex associated to the presentation of G).

5.3(a): With G = F/R and following Kenneth Brown’s proof of Theorem II.5.3, let F = F (S), let Y be a bouquet of circles indexed by S, and let Y˜ be the connected regular covering space of Y corresponding to the normal subgroup R of F = π1Y . Choosing a basepointv ˜ ∈ Y˜ lying over the vertex of Y , we identify G with the group of covering transformations of Y˜ . For any f ∈ F we regard f as a combinatorial path in the CW-complex Y and we denote by f˜ the lifting of f to Y˜ starting atv ˜. ˜ ¯ ¯ This path f ends at the vertex fv˜, where f is the image of f in G. Deﬁne the function d : F → C1Y˜ ˜ by letting df be the sum of the oriented 1-cells which occur in f. Since the lifting of f1f2 is the path ˜ ¯ ˜ f1 ¯ f1f2 ¯ ¯ ¯ v˜ −→ f1v˜ −→ f1f2v˜, we have d(f1f2) = df1 + f1df2 for all f1, f2 ∈ F . Thus, if we regard the G-module C1Y˜ as an F -module via the canonical homomorphism q : F → G, then d is a derivation [since the ¯ F -action is given by restriction of scalars: f1 · df2 = f1df2 = q(f1)df2].

5.3(b): For any free group F we can apply part(a) above with R = {1} to get the desired deriva- ˜ (S) tion d : F → Ω where G = F/1 = F and Ω = C1Y = ZF which is the free module with basis (ds)s∈S [note: ds−1 = −s−1ds]. The above note is a result of d(1) = d(1 · 1) = d(1) + 1d(1) = 2d(1) ⇒ d(1) = 0. P We write the total free derivative df of f as the sum df = s∈S(∂f/∂s)ds, where ∂f/∂s ∈ ZF is the partial derivative of f with respect to s [the coeﬃcient of ds when df is expressed in terms of the basis (ds)]. It is immediate that ∂/∂s : F → ZF is a derivation because d is a derivation: ff 0 7→ d(ff 0) =

14 df + fdf 0 = P[(∂f/∂s) + f(∂f 0/∂s)]ds 7→ (∂f/∂s) + f(∂f 0/∂s). For t ∈ S we have dt = P(∂t/∂s)ds = P s6=t 0ds + 1dt, and so ∂t/∂s = δs,t. Example: S = {s, t} ⇒ ∂(ts−1ts2)/∂s = ∂ts−1/∂s + ts−1(∂ts2/∂s) = [∂t/∂s + t(∂s−1/∂s)] + ts−1[∂t/∂s + t(∂ss/∂s)] = 0 + t[−s−1(∂s/∂s)] + ts−10 + ts−1t[∂s/∂s + s(∂s/∂s)] = −ts−11 + ts−1t[1 + s1] = ts−1ts + ts−1t − ts−1 5.3(c): Consider any free group F = F (S) and derivation d : F → M where M is an F -module. By the ±1 ±1 representation f = s ··· sn and the deﬁnition of a derivation d(gh) = dg + gdh, we have the equation P 1 df = s∈S wsds through trivial induction on n, where ws = ∂f/∂s by deﬁnition of the partial derivative.

θ (S) ∂ ε 5.3(d): Consider θ in the exact sequence 0 → Rab → ZG → ZG → Z → 0 of G-modules, where (S) (S) ZG is free with basis (es)s∈S and ∂es =s ¯ − 1 [bar denotes image in G], and consider ϕ : R → ZG P given by r 7→ s∈S (∂r/∂s)es where (∂r/∂s) is the image of ∂r/∂s under the canonical map ZF → ZG. In order to show that θ is induced by ϕ we must verify exactness of the above sequence, and so we start b1 bn by calculating the partial derivatives of the representation r = s1 ··· sn ∈ Rab (where si 6= sj6=i). Since b Pb−1 j b Pb −j b b ∂s /∂s = j=0 s for b > 0 and ∂s /∂s = − j=1 s for b < 0 and ∂(As )/∂s = A(∂s /∂s) with s - A, we obtain

( b1 bi−1 Pbi−1 j s1 ··· si−1 j=0 si bi > 0 ∂r/∂si = b1 bi−1 Pbi −j −s1 ··· si−1 j=1 si bi < 0

(S) Injectivity of ϕ|Rab follows immediately from the freeness of ZG and the fact that any nontrivial r P has some nontrivial bi (hence ∂r/∂si 6= 0). It suﬃces to show that ∂ϕ(r) = (∂r/∂s)(s − 1) = 0 for ±b1 ±bi−1 P 1±j P ±j ±b1 ±bi−1 ±bi r ∈ Rab. For a particular i,(∂r/∂si)(si −1) = s1 ··· si−1 ( si − si ) = s1 ··· si−1 (si −1), b1 bi−1 b1 bi bi+1 and (∂r/∂si)(si − 1) + (∂r/∂si+1)(si+1 − 1) = −s1 ··· si−1 + s1 ··· si si+1 [suppressing the ±]. Thus, P b1 bn (∂r/∂s)(s − 1) = −1 + 0 + ··· + 0 +s ¯1 ··· s¯n = −1 +r ¯ = −1 + 1 = 0, and exactness is satisﬁed. (T ) If R is the normal closure of a subset T ⊆ F , then the projection ZG → Rab given by g · et 7→ g · [t] and the above exact sequence provides us with a partial free resolution: (T ) ∂2 / (S) ∂1 / ε / / ZG Z;G ZG Z 0

# # - θ Rab

where the matrix of ∂2 is the “Jacobian matrix” (∂t/∂s)t∈T,s∈S.

5.4: Sketch (via Ken Brown): Let G = F/R and use the same notation as in Exercise 5.3(a) above. Consider the following chain map in dimensions ≤ 2 using the bar resolution B∗ ∂2 ∂1 ε B2 / B1 / B0 / Z / 0

Γ2 Γ1 Γ0 (R) ˜ ˜ ZG / C1(Y ) / C0(Y ) / Z / 0 ˜ (S) We have the identification C1(Y ) = ZG = IR where I is the augmentation ideal of ZF (so IR is a (R) free G-module on the images s − 1), and ZG is the free G-module with basis (er)r∈R which maps onto Rab = H1Y˜ ⊂ C1(Y˜ ). The specific chain map is given by Γ2 :[g1|g2] 7→ −g1g2 · r(g1, g2) and Γ1 :[g] 7→ f(g) − 1 and Γ0 :[] 7→ 1, where r(g1, g2) ∈ R and f(g) ∈ F such that f(g) = g ∈ G and f(g)f(h) = f(gh)r(g, h). Applying the coinvariants functor, the group homomorphism C2(G) → Rab given by [g|h] 7→ r(g, h) mod[R,R] induces the isomorphism ϕ : H2G → R ∩ [F,F ]/[F,R] by passage to subquotients. A specific chain map γ in the other direction is given by γ2 : r 7→ hγ1(r − 1) and γ1 : s − 1 7→ [¯s] and γ0 : 1 7→ [ ], where h : B1 → B2 is the contracting homotopy h(g · [h]) = [g|h]. Regarding C2(G) as an ¯ γ1 h F -module via f · [g|h] = [fg|h], the map D : F,→ ZF → I → IR → B1 → B2 → (B2)G ≡ C2(G) is a P derivation such that Ds = [1|s¯]. Moreover, Df = s∈S[∂f/∂s|s¯], where the symbol [·|·] is Z-bilinear. −1 Then D|R : R → C2(G) is a homomorphism (since R acts trivially on C2(G)) which induces ϕ by passage to subquotients.

15 Qn −1 Let a1, . . . , an, b1, . . . , bn ∈ F such that r = i=1[ai, bi] ∈ R. The formula ϕ (r mod[F,R]) = Pn ¯ ¯ ¯ ¯ ¯ −1 ¯ ¯ i=1{[Ii−1|a¯i] + [Ii−1a¯i|bi] − [Ii−1a¯ibia¯i |a¯i] − [Ii−1|bi]}, where Ii = [a1, b1] ··· [ai, bi], is proven in the universal example where F is the free group on a1, . . . , an, b1, . . . , bn and R is the normal closure of r. Using the constructed formula for ϕ−1 and the product rule for derivations, the desired formula arises P −1 −1 −1 from Dr = i Ii−1 · D[ai, bi] and D[a, b] = [1|a] + [a|b] − [aba |a] − [aba b |b].

W 1 5.5(a): With the presentation G = hs1, ··· , sn | r1, ··· , rmi we associate the 2-complex Y = ( s S )∪r1 2 2 ∼ e · · · ∪rm e so that π1Y = G. By computing the Euler characteristic χ(Y ) two diﬀerent ways (by The- P i P i orem 22.2[4]) we obtain the equation (−1) rkZ(HiY ) = (−1) ci, where rkZ is the rank and ci is the number of i-cells. Then 1 − rkZ(Gab) + rkZ(H2Y ) = 1 − n + m, and so rkZ(H2Y ) = m − n + r where r = rkZ(Gab) = dimQ(Q ⊗ Gab). Now H2Y = Ker∂2 is a free abelian group (subgroup of cellular 2-chain group), and by applying Theorem II.5.2[1] we get a surjection H2Y → H2G (from the exact sequence in the theorem). Thus H2G can be generated by m − n + r elements.

5.5(b): Since Gab = 0 and the number of generators equals the number of relations (in the ﬁnite presentation), m − n + r = m − m + 0 = 0. Thus by part(a), H2G can be generated by at most 0 elements, and so H2G = 0.

∼ 5.5(c): Given Gab = 0, H2G = Z2 ⊕ Z2, and n as the number of generators, let m be the number of relations in the presentation of G. Then H2G is generated by the two elements (0, 1) and (1, 0), and r = 0, so by part(a), 2 ≤ m − n + 0 ⇒ m ≥ n + 2. Therefore, any n-generator presentation must involve at least n + 2 relations.

∼ 5.6(a): From the group extension 1 → N,→ G Q → 1 we have G/N = Q, and from Hopf’s formula ∼ ∼ we have H2G = R ∩ [F,F ]/[F,R] and H2Q = S ∩ [F,F ]/[F,S], where G = F/R and Q = F/S with ∼ ∼ −1 −1 R ⊆ S ⊆ F (so N = S/R). (H1N)Q = (Nab)Q = Nab/h{(q −1)·n[N,N]}i = Nab/h{gng n [N,N]}i = ∼ (N/[N,N])/([G, N]/[N,N]) = N/[G, N], where the Q-action on Nab is induced by the conjugation action of G on N, and the latter isomorphism follows from the Third Isomorphism Theorem. Now −1 −1 −1 −1 ∼ [G/N, G/N] = {g1Ng2Ng1 Ng2 N} = {g1g2g1 g2 N} = [G, G]N/N, so we have H1Q = Qab = (G/N)/[G/N, G/N] =∼ G/(N[G, G]), where the latter isomorphism follows from the Third Isomorphism Theorem. Thus, the desired 5-term exact sequence is obtained by showing the exactness of the sequence β γ R ∩ [F,F ]/[F,R] →α S ∩ [F,F ]/[F,S] → N/[G, N] → G/[G, G] →δ G/(N[G, G]) → 0 where γ and δ are induced by the injection and surjection of the group extension. From δ : g[G, G] 7→ g[G, G]N = gN[G, G] we have Kerδ = N/[G, G] and Imδ = G/(N[G, G]). From γ : n[G, N] 7→ n[G, G] we have Kerγ = N ∩ [G, G]/[G, N] and Imγ = N/[G, G] = Kerδ. As deduced above, N/[G, N] = (S/R)/[F/R, S/R] =∼ S/(R[F,S]) and so from β : s[F,S] 7→ s[F,S]R = sR[F,S] we have Kerβ = R ∩ [F,F ]/[F,S] and Imβ = S ∩ [F,F ]/(R[F,S]) =∼ N ∩ [G, G]/[G, N] = Kerγ. Finally, from α : r[F,R] 7→ r[F,S] we have Imβ = R ∩ [F,F ]/[F,S] = Kerβ.

5.6(b): Applying part(a) to the group extension 1 → R,→ F G → 1 we obtain the exact sequence α β γ δ H2F / H2G / (H1R)G / H1F / H1G / 0

∼= ∼= ∼= ∼= 0 R/[F,R] F/[F,F ] G/[G, G] where the ﬁrst vertical isomorphism follows from Example II.4.1[1] and the second vertical isomorphism arises in the solution to part(a). By the First Isomorphism Theorem and exactness of the sequence, ∼ ∼ ∼ H2G = H2G/Kerβ = Imβ = Kerγ = R ∩ [F,F ]/[F,R].

5.7(a): S3 is a closed orientable 3-manifold, and it has a group structure under quaternion multpli- cation (S3 < H as the elements of norm 1). The ﬁnite subgroup G of S3 provides a multiplication action, and it is free because the only solution in H to the equation gx = x for nontrivial x is g = 1. From a result in the solution to Exercise I.4.2, the G-action is a “properly discontinuous” action and so the

16 quotient map ρ : S3 → S3/G is a regular covering space. Following a proof by William Thurston, S3/G is Hausdorff: considering two points α, β of S3 in distinct orbits, form respective neighborhoods Uα and Uβ such that they are disjoint and neither neighborhood contains any translates of α or β (this can be done by the Hausdorff property of S3). Taking the union S K of these two neighborhoods, Int(K − g6=1 gK) yields a neighborhood of α and a neighborhood of β which project to disjoint neighborhoods of Gα and Gβ in S3/G [note: we needed to refine K because if Uα intersects gUβ, then after projecting to the orbit space, ρ(gUβ) = ρ(Uβ) intersects ρ(Uα)]. S3/G is a closed connected 3-manifold since it has S3 as a covering space (Theorem 26.5[6] provides the compactness, Theorem 23.5[6] provides the connectedness, and the property of evenly-covered neighborhoods provides the nonboundary and manifold structure). Since the actions g : S3 → S3 are fixed-point free, we have deg(g) = (−1)3+1 = 1 by Theorem 21.4[4] and hence G acts by orientation-preserving homeomorphisms [note: orientation is in terms of local ori- 3 3 ∼ 3 ∼ entations µx ∈ H3(S ,S − x) = H3(S ) = Z, as defined in [3] on pg234]. Local orientations µGx = µρ(x) 3 3 of S /G = ρ(S ) are given by the images Γx(µx) = Γgx(µgx) under the isomorphisms of local homology 3 3 3 3 groups Γx : H3(S ,S − x) → H3(ρ(S ), ρ(S ) − ρ(x)) which arise from the Excision Theorem and the local-homeomorphism property of covering spaces; the ‘local consistency condition’ follows in the same respect (where a ball B containing Gx and Gy has as preimage under ρ a union of balls, each of which is homeomorphic to B, and such a homeomorphic ball containing g1x and g2y provides the local consistency condition for S3). Therefore, S3/G is orientable. 3 3 ∼ 3 3 ∼ 3 Since S is simply-connected, π1S = 0 and so G = π1(S /G)/π1(S ) = π1(S /G). Applying Poincaré 3 ∼ 1 3 ∼ 3 ∼ Duality and the Universal Coefficient Theorem we obtain H2(S /G) = H (S /G) = Hom(H1(S /G), Z) = Hom(Gab, Z) = 0 [noting that G is finite]. A theorem of Hopf (Theorem II.5.2[1]) gives us an exact se- 3 quence which includes the surjection H2(S /G) → H2G → 0, hence H2G = 0.

5.7(b): The binary icosahedral group G (of order 120) is the preimage in S3 which maps onto the 3 alternating group A5 under S → SO(3) [up to isomorphism with the group of icosahedral-rotational symmetries] as explained in [3] on pg75. By part(a), H2G = 0. Consider the group extension 1 → K → G → A5 → 1 where K is the central kernel of order 2 (corresponding to G mapping onto A5). The ∼ associated 5-term exact sequence becomes 0 → H2(A5) → (H1K)A5 → 0 because H1G = Gab = 0, ∼ ∼ ∼ and thus we obtain the isomorphism H2(A5) = (H1K)A5 . The A5-action on H1K = Kab = K = Z2 ∼ is induced by the conjugation G-action on K = Z2 which is the trivial action (since K ≤ Z(G)), and ∼ ∼ therefore H2(A5) = H1K = Z2.

5.7(c): Consider the abstract group G = hx, y, z ; x2 = y3 = z5 = xyzi which is a finite presentation with the same number of generators as relations. We show that G is perfect (G = [G, G]) so that ∼ H1G = Gab = 0 and H2G = 0 by Exercise 5.5(b) above. Now G/[G, G] is an abelian group with the relations 2¯x = 3¯y = 5¯z =x ¯ +y ¯ +z ¯. From this we see thatx ¯ =y ¯ +z ¯, so we need not look atx ¯. Subsequently, 2¯y + 2¯z = 3¯y ⇒ y¯ = 2¯z and so we need not look aty ¯. Finally, 5¯z = 3(2¯z) = 6¯z ⇒ z¯ = 0 and so all generators vanish, i.e. G/[G, G] = 0. Alternatively, the commutator subgroup is [G, G] ⊆ G, and to prove the opposite inclusion it suffices to show that x, y, z all lie in [G, G] ≡ G0. z }| { x2 = xyz ⇒ xz =}|yz{ ⇒ x2 = yzyz = y3 ⇒ z−1yzy−1 = z−2y ⇒ z2[z−1, y] = y The two overbraced equations allow us to finish by showing that z ∈ G0. yzyz = y3 ⇒ y2 = zyz ⇒ [y, z] = z−1yz−1 ⇒ y = z[y, z]z We then have z5 = xyz ⇒ z3 = xyz−1 = yz · z[y, z]z · z−1 = z[y, z]z · z2[y, z] ⇒ z2 = [y, z]z3[y, z] = z }| { z3z−3 · [y, z]z3[y, z] ⇒ z−1 = g[y, z] ⇒ z = [y, z]−1g−1 ∈ G0 with g = z−3[y, z]z3 ∈ G0 by normality of the commutator subgroup.

With the cyclic subgroup C = hxyzi we have G/C = A5 and hence the group extension 1 → C → G → 2 3 5 A5 → 1 [A5 = hx, y, z ; x = y = z = xyz = 1i of order 60]. The associated 5-term exact sequence ∼ now yields the isomorphism H2(A5) = H1C = C, noting the trivial A5-action (since C is generated by a central element) and noting the cyclicity Cab = C. Thus, by part(b) we deduce that |C| = 2 and hence |G| = 2 · 60 = 120. In fact, G = SL2(F5) is the binary icosahedral group!

17 6.1: Given N/G, let F be a projective resolution of Z over ZG and consider the complex FN . Since a projective ZG-module is also projective as a ZH-module for any subgroup H ⊆ G (by Exercise I.8.2), F is a projective resolution of Z over ZN and so H∗(FN ) = H∗N. By Exercise II.2.3(a), FN is a complex of G/N-modules and so H∗(FN ) inherits a G/N-action, with FN → FN given by x 7→ (gN)x = gx. For Corollary II.6.3 we have the conjugation action α : N → N given by n 7→ gng−1 (for g ∈ G), and we have the augmentation-preserving N-chain map τ : F → F given by x 7→ gx [it commutes with the boundary operator ∂ of F since ∂ is equivariant, and it satisﬁes the condition τ(nx) = gnx = gng−1gx = α(n)τ(x)]. By Proposition II.6.2[1], if α is conjugation by g ∈ N then H∗(α) is the identity (hence trivial action), 0 0 and so τ : FN → FN is given by τ (x) = (gN)x = gx which agrees with the above map.

6.2: For any ﬁnite set A let Σ(A) be the group of permutations of A. For |A| ≤ |B|, choose an injection i : A,→ B and consider the injection Σ(A) ,→ Σ(B) obtained by extending a permutation on A to be the identity on B − iA. In order to show that the induced map H∗Σ(A) → H∗Σ(B) is independent of the choice of i, it suﬃces to show that any two injections i1 and i2 give conjugate maps ¯i1,¯i2 : Σ(A) ,→ Σ(B), because the conjugation map Σ(B) → Σ(B) induces the identity map on homology H∗Σ(B) → H∗Σ(B) by Proposition II.6.2[1]. Let τ be the permutation which takes i1(a) to i2(a) for all a ∈ A and is an arbitrary permutation (B − i1A) → (B − i2A). Then for a permutation ¯i1(˜σ) = σ ∈ Σ(B), the per- −1 mutation τστ is equal to ¯i2(˜σ). Thus ¯i1 and ¯i2 are conjugates of each other by τ, and the result follows.

0 0 6.3(a): Given the homomorphism α : G → G , the n-tuples in Cn(G) are sent to the n-tuples in Cn(G ) coordinate-wise via α, where [gh] 7→ [α(gh)] = [α(g)α(h)]. Thus H1(α) mapsg ¯ to α(g), whereg ¯ denotes the homology class of the cycle [g]. We also have the explicit isomorphism H1G → Gab given byg ¯ 7→ ∼= g mod[G, G]. It is immediate that we have the commutative diagram H1G / Gab

∗ H1(α) α ∼ 0 = 0 H1G / Gab with α∗ : g mod[G, G] 7→ α(g) mod[G0,G0], which is precisely the map obtained from α by passage to the ∼ quotient. Thus, the isomorphism H1() = ()ab is natural.

6.3(b): Suppose G = F/R and G0 = F 0/R0 with F = F (S) and F 0 = F 0(S0) free, and suppose α : G → G0 lifts toα ˜ : F → F 0. Let Y and Y˜ be associated to the presentation of G as in Exercise II.5.3(a), and similarly for Y 0 and Y˜ 0 with G0. Nowα ˜ yields a map Y → Y 0 that sends the combinatorial 1 path s = Ss to the combinatorial pathα ˜(s). ***Incomplete***

7.1: Consider G = G1 ∗A G2 with αk : A → Gk not necessarily injective, and let G¯1 = β1(G1), G¯2 = β2(G2), A¯ = β1α1(A) = β2α2(A) be the images of G1,G2,A in G [where βk : Gk → G arises from ¯ ¯ the amalgamation diagram for G]. Form the amalgam H = G1 ∗A¯ G2 and the commutative diagram: i2 A¯ / G¯2

i1 j2

γ2 G¯1 / H j1 ϕ

γ1 , G where γk is the natural inclusion and ik is the obvious injection [subsequently, we have γ1i1 = γ2i2 and hence the unique map ϕ from the universal mapping property].

18 α2 We also have the commutative diagram: A / G2

r2 α1 β2 G1 / G G¯2 β1 φ j2 r1 G¯1 / H j1 where rk is βk with the codomain restricted to form the inclusion [subsequently, j1r1α1(a) = j1β1α1(a) = j1i1β1α1(a) = j2i2β1α1(a) = j2i2β2α2(a) = j2r2α2(a) and hence we have the unique map φ from the universal mapping property]. From the diagrams (and dropping subscripts), φβ = jr and ϕj = γ. ϕφ(β(g)) = γ(r(g)) = γ(β(g)) = β(g) φϕ(j[β(g)]) = φγ[β(g)] = φ(β(g)) = j(r(g)) = j(β(g)) −1 ∼ Thus ϕφ = idG and φϕ = idH , and so ϕ = φ is an isomorphism (H = G). Consequently, any amalgamated free product is isomorphic to one in which the maps A → Gk are injective.

7.2:

∼ 7.3: Consider the Special Linear Group SL2(Z) = Z4 ∗Z2 Z6 which is the subgroup of all 2x2 integral matrices with determinant 1. Applying the Mayer-Vietoris sequence for groups and using the fact that H∗(Zk) is trivial in positive even dimensions and is isomorphic to Zk in positive odd dimensions, α β γ we get the exact sequence 0 → H2n(SL2(Z)) → Z2 → Z4 ⊕ Z6 → H2n−1(SL2(Z)) → 0. Noting that the only nontrivial map ϕ : Z2 = hti → Z4 = hsi is the canonical embedding deﬁned by 2 t 7→ s , we assert that the induced map under H2n−1 is the same embedding: Considering the two periodic free resolutions of Z, there exists an augmentation-preserving chain map f between them by Theorem I.7.5[1] and we have a commutative diagram 1+t t−1 Z[Z2] / Z[Z2] / Z[Z2]

fn+1 fn 1+s+s2+s3 s−1 Z[Z4] / Z[Z4] / Z[Z4] 2 3 2 The left-side square yields (1+s+s +s )fn+1(1) = fn(1+t) = fn(1)+ϕ(t)fn(1) = (1+s )fn(1), and by 2 3 2 exactness of the bottom row we have 0 = (s − 1)(1 + s )fn(1) = (s − s + s − 1)fn(1) ⇒ fn(1) = 1 + s, 2 3 2 2 3 hence (1 + s + s + s )fn+1(1) = (1 + s )(1 + s) = 1 + s + s + s ⇒ fn+1(1) = 1. Then after moving to quotients, the cycle elements (for odd-dimensional homology) are mapped via ϕ∗(1) = 1 + 1 = 2 while the boundary elements are mapped via ϕ∗(1) = 1, and the result follows [applies to all odd n, and f0(1) = 1]. In general, an injection H,→ G = H × K onto a direct summand will pass to an injection under j any covariant functor T because the composition identity H,→ G H yields the identity T (H) → T (H × K) → T (H) which implies j is injective [this can be applied to H = Z2 and G = Z6 = Z2 × Z3]. 2 3 ∼ Thus we have β(t) = (s1, s2) injective and so H2n(SL2(Z)) = Imα = Kerβ = 0. From the MV-sequence 3 we see that H2n−1(SL2(Z)) is a ﬁnite abelian group of order dividing |Z4 ⊕ Z6| = 24 = 2 · 3, hence contains only 2-torsion and 3-torsion by the Primary Decomposition Theorem. Considering the 3-torsion in the MV-sequence, 0 → 0 → 0 ⊕ Z3 → H2n−1(SL2(Z))(3) → 0, we have H2n−1(SL2(Z))(3) = Z3 by exactness [this row can be extracted from the MV-sequence because ﬁnitely generated abelian groups are direct sums of their Sylow pi-subgroups (by the Primary Decomposition Theorem) and maps between the abelian groups will send primary components (the Sylow pi-subgroups) to respective primary components]. For 2-torsion we consider the 2-torsion subgroup and its MV-sequence, and we obtain a commutative diagram

19 / / / ∼ / / 0 Z2 Z4 ⊕ Z2 H2n−1(Z4 ∗Z2 Z2 = Z4) = Z4 0 0

∼= ∼= ∼= ψ ∼= ∼= 0 / Z2 / Z4 ⊕ Z2 / H2n−1(SL2(Z))(2) / 0 / 0 Then by the Five-Lemma, ψ is an isomorphism and so H2n−1(SL2(Z))(2) = Z4. ∼ ∼ Thus, H2n−1(SL2(Z)) = Z4 ⊕ Z3 = Z12.

 Z i = 0 ∼  ⇒ Hi(SL2(Z)) = Z12 i odd 0 i even

20 3 Chapter III: Homology and Cohomology with Coeﬃcients

0.1: Let F be a flat ZG-module and M a G-module which is Z-torsion-free (ie. Z-flat), and consider the tensor product F ⊗ M with diagonal G-action. Since (F ⊗ M) ⊗G − = (F ⊗ M ⊗ −)G = F ⊗G (M ⊗ −), it suffices to show that X = F ⊗G (M ⊗ −) is an exact functor so that F ⊗ M is ZG-flat (by Corollary 10.5.41[2]). But by the same corollary M ⊗ − is Z-exact (and a G-module) and F ⊗G − is ZG-exact, so X is exact and the result follows.

0.2: Let F be a projective ZG-module and M a Z-free G-module, and consider the tensor product ∼ G ∼ G ∼ F ⊗ M with diagonal G-action. Since HomG(F ⊗ M, −) = Hom(F ⊗ M, −) = Hom(F, Hom(M, −)) = HomG(F, Hom(M, −)) where the second isomorphism is adjoint associativity (Theorem 10.5.43[2]), it suﬃces to show that X = HomG(F, Hom(M, −)) is an exact functor so that F ⊗ M is ZG-projective (by Corollary 10.5.32[2]). But by the same corollary Hom(M, −) is Z-exact (and a G-module) and HomG(F, −) is ZG-exact, so X is exact and the result follows.

G 1.1(a): For a ﬁnite group G and a G-module M we have the norm map N : MG → M induced from P the map M → M which is multiplication by the norm element N = g∈G g. Noting that Nm = |G|m G for both m ∈ MG and m ∈ M , we see that |G| · KerN = 0 (as Nm = Nm/∼ = 0 by deﬁnition of kernel) and |G| · CokerN = 0 (as CokerN = M G/NM and Nm modNM = 0).

1.1(b): Suppose M is an induced module (M = ZG ⊗ A) where A is an abelian group and G acts G G by g · (r ⊗ a) = gr ⊗ a. Then MG = (ZG)G ⊗ A = Z ⊗ A and M = (ZG) ⊗ A = Z · N ⊗ A, where G N is the norm element. The norm map N : MG → M is now given by z ⊗ a 7→ zN ⊗ a (for z ∈ Z) which is clearly a bijection. It is an isomorphism because N[z1 ⊗ a1 + z2 ⊗ a2] = Nz1 ⊗ a1 + Nz2 ⊗ a2 = z1N ⊗ a1 + z2N ⊗ a2 = N[z1 ⊗ a1] + N[z2 ⊗ a2]. L 1.1(c): Let M be a projective ZG-module; it is a direct summand of a free module F = ZG. By L Li application of part(b) above with A = i Z we see that N is an isomorphism for F because F = i ZG = L L ∼ ∼ ∼ i(ZG ⊗ Z) = ZG ⊗ ( i Z). Now (M ⊕ N)G = Z ⊗ZG (M ⊕ N) = (Z ⊗ZG M) ⊕ (Z ⊗ZG N) = MG ⊕ NG, and (M ⊕ N)G = M G ⊕ N G under the coordinate-wise G-action (of F) since g · (m, n) = (g · m, g · n) = ∼ G G (m, n) ⇒ g · m = m , g · n = n. Thus MG ⊕ NG = M ⊕ N , and since the norm map is bilinear we ∼ G have MG = M .

1.2: Using the standard cochain complex, an element of C1(G, M) is a function f : G → M, and under the coboundary map it is sent to (δf)(g, h) = g · f(h) − f(gh) + f(g). The kernel of this map consists of functions which satisfy f(gh) = f(g) + g · f(h), and these are derivations, so Z1(G, M) =∼ Der(G, M). Since an element of C0(G, M) is simply m ∈ M, and under the coboundary map it is sent to (δm)(g) = g · m − m, which is a principal derivation, we have B1(G, M) =∼ PDer(G, M). Thus, H1(G, M) =∼ Der(G, M)/PDer(G, M). If G acts trivially on M then PDer(G, M) = 0 and Der(G, M) = Hom(G, M), so H1(G, M) =∼ Hom(G, M) = Hom(Gab,M) = Hom(H1G, M), where the second-to-last equality comes from the fact that any group homomorphism from G to an abelian group factors through the commutator subgroup [G, G] by Propo- sition 5.4.7[2]. In particular, H1(G) = 0 for any ﬁnite group G.

1.3: Let A be an abelian group with trivial G-action, and let F → Z be a projective resolution ∼ of Z over ZG. Then F ⊗G A = (F ⊗ A)G = Z ⊗G (F ⊗ A), and since the diagonal G-action on F ⊗ A is simply the left G-action on F (since G acts trivially on A), we can apply tensor associa- ∼ ∼ tivity (Theorem 10.4.14[2]) to obtain Z ⊗G (F ⊗ A) = (Z ⊗G F ) ⊗ A = FG ⊗ A. Thus there is a Z universal coeﬃcient sequence 0 → Hn(G) ⊗ A → Hn(G, A) → Tor1 (Hn−1(G),A) → 0 by Proposition G ∼ I.0.8[1]. Also, HomG(F,A) = Hom(F,A) = Hom(FG,A), where the last isomorphism arises because (gu)(m) = g · u(g−1m) = u(g−1m) and so we must have g−1m = m ∈ F for gu = u. Thus there is also a universal coeﬃcient sequence 0 → Ext1 (H (G),A) → Hn(G, A) → Hom(H (G),A) → 0 by Z n−1 n Proposition I.0.8[1].

21 1.4(a): “Let f : C0 → C be a weak equivalence between arbitrary complexes, and let Q be a non- 0 negative cochain complex of injectives. Then the map HomR(f, Q): HomR(C,Q) → HomR(C ,Q) is a weak equivalence.” To prove this, note that the mapping cone C00 = C ⊕ ΣC0 of f is acyclic by Proposition I.0.6[1], where 0 0 00 (ΣC )p = Cp−1 is the 1-fold suspension of C. The mapping cone of HomR(f, Q) is HomR(C ,Q) because 0 Q 0 Q HomR(C ,Q)n ⊕ ΣHomR(C,Q)n = q HomR(Cq,Qq+n) ⊕ q HomR(Cq,Qq+n−1) = Q 0 0 −1 00 q HomR(Cq ⊕ Cq+1,Qq+n) = HomR(C ⊕ Σ C,Q)n = HomR(C ,Q)n

0 −1 0 0 0 00 noting that C ⊕ Σ C = (Cq ⊕ Cq+1) = (Cq−1 ⊕ Cq) = ΣC ⊕ C = C . Thus it suﬃces to show that 00 00 00 HomR(C ,Q) is acyclic (by Proposition I.0.6[1]), i.e., that Hn(HomR(C ,Q)) ≡ [C ,Q]n = 0 ∀ n ∈ Z. 00 n 00 By the uniqueness part of the result of Exercise I.7.4, [C ,Q]n ≡ [Σ C ,Q] is indeed 0, since any map on Q is zero in negative dimensions (so all extensions oﬀ of that zero map are homotopy equivalent).

1.4(b): Let ε : F → Z be a projective resolution and let η : M → Q be an injective resolution. By part(a) above and noting that ε is a weak equivalence (regarded as a chain map with M concentrated in dimension 0), we have a weak equivalence HomR(F,Q) ← HomR(Z,Q). Similarly, by Theorem I.8.5[1] we have a weak equivalence HomR(F,M) → HomR(F,Q). ∗ ∗ G ∗ ∗ ∗ ∼ In particular, H (G, M) = H (Q ) because H (G, M) = H (HomG(F,M)) and H (HomG(F,M)) = ∗ ∗ G ∼ H (HomG(Z,Q)) = H (Hom(Z,Q) ), noting that HomZ(Z,Q) = Q.

2.1: Given projective resolutions F → M and P → N of arbitrary G-modules M and N, there is ∼ = degf·degp an isomorphism of graded modules Γ : F ⊗G P → P ⊗G F given by f ⊗ p 7→ (−1) p ⊗ f, where ∼ we consider diagonal G-action on Fi ⊗ Pj = Pj ⊗ Fi [generally, degx = n for x ∈ Cn]. If we show that Γ (a degree 0 map) is a chain map, then it is a homotopy equivalence (hence a weak equivalence) and so G ∼ G 0 Tor∗ (M,N) = H∗(F ⊗G P ) = H∗(P ⊗G F ) = Tor∗ (N,M). Denote by d and d the boundary operators 0 of F and P , respectively, and denote by D and D the boundary operators of F ⊗G P and P ⊗G F , respectively. Then

D0Γ(f ⊗ p) = D0[(−1)degf·degpp ⊗ f] = (−1)degf·degp(d0p ⊗ f + (−1)degpp ⊗ df) = (−1)degf·degpd0p ⊗ f + (−1)degp·(degf+1)p ⊗ df and ΓD(f ⊗ p) = Γ[df ⊗ p + (−1)degf f ⊗ d0p] = (−1)degp·(degf−1)p ⊗ df + (−1)degf+degf·(degp−1)d0p ⊗ f = (−1)degp·(degf+1)p ⊗ df + (−1)degf·degpd0p ⊗ f Thus D0Γ = ΓD and so Γ is a chain map. (This simultaneously provides a solution to Exercise I.0.5)

3.1: Let P be a projective R-module, and let C be a short exact sequence of S-modules which can be regarded as R-modules via restriction of scalars. Since P is projective, HomR(P, C) is a short exact ∼= sequence, and so it suﬃces to show that the isomorphism of functors HomS(S ⊗R P, −) → HomR(P, −) is natural [because then HomS(S ⊗R P, C) is a short exact sequence which implies that S ⊗R P is a projective S-module]. Given a module homomorphism ψ : M → N, we must check commutativity of the diagram α HomS(S ⊗R P,M) / HomS(S ⊗R P,N)

ϕ1 ϕ2

β HomR(P,M) / HomR(P,N)

where α and β are given by f 7→ ψ ◦ f, and ϕi (i = 1, 2) is given by f 7→ f ◦ i under the universal mapping property with i : P → S ⊗R P , i(p) = 1 ⊗ p. Now ϕ2[α(F )] = ϕ2[ψ ◦ F ] = (ψ ◦ F ) ◦ i and β[ϕ1(F )] = β[F ◦ i] = ψ ◦ (F ◦ i) = (ψ ◦ F ) ◦ i. Therefore, ϕ2α = βϕ1 and the result follows: Extension of scalars takes projective R-modules to projective S-modules.

3.2: Let Q be an injective R-module, and let C be a short exact sequence of S-modules which can be regarded as R-modules via restriction of scalars. Since Q is injective, HomR(C,Q) is a short exact sequence,

22 ∼= and so it suﬃces to show that the isomorphism of functors HomS(−, HomR(S, Q)) → HomR(−,Q) is natural [because then HomS(C, HomR(S, Q)) is a short exact sequence which implies that HomR(S, Q) is an injective S-module]. Given a module homomorphism ψ : M → N, we must check commutativity of the diagram α HomS(N, HomR(S, Q)) / HomS(M, HomR(S, Q))

ϕ1 ϕ2

β HomR(N,Q) / HomR(M,Q)

where α and β are given by f 7→ f ◦ ψ, and ϕi (i = 1, 2) is given by f 7→ π ◦ f under the universal mapping property with π : HomR(S, Q) → Q , π(f) = f(1). Now ϕ2[α(F )] = ϕ2[F ◦ ψ] = π ◦ (F ◦ ψ) and β[ϕ1(F )] = β[π ◦ F ] = (π ◦ F ) ◦ ψ = π ◦ (F ◦ ψ). Therefore, ϕ2α = βϕ1 and the result follows: Co-extension of scalars takes injective R-modules to injective S-modules.

3.3: Given S which is flat as a right R-module, let Q be an injective S-module, let C be a short exact sequence of S-modules, and consider HomR(C,Q) where C and Q are regarded as R-modules via restriction of scalars. Since S is R-flat, S ⊗R C is a short exact sequence, and since Q is S-injective, HomS(S ⊗R C,Q) is a short exact sequence. It suffices to show that the isomorphism of functors ∼= HomS(S ⊗R −,Q) → HomR(−,Q) is natural [because then HomR(C,Q) is a short exact sequence which implies that Q is an injective R-module]. Given a module homomorphism ψ : M → N, we must check commutativity of the diagram α HomS(S ⊗R N,Q) / HomS(S ⊗R M,Q)

ϕ1 ϕ2

β HomR(N,Q) / HomR(N,Q)

where α is given by f 7→ f ◦ (S ⊗R ψ), β is given by f 7→ f ◦ ψ, ϕ1 is given by f 7→ f ◦ iN for the natural map iN : N → S ⊗R N, and ϕ2 is given similarly by f 7→ f ◦ iM . Now ϕ2[α(F )] = ϕ2[F ◦ (S ⊗R ψ)] = (F ◦ (S ⊗R ψ)) ◦ iM = F ◦ ((S ⊗R ψ) ◦ iM ) and β[ϕ1(F )] = β[F ◦ iN ] = (F ◦ iN ) ◦ ψ = F ◦ (iN ◦ ψ). Also, iN [ψ(m)] = 1 ⊗ ψ(m) = (S ⊗R ψ)(1 ⊗ m) = (S ⊗R ψ)[iM (m)]. Therefore, ϕ2α = βϕ1 and the result follows: Restriction of scalars takes injective S-modules to injective R-modules if S is a ﬂat right R-module.

3.4: Given S which is projective as a left R-module, let P be a projective S-module, let C be a short exact sequence of S-modules, and consider HomR(P, C) where C and P are regarded as R-modules via restriction of scalars. Since S is R-projective, HomR(S, C) is a short exact sequence, and since P is S-projective, HomS(P, HomR(S, C)) is a short exact sequence. It suﬃces to show that the isomorphism ∼= of functors HomS(P, HomR(S, −)) → HomR(P, −) is natural [because then HomR(P, C) is a short exact sequence which implies that P is a projective R-module]. Given a module homomorphism ψ : M → N, we must check commutativity of the diagram α HomS(P, HomR(S, M)) / HomS(P, HomR(S, N))

ϕ1 ϕ2

β HomR(P,M) / HomR(P,N)

where α is given by f 7→ φ ◦ f for φ(g) = ψ ◦ g (g : S → M), β is given by f 7→ ψ ◦ f, ϕ1 is given by f 7→ πM ◦ f under the universal mapping property with π : HomR(P,M) → M , πM (f) = f(1), and ϕ2 is given similarly by f 7→ πN ◦ f. Now ϕ2[α(F )] = ϕ2[φ ◦ F ] = πN ◦ (φ ◦ F ) = (πN ◦ φ) ◦ F and β[ϕ1(F )] = β[πM ◦ F ] = ψ ◦ (πM ◦ F ) = (ψ ◦ πM ) ◦ F . Also, πN [φ(f)] = πN [ψ ◦ f] = (ψ ◦ f)(1) = ψ[f(1)] = ψ[πM (f)]. Therefore, ϕ2α = βϕ1 and the result follows: Restriction of scalars takes projective S-modules to projective R-modules if S is a projective left R-module.

th 4.1: Let R = Z/nZ, and note that the ideals of R are the ideals Ix = (x) mod(n) for x|n ∈ Z by the 4 Isomorphism Theorem. It suﬃces to show that every map ϕ : I → R extends to a map R → R so that R is self-injective by Baer’s Criterion (Proposition III.4.1[1]). Given Ix and ϕ(x mod(n)) = r mod(n), we can write n = yx so that ϕ(yx mod(n)) = 0. But ϕ(yx mod(n)) = yϕ(x mod(n)) mod(n) = yr mod(n)

23 and thus yr = mn = myx ⇒ r = xm. Then, since ϕ(x mod(n)) = [x mod(n)][m mod(n)] we can extend ϕ to the domain R by setting ϕ(1 mod(n)) = m mod(n). (a): Let A be an abelian group such that nA = 0, and let C ⊆ A be a cyclic subgroup of order |C| = n. We can regard A as an R-module because xn = x0 ∈ R and R acts on A by xi · a = ia. As C is a [self-injective] subgroup of A, we have an inclusion C,→ A of an injective R-module into an R-module. By definition of “injective module” (pg.782-783 of [5], statement XX.4.I1), every exact sequence of modules 0 → Q → M → M 0 → 0 splits for injective Q, hence Q is a direct summand of M. Therefore, C is a direct summand of A. (b): Given A as above (i.e. an arbitrary abelian group of finite exponent), we regard A as an R- module. Since n is minimal (to annihilate A) there exists an element of order n and hence a cyclic subgroup C =∼ R of order n in A. If we can show that A = A0 ⊕ A00 for A0 of smaller exponent then by induction on n we have that A0 is a direct sum of cyclic groups; thus it remains to show that A00 is a direct sum of modules (each isomorphic to R) and is a direct summand of A. An application of Zorn’s Lemma on the set of direct sums with summands in A isomorphic to R (noting that the set is nonempty because it contains C) provides the maximal element A00; we can use this lemma because an upper bound for any chain would be the direct sum of those elements (the direct sums) in that chain. A ring is Noetherian iff every ideal is finitely generated (by Theorem 15.1.2[2]); thus Z is Noetherian (being a Principal Ideal Domain). Now R = Zn is also Noetherian, because a quotient of a Noetherian ring by an ideal is Noetherian (by Proposition 15.1.1[2]). It is a fact that a ring is Noetherian iff an arbitrary 00 L 00 direct sum of injective modules (over that ring) is injective. Thus A = i R is R-injective, so A is a direct summand of A, and A is a direct sum of cyclic groups. This result is known as Prufer’s Theorem for abelian groups.

G L 5.1: For any H-module M consider the G-module IndH M = g∈G/H gM where this equality follows from Proposition III.5.1[1]. The summand gM is a gHg−1-module and hence gHg−1 is the isotropy G G ∼ G group of this summand in IndH M. By Proposition III.5.3[1], IndH M = IndgHg−1 gM. In particular, by Proposition III.5.6[1] we have the K-isomorphism

−1 0 0 −1 L K gHg ∼ L K gg H(gg ) 0 g∈E IndK∩gHg−1 ResK∩gHg−1 gM = gg0∈E IndK∩gg0H(gg0)−1 ResK∩gg0H(gg0)−1 gg M

K gHg−1 Thus the K-module IndK∩gHg−1 ResK∩gHg−1 gM depends up to isomorphism only on the class of g ∈ E in K\G/H.

G 5.2(a): For any H-module M and G-module N consider the tensor product N ⊗ IndH M which has the diagonal G-action. By Proposition III.5.1[1] and the fact that tensor products commute with direct G ∼ L ∼ L sums, we have N ⊗ IndH M = N ⊗ ( g∈G/H gM) = g∈G/H (N ⊗ gM) which has N ⊗ M as a direct G summand in the underlying abelian group. Treating this as an H-module ResH N ⊗ M with a diago- G ∼ G G nal action so that H is its isotropy group, we have N ⊗ IndH M = IndH (ResH N ⊗ M) by Proposition III.5.3[1]. ∼ G G In particular, for M = Z we have N ⊗ Z[G/H] = IndH ResH N.

G 5.2(b): For any H-module M and G-module N consider U = Hom(IndH M,N) which has the “diagonal” G-action given by (gu)(m) = g · u(g−1m). By Proposition III.5.1[1] and the fact that the Hom-functor “commutes” with direct sums/products, we have U =∼ Hom(L gM, N) =∼ Q Hom(gM, N), where the indices on the sum/product symbols are implicitly the coset representatives in G/H. Thus U admits a direct product decomposition (πg : U Hom(gM, N)). Using the denotation πg(u) = ug, −1 we have [(πgg0)(u)](m) = [πg(g0u)](m) = g0 · ug(g m) = g0 · u −1 (m) = g0 · [π −1 (u)](m), and so 0 g0 g g0 g G πgg0 ∼ π −1 , with m ∈ Ind M. This decomposition has U Hom(M,N) as one of the surjections g0 g H G in the underlying abelian group. Treating this surjection as an H-module π1 : U Hom(M, ResH N) G ∼ G G so that H is its isotropy group, we have Hom(IndH M,N) = CoindH Hom(M, ResH N) by Proposition III.5.8[1]. G ∼ G G An analogous proof will provide Hom(N, CoindH M) = CoindH Hom(ResH N,M).

5.3: Let F be a projective G-module and M a Z-free G-module, and consider the tensor product F ⊗ M with diagonal G-action. By Corollary III.5.7[1], ZG⊗M is a free G-module since M is free as a Z-module.

24 L Since F is projective, it is a direct summand of a free G-module F = ZG, so that F = F ⊕ K. As i L the tensor product commutes with arbitrary direct sums (Corollary XVI.2.2[5]), F ⊗ M = i(ZG ⊗ M) and hence is a free G-module. Finally, F ⊗ M = (F ⊕ K) ⊗ M = (F ⊗ M) ⊕ (K ⊗ M) and so F ⊗ M is a direct summand of F ⊗ M, hence G-projective. Note that this gives a new proof of the result of Exercise III.0.2 above.

5.4(a): If |G : H| = ∞ then there are infinitely many distinct coset representatives in G/H. Now G L IndH M = g∈G/H gM by Proposition III.5.1[1], and it has a transitive G-action which permutes the PN summands. Consider an arbitrary [nontrivial] element x = i=1 gimi with all mi 6= 0 (this also refers to the sum over all coset representatives with cofinitely many mi = 0). We may take the summand 0 0 entry g1m1 of x and a summand g M which doesn’t appear in the representation of x (i.e. g 6= gj for 00 00 0 1 ≤ j ≤ N), and there then exists g ∈ G such that g · g1m1 = g m because the G-action is transitive 00 G G on the summands. Thus x is not fixed by G (since g · x 6= x), and so (IndH M) = 0. Note that if the index is finite, then this result does not hold. For instance, take 2Z ⊂ Z = hxi which has index | : 2 | = | | = 2. Then IndZ M = M ⊕ xM where x is the coset representative of x(2 ) Z Z Z2 2Z Z which generates . Since x2i ·(m , xm ) = (x2im , x(x2im )), we must have (IndZ M)Z ⊆ M 2Z ⊕xM 2Z. Z2 1 2 1 2 2Z Since x2j+1 · (m ¯ , xm¯ ) = x · (m ¯ , xm¯ ) = (m ¯ , xm¯ ), we must havem ¯ =m ¯ and hence (IndZ M)Z = 1 2 1 2 2 1 1 2 2Z 2 2i i {(m, ¯ xm¯ ) | m¯ ∈ M Z}. For M = Z4 with 2Z-action defined as x · m = (−1) m, the largest submodule on which 2 acts trivially [so that −m = m] is . Thus (IndZ )Z 6= 0 is our desired example. Z Z2 2ZZ4 5.4(b):

5.4(c): Assume statement (i), so that there is a finitely generated subgroup G0 ⊆ G such that |G0 : G0 ∩ gHg−1| = ∞ for all g ∈ G (with H ⊆ G). Using the analogue of Proposition III.5.6[1] for coinduction and −1 0 G G ∼ L G0 gHg ∼ passing to G -coinvariants, we obtain (ResG0 CoindH M)G0 = ( g∈E CoindG0∩gHg−1 ResG0∩gHg−1 gM)G0 = L G0 0 g∈E(CoindG0∩gHg−1 M )G0 where this latter isomorphism follows from commutativity of the tensor product with direct sums (and E is the set of representatives for the double cosets KgH). It is a G fact that if G is finitely generated and |G : H| = ∞ then (CoindH M)G = 0 for any H-module M G G L [part(b) above], so by statement (i) we have (ResG0 CoindH M)G0 = 0 = 0 for any H-module M. Noting that restriction of scalars gives the action g0 · x = ϕ(g0)x where ϕ is a map G0 → G, we have G G G G (CoindH M)G ⊆ (ResG0 CoindH M)G0 . Thus (CoindH M)G = 0 and (i) implies (ii). G Assume statement (ii), so that (CoindH M)G = 0 for all H-modules M. Then in particular (for M = Z) G there is only one element of (CoindH Z)G, and that element must be zero, so (ii) implies (iii). G Assume statement (iii), so that the element of (CoindH Z)G represented by the augmentation map G ε ∈ CoindH Z is zero. We first note that [in general] ifn ¯0 = 0 ∈ NG for some n0 ∈ N then it is 0 also zero in NG0 for some finitely generated subgroup G ⊆ G; this is because NG = N/hgn − ni and so n0 can be written as a finite Z-linear combination of elements of the form gn − n, which implies that 0 G we can take G to be the subgroup generated by those specific g’s. In particular,ε ¯ = 0 ∈ (CoindH Z)G0 for some finitely generated subgroup G0 of G. Using the double coset formula (analogue of Proposition III.5.6[1]) and treating Z and other modules appropriately over specific groups (to ignore restriction), G0 we must haveε ¯g = 0 ∈ (CoindG0∩gHg−1 Z)G0 for all g ∈ G, where εg denotes the component of the augmentation map in the specific summand of coinduction. Now if |G0 : G0 ∩ gHg−1| < ∞ then G0 ∼ G0 ∼ 0 0 −1 P 0 0 P 0 CoindG0∩gHg−1 Z = IndG0∩gHg−1 Z = Z[G /G ∩ gHg ], giving εg = g0∈K g ⊗ εg(g ) = ( g ) ⊗ 1 0 0 −1 00 00 0 where K is the set of coset representatives for the quotient G /G ∩ gHg . Then g εg = εg ∀ g ∈ G , G0 0 0 −1 soε ¯G 6= 0 ∈ (CoindG0∩gHg−1 Z)G0 . Thus we must have |G : G ∩ gHg | = ∞, so (iii) implies (i).

5.5: Let G be a ﬁnite group and let k be a ﬁeld, and consider the free module kG. We have kG =∼ G ∼ G kG⊗k k = Ind{1}k = Coind{1}k = Homk(kG, k), where the second-to-last equation follows from the ana- ∼ ∼ logue over k of Proposition III.5.9[1] since |G| < ∞. Then HomkG(−, kG) = HomkG(−, Homk(kG, k)) = Homk(−, k) where the last isomorphism follows from the universal property of co-induction. It is a fact that every k-vector space is an injective k-module [if the vector space W with basis B is a subspace of a vector space V , then we can extend B to a basis of V and then V = W ⊕ U where U is the vector space spanned by the additional basis vectors extended from B]; thus k is injective ⇒ Homk(−, k) is exact ⇒ HomkG(−, kG) is exact ⇒ kG is self-injective as a kG-module.

25 A Noetherian ring is a commutative ring which satisfies the Ascending Chain Condition on ideals (i.e. no infinite increasing chain of ideals), and any field k is Noetherian because the only ideals are {0} and k (giving {0} ⊆ k) by Proposition 7.4.9[2]. Now kG is Noetherian (for G finite) because it is a finite- dimensional k-vector space and so any infinite ascending chain of subspaces would require cofinitely many of those subspaces to have dimension greater than |G|, a contradiction; since ideals of kG are necessarily k-subspaces, the result follows. It is a fact that a ring R is Noetherian iff an arbitrary direct sum of L injective R-modules is injective. Thus the free module F = i kG is injective and so any projective kG-module is kG-injective (because a projective module is a direct summand of a free module, and a direct summand of an injective module is injective). Assuming the claim is true that any kG-module is a submodule of a kG-projective module, then by definition of “injective” it follows that any injective kG-module is a direct summand of a kG-projective module, hence kG-projective; it remains to prove this claim. We have a canonical kG-module injective map M → HomkH (kG, M) where the kG-module M can be regarded as a kH-module by restriction of G ∼ G scalars (see pg64 of [1]). But HomkH (kG, M) = CoindH M = IndH M = kG ⊗kH M, where this second- to-last equation follows from the analogue over k of Proposition III.5.9[1] since |G| < ∞. Using H = {1}, ∼ L this says that M is a submodule of kG ⊗k M. But M is treated as a k-vector space (M = i k), so ∼ L ∼ L kG ⊗k M = i(kG ⊗k k) = i kG; i.e. kG ⊗k M is a free [hence projective] kG-module.

6.1(a): We ﬁrst note that an arbitrary direct sum of projective resolutions is projective, which follows from the fact that an arbitrary direct sum of projective modules is projective and from the exactness of the row for each summand. We then note that homology commutes with direct sums, and this follows L L L L from the obvious facts Ker( i di) = i Ker(di) and Im( i di) = i Im(di). From these two facts and noting that the tensor product commutes with direct sums, we see that the Tor-functor commutes with direct sums, R L L ∼ L ∼ L L R Tor∗ (M, i Ni) = H∗(M ⊗R ( i Ni)) = H∗( i(M ⊗R Ni)) = i H∗(M ⊗R Ni) = i Tor∗ (M,Ni).

For the amalgamation G = G1 ∗A G2 consider the short exact sequence of permutation modules 0 → Z[G/A] → Z[G/G1] ⊕ Z[G/G2] → Z → 0. By Proposition III.6.1[1] we have the long exact sequence / Hn(G, Z[G/A]) / Hn(G, Z[G/G1] ⊕ Z[G/G2]) / Hn(G, Z) /

∼= ∼= ∼= Hn(A) Hn(G1) ⊕ Hn(G2) Hn(G) where the vertical isomorphisms follow from the fact H∗(H) = H∗(G, Z[G/H]), and the middle isomor- G phim utilizes commutativity of direct sums which follows from above because H∗(G, −) = Tor∗ (Z, −). This long exact sequence is the Mayer-Vietoris sequence for the amalgam G.

6.1(b): For the amalgamation G = G1 ∗A G2 consider the short exact sequence of permutation modules 0 → Z[G/A] → Z[G/G1]⊕Z[G/G2] → Z → 0. Applying −⊗M still yields a short exact sequence because all of the modules are free Z-modules (we consider the sequence of permutation modules as a free resolution of Z, hence a homotopy equivalence by Corollary I.7.6[1]); this homotopy equivalence for Z gives a homotopy equivalence for Z ⊗ M = M (since functors preserve identities), hence a weak equivalence ∼ G G (which can be stated as a resolution). By Proposition III.5.6[1] we have Z[G/A] ⊗ M = IndAResAM, G G ∼ G and by Shapiro’s Lemma we apply H∗(G, −) to obtain H∗(G, IndAResAM) = H∗(A, ResAM). Similar results follow for the other modules, and so by Proposition III.6.1[1] the exact sequence (which resulted from the sequence of permutation modules after application of − ⊗ M) yields a long exact sequence [the Mayer-Vietoris sequence for homology with coeﬃcients] H (A, ResGM) / H (G , ResG M) ⊕ H (G , ResG M) / H (G, M) n A n 1 G1 n 2 G2 n Now consider the original sequence of permutation modules, but instead apply Hom(−,M) which still yields a short exact sequence (same reason as mentioned above). By the analogue over co-induction of ∼ G G Proposition III.5.6[1] (or the result of Exercise III.5.2(a) above) we have Hom(Z[G/A],M) = CoindAResAM, ∗ ∗ G G ∼ ∗ G and by Shapiro’s Lemma we apply H (G, −) to obtain H (G, CoindAResAM) = H (A, ResAM). Similar results follow for the other modules, and so by Proposition III.6.1[1] the exact sequence (which resulted from the sequence of permutation modules after application of Hom(−,M) yields a long exact sequence [the Mayer-Vietoris sequence for cohomology with coeﬃcients]

26 Hn(G, M) / Hn(G , ResG M) ⊕ Hn(G , ResG M) / Hn(A, ResGM) 1 G1 2 G2 A

∂1 ε 7.1(a): Consider the exact sequence · · · → C1 → C0 → M → 0, where each Ci is H∗-acyclic. We can apply the dimension-shifting technique using the short exact sequences 0 → Kerε → C0 → M → 0 ∼ ∼ and 0 → Ker∂i → Ci → Ker∂i−1 → 0 to obtain the isomorphism Hn(G, M) = H1(G, Ker∂n−2) = Ker{H0(G, Ker∂n−1) → H0(G, Cn−1)} = Ker{(Ker∂n−1)G → (Cn−1)G}. Now consider the diagram below concerning CG ¯ ∂n+1 ∂¯ (C ) / (C ) n / (C ) n+1 G : n G 9 n−1 G γ1 α β γ2 % % % % (Zn)G (Zn−1)G with Zi = Ker∂i, noting that the composition βα is exact [right-exactness of (−)G on 0 → Ker∂n → ∼ ¯ Cn → Ker∂n−1 → 0] and Kerγ2 = Hn(G, M). Thus Im∂n+1 = Imα = Kerβ (since γ1 is surjective) and so ¯ Hn(CG) = Ker∂n/Kerβ. Noting that (Zn−1)G = (Cn)G/Kerβ by the First Isomorphism Theorem, take the kernel of γ2 [denoted K/Kerβ] and take its preimage under β to obtain K in (Cn)G. Since K maps ¯ ¯ ¯ to zero under γ2β = ∂n we have K ⊆ Ker∂n. We also have Ker∂n ⊆ K [hence they are equal] because ¯ ¯ by commutativity of the maps Ker∂n maps to K/Kerβ and hence lies in K. Thus Ker∂n/Kerβ = Kerγ2 ∼ and so H∗(G, M) = H∗(CG).

7.1(b): Consider the exact sequence 0 → M → C0 →δ0 C1 →δ1 · · · , where each Ci is H∗-acyclic. We can apply the dimension-shifting technique using the short exact sequences 0 → M → C0 → Kerδ0 → 0 n ∼ 1 ∼ and 0 → Kerδi−1 → Ci → Kerδi → 0 to obtain the isomorphism H (G, M) = H (G, Kerδn−2) = 0 0 G G Coker{H (G, Cn−1) → H (G, Kerδn−1)} = Coker{(Cn−1) → (Kerδn−1) }. Using a similar approach as in part(a) above, we see that H∗(G, M) =∼ H∗(CG).

7.2: This will reprove Proposition III.2.2[1] on isomorphic functors. Method 1 : Let M be Z-torsion-free, so that M ⊗ − is an exact functor (M is Z-flat). Then H∗(G, M ⊗ −) is a homological functor because given a short exact sequence of modules C, M ⊗ C is a short exact sequence and F ⊗G (M ⊗ C) is a short exact sequence of chain complexes (F is projective, hence flat), so the corresponding long exact homology sequence gives us the desired property (by G 0 Lemma 24.1[4] and Theorem 24.2[4]). Similarly, Tor∗ (M, −) = H∗(F ⊗G −) is a homological functor, where F 0 → M is a projective resolution. Both functors are effaceable [erasible] in positive dimensions, 0 since the chain complexes F ⊗G (M ⊗ P ) and F ⊗G P are exact for P projective. In dimension 0, ∼ ∼ ∼ G H0(G, M ⊗ N) = Z ⊗G (M ⊗ N) = (M ⊗ N)G = M ⊗G N = Tor0 (M,N). Therefore, by Theorem ∼ G III.7.3[1] we have an isomorphism of ∂-functors H∗(G, M ⊗ −) = Tor∗ (M, −). [The case for cohomology is similar]. G 0 Method 2 : The chain complex associated to the group Tor∗ (M,N) is given by · · · → F0 ⊗G N → 0 0 M⊗GN → 0, where F → M is a projective resolution. This can be rewritten as · · · → (F0⊗N)G → (M⊗ G 0 N)G → 0 which yields Tor∗ (M,N) = H∗(CG), where C is the chain complex (Fi ⊗N). This is indeed an 0 0 exact sequence because the universal coefficient theorem yields H∗(F ⊗N) = H∗(F )⊗N = N (nontrivial 0 0 L only in dimension 0). Now Fi is projective and hence a summand of a free module F = Fi ⊕K = j ZG. ∼ 0 ∼ L Then H∗(G, F⊗N) = H∗(G, Fi ⊗N)⊕H∗(G, K ⊗N) and H∗(G, F⊗N) = j H∗(G, ZG⊗N) = 0, noting 0 0 that induced modules are H∗-acyclic by Corollary III.6.6[1]. Thus H∗(G, Fi ⊗ N) = 0 and each Fi ⊗ N is G ∼ H∗-acyclic. We can now apply Exercise III.7.1(a) which implies Tor∗ (M,N) = H∗(CG) = H∗(G, M ⊗N). [The case for cohomology is similar].

7.3: For dimension-shifting in homology, we can choose the induced module M = ZG ⊗Z M which maps onto M by ϕ(r ⊗ m) = rm; it is an H∗-acyclic module by Corollary III.6.6[1]. This map is Z-split because it composes with the natural map i : M → M to give the identity, m 7→ 1 ⊗ m 7→ 1m = m.

For dimension-shifting in cohomology, we can choose the coinduced module M = HomZ(ZG, M) which provides the embedding M → M given by m 7→ (r 7→ rm); it is an H∗-acyclic module by Corollary III.6.6[1]. This map is Z-split because it composes with the natural map π : M → M to give the identity, m 7→ (r 7→ rm) 7→ [r 7→ rm](1) = 1m = m.

27 8.1: Let H be a central subgroup of G and let M be an abelian group with trivial G-action. Then the isomorphism c(g):(H,M) → (gHg−1,M) becomes the identity on (H,M) given by (h 7→ ghg−1 = h, m 7→ gm = m). By Corollary III.8.2[1], this conjugation action of G on (H,M) induces an action of −1 G/H on H∗(H,M) given by gH · z = c(g)∗z. Letting α : H → gHg = H denote the group map of c(g), and letting F be a projective resolution of Z over ZG, we can choose the chain map τ : F → F to be the identity, since it satisﬁes the condition τ(hx) = hx = α(h)x = α(h)τ(x). Thus the chain map F ⊗H M → F ⊗H M given by x ⊗ m 7→ x ⊗ gm = x ⊗ m is the identity, and so the induced map c(g)∗ is the identity on H∗(H,M) which gives the trivial G/H-action gH ·z = c(g)∗z = z. Similarly, by Corollary III.8.4[1], we have an induced action of G/H on H∗(H,M) given by gH ·z = (c(g)∗)−1z. Using the same chain map τ, we have the cochain map HomH (F,M) → HomH (F,M) given by f 7→ [x 7→ gf(x) = f(x)] which is the identity. Thus the induced map c(g)∗ is the identity on H∗(H,M) which gives the trivial G/H-action gH · z = (c(g)∗)−1z = z. Alternatively, the trivial G/H-action follows immediately from the fact that functors preserve identities, ∗ where H∗ and H are the functors in question and c(g) is the identity map in question.

G 8.2: Let α : H,→ G be an inclusion, let M be an H-module, let i : M → IndH M be the canon- G ical H-map i(m) = 1 ⊗ m, and let π : CoindH M → M be the canonical H-map π(f) = f(1). We can take the chain map τ : F → F to be the identity (F is a free resolution of Z over ZG) since τ(hx) = hx = α(h)x = α(h)τ(x) and F can be regarded as a free resolution over ZH. G Consider (α, i)∗ on the chain level, induced by the map F ⊗H M → F ⊗G IndH M given by x ⊗ m 7→ x ⊗ (1 ⊗ m). This map is a homotopy equivalence because we can use the universal property of tensor products to deﬁne its inverse x ⊗ (g ⊗ m) = xg ⊗ (1 ⊗ m) 7→ xg ⊗ m, the composite map being x ⊗ (g ⊗ m) 7→ xg ⊗ m 7→ xg ⊗ (1 ⊗ m) = x ⊗ g · (1 ⊗ m) = x ⊗ (g ⊗ m). In particular we have a ∼ G weak equivalence which yields the isomorphism H∗(H,M) = H∗(G, IndH M) of Shapiro’s Lemma given by (α, i)∗. ∗ G Now consider (α, π) on the cochain level, induced by the map HomH (F,M) → HomG(F, CoindH M) given by [x 7→ f(x)] 7→ [x 7→ (g 7→ gf(x))] with g ∈ ZG. This map is a homotopy equivalence because we can use the universal property of co-induction to deﬁne its inverse [x 7→ (g 7→ gf(x))] 7→ [x 7→ (1 7→ f(x))] = [x 7→ f(x)]. In particular we have a weak equivalence which yields the isomorphism ∗ ∼ ∗ G ∗ H (H,M) = H (G, CoindH M) of Shapiro’s Lemma given by (α, π) .

G 9.1: Considering homology, let x ⊗H m represent z ∈ H∗(H,M). Computing corH z on the chain G level yields x ⊗H m 7→ x ⊗G m, while computing corgHg−1 gz on the chain level yields gx ⊗gHg−1 gm 7→ G G gx ⊗G gm = x ⊗G m. Since the images are equal, corgHg−1 gz = corH z. G Considering homology, let x ⊗G m represent z ∈ H∗(G, M). Computing resgHg−1 z on the chain level P 0 G yields x ⊗G m 7→ g0∈gHg−1\G g · (x ⊗gHg−1 m). Computing g · resH z on the chain level yields P 0 P 0 P 00 x⊗G m 7→ g0∈H\G g ·(x⊗H m) 7→ g0∈H\G(gg )·(x⊗gHg−1 m) = g00∈gHg−1\G g ·(x⊗gHg−1 m), where this last equality arises from g(Hg0) = gHg−1gg0 = (gHg−1)g00 with g00 = gg0 being the coset representa- G G tive. Since the images are equal (the sums are the same for all coset representatives), g·resH z = resgHg−1 z. ∗ G Considering cohomology, let fG represent z ∈ H (G, M). Computing g · resH z on the chain level yields −1 −1 −1 −1 [x 7→ fG(x)] = [x 7→ g fG(gx)] 7→ [x 7→ g fH (gx)] 7→ [x 7→ gg fgHg−1 (g gx)] = [x 7→ fgHg−1 (x)], G while computing resgHg−1 z on the chain level yields [x 7→ fG(x)] 7→ [x 7→ fgHg−1 (x)]. Since the images G G are equal, g · resH z = resgHg−1 z. ∗ G Considering cohomology, let f represent z ∈ H (H,M). Computing corH z on the chain level yields P 0 G f 7→ g0∈G/H g · f, where G acts diagonally on f ∈ Hom(F,M). Computing corgHg−1 gz on the P 0 P 00 chain level yields g · f 7→ g0∈G/gHg−1 (g g) · f = g00∈G/H g · f, where this last equality arises from g0(gHg−1)g = g0gH = g00H with g00 = g0g being the coset representative. Since the images are equal G G (the sums are the same for all coset representatives), corgHg−1 gz = corH z.

9.2: The transfer map H1(G) → H1(H) can be regarded as a map of abelian groups Gab → Hab. Ifg ¯ denotes the representative of Hg then ρg = gg¯−1, where ρ : G → H is the unique map of left H-sets which sends ever coset representative to 1. Now the transfer map is induced by the composite chain map tr F (G)G → F (G)H → F (H)H , where the latter map concerns the chain map τ : F (G) → F (H) given P 0 by (g0, g1) 7→ (ρg0, ρg1). Using bar notation, this composite chain map is given by [g] 7→ g0∈E g [g] 7→

28 P 0 0 P 0 0−1 0 P 0 P 0 g0∈E τ(g , g g) = g0∈E(g g , ρ(g g)) = g0∈E(1, ρ(g g)) = g0∈E[ρ(g g)], where E is a set of representatives for the right cosets Hg0 and we note then that g0 = g0. Note that for homology classes, [g1]+[g2] = [g1g2] because of the boundary map ∂2[g1|g2] = [g2]−[g1g2]+[g1]; thus (as a homology class) P 0 Q 0 g0∈E[ρ(g g)] = [ g0∈E ρ(g g)]. Using the isomorphism H1(G) → Gab given by [g] 7→ g mod [G, G], the Q 0 0 −1 transfer map Gab → Hab is computed as g mod [G, G] 7→ g0∈E g g(g g) mod [H,H]. Example: The transfer map Z → nZ is multiplication by n, since x 7→ [1x(1x)−1][xx(xx)−1] ··· [xn−2x(xn−2x)−1][xn−1x(xn−1x)−1] = 1 · 1 ··· 1 · xn(1)−1 = xn, where x is the generator of Z.

10.1: The symmetric group G = S3 on three letters is the group of order 3! = 6 whose elements are the permutations of the set {1, 2, 3}. The Sylow 3-subgroup is generated by the cycle (1 2 3), and a Sylow 2-subgroup is generated by the cycle (1 2). Noting the semi-direct product representation S3 = Z3 o Z2 ∗ ∗ ∗ ∼ ∗ ∗ 2 where Z2 acts on Z3 by conjugation, we have H (S3) = H (S3)(2) ⊕ H (S3)(3) = H (S3)(2) ⊕ H (Z3)Z ∼ by Theorem III.10.3[1]. Now S3 is the unique nonabelian group of order 6, so D6 = S3 and we can use ∼ 2i i Exercise AE.9 which implies that the Z2-action on H2i−1(Z3) = H (Z3) is multiplication by (−1) (we n 2 can pass this action to cohomology by naturality of the UCT). Thus H (Z3)Z is isomorphic to Z3 for n = 2i where i is even, and is trivial for n odd and n = 2i where i is odd. Taking any Sylow 2-subgroup ∼ ∗ H = Z2, Theorem III.10.3[1] states that H (S3)(2) is isomorphic to the set of S3-invariant elements of ∗ 2i−1 2i−1 2i−1 H (H). In particular we have the monomorphism H (S3)(2) ,→ H (H) = 0, so H (S3)(2) = 0. −1 2i ∼ H gHg An S3-invariant element z ∈ H (H) = Z2 must satisfy the equation resK z = resK gz, where K denotes H ∩ gHg−1. If g ∈ H then gHg−1 = H and the above condition is trivially satisfied for all z (hz = z by Proposition III.8.1[1]). If g∈ / H then K = {1} because H is not normal in S3 and only contains two elements, so the intersection must only contain the trivial element. But then the image of 2i both restriction maps is zero, so the condition is satisfied for all z; thus H (S3)(2) = Z2. Alternatively, a theorem of Richard Swan states that if G is a finite group such that Sylp(G) is abelian and M is a G ∗ NG(Sylp(G)) trivial G-module, then Im(res ) = H (Syl (G),M) . It is a fact that NS3 ( 2) = 2 (refer Sylp(G) p Z Z ∼ S3 Z2 to pg51[2]), so taking G = S3 and H = Syl2(S3) = Z2 and M = Z we have Im(resH ) = (Z2) = Z2 in the even-dimensional case. Since any invariant is in the image of the above restriction map (by Theorem 2i III.10.3[1]), the result H (S3)(2) = Z2 follows.  Z n = 0   n ∼ Z6 n ≡ 0 mod 4 , n 6= 0 ⇒ H (S3) = Z2 n ≡ 2 mod 4  0 otherwise

10.2(a): Let H be a subgroup of G of ﬁnite index, let C be the double coset HgH, and let T (C) be the en- ∗ ∗ G domorphism H (G, f(C)) of H (H,M) where f(C) is the G-endomorphism of IndH M given by 1⊗m 7→ −1 P −1 H gHg c∈C/H c ⊗ cm. To show that T (C)z = corH∩gHg−1 resH∩gHg−1 gz, it suﬃces to check this equation H P in dimension 0 (by Theorem III.7.5[1]). The right side maps m ∈ M to h∈H/H∩gHg−1 h(gm) = P P 0 H 0 hg∈HgH/H (hg)m = g0∈C/H g m ∈ M , where g = hg as a coset representative. Now T (C) in dimension 0 is given by the composite map

∗ −1 −1 0 α 0 G β 0 G f 0 G β 0 G α H (H,M) → H (G, CoindH M) → H (G, IndH M) → H (G, IndH M) → H (G, CoindH M) → H0(H,M) P where α is the Shapiro isomorphism m 7→ (s 7→ m), and β is the canonical isomorphism F 7→ x∈G/H x⊗ F (x−1), and f ∗ is induced by f(C), and α−1 is the inverse F 7→ F (1), and β−1 is the inverse x ⊗ m 7→ β−1[x ⊗ m](s · x) which is sxm if sx ∈ H and is 0 if sx∈ / H. This composite is given by

P P P −1 P P −1 T (C): m 7→ (s 7→ m) 7→ x∈G/H x ⊗ m 7→ x c∈C/H x(c ⊗ cm) = x c xc ⊗ cm 7→ P P −1 −1 −1 P P −1 −1 −1 H x c β [xc ⊗ cm](s · xc ) 7→ x c β [xc ⊗ cm](xc ) ∈ M To simplify this last term, note that the image of β−1[xc−1 ⊗ cm] is nontrivial iﬀ xc−1 ∈ H ⇔ x ∈ C, and for each x there is at most one c such that xc−1 ∈ H. Thus the double sum reduces to

29 P −1 P x∈G/H , x∈C xc · cm = x∈C/H xm, and this is precisely the image of the right-side map. G G Note: α was determined by noting that any element f of (CoindH M) must satisfy f(xg) = f(x) and hence f is determined by f(1) = m. So f is given by g 7→ m, but it must also commute with the H-action which means that hg 7→ hm and hence hm = m, i.e. m ∈ M H . Thus α(m) = (s 7→ m), s ∈ G.

10.2(b): If z ∈ H∗(H,M) is G-invariant where H ⊆ G is of ﬁnite index as above, then T (C)z = −1 H gHg H H −1 corH∩gHg−1 resH∩gHg−1 gz = corH∩gHg−1 resH∩gHg−1 z = |H : H ∩ gHg | = a(C)z, where the second-to- last equality follows from Proposition III.9.5[1].

10.2(c): Let X = {z ∈ H∗(H,M) | T (C)z = a(C)z ∀ C} where C is any H-H double coset −1 G and a(C) = |C/H| = |H : H ∩ gHg |. Since the image of the restriction map resH lies in the set of G-invariant elements of H∗(H,M), and such elements lie in X by part(b) above, we have G Im(resH ) ⊆ X. In the situation of Theorem III.10.3[1] and Proposition III.10.4[1], consider the ele- G n n ment w = corH z ∈ H (G, M) where z is an arbitrary element of X. Then either H (H,M) is anni- n hilated by |H| [H = Sylp(G)] in which case w ∈ H (G, M)(p), or |G : H| is invertible in M [hence −1 n G P H gHg in H (G, M)]. Using Proposition III.9.5[1] we obtain resH w = g∈H\G/H corH∩gHg−1 resH∩gHg−1 gz = P P P −1 g∈H\G/H T (C)z = g∈H\G/H a(C)z = ( g∈H\G/H |H : H ∩ gHg |)z = |G : H|z. Since either G 0 0 |G : H| is prime to p or is invertible in M, it follows that z = resH w where w = w/|G : H|. Thus G G X ⊆ Im(resH ) ⇒ X = Im(resH ).

30 4 Chapter IV: Low-Dimensional Cohomology and Group Extensions

2.1: If d is a derivation [crossed homomorphism], then d(1) = d(1 · 1) = d(1) + 1 · d(1) = 2 · d(1) and so d(1) = 0.

2.2: Let I be the augmentation ideal of ZG and let D : G → I be the derivation deﬁned by g 7→ g−1 (this is the principal derivation G → ZG corresponding to 1 ∈ ZG). Given any G-module A and any derivation d : G → A, we can extend d to an additive map d¯: ZG → A such that d¯(rs) = dr¯ · ε(s) + r · ds¯ , where ε is the augmentation map and r, s ∈ ZG. This map is well-deﬁned because d¯(rs · t) = d¯(rs) · ε(t) + rs · dt¯ = dr¯ · ε(s) · ε(t) + r · ds¯ · ε(t) + rs · dt¯ = dr¯ · ε(st) + r · (ds¯ · ε(t) + s · dt¯ ) = dr¯ · ε(st) + r · d¯(st) = d¯(r · st), and dg¯ = d¯(g1) = dg¯ · ε(1) + g · d¯1 = dg · 1 + g · d1 = dg + g · 0 = dg. The restriction f of d¯ to I is a G-module homomorphism since f(r · s) = f(r) · ε(s) + r · f(s) = f(r) · 0 + r · f(s) = r · f(s), and f(g − 1) = dg¯ − d¯1 = dg − d1 = dg − 0 = dg, so f is the unique module map I → A such that d = fD. ∼ This means D is the universal derivation on G, and Der(G, A) = HomZG(I,A).

2.3(a): Let F = F (S) be the free group generated by the set S, and consider the F -module A with a family of elements (as)s∈S. For the set map S → A o F deﬁned by s 7→ (as, s), there is a unique extension to a homomorphism ϕ : F → A F by the universal mapping property of F . This is a splitting π o of 1 → A → A o F → F → 1 because πϕ(f) = π(df, f) = f, where d : F → A is some function which maps s ∈ S to as ∈ A. Since derivations F → A correspond to splittings of the above group extension, d is the unique derivation such that ds = as ∀ s ∈ S.

2.3(b): Given any function φ : S → A where A is a ZF -module, there is a unique map d : F → A by part(b) above, hence a unique ZF -module homomorphism f : I → A such that fD = d by Exercise IV.2.2 above. In particular, I satisﬁes the universal property of free modules, φ = fD|S, and so the augmentation ideal I of ZF is a free ZF -module with basis (Ds)s∈S = (s − 1)s∈S. Note that this reproves Exercise II.5.3(b).

P 2.3(c): By part(b) above the universal derivation D : F → I satisﬁes Df = s∈S(∂f/∂s)ds, where I is the augmentation ideal of ZF . By Exercise IV.2.2 any derivation d : F → M (where M is an F -module) corresponds to a unique F -module map ϕ : I → M and hence satisﬁes df = ϕ(Df) = P P s∈S(∂f/∂s)ϕ(Ds) = s∈S(∂f/∂s)ds, where ∂f/∂s lies in ZF . Note that this reproves Exercise II.5.3(c).

2.4(a): Let G = F/R where F = F (S) and R is the normal closure of some subset T ⊆ F . For any G-module A, derivations d : G → A correspond to splittings of 1 → A → A o G → G → 1; they are of the form s(g) 7→ (dg, g) ∈ A o G. Consider the homomorphism ϕ : F → A o G given by f 7→ (df, g) where g is the image of f under the projection map p : F → F/R, which is the extension of the set map S → A o G, s 7→ (ds, p(s)), by the universal mapping property of F . Now r = ftf −1 ∈ R is mapped to ϕ(r) = df + f · (dt + t · df −1) = df + f · dt − ftf −1 · df = (1 − r) · df + f · dt = dt, so ϕ induces a homomorphism G → A o G iﬀ dt = 0 ∀ t ∈ T [note: 1 − r = 1 − 1 = 0 when computing the G-action on df ∈ A]. This homomorphism is a splitting iﬀ d is a derivation, and so derivations G → A correspond to derivations d : F → A such that d(T ) = 0.

2.4(b): From part(a) above and Exercise IV.2.3(c) we see that derivations G → A correspond to P derivations d : F → A such that dt = s∈S(∂t/∂s)ds = 0 for all t ∈ T , where ∂t/∂s is the image of ∂t/∂s under ZF → ZG due to the G-action on A (restriction of scalars). By Exercise IV.2.3(a) these P correspond to families (as)s∈S of elements of A such that s∈S(∂t/∂s)as = 0 for all t ∈ T .

2.4(c): The identity map idI : I → I [where I is the augmentation ideal of ZG] corresponds to a derivation d : G → I such that ds¯ = idI D(¯s) = idI (¯s − 1) =s ¯ − 1 by Exercise IV.2.2, and this cor- P responds to a family (¯s − 1)s∈S of elements of I such that s∈S(∂t/∂s)(¯s − 1) = 0 for all t ∈ T by (T ) ∂2 (S) ∂1 P part(b) above. Thus there is an exact sequence ZG → ZG → I where ∂2et = s∈S(∂t/∂s)es and ∂1es =s ¯ − 1 (et and es are basis elements of their respective groups). Now ∂1 is surjective because

31 the elementss ¯ − 1 generate I as a left ZG-ideal by Exercise I.2.1(b), where S¯ is a set of generators for ∂2 ∂1 G = F (S)/R, and so we have the desired exact sequence ZG(T ) → ZG(S) → I → 0. Note that this reproves the second part of Exercise II.5.3(d).

2.4(d): Since the augmentation ideal I of ZF is free (by Exercise IV.2.3(b)), we have a free resolution ε ∼ 0 → I → ZF → Z → 0 of Z over ZF . Taking R-coinvariants and noting that (ZF )R = Z[F/R] = ZG by ∼ Exercise II.2.1, we obtain a complex IR → ZG → Z → 0 whose homology is H∗R because H1(R, Z) = Ker(IR → ZG) by the dimension-shifting technique; ZF is an H∗-acyclic module by Proposition III.6.1[1] since it is free as a ZR-module by Exercise I.8.2, and it maps onto Z with kernel I. Since I is free with ∼ (S) ϕ (S) ∂1 basis (s − 1)s∈S, I = ZF and we have the exact sequence 0 → H1R → ZG → ZG → Z → 0, where we note that taking coinvariants is a right-exact functor. Now we can map the standard (bar) resolution of Z over ZR to the aforementioned free resolution: ∂2 ∂1 ∂0 ··· / F2 / F1 / F0 / Z / 0

φ2 φ1 φ0 id i ε 0 / I / ZF / Z / 0 where φn>1 = 0, φ1[r] = r − 1 = Dr, and φ0[ ] = 1. This is a commutative diagram because φ0∂1[r] = φ0(r[] − 1) = r − 1 = i(r − 1) = iφ1[r] and φ1∂2[r1|r2] = φ1(r1[r2] − [r1r2] + [r1]) = r1(r2 − 1) − (r1r2 − 1) + r1 − 1 = r1r2 − r1 − r1r2 + 1 + r1 − 1 = 0. By applying the coinvariants functor ∼ and noting that φ1|R = D|R and H1R = Ker(∂1)R/Im(∂2)R = Rab, we have a commutative diagram D| R R / I

ϕ (S) H1R / ZG D where the vertical arrows are quotient maps. The composite F → I → ZG(S) is a derivation such that

s 7→ es because s1s2 7→ (s1 − 1) + s1 · (s2 − 1) 7→ es1 +s ¯1es2 . Thus the map ϕ is given by r mod [R,R] 7→ P θ (S) ∂ ε s∈S(∂r/∂s)es, and we have the desired exact sequence 0 → Rab → ZG → ZG → Z → 0 where P ∂es =s ¯ − 1 and θ(r mod [R,R]) = s∈S(∂r/∂s)es. Note that this reproves the ﬁrst part of Exercise II.5.3(d).

3.1(a): Let 0 → A → E →π G → 1 be an extension and let α : G0 → G be a group homomorphism, and 0 0 0 0 consider the pull-back (fiber-product) E ×G G = {(e, g ) ∈ E × G | π(e) = α(g )}. The kernel of the 0 0 0 0 ∼ canonical projection p : E ×G G → G corresponds to g = 0 ⇒ α(g ) = 0 = π(e) ⇒ Kerπ = A, and 0 p 0 thus we have an extension 0 → A → E ×G G → G → 1 which by definition fits into the commutative diagram / / π / / 0 A EO GO 1 α

0 p 0 0 / A / E ×G G / G / 1 This extension is classified up to equivalence (by fitting into the above commutative diagram) because given another such extension [corresponding to E0] of G0 by A, commutativity of the right-hand square 0 0 implies there is a unique map φ : E → E ×G G by the universal property of the pull-back, and this gives commutativity of the right-half of the diagram below: / i1 / π / / 0 A EO GO X 1 ϕ α i p 0 / A 2 / E × G0 / G0 / 1 GO φ i 0 / A 3 / E0 / G0 / 1 Note that α for the E0-extension yields the identity map G0 → G0. It suffices to show that the left-side

32 of the diagram also commutes, for then we can apply the Five-Lemma which states φ is an isomorphism 0 ∼ 0 0 (E = E ×G G ). Now φi3(a) = i2(b) for some b ∈ A because i3(a) maps to 0 ∈ G by exactness of the 0 bottom row and hence lies in the kernel of E ×G G which is contained in i2(A). Then ϕφi3(a) = ϕi2(b), and ϕφi3(a) = i1(a) by commutativity of the outer left-hand square while ϕi2(b) = i1(b) by commutativity of the top left-hand square. Thus i1(a) = i1(b) ⇒ a = b by injectivity of the inclusion, and this yields φi3(a) = i2(a) which gives commutativity of the bottom left-hand square and completes the proof. Therefore, α induces a map E(G, A) → E(G0,A), and this corresponds to H2(α, A): H2(G, A) → H2(G0,A) under the bijection of Theorem IV.3.12[1].

p 3.1(b): Let 0 → A →i E → G → 1 be an extension and let f : A → A0 be a G-module homomorphism, and consider the largest quotient E0 of A0 o E such that the left-hand square in the following diagram commutes: p 0 / A i / E / G / 1

f φ 0 p0 0 / A0 i / E0 / G / 1 0 0 0 0 0 0 Explicitly, E = A o E/∼ with the equivalence relation (a1, e1) ∼ (a2, e2) iff a1 + f(a1) = a2 + f(a2) and e1 − i(a1) = e2 − i(a2) for some a1, a2 ∈ A; this relation is obviously reflexive and symmetric. It is tran- 0 0 0 0 0 0 sitive because if a1 + f(a1) = a2 + f(a2) and a2 + f(c) = a3 + f(a3), then a1 + f(a1) = a3 + f(a3 + a2 − c) 0 0 0 0 0 and e1 − i(a1) = [e3 − i(a3) + i(c)] − i(a2) = e3 − i(a3 + a2 − c). Define i by i (a ) = (a , 0) and define p 0 0 0 0 0 by p (a , e) = p(e) and define φ by φ(e) = (0, e). The map p is well-defined because for (a1, e1) ∼ (a2, e2) 0 0 0 0 0 0 we have p (a2, e2) = p(e2) = p(e1) + p[i(a2 − a1)] = p(e1) + 0 = p(e1) = p (a1, e1). Now p (a , e) = 0 ⇒ p(e) = 0 ⇒ ∃ a | i(a) = e ⇒ (a0, e) = (a0, i(a)) ∼ (a0 + f(a), 0) = i0(a0 + f(a)) ⇒ Kerp0 ⊆ Imi0, and p0[i0(a0)] = p0(a0, 0) = p(0) = 0 ⇒ Imi0 ⊆ Kerp0, so Kerp0 = Imi0 and the bottom row is an E0-extension. This extension is classified up to equivalence (by fitting into the above commutative diagram) because given another such extension [corresponding to E00] of G by A0, we get a diagram p 0 / A i / E / G / 1

f φ 0 p0 0 / A0 i / E0 / G / 1

ϕ 00 p00 0 / A0 i / E00 / G / 1 where E → E00 is the map Φ and the identity map A0 → A0 is induced from the f for the E00-extension. We obtain an induced map ϕ : E0 → E00 given by ϕ(a0, e) = i00(a0) + Φ(e) which is well-deﬁned be- 0 0 0 00 0 00 00 cause if (a1, e1) ∼ (a2, e2) then ϕ(a2, e2) = i (a1) + i f(a1) − i f(a2) + Φ(e1) − Φi(a1) + Φi(a2) = 00 0 00 00 00 00 0 0 [i (a1) + Φ(e1)] + i f(a1) − i f(a2) − i f(a1) + i f(a2) = ϕ(a2, e2) + 0 + 0 = ϕ(a1, e1), noting that Φi = i00f by commutativity of the outer left-hand square. The bottom right-hand square is commutative because p00ϕ(a0, e) = p00i00(a0) + p00Φ(e) = 0 + p(e) = p(e) = p0(a0, e). The bottom left-hand square [hence the whole diagram] is also commutative because ϕi0(a0) = ϕ(a0, 0) = i00(a0) + Φ(0) = i00(a0) + 0 = i00(a0). We can now apply the Five-Lemma which states ϕ is an isomorphism (E00 =∼ E0). Therefore, f induces a map E(G, A) → E(G, A0), and this corresponds to H2(G, f): H2(G, A) → H2(G, A0) under the bijection of Theorem IV.3.12[1].

p 3.2(a): Let 0 → A0 →i A → A00 → 0 be a short exact sequence of G-modules and let d : G → A00 be a derivation, and consider the set-theoretic pull-back E = {(a, g) ∈ A o G | p(a) = d(g)} where we note that A o G = A × G as sets. If d¯ : E → A and π : E → G are the canonical projections, then dπ[(a1, g1)(a2, g2)] = d(g1g2) = d(g1) + g1 · d(g2) = p(a1) + g1 · p(a2) and the group law on E can be that ¯ of the semi-direct product due to the agreement pd(a1 +g1 ·a2, g1g2) = p(a1 +g1 ·a2) = p(a1)+g1 ·p(a2) = ¯ dπ(a1 + g1 · a2, g1g2). Thus E can be regarded as a subgroup of A o G, and d is a derivation because ¯ ¯ ¯ ¯ d[(a1, g1)(a2, g2)] = d(a1 + g1 · a2, g1g2) = a1 + g1 · a2 = d(a1, g1) + g1 · d(a2, g2). Mimicking the proof of Exercise IV.3.1(a) verbatim, for each derivation d there is an extension 0 → A0 → E → G → 1 characterized by the fact that it ﬁts into a commutative diagram with derivation d¯

33 / 0 i / p / 00 / 0 A AO AO 0

d¯ d j 0 / A0 / E π / G / 1 This construction gives a map Der(G, A00) → E(G, A0).

3.2(b): What follows will be set-theoretic, and we use the same notation/maps as in part(a). A lifting of d : G → A00 to a function l : G → A is given by l(g) = p−1d(g), where p−1(x) = 0 if x∈ / Imp. This yields a cross-section s : G → E of π given by s(g) = d¯−1l(g), because πs = πd¯−1l = πd¯−1p−1d = π(pd¯)−1d = −1 −1 −1 π(dπ) d = ππ d d = idG. Consider the diagram Der(G, A00) / E(G, A0)

∼= H1(G, A00) δ / H2(G, A0) where the top-horizontal map is the constructed map of part(a) and the left-vertical map is the natural quotient; the isomorphism (right-vertical map) sends an extension to a factor set f : G × G → A0. The 1 00 00 connecting homomorphism δ([d]) = [w2] is defined by lifting the element d of Z (G, A ) = Der(G, A ) 1 [equality comes from Exercise III.1.2] to an element w1 in C (G, A) and then taking its image under ∗ 2 0 the coboundary map ∂ w2 and noting that it must be the image of some w2 ∈ Z (G, A ) under the ∗ 2 0 2 injection i : C (G, A ) → C (G, A); all of this is well-defined. From above, w1 in this scenario is l and so ∂∗l = L : G × G → A is defined by L(g, h) = gl(h) − l(gh) + l(g) via the equation on pg59[1]. ∗ 0 Then we have i (w2) = L, and w2 : G × G → A is a factor set which corresponds to a section S via S(g)S(h) = j(w2(g, h))S(gh) by the normalization condition on pg91[1]. I claim that S is the section s defined above and that w2 is the factor set corresponding to s. Assuming this claim holds for the moment, the connecting homomorphism sends the [class of the] derivation d to the factor set w2 which is precisely the image of the extension under the vertical-isomorphism, and that extension is the image of d under the top-horizontal map, so commutativity of the diagram is satisfied. Therefore, it suffices to prove the claim, i.e. we must check the validity of the equation s(g)s(h) = j(w2(g, h))s(gh). Note that ¯−1 ¯−1 j ◦ w2 = d L by commutativity of the left-side of the diagram in part(a), and s = d l, so the equation reduces to l(g) + [d¯−1l(g)] · l(h) = L(g, h) + [d¯−1L(g, h)] · l(gh) after we apply the derivation d¯ to both sides. The action of E on A is given by e · a = π(e)a, so [d¯−1l(g)] · l(h) = π[s(g)]l(h) = gl(h). Thus the equation becomes l(g)+gl(h) = [gl(h)−l(gh)+l(g)]+[d¯−1L(g, h)]·l(gh) ⇒ l(gh) = [d¯−1L(g, h)]·l(gh). It now suffices to show that the element π[d¯−1L(g, h)] fixes l(gh), i.e. d¯−1L(g, h) ∈ A0 ⊂ E. But this is ¯−1 0 0 0 immediate because d L(g, h) = j(w2(g, h)) ∈ A where A has lazily been identified with j(A ).

3.3: Let G be a ﬁnite group which acts trivially on Z. For any homomorphism G → Q/Z we can construct a central extension of G by Z by pulling back the canonical extension (the top row) 0 / / / / / 0 Z QO Q O Z

0 / Z / E / G / 1 Thus we have a map ϕ : Hom(G, Q/Z) → E(G, Z), which by Exercise IV.3.2 can be identified with the boundary map δ : H1(G, Q/Z) → H2(G, Z) since H1(G, Q/Z) =∼ Hom(G, Q/Z) by Exercise III.1.2. Now H1G = Gab is finite, and so is H2G by Exercise AE.16 [G is finite and Z is finitely generated]; thus Hom(H2G, Q) = 0 = Hom(H1G, Q) because there are no nontrivial homomorphisms from a finite group into the infinite group . Also, is -injective and hence Ext1 (H G, ) = 0 = Ext1 (H G, ) Q Q Z Z 0 Q Z 1 Q by Proposition 17.1.9[2]. The universal coefficient sequence of Exercise III.1.3 now yields H1(G, Q) = 0 = H2(G, Q); alternatively, we could have simply noted that Hn(G, Q) = 0 ∀ n > 0 by Corollary III.10.2[1] since |G| is invertible in Q. By the long exact cohomology sequence (Proposition III.6.1[1]), δ is an isomorphism. Thus ϕ is a bijection.

3.4(a): Let E be a group which contains a central subgroup C ⊆ Z(E) of ﬁnite index n. Method 1 : We have a central extension 1 → C → E → G → 1 by taking G = E/C. The nth power map C → C (where C is considered as an abelian group) induces multiplication by n on H2(G, C)

34 because the cochain f : G2 → C is sent to nf : G2 → nC ⊆ C, and this is the zero-map because n = |G| annihilates H2(G, C) by Corollary III.10.2[1]. Now Exercise IV.3.1(b) implies the above central extension ﬁts uniquely into a commutative diagram with the split extension [the trivial element in E(G, C)] 1 / C / E / G / 1

n φ 1 / C / C × G / G / 1 where we note that C o G =∼ C × G because C is central. Looking at the left-hand square with E restricted to C, commutativity implies φ(c) = (cn, 1), so the C-component of E → C × G gives us the desired homomorphism E → C whose restriction to C is the nth power map. Method 2 : The abelianization map ρ : E → Eab composed with the transfer map tr : Eab → Cab = C Q −1 is given by ϕ : e 7→ e mod [E,E] 7→ g∈C\E ge(ge) . Now |C\E| = n and c ∈ C commutes with all Q −1 n Q −1 n Q −1 n elements of E, so ϕ(c) = g∈C\E gc(gc) = c g(cg) = c gg = c . Thus ϕ = tr ◦ ρ : E → C is the desired homomorphism whose restriction to C is the nth power map.

3.4(b): Given a finitely generated group E, suppose the commutator subgroup [E,E] of E is finite. Then there are finitely many nontrivial elements g−1(ege−1) where g is a generator of E (and e ∈ E is arbitrary), so there are only finitely many conjugates ege−1 of each generator g of E. Consider the inner automorphism group Inn(E) =∼ E/C, where C = Z(E) is the center of E. An arbitrary element is −1 a function fe(x) = exe [with e ∈ E fixed] which is determined by where it sends the generators of E. Since there are finitely many generators and finitely many conjugates of each generator, there are only finitely many non-identity maps fe ∈ Inn(G). Thus |Inn(G)| is finite, and so the center C of E has finite index. Conversely, suppose the center C of the finitely generated group E has finite index. Then part(a) above n gives us a homomorphism ϕ : E → C such that ϕ|C is the nth power map C → C. It suffices to show that Kerϕ is finite, for then E/Kerϕ is abelian (being isomorphic to a subgroup of the abelian group C by the First Isomorphism Theorem) and hence [E,E] ⊆ Kerϕ by Proposition 5.4.7[2], so the commutator subgroup [E,E] of E is finite. Now ϕ is the C-component of the map φ : E → C × G given in part(a), and the kernel of φ is contained in the kernel of the projection π : E → G by commutativity of that diagram in part(a), so Kerϕ ⊆ Kerφ ⊆ Kerπ = C and hence Kerϕ ⊆ Ker(C →n C). It suffices to show that C ⊆ E is a finitely generated abelian group, for then Ker(C →n C) is a subgroup of finite exponent in a finitely generated abelian group (hence finite by the Fundamental Theorem of Finitely Generated Abelian Groups). The Nielsen-Schreier Theorem (Theorem 85.1[6]) states that every subgroup of a free group is free. The Schreier Index Formula (Theorem 85.3[6]) states that for a free group F of finite

rank with a subgroup H of finite index, rkZH = |F : H|(rkZF − 1) + 1. The finitely generated group E has the presentation F/R with F free of finite rank, and C ⊆ F corresponds bijectively to some H ⊆ F with H/R = C by the Lattice [4th] Isomorphism Theorem; H is free and finitely generated by the Nielsen-Schreier Theorem and the Schreier index formula. Since quotients of finitely generated groups are finitely generated (the generators of the quotient are the images of the generators under the projection), C is finitely generated (and abelian since it lies in the center of E).

3.5(a): Let E be a group which contains an infinite cyclic central subgroup of finite index. Method 1 : The group E of an extension in E(G, Z) gives us a homomorphism ϕ : G → Q/Z by Exercise IV.3.3 (where G = E/Z), along with a map φ : E → Q which is injective on Z ⊆ E. Now the nontrivial m m ∼ finitely generated subgroups of Q are of the form n Z with m, n ∈ N, so Imφ = n Z = Z [note: E is finitely generated because both Z and E/Z are]. Thus we have a surjective map φ˜ : E → Imφ =∼ Z which is simply φ rephrased. Its kernel F := Kerφ˜ = Kerφ is finite because |F ∩Z| = 1 and |F Z/Z| = |F : F ∩Z| = |F | by the 2nd Isomorphism Theorem, where we note that F Z/Z ⊆ G is finite [F Z is a subgroup of E because φ˜ F/E]. We then have an extension 0 → Kerφ˜ → E → Z → 0 which must split (by Exercise AE.27) because Z is free. Thus E =∼ F o Z with F finite. Method 2 : By Exercise IV.3.4(a) we have a map φ : E → Z which has a finite kernel as proved in Ex- ercise IV.3.4(b). Letting φ˜ : E → φ(E) be the surjective map induced from φ, we have φ(E) = mZ =∼ Z for some m ∈ N and so φ˜ : E → Z is a surjective map with finite kernel. We then have an extension 0 → Kerφ˜ → E → Z → 0 which must split (by Exercise AE.27) because Z is free. Thus E =∼ F o Z with

35 F ﬁnite (taking F = Kerφ˜ = Kerφ).

3.5(b): Let E be a torsion-free group which has an infinite cyclic subgroup Z of finite index. Let E act by left multiplication on the finite set T of left cosets of Z in E and let πZ : E → S|T | be the associated permutation representation afforded by this action. By Theorem 4.2.3[2], the kernel of the action T −1 is the core subgroup KerπZ = e∈E eZe ≡ A and so A ⊆ Z is a normal subgroup of E of finite index (A is necessarily nontrivial and hence infinite cyclic). The group action of E on A =∼ Z by conjugation is a homomorphism E → Aut(Z) = {±1}, and the kernel is a subgroup E0 ⊆ E of index 1 or 2 such that A ⊆ Z(E0). By part(a) we have E0 =∼ F o A with F finite, but since E is torsion-free, F = {1} and hence E0 =∼ Z. We therefore have an extension 0 → Z → E → G → 1 with |G| ≤ 2 (take G = E/E0). If 2 2 G = {1} we are done. If G = Z2 acts non-trivially on Z then H (G = Z2, Z) = ZZ /NZ = 0/NZ = 0, where N ∈ ZG is the norm element. In view of Theorem IV.3.12[1] we see that the extension splits (so E =∼ Z o G), contradicting the assumption that E is torsion-free. Hence G acts trivially, so Z is central in E and E =∼ Z by part(a).

3.6: Let E be a finitely generated torsion-free group which contains an abelian subgroup of finite index. This subgroup is isomorphic to Zn for some n by the Fundamental Theorem of Finitely Gen- erated Abelian Groups. Note that Zn is a polycyclic group because it is solvable (it has the series 1 / Z / Zn with Zn/Z = Zn−1 abelian) and every subgroup is finitely generated; an equivalent definition of a polycyclic group is that it has a subnormal series with each quotient cyclic (so for Zn we have the series 1 / Z / Z2 / ··· / Zn−1 / Zn with each quotient Zi/Zi−1 = Z cyclic). Thus E is a virtually polycyclic group because it has a polycyclic subgroup of finite index [note: E is also virtually abelian]. Let E act by left multiplication on the finite set T of left cosets of Zn in E and let n πZ : E → S|T | be the associated permutation representation afforded by this action. By Theorem n −1 n n T 4.2.3[2], the kernel of the action is KerπZ = e∈E eZ e =: A and so A ⊆ Z is a normal subgroup of E of finite index; A is necessarily nontrivial and hence isomorphic to Zn because it is a subgroup of Zn of finite index (|E : Zn| = |(E/A)/(Zn/A)| is finite and |E : A| is finite). We therefore have an p extension 0 → Zn → E → G → 1 with G finite. The group action of G on Zn is a homomorphism n ∼ ρ : G → Aut(Z ) = GLn(Z), and GLn(Z) contains only integral matrices with determinant ±1 as de- × duced from the surjective map det : GLn(Z) → Z = {±1} or from the fact that an integral matrix is invertible iff its determinant is a unit in Z. Consider the finite kernel K := Kerρ and its preimage E0 := p−1(K) ⊆ E under p, and note that we have E0 /E by the Lattice Isomorphism Theorem because E0/Zn = K/G = E/Zn. Now this torsion-free group E0 is finitely generated because its subgroup Zn and its quotient K are both finitely generated [if e ∈ E0 − Zn then it can be written as a finite sum with generators of K, and if e ∈ Zn then it can be written as a finite sum with generators of Zn]. Also, the corresponding K-action on Zn is trivial (as K is the kernel of the G-action) and hence Zn lies in the center of E0, so by Exercise IV.3.4(b) the commutator subgroup [E0,E0] is finite. But the only finite 0 0 0 0 0 0 0 0 subgroup of a torsion-free group is the trivial group, so [E ,E ] = 0 and Eab = E /[E ,E ] = E (i.e. E is abelian, hence isomorphic to Zn by the Fundamental Theorem of Finitely Generated Abelian Groups, noting that |E0 : Zn| = |K| is finite). Letting F = E/E0 =∼ (E/Zn)/(E0/Zn) = G/K where the isomorphism follows from the 3rd Isomorphism Theorem, we have a group extension 0 → Zn → E → F → 1 coupled with the faithful group action F,→ GLn(Z); this action is faithful because we modded the map ρ by its kernel and injected A =∼ Zn into E0 =∼ Zn. Since |F | = r annihilates H2(F, Zn) by Corollary III.10.2[1], the rth power map Zn → Zn induces the zero-map on H2(F, Zn). Exercise IV.3.1(b) then implies the above extension fits uniquely into a commutative diagram with the split extension [the trivial element in E(F, Zn)] i1 p1 0 / Zn / E / F / 1

r φ

i2 p2 0 / Zn / Zn o F / F / 1 The map φ is injective because if φ(e) = 0 then p1e = 0 ⇒ ∃ z | i1z = e ⇒ i2(rz) = φ(i1z) = 0 ⇒ rz = 0 ⇒ z = 0 ⇒ e = i1z = i10 = 0. Alternatively, we could simply note that since the r-map is injective, Kerφ ∩ Zn = {1} and hence Kerφ injects into F , which means Kerφ is trivial by commutativity of the right-side diagram. n A crystallographic group is a discrete cocompact subgroup of the group R o On of isometries of some

36 Euclidean space; in general, V ⊆ W is cocompact if W/V is compact. Note that Rn has the usual topology (its basis consists of the open n-balls), and the relative topology for the subspace Zn is the n n discrete topology (any point x ∈ Z is equal to the intersection Z ∩ Bx where Bx is the ball of radius 1 T 3 centered at x). The orthogonal group On = {M ∈ GLn(R) | M M = 1} has the relative topology ∼ n2 induced from the matrix group Mn(R) = R . It is a fact that a faithful action F,→ GLn(R) can also be considered as a faithful action F,→ On, so in our case we have an injection F,→ On since GLn(Z) ⊂ GLn(R). The product topology on n n n n Z o F ⊂ R o On is the discrete topology [thus Z o F is discrete] because Z has the discrete topology as mentioned above and the relative topology on F is discrete since F is ﬁnite. Since a subgroup of a discrete group is discrete (the relative topology induced from the discrete topology is discrete), E can n be embedded (by φ) as a discrete subgroup of R o On. The Zn-action on Rn given by translation x 7→ x + z is properly discontinuous, so the quotient map p : Rn → Rn/Zn =∼ Tn is a regular covering map (see Exercise I.4.2) where Tn ≡ S1 × · · · × S1 is the compact n-torus. Now On is a compact space by the Heine-Borel Theorem because it is a closed bounded n2 n n n subspace of R (see pg292[3]), so Z is a cocompact subgroup of Isom(R ) := R o On because the n n quotient T o On is compact. But Z is a subgroup of E of ﬁnite index, so E is also cocompact. This completes the proof that E is a crystallographic group.

3.7(a): Let G be a perfect group (so H1G = 0) and let A be an abelian group with trivial G-action. 2 ∼ The universal coeﬃcient sequence of Exercise III.1.3 then implies H (G, A) = Hom(H2G, A).

3.7(b): Let G be a perfect group and let A be any abelian group with trivial G-action. Yoneda’s lemma 2 from Exercise I.7.3(a) states that a natural transformation ϕ : Hom(H2G, −) → H (G, −) is determined

by where it sends idH2G ∈ Hom(H2G, H2G), and so for the isomorphism ϕ of part(a) there is an element 2 2 u ∈ H (G, H2G) such that ϕ(idH2G) = u. Now for any v ∈ H (G, A) there is a unique map f : H2G → A such that ϕ(f) = v because ϕ is an isomorphism, and Yoneda’s lemma gives ϕ(f) = H2(G, f)u. Thus 2 2 u is the “universal” cohomology class of H (G, H2G), in the sense that for any v ∈ H (G, A) there is a 2 unique map f : H2G → A such that v = H (G, f)u.

3.7(c): In view of Theorem IV.3.12[1], part(b) can be reinterpreted as saying that the [perfect] group π G admits a “universal central extension” 0 → H2G → E → G → 1 characterized by a certain property. 0 This property states that given any abelian group A and any central extension 0 → A → E0 →π G → 1, there is a unique map f : H2G → A such that the extension is the image of the universal extension under H2(G, f) = E(G, f). By Exercise IV.3.1(b) this latter part means there is a map E → E0 making the following diagram commute π / E > G

π0 E0 0 and we assert that this map is unique. Indeed, two such maps h1, h2 : E ⇒ E which induce the same 0 0 map f must diﬀer by a homomorphism φ : E → A because π h1(e) = π(e) = π h2(e) ⇒ h1e = h2e · a 0 with a ∈ Kerπ = A, giving φ(e) = a [it is obviously a homomorphism since h1 and h2 are]; φ must ∼ also factor through G (i.e. φ is equal to E → E/H2G = G → A) because h1 and h2 must agree on where it sends H2G by commutativity of the diagram in Exercise IV.3.1(b). Now any homomorphism ψ : G → A satisﬁes [G, G] ⊆ Kerψ by Proposition 5.4.7[2]; but G is perfect, so G = Kerψ and there are no non-trivial maps G → A (hence φ = 0 ⇒ h1 = h2). Note: It happens to be true that E in the universal central extension is necessarily perfect, so there is another way to show uniqueness of the map E → E0 [this is presented in John Milnor’s Introduction to 0 0 0 0 Algebraic K-Theory]. For any y, z ∈ E we have h1y = h2y·c and h1z = h2z·c with c, c ∈ Kerπ ⊆ Z(E ). −1 −1 −1 −1 Thus h1(yzy z ) = h2(yzy z ) by basic rearrangements of the two previous equations, noting that we can move c and c0 around as they lie in the center of E0. Since E = [E,E], it is generated by commutators and hence h1 = h2.

3.8(a): Let 0 → A →i E →π G → 1 be a central extension with G abelian. The commutator pairing associated to the extension is the map c : G×G → A deﬁned by c(g, h) = i−1([˜g, h˜]) = i−1(˜gh˜g˜−1h˜−1), whereg ˜

37 and h˜ are lifts of g and h to E. Since A/E and G = E/A is abelian, [E,E] ⊆ A by Proposition 5.4.7[2] and so i−1([˜g, h˜]) is deﬁned. Any other lift of g to E is of the formga ˜ = ag˜, and [ag,˜ a0h˜] = [˜g, h˜] because A lies in the center of E. Thus c is well-deﬁned, and it is alternating because c(g, g) = i−1([˜g, g˜]) = i−1(1) = 0. Nowg ˜h˜ is a lifting of g + h because π(˜gh˜) = π(˜g) + π(h˜) = g + h, and [˜gh,˜ k˜] =g ˜h˜k˜h˜−1g˜−1k˜−1 = g˜[h,˜ k˜]˜g−1[˜g, k˜] = [˜g, k˜][h,˜ k˜] where the last equality follows from [E,E] ⊆ A ⊆ Z(E). We then have c(g + h, k) = c(g, k) + c(h, k) because i is injective, giving i−1([˜g, k˜][h,˜ k˜]) = i−1([˜g, k˜]) + i−1([h,˜ k˜]). An analogous computation gives c(k, g + h) = c(k, g) + c(k, h), so c is Z-bilinear. Since c is alternating and bilinear, c(g, h) = −c(h, g) and hence c can be viewed as a map V2 G → A, where V2 G = G⊗G/h{g⊗g}i is the second exterior power of G.

3.8(b): Let f be a factor set to the central extension in part(a). To show that c(g, h) = f(g, h) − f(h, g) it suﬃces to show that [˜g, h˜] = i[f(g, h)]i[f(h, g)]−1 because i is injective. Given the section s : G → E, f satisﬁes s(g)s(h) = i[f(g, h)]s(gh) and s(h)s(g) = i[f(h, g)]s(gh). Thus we have [s(g), s(h)] = s(g)s(h)s(g)−1s(h)−1 = i[f(g, h)]i[f(h, g)]−1. Since s(g) is a lifting of g for all g ∈ G, we have the desired [s(g), s(h)] = [˜g, h˜].

3.8(c): Let θ : H2(G, A) → Hom(V2 G, A) be the map which sends the class of a cocycle f to the alternating map f(g, h) − f(h, g). This map is well-defined because [δc](g, h) − [δc](h, g) = c(g) + gc(h) − c(gh) − c(h) − hc(g) + c(hg) = c(g) + c(h) − c(gh) − c(h) − c(g) + c(gh) = 0, where we note that G is abelian and the G-action on the cochain c is trivial since A is central. In view of Theorem IV.3.12[1], part(b) implies this image is the commutator pairing c(g, h), and f ∈ Kerθ iff θf = c(g, h) = 0. Now c(g, h) = i−1([˜g, h˜]) = 0 iff [˜g, h˜] = 0 which is equivalent to E being abelian, i.e. [E,E] = 0. Thus there ∼ is a bijection Kerθ = Eab(G, A), where Eab(G, A) is the set of equivalence classes of abelian extensions of G by A.

4.1: Let Q2n be a generalized quaternion group. It is a fact that Q2n has a unique element of order 2, hence a unique subgroup of order 2. Any subgroup of Q2n is also a 2-group (by Lagrange’s Theorem) and so it has an element of order 2 by Cauchy’s Theorem; this element is then unique in each subgroup (giving a unique Z2 subgroup). Therefore, by Theorem IV.4.3[1] every subgroup of Q2n is either a cyclic group or a generalized quaternion group. 2n−1 4 −1 −1 Alternatively, Q2n = hx, y | x = y = 1, yxy = x i has the property that any subgroup G is a 2-group with a unique Z2 subgroup (as mentioned above). If G is abelian, then by the Fundamental ∼ Theorem of Finitely Generated Abelian Groups, G = Z2α1 × · · · × Z2αi which has more than one Z2 sub- ∼ group if i > 1, and so G = Z2α is a cyclic subgroup. Suppose G is nonabelian, which means G contains elements of the form xk (they necessarily form a cyclic subgroup H ⊂ hxi) and elements of the form xky i j n−1 [note: each element of Q2n can be written uniquely in the form x y for 0 ≤ i ≤ 2 − 1 and 0 ≤ j ≤ 1, k 4 k k 2 k −k 2 2 2 4 and any element in Q2n − hxi has order 4 because (x y) = (x yx y) = (x x yy) = (y ) = y = 1]. Let X = {xk1 y, xk2 y} ⊂ G be the elements of the form xky and let xi be the generator of H. Now n−2 xk2 y · xk1 y = xk2−k1 y2 = xMk+2 and (xk1 y)2 = y2, so xMky2 · y2 = xMk and xMk · xk1 y = xk2 y. Thus xk1 y is a generator of G (of order 4) and the other generator is xr = xgcd(i,Mk) which is cyclic [note: if X km includes other elements (up to x y) then the above still applies with r = gcd(i, km −k1, . . . , km −km−1), and if either i = 0 or k1 = ··· = km = 0 then omit those integers/diﬀerences in the gcd-term]. Since (xk1 y)xr(xk1 y)−1 = xk1 (yxry−1)x−k1 = xk1 x−rx−k1 = x−r, the presentation is complete and G is a generalized quaternion group.

2n−1 2 −1 4.2: The dihedral group D2n = Z2n−1 o Z2 = hx, y | x = y = 1, xy = yx i has the property that each element can be written uniquely in the form yixj for 0 ≤ i ≤ 1 and 0 ≤ j ≤ 2n−1 − 1. k 2 k k k k −k k Any element not in Z2n−1 = hxi has order 2 because (yx ) = yx yx = (yx y)x = x x = 1. If a subgroup contains only elements of hxi then it is cyclic, so we consider the only other subgroups G, and these contain elements of the form yxk and elements in hxi (which necessarily form a subgroup H ⊆ hxi). Let X = {yxk1 , . . . , yxkm } ⊂ G be the elements of the form yxk and let xi be the generator of H. The given subset X implies G contains the elements xkm−k1 , . . . , xkm−km−1 . Then xr = xgcd(i,km−k1,...,km−km−1) generates the elements in G of the form xk (because k would be a multiple of r), and yxk1 · x(km−k1)−(km−kj ) = yxkj for any j ≤ m [note: if i = 0 (meaning the only elements of k G are of the form yx ) then r = gcd(km − k1, . . . , km − km−1), and if k1 = ··· = km = 0 (meaning y is

38 the only element in G of the form yxk) then r = i]. Thus G = hxr, yxk1 i where yxk1 is of order 2 and (yxk1 )xr(yxk1 ) = (yxk1+ry)xk1 = x−k1−rxk1 = x−r, i.e. G is a dihedral group. The non-cyclic abelian subgroups of D2n are isomorphic to Z2 × Z2 = D4. The group D8 contains two non-cyclic abelian normal subgroups, hx2, yi and hx2, yxi. We assert that such subgroups are not normal if n > 3 (i.e. if |D2n | > 8). From the subgroup presentations above we see that the non-cyclic abelian 2n−2 i n−1 subgroups are Hi = hx , yx i for all i ∈ {0,..., 2 − 1}. It suﬃces to show that there is an integer j i −j i−2j n−2 j such that the element x (yx )x = yx does not lie in Hi (i.e. i − 2j is neither i nor i + 2 modulo 2n−1). The ﬁrst condition implies 2j 6≡ 0 mod 2n−1 ⇒ j 6= 2n−2m for any m ∈ Z, and the latter condition implies 2j + 2n−2 6≡ 0 mod 2n−1 ⇒ j 6= 2n−2m − 2n−3. Thus we can take j = 2n−2 − 1, and 2n−2−1 2n−2−1 1−2n−2 x ∈ D2n will yield the non-normality (x )Hi(x ) * Hi [note: if n = 3 then j cannot be an even integer 2m nor can it be an odd integer 2m − 1, which means j does not exist].

n 4.3: Let G = Zq o Z2 with q = 2 (n ≥ 3) and let A ⊂ Zq be the subgroup of order 2. Note that A o Z2 = A × Z2 is a non-cyclic abelian subgroup of G since Z2 acts trivially on A [multiplication by −1 + 2n−1 is the action, and (−1 + 2n−1)2n−1 = −2n−1 + 2n−22n = 2n−1 + 0 = 2n−1 for the generator n−1 2 ∈ A]. If H ⊂ G is a non-cyclic proper subgroup, then H *Zq and HZq is a subgroup of G by Corollary 3.2.15[2], so HZq = G. Thus 2q = |G| = |HZq| = |H|q/|H ∩ Zq| ⇒ |H : H ∩ Zq| = 2, and ∼ since H ∩ Zq /H we have H/(H ∩ Zq) = Z2 and this yields the extension 0 → H ∩ Zq → H → Z2 → 0. n−1 m The generator of Z2 acts as multiplication by −1 on H ∩ Zq because −1 + 2 ≡ −1 mod 2 with m m < n [note: 2 = |H ∩ Zq| with m < n because H ∩ Zq is a subgroup of Zq and hence has m order 2 , and if m = n then Zq ⊆ H which means H = G, contradicting the assumption that H is proper]. If H were abelian then H ∩ Zq would be central in H, and if H ∩ Zq were cen- ∼ tral in H then H would be abelian since H/(H ∩ Zq) = Z2 is cyclic; this latter fact holds because a H/(H ∩ Zq) = {1(H ∩ Zq), x(H ∩ Zq)} and so any h ∈ H has a representation h = x z for z ∈ H ∩ Zq, a1 a2 a1 a2 a2 a1 a2 a1 giving h1h2 = x z1x z2 = x x z1z2 = x x z2z1 = x z2x z1 = h2h1. Now H ∩ Zq is central in H iff the Z2-action is trivial (−z = z) iff H ∩ Zq = A, and so H is abelian iff H ∩ Zq = A. 1 Z2 We have H (Z2, Zq) = KerN where N :(Zq)Z2 → (Zq) is the norm map. But (Zq)Z2 is a quotient of a cyclic group (it is then necessarily cyclic of order m = 2r where m|q), and in particular we must have (−1 + 2n−1)1 = 1 mod m ⇒ 2n−2 − 1 = ms = 2(2r−1s) which is impossible because an even number cannot equal an odd number, so (Zq)Z2 = 0 and the kernel of the norm map is trivial. Thus 1 H (Z2, Zq) = 0, and by Proposition IV.2.3[1] this means that the extension 0 → Zq → G → Z2 → 0 has a unique splitting (up to conjugacy) and hence that G contains only two conjugacy classes of subgroups of order 2 (specifically, 0 × Z2 lies in one class, and A × 0 is the other class because the number of conjugates of A is |G : NG(A)| = |G : G| = 1 by Proposition 4.3.6[2]). As an abelian non-cyclic subgroup, H contains at least two subgroups of order 2 (one necessarily being A) and hence H can be A × Z2 and its conjugates. We assert that H = A × Z2 is not normal in G. Indeed, the element (1, x) ∈ G where x is the genera- n−1 n−1 −1 n−1 tor of Z2 can be used with (2 , 1) ∈ H to obtain (1, x)(2 , 1)(1, x) = (1 + x · 2 , x)(−1, x) = (1 + 2n−1 + x · (−1), x2) = (2 + 2n−1, 1) which is not in H since 2 + 2n−1 is neither 0 nor 2n−1 modulo 2n for n ≥ 3.

4.4: Let G be a p-group such that every abelian normal subgroup is cyclic and choose a maximal abelian n normal subgroup Zq ⊂ G with q = p , so we have the corresponding extension 0 → Zq → G → H → 1. ∼ If |H| = 1 then G = Zq is cyclic, hence of type (A). If |H| = p then Theorem IV.4.1[1] states G is of type (A),(D),(E), or (F) because groups of type (C) have a non-cyclic abelian subgroup of index p by Proposition IV.4.4[1] (such subgroups are necessarily normal by Corollary 4.2.5[2]) and groups of type (B) are non-cyclic abelian normal subgroups of themselves [note: if G is of type (D) then we must have 4 ∼ |G| ≥ 16 = 2 because G = D8 contains two non-cyclic abelian normal subgroups and G = D4 = Z2 ×Z2 is a non-cyclic abelian normal subgroup of itself]. Suppose, then, that |H| ≥ p2, and consider the normal 0 0 0 subgroups H ⊆ H of order p. If such an H acted trivially on Zq, then the inverse image G ⊂ G 0 0 0 ∼ of H = G /Zq would be a central extension of H = Zp by Zq, hence an abelian subgroup of G [as 0 explained in Exercise IV.4.3] bigger than Zq. But G /G by the Lattice Isomorphism Theorem (since 0 0 0 H /H), so it contradicts the maximality of Zq and hence H cannot act trivially on Zq. Now G is a p-group with a cyclic subgroup Zq of index p, so by Theorem IV.4.1[1] it is of type (C),(D),(E), or (F) because types (A) and (B) were eliminated by the previous statement. Also, if H0 acted as in (C)

39 then we would have a unique non-cyclic abelian subgroup G00 of G0 of index p by Proposition IV.4.4[1]; this applies to all p and n ≥ 2 except for the case p = 2 = n. Now G acts on G0 by conjugation (since G0 /G) and hence maps G00 to another non-cyclic p-index subgroup. But G00 is the only such subgroup, so conjugation sends G00 to itself (i.e. G00 /G) and G is a non-cyclic abelian normal subgroup of G, contradicting the hypothesis. Thus the only possibility is that p = 2 since groups of type (D),(E), and (F) are 2-groups, and the non-trivial element of H0 acts as −1 [corresponding to (D),(E), and (C)] or n−1 0 ∗ ∼ −1 + 2 with n ≥ 3 [corresponding to (F)]. This means that H embeds in Z2n = Z2 × Z2n−2 as n−3 ∗ either Z2 × {0} or {(0, 0), (1, 2 )}. But the composite H → Z2n → {±1} [where the ﬁrst map is the action-representation and the latter map is the projection] has a non-trivial kernel, and we can simply 0 0 n−3 n−2 ∗ take H to be contained in the kernel; this implies H is embedded as {0} × 2 Z/2 Z ⊂ Z2n which is not any of the aforementioned subgroups. Thus we do not have |H| ≥ p2 and the proof is complete.

6.1: Suppose N is a group with trivial center. Method 1 : The center of N is C = {1}, and so H2(G, C) = 0 = H3(G, C) for any group G. Then any homomorphism ψ : G → Out(N) gives rise to an obstruction in H3(G, C) = 0 which necessarily vanishes, thus E(G, N, ψ) 6= ∅ by Theorem IV.6.7[1]. Therefore, by Theorem IV.6.6[1], E(G, N, ψ) =∼ H2(G, C) = 0 and hence there is exactly one extension of G by N (up to equivalence) corresponding to any homomorphism G → Out(N). Method 2 : Note that N is an Aut(N)-crossed module via the canonical map α : N → Aut(N) and the canonical action of Aut(N) on N, and Kerα = Z(N) = {1}. Thus any extension of G by N ﬁts into a commutative diagram with exact rows 1 / N i / E π / G / 1

ψ 1 / N α / Aut(N) / Out(N) / 1 where ψ is determined by the E-extension. By Exercise IV.3.1(a), this is a pull-back diagram and hence all such extensions for a given ψ are equivalent (under the above diagram). Thus there is exactly one extension of G by N (up to equivalence) corresponding to any homomorphism G → Out(N).

6.2: Let 1 → N → E → G → 1 be an extension of ﬁnite groups such that gcd(|N|, |G|) = 1. Corol- lary IV.3.13[1] states that such an extension must split if N is abelian, and it is indeed true that this result could be generalized to the non-abelian case. Now one might hope to deduce this generalization directly from Theorem IV.6.6[1] in view of the vanishing of H2(G, C) [note that H2(G, C) = 0 by the cohomology analogue of Exercise AE.16 because gcd(|C|, |G|) = 1, where C is the center of N]. However, this does not work because E(G, N, ψ) doesn’t contain the semidirect product (hence the split extension) N o G unless ψ : G → Out(N) [which is determined by the above extension] lifts to a homomorphism ϕ : G → Aut(N).

40 5 Chapter V: Products

t−1 1.1: Let G = hti be a ﬁnite cyclic group of order m and let F be the periodic resolution · · · → ZG → N t−1 ε Pm−1 i ZG → ZG → ZG → Z → 0, where N = i=0 t is the norm element (note that Fn = ZG for all n ≥ 0). Let ∆ : F → F ⊗ F be the map whose (p, q)-component ∆pq : Fp+q → Fp ⊗ Fq is given by  1 ⊗ 1 p even  ∆pq(1) = 1 ⊗ t p odd, q even P i j 0≤i

2.1: Let M (resp. M 0) be an arbitrary G-module (resp. G0-module), let F (resp. F 0) be a projective reso- 0 0 0 0 0 lution of Z over ZG (resp. ZG ), and consider the map (F ⊗GM)⊗(F ⊗G0 M ) → (F ⊗F )⊗G×G0 (M⊗M ) 0 0 0 0 0 0 given by (x ⊗ m) ⊗ (x ⊗ m ) 7→ (x ⊗ x ) ⊗ (m ⊗ m ). Note that (F ⊗G M) ⊗ (F ⊗G0 M ) = 0 0 0 0 (F ⊗ M)G ⊗ (F ⊗ M )G0 , which is the quotient of F ⊗ M ⊗ F ⊗ M by the subgroup generated ∼ by elements of the form gx ⊗ gm ⊗ g0x0 ⊗ g0m0. The isomorphism F ⊗ M →= M ⊗ F of chain complexes (M in dimension 0) is given by x ⊗ m 7→ (−1)degm·degxm ⊗ x = m ⊗ x, and so the aforementioned quotient is isomorphic to F ⊗ F 0 ⊗ M ⊗ M 0 modulo the subgroup generated by elements of the form (gx⊗g0x0 ⊗gm⊗g0m0) = (g, g0)·(x⊗x0 ⊗m⊗m0) where this latter action is the diagonal (G×G0)-action. 0 0 0 0 Now this is precisely (F ⊗ F ⊗ M ⊗ M )G×G0 = (F ⊗ F ) ⊗G×G0 (M ⊗ M ) and hence the considered map is an isomorphism. Assuming now that either M or M 0 is Z-free, we have a corresponding Künnethformula Ln 0 0 0 0 Ln−1 Z 0 0 p=0 Hp(G, M)⊗Hn−p(G ,M ) ,→ Hn(G×G ,M ⊗M ) p=0 Tor1 (Hp(G, M),Hn−p−1(G ,M )) by Proposition I.0.8[2]. Note that in order to apply the proposition we needed one of the chain complexes (say, F ⊗G M) to be dimension-wise Z-free (and so with a free resolution F this means we needed M to be Z-free). Actually, the general Künneththeorem has a more relaxed condition and it suffices to choose M (or M 0) as a Z-torsion-free module. *Must require some conditions on groups/actions: We have assumed G acts trivially on M 0 while G0 acts trivially on M. Since these actions don’t mix, we cannot deduce a Universal Coefficients theorem (UCT) with nontrivial actions.

41 2.2: [[no proofs, just notes]] (for more info, see topological analog in §60[4]) Let M (resp. M 0) be an arbitrary G-module (resp. G0-module), let F (resp. F 0) be a projective resolu- 0 0 0 tion of Z over ZG (resp. ZG ), and consider the cochain cross-product HomG(F,M)⊗HomG0 (F ,M ) → 0 0 0 0 0 0 HomG×G0 (F ⊗ F ,M ⊗ M ) which maps the cochains u and u to u × u given by hu × u , x ⊗ x i = degu0·degx 0 0 (−1) hu, xi⊗ hu , x i. This map is an isomorphism under the hypothesis that either H∗(G, M) or 0 0 H∗(G ,M ) is of finite type, that is, the ith-homology group is finitely generated for all i (alternatively, we could simply require the projective resolution F or F 0 to be finitely generated). For example, if M = Z then HomG(−, Z) commutes with finite direct sums, so we need only consider the case F = ZG. An inverse to the above map is given by t 7→ ε ⊗ φ, where ε is the augmentation map and φ : F 0 → M 0 0 0 L∞ is given by φ(f ) = t(1 ⊗ f ). Note that this does not hold for infinitely generated P = ZG because ∼ Q∞ HomG(P, Z) = Z is not Z-projective (i.e. free abelian). Assuming now that either M or M 0 is Z-free, we have a corresponding Künnethformula Ln p n−p 0 0 n 0 0 Ln+1 Z p n−p+1 0 0 p=0 H (G, M)⊗H (G ,M ) ,→ H (G×G ,M ⊗M ) p=0 Tor1 (H (G, M),H (G ,M )) by Proposition I.0.8[2]. *Must require some conditions on groups/actions: We have assumed G acts trivially on M 0 while G0 acts trivially on M. Since these actions don’t mix, we cannot deduce a Universal Coefficients theorem (UCT) with nontrivial actions.

0 G q ∗ ∗ 3.1: Let m ∈ H (G, M) = M and u ∈ H (G, N), and let fm : H (G, N) → H (G, M ⊗ N) denote the map induced by the coefficient homomorphism n 7→ m ⊗ n. This homomorphism is also given ∼= by n 7→ 1⊗n 7→ m⊗n where the former map in the composition is the canonical isomorphism N → Z⊗N, the latter map in the composition is Fm ⊗idN : Z⊗N → M ⊗N, and Fm : Z → M is given by F (1) = m. Then using two properties of the cup product (existence of identity element and naturality with respect ∗ ∗ ∗ to coefficient homomorphisms) we obtain fm(u) = (Fm ⊗ idN ) (1 ` u) = Fm(1) ` idN (u) = m ` u. 0 G Now let m ∈ H (G, M) = M and z ∈ Hq(G, N), and let fm : H∗(G, N) → H∗(G, M ⊗ N) denote the map induced by the same coefficient homomorphism n 7→ m ⊗ n. Then using two properties of the cap product (existence of identity element and naturality with respect to coefficient homomorphisms) ∗ ∗ ∗ we obtain fm(z) = (Fm ⊗ idN ) (1 a z) = Fm(1) a idN (z) = m a z.

3.2: Consider the diagonal transformation ∆ presented in Exercise V.1.1, along with the cohomology groups H2r(G, M) =∼ M G/NM and H2r+1(G, M 0) =∼ Ker(N : M 0 → M 0)/IM 0 of the ﬁnite cyclic group G = hσi of order n, where I = hσ − 1i is the augmentation ideal of G. The cup product in ∗ 0 0 degv·degx 0 H (G, M ⊗M ) is given by u ` v = (u×v)◦∆ with hu×v, x⊗x i = (−1) hu, xi⊗hv, x i. Choose representatives hu, xi = m ∈ M of Hi(G, M) with m ∈ M G for i even and m ∈ Ker(N : M → M) for i odd, and choose representatives hv, x0i = m0 ∈ M 0 of Hj(G, M 0) with m0 ∈ M 0G for j even and m0 ∈ Ker(N : M 0 → M 0) for j odd. If i is even then (−1)degv·degx = (−1)degv·i = 1, and if j is even then (−1)degv·degx = (−1)−j·degx = 1, and if both i and j are odd then (−1)degv·degx = (−1)−j·i = −1. Thus the cup product element of Hi+j(G, M ⊗ M 0) is represented by m ⊗ m0 for i or j even and is represented P p q 0 by − 0≤p

3.3(a): Let G be a ﬁnite group which acts freely on S2k−1, and consider the exact sequence of G- 2k−1 modules from pg20[1], 0 → Z → C2k−1 → · · · → C1 → C0 → Z → 0 where each Ci = Ci(S ) is free. Tensoring the sequence with an arbitrary G-module M gives an exact sequence (as explained on

∂2k−1 ∂1 ∂0 pg61[1]) 0 → M → C2k−1 ⊗ M → · · · → C1 ⊗ M → C0 ⊗ M → M → 0. We can break this up into short exact sequences 0 → Ker∂0 → C0 ⊗ M → M → 0 and 0 → Ker∂i → Ci ⊗ M → Ker∂i−1 → 0 for all i, and we can then apply the H∗(G, −) functor to obtain corresponding long exact cohomology sequences. First note that HomG(F,C∗ ⊗ M) → HomG(F,M) is a weak equivalence by Theorem n ∼ n I.8.5[1], so H (G, Ci ⊗ M) = H (HomG(F,M)) = 0 for all i with n > 0 (M is considered a chain complex concentrated in dimension 0). Thus we can apply the dimension-shifting argument to the above i ∼ i+1 ∼ i+2 ∼ ∼ i+2k short exact sequences and obtain H (G, M) = H (G, Ker∂0) = H (G, Ker∂1) = ··· = H (G, M) for i > 0 where we note that Ker∂2k−1 = M. For i = 0 we use the ﬁrst short exact sequence men- 0 1 1 tioned above to obtain the exact sequence H (G, M) → H (G, Ker∂0) → H (G, C0 ⊗ M) = 0, with 1 ∼ 2k i i+2k H (G, Ker∂0) = H (G, M). Thus there is an iterated coboundary map d : H (G, M) → H (G, M) which is an isomorphism for i > 0 and an epimorphism for i = 0.

42 3.3(b): Consider the “periodicity map” d from part(a) above with the ﬁnite group G which acts freely on the sphere S2k−1. Since d is simply an iteration of coboundary maps δ, we can use Property V.3.3[1] ∗ ∗ 0 which states d(w ` v) = d(w) ` v for any w ∈ H (G, Z) and v ∈ H (G, M). Choosing w = 1 ∈ H (G, Z) and noting that 1 ` v = v by Property V.3.4[1], d(v) = d(1 ` v) = d(1) ` v. But d is an isomorphism from H0(G, Z) to H2k(G, Z) by part(a) above, so there exists an element u ∈ H2k(G, Z) satisfying d(1) = u. Thus there is an element u ∈ H2k(G, Z) such that the “periodicity map” d of H∗(G, M) is ∗ given by d(v) = u ` v for all v ∈ H (G, M).

3.3(c): Let G be a ﬁnite cyclic group of order |G| = n, and note that it acts freely on the circle S1 by rotations. By part(b) above, the periodicity isomorphism d maps 1 ∈ H0(G, Z) =∼ Z to a generator 2 ∼ ∗ α ∈ H (G, Z) = Zn. For the ring structure on H (G, Z) with multiplication being the cup product, 2 4 ∼ 2 α is a generator of H (G, Z) = Zn since d is an isomorphism and d(α) = α ` α = α . General- m 2m ∼ izing, α is a generator of H (G, Z) = Zn and hence the cohomology ring is the polynomial ring H∗(G, Z) =∼ Z[α]/(nα) with |α| = 2. Alternatively, we could note that the inﬁnite-dimensional lens space L∞ is a K(G, 1)-complex, so H∗(G, Z) =∼ H∗(L∞) by the cohomological analog of Proposition II.4.1[1]. The ring structure was calculated in Example 3.41 on pg251[3], giving H∗(L∞) =∼ Z[α]/(nα) with |α| = 2.

3.4: Let G be cyclic of order n and consider the endomorpism α(m) of G given by α(m)g = gm, ∗ ∗ ∗ ∗ for any m ∈ Zn. Since α(m) (u ` v) = α(m) u ` α(m) v and H (G, Z) consists of elements of the form P i ∗ 2 i ziβ with zi ∈ Z and |β| = 2 by Exercise V.3.3(c), it suﬃces to calculate α(m) on H (G, Z) in order to calculate α(m)∗ : H∗(G, Z) → H∗(G, Z). By Exercise III.1.3 we have the universal coeﬃcient isomor- 2 ∼ phism H (G, Z) = Ext(H1G, Z) since Hom(H1G, Z) = Hom(Zn, Z) = 0. The map α(m) is multiplication by m on Gab = G and hence is multiplication by m on H1G by Exercise II.6.3(a). Now f(mg) = mf(g) for any group homomorphism f, so given a free resolution F of the abelian group Gab = H1G = G (namely, its presentation) we have the commutative diagram φ 0o Hom(F1, Z) o Hom(F0, Z) o Hom(G, Z)

m m m

φ 0o Hom(F1, Z) o Hom(F0, Z) o Hom(G, Z) ∗ Thus α(m) induces multiplication by m on Cokerφ ≡ Ext(H1G, Z) and hence α(m) is multiplication by m on H2(G, Z) by naturality of the universal coeﬃcient formula. Alternatively, we could have used the interpretation of H2 in terms of group extensions (Exercise IV.3.1(a)) which states that α(m) induces a map E(G, Z) → E(G, Z), sending the extension E to the ﬁber-product E ×G G which correponds to mE (i.e. the elements em ∈ E) and hence gives the m-multiplication map in cohomology by Theorem IV.3.12[1]. Referring back to the cohomology ring, α(m)∗ is multiplication by mi for the (2i)th-dimension P i P ∗ P ∗ ∗ P since i ziβ 7→ i ziα(m) (β ` ··· ` β) = i zi[α(m) β] ` ··· ` [α(m) β] = i zi(mβ) ` ··· ` P i i (mβ) = i zim β .

3.5: The symmetric group S3 on three letters has a semi-direct product representation S3 = Z3 o Z2 ∗ ∗ ∗ ∼ ∗ ∗ 2 where Z2 acts on Z3 by conjugation. Thus H (S3) = H (S3)(2) ⊕ H (S3)(3) = H (S3)(2) ⊕ H (Z3)Z by ∗ ∗ Theorem III.10.3[1], with H (S3)(2) isomorphic to the set of S3-invariant elements of H (Z2). Exercise ∗ ∼ ∗ 2 III.10.1 showed that H (S3)(2) = Z2, so it suﬃces to compute H (Z3)Z where we know that Z2 acts by 2 2 conjugation on Z3 (1 7→ 1, x 7→ x , x 7→ x). But this action can be considered as the endomorphism α(2) from Exercise V.3.4 above since (1)2 = 1 and (x)2 = x2 and (x2)2 = x4 = x, and that exercise implies that the induced map on the (2i)th-cohomology is multiplication by 2i [we know that the cohomology is trivial in odd dimensions]. Now 21 ≡ 2 mod3 and 22 ≡ 1 mod3, so by multiplying both of those statements by 2 repeatedly we see that 2i ≡ 1 mod3 for i even and 2i ≡ 2 mod3 for i odd. Thus 2i ∼ the largest Z2-submodule of H (Z3) = Z3 on which Z2 acts trivially is Z3 (itself) for i even, and is 0 ∗ for i odd. It now follows that the integral cohomology H (S3) is the same as that which was deduced in Exercise III.10.1, namely, it is Z2 in the 2 mod4 dimensions and is Z6 in the 0 mod4 dimensions and is 0 otherwise (besides the 0th-dimension in which it is Z).

4.1:

43 4.2:

5.1(a): Let G be an abelian group so that ZG is a commutative ring, and let M ⊗ZG N be the tensor product with ZG-module structure deﬁned by r·(m⊗n) = rm⊗n = m⊗rn, where r ∈ ZG. There exists × µ∗ a Pontryagin product given by H∗(G, M) ⊗ H∗(G, N) → H∗(G × G, M ⊗ N) → H∗(G, M ⊗ZG N). The former map is the homology cross product (see pg109[1]), where G×G acts on M ⊗N via (g, g0)·(m⊗n) = 0 gm⊗g n. The latter map in the composition is induced by µ = (α, f):(G×G, M ⊗N) → (G, M ⊗ZG N), 0 0 where α(g, g ) = gg and f(m ⊗ n) = m ⊗ n. As explained on pg79[1], µ∗ is a well-deﬁned map as long as f(rm) = α(r)f(m) for r ∈ ZG. This is indeed the case because α(g, g0)f(m ⊗ n) = gg0 · (m ⊗ n) = g · (m ⊗ g0n) = gm ⊗ g0n = f(gm ⊗ g0n) = f[(g, g0) · (m ⊗ n)].

i 2i 5.1(b): Take G = Z3 = hgi and take the modules M = N = Z7 with nontrivial Z3-action, i.e. g·x = x and g2 · xi = x4i with g3 = 1 [consistency check: g3 · xi = x8i = (x8)i = xi]. Note that there is only one ∼ nontrivial action in this scenario because an action is a homomorphism Z3 → Aut(Z7) = Z6 and there are two such maps, the trivial map and the inclusion. The tensor product Z7 ⊗Z3 Z7 can be interpreted as the group (Z7 ⊗ Z7)Z3 where Z3 acts diagonally, g · (m ⊗ n) = gm ⊗ gn. This group contains only idempotent elements since (xi ⊗ xj)2 = x2i ⊗ x2j = gxi ⊗ gxj = g · (xi ⊗ xj) ∼ xi ⊗ xj. Now the tensor product ⊗ can be interpreted as the group ( ⊗ ) where acts anti-diagonally, Z7 Z[Z3] Z7 Z7 Z7 Z3 Z3 g · (m ⊗ n) = g−1m ⊗ gn = g2m ⊗ gn. This group, however, does contain elements which are not idempotent. For example, (x ⊗ x)2 = x2 ⊗ x2 = g2x4 ⊗ gx ∼ x4 ⊗ x = (x ⊗ x)(x3 ⊗ 1) and (x3 ⊗ 1) 1 ⊗ 1. Thus ⊗ ⊗ . Z7 Z[Z3] Z7 Z7 Z3 Z7

0 ∗ 5.2: If G and G are abelian and k is a commutative ring, then the cohomology cross-product H (G, k)⊗k ∗ 0 × ∗ 0 0 0 0 H (G , k) → H (G × G , k ⊗ k) is a k-algebra homomorphism. Moreover, z × z = p∗z ` p∗z for any z ∈ H∗(G, k) and z0 ∈ H∗(G0, k), where p : G × G0 → G and p0 : G × G0 → G0 are the projections. 0 To prove these two statements, note ﬁrst that the cross-product is given by hz×z0, x⊗x0i = (−1)|z ||x|z(x)⊗ z0(x0) for x ∈ F and x0 ∈ F 0, with F and F 0 being the resolutions for G and G0, respectively. Then

0 0 0 0 0 0 0 |z ||z2| 0 0 0 |z ||z2| |z z ||x| h(z1 ⊗ z1)(z2 ⊗ z2), x ⊗ x i = h(−1) 1 z1z2 × z1z2, x ⊗ x i = (−1) 1 (−1) 1 2 z1(x)z2(x) ⊗ 0 0 0 0 |z0 ||x| |z0 ||x| 0 0 0 0 0 0 0 0 z1(x )z2(x ) = (−1) 1 (−1) 1 [z1(x) ⊗ z1(x )][z2(x) ⊗ z2(x )] = hz1 × z1, x ⊗ x i · hz2 × z2, x ⊗ x i and so the cross-product is a k-algebra homomorphism (it obviously commutes with addition). Since 0 0 0 0 0 z ⊗ z = (z ⊗ 1) · (1 ⊗ z ) with 1 ∈ H (G, Z) = Z, it follows that z × z = (z × 1) ` (1 × z ), where we make the identiﬁcations Z ⊗ k = k = k ⊗ Z to switch from tensor-multiplication to cup product. I claim 0 that z × 1 = p∗z; indeed, by naturality of the cross-product it suﬃces to check this for G = 1. But ∗ ∗ 0 this is obvious because p∗ : H (G, k) → H (G × 1, k) maps z to z × 1, noting that F = 1. Similarly, 0 0 0 1 × z = p∗z , whence the proposition.

5.3(a):A directed set D is a partially-ordered set having the property that for each pair α, β ∈ D there exists γ ∈ D such that α, β ≤ γ.A directed system of groups is a family of groups {Gα} indexed by a directed set D along with a family of homomorphisms {fαβ : Gα → Gβ} such that fαα = idG and fαγ = fβγ ◦ fαβ for α ≤ β ≤ γ. The direct limit G = lim Gα of this directed system α ` −→ is defined to be α Gα/∼ where gα ∼ gβ if there exists some γ ∈ D such that fαγ (gα) = fβγ (gβ). Now for any G-module M, we have a compatible family of maps H∗(Gα,M) → H∗(G, M), hence a map ϕ : lim H (G ,M) → H (G, M). It is true that homology commutes with direct limits of −→ ∗ α ∗ chain complexes (see Albrecht Dold’s Lectures on Algebraic Topology, Proposition VIII.5.20), so to prove that ϕ is an isomorphism it suffices to show that lim(F ⊗ M) = F ⊗ M where F is the −→ α Gα G α standard resolution for Gα and F is the standard resolution for G. The obvious maps Fα → Fβ are given by (g1, . . . , gn) 7→ (fαβ(g1), . . . , fαβ(gn)), and F is obviously the direct limit of Fα. Thus ` F ⊗G M = ( Fα/∼) ⊗G M, and since we can switch actions from G to Gα via restriction of scalars, ` α ` ( α Fα/∼)⊗G M = α(Fα ⊗Gα M)/∼ where ∼ is simply altered by tensoring each tuple with an element of M. But ` (F ⊗ M)/∼ is by definition the direct limit of F ⊗ M, so lim(F ⊗ M) = F ⊗ M α α Gα α Gα −→ α Gα G and hence group homology commutes with direct limits of groups.

5.3(b): It is a fact that any group is the direct limit of its ﬁnitely generated subgroups. So for any abelian

44 group G = lim G and commutative ring k, the homology ring H (G, k) is isomorphic to lim H (G , k) −→ α ∗ −→ ∗ α by part(a) above. If each of those rings H∗(Gα, k) is strictly anti-commutative, then H∗(G, k) will obviously be strictly anti-commutative since it is a quotient of the direct sum of those rings. Thus we can reduce to the case where G is a finitely generated abelian group, hence isomorphic to a finite product of cyclic groups. Let’s argue by induction on the number of cyclic factors. The infinite cyclic group has resolution F = V(x) which is strictly anti-commutative as explained in subsection 5.2 on pg118[1], and V the finite cyclic group has resolution F = (e1)⊗ZG Γ(e2) which is strictly anti-commutative by property (iii) of subsection 5.3 on pg119[1]. Since the admissible product on F induces a k-bilinear product on 0 0 0 0 F ⊗G k via (f ⊗k)(f ⊗k ) = ff ⊗kk , the complex F ⊗G k is strictly anti-commutative; thus H∗(G, k) is strictly anti-commutative for G cyclic. Applying the inductive hypothesis, we can attach another cyclic factor by the method of subsection 5.4 on pg119[1]: F and F 0 are resolutions with admissible product for G (cyclic) and G0 (inductive group), so F ⊗ F 0 is a resolution with admissible product for G × G0. This resolution is strictly anti-commutative because (x ⊗ y)(x ⊗ y) = (−1)|x|·|y|x2 ⊗ y2 = 0 if either x2 or y2 is 0, so for x (resp. y) of odd degree and y (resp. x) of even degree we have x ⊗ y of odd degree 2 0 0 which satisfies (x ⊗ y) = 0. Then (F ⊗ F ) ⊗G×G0 k is strictly anti-commutative and so is H∗(G × G , k), completing the inductive process. Thus the ring H∗(G, k) is strictly anti-commutative for any abelian group G and commutative ring k.

5.4: Let n = p + q + r, and let σ ∈ Sn be a permutation with signature sgn(σ) being the number of inversions of σ; an inversion of σ is a pair of elements (i, j) such that i < j and σ(i) > σ(j) [it also indicates the number of swaps needed to give the original sequence ordering]. A permutation σ is called a (p, q, r)-shuffle if σ(i) < σ(j) for 1 ≤ i < j ≤ p and for p + 1 ≤ i < j ≤ p + q and for p + q + 1 ≤ i < j ≤ p + q + r. But this permutation is clearly the composition of a (p, q)-shuffle τ1 and a (p + q, r)-shuffle τ2 since the former shuffle will give τ1(i) < τ1(j) for 1 ≤ i < j ≤ p and for p + 1 ≤ i < j ≤ p + q, and the latter shuffle will give τ2(i) < τ2(j) for p + q + 1 ≤ i < j ≤ p + q + r and will preserve the ordering of the former shuffle via 1 ≤ i < j ≤ p + q. Then sgn(σ) = sgn(τ1) + sgn(τ2) because the inversions of τ2 give the original ordering of the sequence up to changes in {1, . . . , p + q} and the inversions of τ1 give the original ordering of that set. Therefore (in the bar resolution), P [g1| · · · |gp] · [gp+1| · · · |gp+q] · [gp+q+1| · · · |gp+q+r] = σ σ[g1| · · · |gp+q+r] where σ ranges over the (p, q, r)-shuffles. P The notation is [g1| · · · |gn] · [gn+1| · · · |gn+m] = τ τ[g1| · · · |gn+m] where τ ranges over all (p, q)-shuffles, sgnσ and σ[g1| · · · |gn] = (−1) [gσ−1(1)| · · · |gσ−1(n)]. Generalizing, we have P [g1| · · · |ga1 ] · [ga1+1| · · · |ga1+a2 ] ··· [ga1+···+an−1+1| · · · |ga1+···+an ] = σ σ[g1| · · · |ga1+···+an ] where σ ranges over the (a1, . . . , an)-shuffles.

6.1: Let G be an abelian group (written additively) with n ∈ Z, and consider the endomorphism g 7→ ng of G. To see what it induces on the rational homology ring H∗(G, Q) it suﬃces to ﬁgure out what the V∗ endomorphism induces on the exterior algebra (G ⊗ Q) by Theorem V.6.4[1] (they’re isomorphic), V∗ noting that the isomorphism in said theorem is natural. Now the induced map on (G ⊗ Q) is uniquely V1 determined by the induced map ϕ : G ⊗ Q → (G ⊗ Q) = G ⊗ Q by the universal mapping property of exterior algebras (pg122[1]), and this map ϕ is given by g ⊗ q 7→ ng ⊗ q = n(g ⊗ q), i.e. multiplication by n. Then on the i-fold tensor product of G ⊗ Q with itself (hence the exterior algebra) we have the i induced map as f1 ⊗ · · · ⊗ fi 7→ nf1 ⊗ · · · ⊗ nfi = n (f1 ⊗ · · · ⊗ fi) where f1, . . . , fi ∈ G ⊗ Q. Thus the i th original endomorphism on G induces multiplication by n on H∗(G, Q) for the i -dimension.

6.2: Let A and B be strictly anti-commutative graded k-algebras, where k is a commutative ring. We assert that A⊗k B is the sum (i.e. coproduct) of A and B in the category of strictly anti-commutative graded k-algebras, via the maps fA : A → A⊗k B and fB : B → A⊗k B with a 7→ a⊗1 and b 7→ 1⊗b. The coproduct refers to the pair (A⊗k B, {fA, fB}) satisfying the universal property that given a family of algebra homomorphisms {gA : A → C, gB : B → C}, there exists a unique algebra homomorphism h : A ⊗k B → C such that hfA = gA and hfB = gB. We deﬁne a k-bilinear map A×B → C given by a×b 7→ gA(a)gB(b). Then Corollary 10.4.16[2] gives us a unique homomorphism h : A⊗k B → C given by a⊗b 7→ gA(a)gB(b),

45 |a2||b1| and this is clearly an algebra homomorphism since h[(a1 ⊗ b1)(a2 ⊗ b2)] = h[(−1) a1a2 ⊗ b1b2] = |a2||b1| |a2||b1| |a2||b1| (−1) gA(a1a2)gB(b1b2) = (−1) gA(a1)[(−1) gB(b1)gA(a2)]gB(b2) = h(a1 ⊗ b1)h(a2 ⊗ b2). Now hfA(a) = h(a ⊗ 1) = gA(a)gB(1) = gA(a), with a similar calculation for B, and so the universal property of tensor products (Theorem 10.4.10[2]) guarantees that A ⊗k B is the categorical sum A + B.

6.3(a): Let A be a strictly anti-commutative graded ring with a diﬀerential ∂ and a system of di- (i) vided powers, so that there is a family of functions ϕi : A2n → A2ni denoted x 7→ x satisfying the properties on pg124[1]. Note that x(i) is a cycle if x is a cycle because ∂x(i) = x(i−1)∂x = x(i−1) · 0 = 0. Thus ϕi restricted to the kernel Z2n ⊆ A2n gives a function Z2n → Z2ni. These functions inherit the same properties associated with A because every term in all of the properties are cycles xi and x(j), (i) assuming x is a cycle. Now suppose x is a boundary whenever x is a boundary. Then ϕi induces a (i) (i) (i) (i) function H2nA → H2niA because x = x + ∂y 7→ x + (∂y) = x + ∂w = x , and the properties remain untouched. Therefore, we have an induced system of divided powers on H∗A.

6.3(b): Consider the divided polynomial algebra Γ(y) with degy = 2, and assume that y is a boundary, y = ∂x for some x. I claim that y(i) is not a boundary. Indeed, suppose y(i) = ∂f for some P (j) i P j P j−1 element f = j zjy ∈ Γ(y), where zj ∈ Z. Then y /i! = j zj∂(y )/j! = j zj[jy ∂y]/j! = P j−1 2 i ∂y j zjy /(j − 1)! = ∂ x · w = 0 · w = 0. But this implies y = 0, a contradiction. V∗ 6.4(a): By Theorem V.6.4[1] we have an injection ψ : (G ⊗ k) → H∗(G, k) for G abelian and k a PID. This k-algebra map was the unique extension of the isomorphism G ⊗ k → H1(G, k) in dimension 1. Now ψ[(g ⊗ 1) ∧ (h ⊗ 1)] = ψ(g ⊗ 1) · ψ(h ⊗ 1) by definition of an algebra map, where · is the Pontryagin product. On the bar resolution this product is given by the shuffle product, and the P isomorphism ψ : G ⊗ k → H1(G, k) sends g ⊗ 1 to [g]. Thus (g ⊗ 1) ∧ (h ⊗ 1) 7→ [g] · [h] = σ[g|h] = [g|h] − [h|g], where σ ran over the two possible (1, 1)-shuffles. Remark: This map is well-defined because for (g ⊗ 1) ∧ (g ⊗ 1) = 0 ∈ V2(G ⊗ k), the image is [g|g] − [g|g] = 0.

6.4(b): Let k be a PID in which 2 is invertible, let G be an abelian group, and consider the map V∗ V∗ C2(G, k) → (G⊗k) given by [g|h] 7→ (g⊗1)∧(h⊗1)/2. This induces a map ϕ : H2(G, k) → (G⊗k) because any 3-coboundary ∂[r|s|t] = [s|t] − [rs|t] + [r|st] − [r|s] is mapped to the trivial element (s⊗1)∧(t⊗1)/2−(r +s⊗1)∧(t⊗1)/2+(r ⊗1)∧(s+t⊗1)/2−(r ⊗1)∧(s⊗1)/2 = [(s⊗1)∧(t⊗1)/2]− [(r ⊗1)∧(t⊗1)/2]−[(s⊗1)∧(t⊗1)/2]+[(r ⊗1)∧(s⊗1)/2]+[(r ⊗1)∧(t⊗1)/2]−[(r ⊗1)∧(s⊗1)/2] = 0. Using ψ from part(a) above, ϕ is its left-inverse because (g ⊗ 1) ∧ (h ⊗ 1) 7→ [g|h] − [h|g] 7→ (g ⊗ 1) ∧ (h ⊗ 1)/2 − (h ⊗ 1) ∧ (g ⊗ 1)/2 = (g ⊗ 1) ∧ (h ⊗ 1)/2 + (g ⊗ 1) ∧ (h ⊗ 1)/2 = (g ⊗ 1) ∧ (h ⊗ 1), where we note in the last equality that the exterior algebra is strictly anti-commutative.

6.5: Let G be abelian and let A be a G-module with trivial G-action. In view of the isomorphism ∼ V2 H2G = G, the universal coeﬃcient theorem gives us a split exact sequence 0 → Ext(G, A) → 2 θ V2 V2 H (G, A) → Hom( G, A) → 0. The isomorphism ψ : G → H2G is given by g ∧ h 7→ [g|h] − [h|g] 2 by Exercise V.6.4(a). Now the map β : H (G, A) → Hom(H2G, A) in the universal coeﬃcient sequence sends the class kfk of the cocycle f to (βkfk)([g1|g2]) = f(g1, g2). Then θ is given by kfk 7→ βkfk 7→ (βkfk ◦ ψ)(g ∧ h) = (βkfk)([g|h] − [h|g]) = f(g, h) − f(h, g), and this element is an alternating map. Thus θ coincides with the map θ in Exercise IV.3.8(c), and so we see that every alternating map comes ∼ from a 2-cocycle. It also follows that Ext(G, A) = Eab(G, A), whence the name “Ext.”

46 6 Chapter VI: Cohomology Theory of Finite Groups

2.1: Suppose |G : H| < ∞ and that M is a ZG-module with a relative injective resolution Q, and η : M → 0 0 G G 0 Q is the canonical admissible injection (i.e. Q = CoindH ResH M). If M is free as a ZH-module then Q η is ZG-free by Corollary VI.2.2[1]. Since η is H-split, the exact sequence 0 → M → Q0 → Cokerη → 0 is split-exact and hence Q0 =∼ M ⊕Cokerη. Since M and Q0 are both ZH-free (a free ZG-module is ZH-free by Exercise I.3.1), Cokerη is by definition stably free. Let us find this cokernel explicitly. The canonical G ∼ G G G injection is given by M,→ HomH (ZG, ResH M) = IndH ResH M = ZG ⊗ZH ResH M, m 7→ [s 7→ sm] 7→ P −1 G G ∼ g∈G/H g ⊗g m. But IndH ResH M = Z[G/H] ⊗ M with the mapping g ⊗m 7→ g ⊗gm, by Proposition P III.5.6[1]. Thus the canonical injection is given by m 7→ g∈G/H g ⊗ m. We can now consider M as a free ZH-module, and without loss of generalization we can assume M = ZH (since direct sums commute L with tensor products). We can also regard Z[G/H] = g∈G/H Z[g] as a direct sum of integers via the L|G:H| L|G:H| isomorphism Z[g] =∼ Z, g 7→ 1. Thus the H-split injection is η : ZH,→ Z ⊗ ZH =∼ ZH with the mapping m 7→ (g1, . . . , g|G:H|) ⊗ m 7→ (1,..., 1) ⊗ m 7→ (m, . . . , m). The cokernel of this map L|G:H| L|G:H|−1 Ln is ZH/ZH =∼ ZH, a free ZH-module. Indeed, For any finite direct sum X, the quotient of this group by its diagonal subgroup X = {(x, . . . , x)} is the direct sum Ln−1 X because −1 −1 any element (x1, . . . , xn−1, xn)X is equivalent in the quotient group to (x1xn , . . . , xn−1xn , 1)X via the −1 −1 −1 element (xn , . . . , xn , xn ). Since Cokerη is free as a ZH-module, we can apply Corollary VI.2.2[1] to get an admissible injection Cokerη ,→ Q1 with Q1 free as a ZG-module. Continuing in this way, we obtain the resolution Q as a complex of free ZG-modules.

3.1: Let G act freely on S2k−1, as on pg20[1]. From this we have a free resolution of Z over ZG which is periodic of period 2k: ε · · · → C1 → C0 → C2k−1 → · · · → C1 → C0 → Z → 0 2k−1 where C∗ = C∗(X) and X ≈ S is a free G-complex that makes each Ci ﬁnitely generated. Then by Proposition VI.3.5[1] we can take the above resolution ε : C → Z, form the dual (backwards resolution) of its suspension ε∗ : Z∗ = Z → ΣC, and then splice together C and ΣC to form a complete resolution F for G. This resolution is obviously periodic of period 2k because ε is periodic of period 2k and the suspension just shifts the resolution (leaving the period unaltered) and the dual functor forms a periodic resolution of the same period. If G = hti is ﬁnite cyclic of order n, and k = 1, then G acts by rotations on the circle (n vertices/edges) and we have a periodic resolution of period 2: t−1 N t−1 ε · · · → ZG → ZG → ZG → ZG → Z → 0 where N = 1 + t + ··· + tn−1 is the norm element. Now ε(1) = 1, so by Proposition VI.3.4[1] the dual ∗ ∗ P t−1 N ε : Z → ZG is given by ε (1) = g∈G g = N. The maps ZG → ZG and ZG → ZG are invariant under ∼ the dual functor because HomG(ZG, ZG) = ZG. Therefore, the explicit complete resolution is: t−1 N t−1 N t−1 N t−1 · · · → ZG → ZG → ZG → ZG → ZG → ZG → ZG → ZG → · · · .

5.1: We have a natural map H∗ → Hb ∗ which is an isomorphism in positive dimensions and an epimorphism in dimension 0. The cup product for both functors agrees in dimension 0 because `: H0(G, M) ⊗ H0(G, N) → H0(G, M ⊗ N) is the map M G ⊗ N G → (M ⊗ N)G induced by the in- G G 0 0 0 clusions M ,→ M and N ,→ N, and `: Hb (G, M) ⊗ Hb (G, N) → Hb (G, M ⊗ N) is induced by G G G 0 0 M ⊗ N → (M ⊗ N) via the surjection H Hb . From this compatibilism in dimension 0 we can deduce that the diagram Hp(G, M) ⊗ Hq(G, M) ` / Hp+q(G, M ⊗ N)

Hb p(G, M) ⊗ Hb q(G, M) ` / Hb p+q(G, M ⊗ N) commutes for all p, q ∈ Z. Indeed, embed M in the (co)induced module M¯ = Hom(ZG, M) =∼ Z ⊗ M (noting that G is ﬁnite for our purposes) and let 0 → M → M¯ → C → 0 be the canonical Z-split exact sequence (see Exercise III.7.3). For any G-module N the sequence 0 → M ⊗N → M¯ ⊗N → C ⊗N → 0 is

47 exact, and the module M¯ ⊗ N is induced (see Exercise III.5.2(b)). We therefore have dimension-shifting isomorphisms δ : Hp(G, C) → Hp+1(G, M) and δ : Hp(G, C ⊗ N) → Hp+1(G, M ⊗ N) which commutes with the cup product (see pg110[1]). The compatibilism for p = 0, q = 0 allows us to prove by ascending induction on p that the above diagram is commutative for p ≥ 0 and q = 0. Embedding N in an induced module, we then see that it commutes for q ≥ 0. The scenario p, q < 0 is trivial because Hp = 0 for p < 0. Thus the cup product on Hb ∗ is compatible with that deﬁned originally on H∗, and the natural map H∗ → Hb ∗ preserves products.

∗ ∗ ∗ ∗ 5.2(a): Let Hb (G)(p) be the p-primary component of Hb (G) = Hb (G, Z), so that we have Hb (G) = L ∗ ∗ ∗ ∗ p | |G| Hb (G)(p). Since Hb (G)(p) is a subgroup of Hb (G), in order to show that it is an ideal in Hb (G) ∗ ∗ ∗ it suffices to show that α ` β ∈ Hb (G)(p) for α ∈ Hb (G)(p) and β ∈ Hb (G). But this is trivial because s s s L ∗ if p α = 0 then p (α ` β) = (p α) ` β = 0 ` β = 0. Now q6=p Hb (G)(q) is also an ideal because ∗ ∗ a sum of ideals is an ideal. Consequently, Hb (G)(p) is a quotient ring of Hb (G) via the projection ∗ ∗ Hb (G) Hb (G)(p). ∗ ∗ ∗ Note: The p-primary ring Hb (G)(p) is not a subring of Hb (G) because the inclusion Hb (G)(p) ,→ ∗ ∗ 0 Hb (G) does not preserve identity elements. The identity of Hb (G)(p) is 1 ∈ Zpr = Hb (G)(p) while the ∗ 0 identity of Hb (G) is 1 = (1,..., 1) ∈ Z|G| = Hb (G), where we have used the factorization of Z|G| into the direct sum of its p-primary components and |G| = prm with p - m. The inclusion will send the identity 1 to (1, 0, 0,..., 0) 6= 1, where we take the first summand of Z|G| to be Zpr . ∼ ∗ = Q ∗ 5.2(b): Consider the group isomorphism ϕ : Hb (G) → p | |G| Hb (G)(p), where each factor on the right is a ring via part(a), and multiplication in the product is done componentwise [note: we switch from direct sum to direct product notation in order to emphasize the fact that we are dealing cat- P egorically with rings]. The map is given by ϕ(α) = (. . . , αp,...), where α = p αp in the decom- ∗ L ∗ position Hb (G) = p | |G| Hb (G)(p). In order to show that this map is a ring isomorphism it suffices to show that (αβ)p = αpβp, where αβ = α ` β, for then ϕ(α ` β) = (..., (αβ)p,...) = P P (. . . , αpβp,...) = (. . . , αp,...)(. . . , βp,...) = ϕ(α) ` ϕ(β). Writing α = p αp and β = p0 βp0 we P P 0 have αβ = p,p0 αpβp0 = p αpβp, because αpβp06=p is annihilated by both p and p 6= p, and hence annihilated by 1 = gcd(p, p0) [note that gcd(p, p0) = mp + np0 for some m, n ∈ Z by the Euclidean P Algorithm]. It is now obvious that (αβ)p = ( q αqβq)p = αpβp.

6.1:

6.2: Let R be a ring, let C be a chain complex of finitely generated projective R-modules, and let C¯ be the dual complex HomR(C,R) of finitely generated projective right R-modules. For any ¯ 0 0 ¯ 0 z ∈ (C⊗b RC)n any any chain complex C , there is a graded map ψz : HomR(C,C ) → C⊗b RC of ¯ degree n, given by ψz(u) = (idC¯ ⊗bu)(z). Let z ∈ (C⊗b RC)0 correspond to idC under the isomor- ¯ phism ϕ : C⊗b RC → HomR(C,C) from Exercise VI.6.1. This element is indeed a cycle, because 0 ϕ−1(∂z) = D0ϕ0(z) = D0(idC ) = d ◦ idC − (−1) idC ◦ d = d − d = 0 implies that ∂z = 0 since ϕ ¯ ∗ p is an isomorphism. Then z = (zp)p∈Z, where zp ∈ C−p ⊗R Cp = (Cp) ⊗R Cp corresponds to (−1) idCp ∗ ∼ under the canonical isomorphism φ :(Cp) ⊗R Cp = HomR(Cp,Cp) of Proposition I.8.3[1] given by p c¯⊗c 7→ hc,¯ xic. Indeed, denoting zp =c ¯⊗c, the canonical isomorphism gives hc,¯ xic = (−1) idCp and the −p·p p2 p p(p+1) isomorphism ϕ0 = (ϕ−pp)p∈Z then becomes (−1) hc,¯ xic = (−1) (−1) idCp = (−1) idCp = idCp , 0 ¯ 0 agreeing with our choice for z. Now ψz is induced by maps ψpq : HomR(C−p,Cq) → Cp ⊗R Cq. Since 0 u ∈ HomR(C−p,Cq) is of degree p + q and idC¯p is in dimension p, these maps are clearly given by (q+p)p ψpq(u) = (−1) (idC¯p ⊗ u)(z−p); see the definition of a map between completed tensor products on −1 0 0 pg137[1]. Then Exercise I.8.7 states that ϕ = ψ −1 = ψ , where ψ : Hom (C ,C ) → C¯ ⊗ C pq ϕ (idCp ) r r R −p q p R q −1 pq −1 pq p is defined by ψr(u) = (idC¯p ⊗R u)(r). Since r = ϕ (idCp ) = (−1) φ (idCp ) = (−1) (−1) z−p, we −1 pq p (pq+p2) have ϕpq (u) = ψr(u) = (idC¯p ⊗R u)((−1) (−1) z−p) = (−1) (idC¯p ⊗ u)(z−p) = ψpq(u), noting p p2 ¯ 0 0 that (−1) = (−1) . Therefore, ψz is the inverse of the isomorphism ϕ : C⊗b RC → HomR(C,C ) of Exercise VI.6.1.

48 6.3: Let G be a finite group, let F be an acyclic chain complex of projective ZG-modules, and let ε0 : F 0 → Z be a complete resolution. Part(b) of Proposition VI.6.1[1] states that if F is of finite type 0 0 then ε ⊗ M induces a weak equivalence F ⊗b G(F ⊗ M) → F ⊗b GM = F ⊗G M. The proof of the propo- ¯ sition considers the dual F = HomG(F, ZG) and utilizes the fact that it is projective. However, if we did not impose the finiteness hypothesis on F then its dual would not necessarily by projective. Indeed, L∞ ∼ Q∞ the dual of an infinite direct sum is an infinite direct product, and HomG( ZG, ZG) = ZG is not ZG-projective. If it were ZG-projective, then by Exercise I.8.2 it would also be Z-projective (i.e. Q∞ free abelian). But any subgroup of a free abelian group is free abelian by Theorem I.7.3[5], and Z Q∞ is a subgroup of ZG which is not free abelian, giving the desired contradiction. Thus the finiteness hypothesis is necessary for the given proof – this does not guarantee that the finiteness hypothesis is necessary for the statement of the proposition.

6.4:

∼ i = 7.1(a): Let ϕ : Hb (G, M) → Hb−1−i(G, M) be the isomorphism established in the proof of Proposi- −1 tion VI.7.2[1], which on the chain level has the inverse ϕ : HomG(F, ZG) ⊗G M → HomG(F,M) given by u ⊗ m 7→ [x 7→ u(x) · m], and let z = ϕ(1) ∈ Hb−1(G, Z). For an arbitrary G-module coeﬃcient homomorphism h : M → N we have a commutative diagram ϕ−1 HomG(F, ZG) ⊗G M / HomG(F,M)

α β ϕ−1 HomG(F, ZG) ⊗G N / HomG(F,N) where α(u ⊗ m) = u ⊗ h(m) and β(f) = h ◦ f, because βϕ−1(u ⊗ m) = β[u(x) · m] = h(u(x) · m) = u(x) · h(m) = ϕ−1(u ⊗ h(m)) = ϕ−1α(u ⊗ m). Thus ϕ−1 is natural and hence so is ϕ. Since ϕ is natural the following diagram with short exact rows is commutative (suppressing the end 0’s) 0 00 HomG(F,M ) / HomG(F,M) / HomG(F,M )

ϕ ϕ ϕ 0 00 HomG(F, ZG) ⊗G M / HomG(F, ZG) ⊗G M / HomG(F, ZG) ⊗G M and so ϕ is compatible with connecting homomorphisms in long exact sequences by Proposition I.0.4[1].

0 7.1(b): By definition of z, ϕ and a z agree on 1 ∈ Hb (G, Z) since 1 a z = z. If now M and u ∈ Hb 0(G, M) are arbitrary, there is a coefficient homomorphism Z → M such that 1 7→ u under the induced map α : Hb 0(G, Z) → Hb 0(G, M), noting that this cohomology map is induced from H0(G, Z) = ZG → M G = H0(G, M). Since the cap product is natural with respect to coefficient homomorphisms we have a commutative diagram 0 az Hb (G, Z) / Hb−1(G, Z)

α β 0 az Hb (G, M) / Hb−1(G, M) which deﬁnes β(z) = β(1 a z) = α(1) a z = u a z. Thus by naturality of ϕ from part(a) we have an analogous commutative diagram as above (replacing a z with ϕ), and this yields ϕ(u) = ϕα(1) = βϕ(1) = β(z) = u a z.

7.1(c): The maps ϕ and a z agree in dimension 0 (referring to the domain) by part(b), and ϕ is δ-compatible by part(a). Thus we can use dimension-shifting [the δ boundary isomorphisms] to deduce that ϕ and a z agree in all dimensions, up to sign. Indeed, we have the commutative diagram 0 ϕ Hb (G, M) / Hb−1(G, M)

∼= δ ∼= δ n ϕ Hb (G, K) / Hb−1−n(G, K)

49 where the vertical maps are due to iterations of the dimension-shifting technique 5.4 on pg136[1]. These isomorphisms provide ambiguity in the sign, so ϕ(u) = ±u a z in any dimension and hence Proposition VI.7.2[1] has been reproved (the isomorphism is given by the cap product with the fundamental class z).

7.2: Let A be an abelian torsion group, and consider the injective resolution 0 → Z → Q → Q/Z → 0 of δ0 δ1 Z. Applying Hom(A, −) gives the cochain complex 0 → Hom(A, Z) → Hom(A, Q) → Hom(A, Q/Z) → 0, and we have Ext(A, ) ≡ Ext1 (A, ) = Kerδ1/Imδ0 = Hom(A, / )/Imδ0 = A0/Imδ0. But is torsion- Z Z Z Q Z Q free, so Hom(A, Q) = 0 and Imδ0 = 0. Thus Ext(A, Z) = A0. Equivalently, Theorem 17.1.10[2] provides us with a long exact sequence 0 → Hom(A, Z) → Hom(A, Q) → Hom(A, Q/Z) → Ext(A, Z) → Ext(A, Q). But Hom(A, Q) = 0 as mentioned above, and Ext(A, Q) = 0 ∼= by Proposition 17.1.9[2] since Q is Z-injective. Thus we have a desired isomorphism A0 = Hom(A, Q/Z) → Ext(A, Z).

7.3: Let G be a finite group, let M be a G-module which is free as an abelian group, and let F be a projective resolution of Z over ZG. Note that M ∗ = Hom(M, Z) by Proposition VI.3.4[1]. Consider the split exact coefficient sequence 0 → Z → Q → Q/Z → 0. Since M is Z-free, applying the functor Hom(M, −) yields the exact sequence 0 → M ∗ → Hom(M, Q) → M 0 → 0 where M 0 = Hom(M, Q/Z); this sequence is Z-split exact because the original coefficient sequence is split exact (Hom commutes with direct sums). I claim that Hb ∗(G, Hom(M, Q)) = 0 and Hb ∗(G, Hom(M, Q) ⊗ M) = 0. Assuming this for the moment, we then have dimension-shifting isomorphisms δ : Hb j(G, M 0) → Hb j+1(G, M ∗) for all j. It is a fact that the tensor product of a G-module with a Z-split exact sequence is exact, so 0 → M ∗ ⊗ M → Hom(M, Q) ⊗ M → M 0 ⊗ M → 0 is an exact sequence. Thus we also have dimension- shifting isomorphisms δ : Hb j(G, M 0 ⊗ M) → Hb j+1(G, M ∗ ⊗ M). Moreover, we have a commutative diagram ` α Hb i−1(G, M 0) ⊗ Hb −i(G, M) / Hb −1(G, M 0 ⊗ M) / Hb −1(G, Q/Z)

∼= δ⊗id ∼= δ ∼= δ ` β Hb i(G, M ∗) ⊗ Hb −i(G, M) / Hb 0(G, M ∗ ⊗ M) / Hb 0(G, Z) where the left-side square follows from compatibility with δ (see pg110[1]) and the right-side square follows from naturality of the long exact cohomology sequence (see pg72[1]); α and β are induced by the evaluation maps. Since the top row is a duality pairing by Corollary VI.7.3[1], so is the bottom row. It suﬃces to prove the claim. The analog of Proposition III.10.1[1] for Tate cohomology states that if Hb n(H,M) = 0 for some n with H ⊆ G, then Hn(G, M) is annihilated by |G : H|. Taking H = {1} H and M a rational vector space, the norm map N : MH → M is an isomorphism (N = idM ). Then Hb −1({1},M) = KerN = 0 = CokerN = Hb 0({1},M), so Hb −1(G, M) and Hb 0(G, M) are annihilated by |G| and are thus trivial groups since |G| is invertible in M. The claim is now justiﬁed since Hom(M, Q) and Hom(M, Q) ⊗ M are both rational vector spaces.

0 7.4: Let k be an arbitrary commutative ring and Q an injective k-module. Let A = Homk(A, Q) for any k-module A. If M is a kG-module, then the pairing Hb i(G, M 0) ⊗ Hb −1−i(G, M) →` Hb −1(G, M 0 ⊗ M) → Hb −1(G, Q) ,→ Q induces an isomorphism Hb i(G, M 0) =∼ Hb −1−i(G, M)0. Indeed, this is simply the analog of Corollary VI.7.3[1], and the proof of that corollary goes through untouched if we replace Q/Z by Q (as both are injective) and Z by k (as both are commutative ring coeﬃcients).

8.1: Let G be a group and M a ZG-module such that Hb ∗(G, M) = 0 but M is not cohomologically ∼ trivial. If G is cyclic, then by Theorem VI.8.7[1] it cannot be a p-group, so G = Zp ×Zq = Zpq for distinct primes p and q. The complete resolution from Exercise VI.3.1 then implies that Hb n(G, M) = CokerN n G for n even and Hb (G, M) = KerN for n odd (see pg58[1]), so N : MG → M is an isomorphism. Also, i j Proposition VI.8.8[1] implies that either Hb (Zp,M) 6= 0 for some i or Hb (Zq,M) 6= 0 for some j (or i both). As mentioned on pg150[1] as a consequence of Theorem VI.8.5[1], Hb (Zp, Zp) 6= 0 for all i > 0. Zpq Therefore, let us consider M = Zp. It suﬃces to ﬁnd a Zpq-action on Zp such that N :(Zp)Zpq → (Zp) Zpq is an isomorphism. But Zp is simple, so either Zpq acts trivially on Zp or (Zp)Zpq = (Zp) = 0. But if

50 Zpq acts trivially on Zp then N : Zp → Zp is the zero map (hence has nontrivial kernel) since multipli- Zpq cation by |Zpq| = pq annihilates Zp. Thus we must have (Zp)Zpq = (Zp) = 0. Taking p to be an odd i 2i prime, this condition is satisﬁed by the pq-action x · m = m , where p = hmi and pq = hxi, as long Z pq Z Z as 2pq ≡ 1 mod p (because we must have m = xpq · m = m2 ). For example, {p = 3, q = 2} works, as does {p = 7, q = 3}. As a result, a desired example is the group G = Z6 = hxi and the Z6-module M = Z3 = hmi coupled with the action x · mi = m2i.

8.2: Suppose M is a G-module which is Z-free and cohomologically trivial (G is of course ﬁnite). Then M is ZG-projective by Theorem VI.8.10[1], so F = M ⊕ N for some projective module N and free module F. For any G-module K, the module Hom(F,K) is an induced module by Exercise III.5.2(b) ∼ 0 G 0 0 (since F = ZG ⊗ F = Ind{1}F where F is a free Z-module of the same rank) and hence cohomologically trivial. Since the Hom-functor commutes with direct sums, Hom(M,K) is also cohomologically trivial for any G-module K. Alternatively, for any G-module K choose an exact sequence 0 → L → F → K → 0 with F free (such sequences exist because every module is a quotient of a free module). Since M is Z-free, we can apply Hom(M, −) to get the exact sequence 0 → Hom(M,L) → Hom(M,F ) → Hom(M,K) → 0. Since F and L are also Z-free, Hom(M,L) and Hom(M,F ) are cohomologically trivial by Lemma VI.8.11[1]; L is free because it embeds in the free Z-module F and any submodule of a Z-free module is free. Thus Hom(M,K) is cohomologically trivial by the long exact Tate cohomology sequence.

8.3(a): Let M and P be ZG-modules such that M is Z-free and P is ZP -projective, and consider ϕ any exact sequence 0 → P →i E → M → 0. The obstruction to splitting the sequence lies in H1(G, Hom(M,P )) =∼ Ext1 (M,P ), where the isomorphism follows from Proposition III.2.2[1]. More ZG precisely, we have a short exact sequence of G-modules 0 → Hom(M,P ) → Hom(M,E) → Hom(M,M) → δ 1 0 since M is Z-free, and this yields the sequence HomG(M,E) → HomG(M,M) → H (G, Hom(M,P )) G 0 via the long exact cohomology sequence, where we recall that HomG(·, ·) = Hom(·, ·) = H (G, Hom(·, ·)). Hence the extension splits iff δ(idM ) = 0, because if the extension splits then there is a section s : M → E which maps onto idM (i.e. s 7→ ϕ ◦ s = idM ) and so idM ∈ Kerδ by exactness, and if δ(idM ) = 0 then by exactness there exists a map M → E which maps onto idM and that map is then the desired section. It suffices to show that Hom(M,P ) is cohomologically trivial, for then H1(G, Hom(M,P )) = 0 and δ = 0. By additivity [Hom commutes with direct sums], it suffices to show that Hom(M, F) is cohomologically trivial for any free ZG-module F, since the projective P is a direct summand of some F. But ∼ G 0 0 F = Coind{1}F where F is a free Z-module of the same rank (by Corollary VI.2.3[1]), so Hom(M, F) is induced (by Exercise III.5.2(b)) and hence cohomologically trivial. Alternatively, since M is Z-free the original exact sequence in consideration is Z-split (see Exercise AE.27), so the injection i : P → E of G-modules is a Z-split injection, hence admissible. Since P is G-projective, it is relatively injective by Corollary VI.2.3[1] and so the mapping problem P i / E

idP f P can be solved (i.e. there exists a map f : E → P such that f ◦ i = idP ). But this just means that i ϕ f : E → P is a ZG-splitting homomorphism for the sequence 0 → P → E → M → 0, and so this sequence splits.

8.3(b): Let M be a ZG-module such that proj dim M < ∞, and consider the projective resolution 0 → Pn → · · · → P0 → M → 0. We can break this up into short exact sequences Zi → Pi → Zi−1 → 0, where Zi is the kernel of Pi → Pi−1. Now Zi (i ≥ 0) is Z-free because it is a submodule of a ZG-projective module which is a submodule of a ZG-free module F, and F is also necessarily Z- free and any subgroup (in particular, Zi) of a Z-free group is Z-free. Therefore, for the sequence 0 → Pn = Zn−1 → Pn−1 → Zn−2 → 0 with Pn G-projective and Zn−2 Z-free, part(a) above implies that ∼ this sequence splits and hence Pn−1 = Pn ⊕ Zn−2. Since Pn−1 is G-projective, so is Zn−2. One now sees by descending induction on i that Zi is G-projective for i ≥ 0, so that 0 → Z0 → P0 → Z−1 = M → 0

51 is a projective resolution of length 1. Thus proj dim M ≤ 1.

∼ 2k−1 8.4: Let G be a group such that there exists a free, finite G-CW-complex X with H∗(X) = H∗(S ). 2k−1 Since Hi(S ) = 0 for i 6= 0 and i 6= 2k −1, the augmented cellular chain complex C∗ = C∗(X) is a free ∂2k−1 ∂2k−2 resolution of Z over ZG up to dimension 2k − 1. From the chain sequence C2k−1 → C2k−2 → C2k−3 ∼ st we have Ker∂2k−2 = Im∂2k−1 = C2k−1/Ker∂2k−1 [the isomorphism is due to the 1 Isomorphism ∂2k−2 Theorem] and hence we have an exact sequence 0 → C2k−1/Ker∂2k−1 ,→ C2k−2 → C2k−3 → · · · . Now Im∂2k = B is the module of (2k − 1)-boundaries of C∗, and we have a surjection C2k−1/B ∼ ∼ C2k−1/Ker∂2k−1 with kernel Ker∂2k−1/B = Z [note: this isomorphism comes from the fact that Z = 2k−1 ∂2k−2 H2k−1(S )]. Thus we have an exact sequence S of G-modules 0 → Z → C2k−1/B → C2k−2 → ε C2k−3 → · · · → C0 → Z → 0 where each Ci is G-free. I claim that C2k−1/B is Z-free and has finite projective dimension. Assuming this claim holds, C2k−1/B is cohomologically trivial by Theorem VI.8.12[1] and hence is ZG-projective by Theorem VI.8.10[1]. Thus we can splice together an infinite number of copies of S (which forms an acyclic complex of projective G-modules) and we can then apply Proposition VI.3.5[1] to obtain a complete resolution which is periodic of period 2k. It suffices to prove the claim. Since C2k−1 is ZG-free, it is necessarily Z-free and hence any subgroup (in particular, B) must also be Z-free; thus C2k−1/B is Z-free. As X is a finite complex, C∗(X) stops after Cn(X) for some 2k − 1 < n < ∞. ∂n ∂2k Thus we have a finite projective resolution 0 → Cn → · · · → C2k → C2k−1 → C2k−1/B → 0 and proj dim C2k−1/B < ∞.

9.1: Let G be a nontrivial ﬁnite group which has periodic cohomology of period d. Then there is an element u ∈ Hb d(G) which is invertible in the ring Hb ∗(G), so cup product with u gives a periodicity isomorphism v 7→ u ` v. Taking v = u and using anti-commutativity of the cup product, d2 2 u ` u = (−1) (u ` u). If d is not even, then 2u = 0. If |G| = 2 then G is cyclic (of order 2) and hence 2 2 has period d = 2 (which is even), so we must have |G| ≥ 3. But then 2u = 0 implies that u ` u = u = 0 and hence u = 0 by the periodicity isomorphism. This contradicts the fact that u is nontrivial (it is invertible), so d must be even.

9.2: If G is cyclic then Hb ∗(G) is periodic of period 2 because G admits a 2-dimensional ﬁxed-point- i free representation as a group of rotations (see pg154[1]); we could also just note that Hb (G) is Z|G| for i even and is 0 for i odd. Conversely, if a group G has periodic cohomology of period 2, then −2 ∼ 0 Gab = H1G = Hb (G) = Hb (G) = Z|G|. Now |G| = |Gab| = |G|/|[G, G]| ⇒ |[G, G]| = 1 and hence ∼ G = Gab = Z|G|, which is cyclic.

∗ −1 0 9.3: Suppose Hb (G) is periodic of period 4. We have Hb (G) = 0 and Hb (G) = Z|G|, as explained on −2 1 1 pg135[1]. We also have Hb (G) = H1G = Gab, and Hb (G) = H G = Hom(G, Z) = 0 by Exercise III.1.2 (noting that G is ﬁnite by hypothesis). Thus, since Hb n(G) =∼ Hb n+4(G) for all n,   Z n = 0 Z n = 0     ∼ Gab n ≡ 1 mod 4 n ∼ Z|G| n ≡ 0 mod 4 , n 6= 0 Hn(G) = H (G) = Z|G| n ≡ 3 mod 4 Gab n ≡ 2 mod 4   0 otherwise 0 otherwise Note that this reproves Exercise II.5.7(a), because the ﬁnite subgroup G ⊂ S3 ⊂ H∗ has Hb ∗(G) periodic of period 4 (see pg155[1]) and hence H2G = 0.

∗ ∼ Q ∗ 9.4: Suppose the finite group G has periodic cohomology. Now Hb (G) = p | |G| Hb (G)(p) by Exer- ∗ ∗ cise VI.5.2(b), and each factor Hb (G)(p) embeds in Hb (Sylp(G)) by the Tate cohomology version of i i Theorem III.10.3[1], so Hb (G) = 0 for i odd if Hb (Sylp(G)) = 0 for each prime p. Thus to show that Hb i(G) = 0 for i odd it suffices to do this when G is a p-group. For the p-group G (of order pr) with periodic cohomology, G is either a cyclic group Zpr or a generalized quaternion group Q2r by Proposition VI.9.3[1]. If G is cyclic then it has period 2 by Exercise VI.9.2, so since Hb −1(G) = 0, Hb i(G) = 0 for i odd. If G is generalized quaternion then it has period 4 as explained on pg155[1] (it is a finite subgroup of H∗), so by Exercise VI.9.3, Hb i(G) = 0 for i odd.

52 9.5: Suppose G is a p-group which has a unique subgroup C of order p; note that C is necessarily cyclic. Choose a fixed-point-free representation of C on a 2-dimensional vector space W (as a group of rotations), and form the induced module V = ZG ⊗ZC W . Since C is unique, it is normal in G and G ∼ L hence ResC V = g∈G/C gW by Proposition III.5.6[1]; each gW is clearly a fixed-point-free representation of C (c · gw = (cg)w 6= gw). Consequently, V is a fixed-point-free represenation of C. But then V is also a fixed-point-free representation of G, for a nontrivial isotropy group Gv(v ∈ V − {0}) would contain an element x of order p (by Cauchy’s Theorem, Theorem 3.2.11[2]) and hence would contain C (uniqueness implies C = hxi), contradicting the fact that C acts freely on V − {0}. Thus G admits a periodic complete resolution of period 2|G : C| as explained on pg154[1], so G has periodic cohomology.

9.6: Let G = Zm o Zn, where m and n are relatively prime and Zn acts on Zm via a homomor- ∗ phism Zn → Zm whose image has order k. If a prime q divides n, then a Sylow q-subgroup H lies in Zn and is necessarily central (in Zn) because cyclic groups are abelian. By Theorem III.10.3[1] we ∗ ∼ ∗ n/H ∗ have Hb (Zn)(q) = Hb (H)Z and by Exercise III.8.1 we know that Zn/H acts trivially on Hb (H), ∗ ∼ ∗ ∗ ∗ ∼ ∗ so Hb (H) = Hb (Zn)(q). By Theorem III.10.3[1] we also have Hb (G)(q) ⊆ Hb (H) = Hb (Zn)(q). Since ∗ ∗ ∗ ∼ ∗ Hb (Zn)(q) ⊆ Hb (G)(q) by Exercise AE.55, we must have Hb (G)(q) = Hb (Zn)(q). Now if a prime p divides ∗ ∼ ∗ m, then the same argument yields Hb (H) = Hb (Zm)(p) where H is now a Sylow p-subgroup of Zm ⊂ G. −1 ∼ We have H/G because H is the unique subgroup of Zm /G (hence gHg = H for all g ∈ G), so Theorem ∗ ∼ ∗ G/H ∼ ∗ G/H ∗ (Zm/H)oZn III.10.3[1] implies Hb (G)(p) = Hb (H) = Hb (Zm)(p) = Hb (Zm)(p) . We have a Zn-action ∗ on Zm given by ϕ : Zn → Aut(Zm) = Zm, and we have a trivial Zm/H-action ψ : Zm/H → Aut(Zm) given by ψ(g) = idZm ≡ 1. We then have a (Zm/H) o Zn-action ψ o ϕ which is precisely the action ∗ ϕ : Zn → Zm because (ψ oϕ)(g, z) = ψ(g)·ϕ(z) = 1·ϕ(z) = ϕ(z). Thus we can consider the G/H-action ∗ ∼ ∗ Zn ∗ on Zm (hence on its Tate cohomology) as the Zn-action, so Hb (G)(p) = Hb (Zm)(p). Since Hb (G) is the direct sum of its primary components and p (resp. q) ranges over prime divisors of m (resp. n), we have ∗ ∼ ∗ ∗ n the isomorphism Hb (G) = Hb (Zn) ⊕ Hb (Zm)Z . Let us examine the Zn-action a little more carefully. The image of Zn under the action-homomorphism k 2 consists solely of automorphisms f : Zm → Zm such that f = idZm . Such a map induces f∗ = Hb (f) on 2 ∼ nd Hb (Zm) = Zm. If α generates the 2 -dimension cohomology, then in the cohomology ring, f∗(α) = λα k k k i for some λ ∈ Zm. But α = id∗(α) = f∗ (α) = λ α, so λ ≡ 1 mod m. Noting that Hb (Zm) = 0 for i i odd, but is nontrivial for i even. Now f∗(α ) = f∗(α ` ··· ` α) = f∗(α) ` ··· ` f∗(α) = (λα) ` ··· ` i i i n (λα) = λ α which is the identity only when i is a multiple of 2k. Thus Hb (Zm)Z is nontrivial only when i ∈ 2kZ (in which case it is Zm since the action is trivial). Therefore,  i ≡ 0 mod 2k Zmn i ∼ i i n ∼ Hb (Zm o Zn) = Hb (Zn) ⊕ Hb (Zm)Z = 0 i odd  Zn otherwise for the case gcd(m, n) = 1. This means that the period of H∗(G) is 2k.

9.7: Let Fq be a ﬁeld with q elements, where q is a prime power. The special linear group G = SLn(Fq) is ∗ the kernel of the surjective determinant homomorphism det : GLn(Fq) → Fq , i.e. it is the group of matrices with determinant 1. Let us ﬁrst assume that n ≥ 3. Then the cyclic groups A = hdiag(a, a−1, 1,..., 1)i −1 and B = hdiag(1,..., 1, b, b )i form a non-cyclic abelian subgroup A×B ⊆ G, noting that Fq is commu- 1 a tative. If we now let n = 2 then we will assume that q is not prime. Then the cyclic groups A = h 0 1 i 1 b 1 a 1 b 1 a+b and B = h 0 1 i form a non-cyclic abelian subgroup A × B ⊆ G, since 0 1 0 1 = 0 1 with b not equal to any multiple of a (and vice versa). By Theorem VI.9.5[1], G = SLn(Fq) does not have periodic cohomology if n ≥ 3 or if q is not prime. Note that if n = 2 and q is prime then SL2(Fq) does have periodic cohomology, as explained on pg157[1].

9.8:

9.9: Suppose that G has p-periodic cohomology. Let P ⊆ G be a subgroup of order p, let N(P ) (resp. C(P )) be the normalizer (resp. centralizer) of P in G, and let W = N(P )/C(P ); note that

53 if C(P ) = P then W is called the Weyl group. Choose a Sylow p-subgroup H containing P . Since G has p-periodic cohomology, H is either cyclic or generalized quaternion by Theorem VI.9.7[1]. I assert that H ⊆ C(P ) ⊆ N(P ). Indeed, if H is cyclic then it is necessarily abelian so the result follows, and if H is generalized quaternion (must have p = 2) then it has a unique element of order 2 ∼ (as stated on pg98[1]) and this is then the generator for P = Z2 which must be central in H, so the ∗ result follows. Denote by XG the G-invariant elements of Hb (H,M) and similarly for XN(P ), where −1 ∗ H gHg an element z ∈ Hb (H,M) is G-invariant if resH∩gHg−1 z = resH∩gHg−1 gz for all g ∈ G. Theorem ∗ ∼ ∗ ∼ III.10.3[1] states that Hb (G, M)(p) = XG and Hb (N(P ),M)(p) = XN(P ). Now trivially, XG ⊆ XN(P ), ∗ ∼ ∼ ∗ so it suﬃces to show that XN(P ) ⊆ XG, for then Hb (G, M)(p) = XG = XN(P ) = Hb (N(P ),M)(p). Note that P is the unique subgroup in H of order p because if H is generalized quaternion then the reasoning is as stated above and if H is cyclic then every subgroup has unique order (by Theorem 2.3.7[2]). If H ∩ gHg−1 is trivial then every element in H∗(H,M) is clearly invariant for such g ∈ G. If H ∩ gHg−1 is not trivial then its order is at least p and the intersection contains P . This implies P ⊆ gHg−1 ⇒ g−1P g ⊆ H ⇒ g−1P g = P and hence g ∈ N(P ), so the question of G-invariance reduces to the question of N(P )-invariance, i.e. XN(P ) ⊆ XG. Also, XN(P ) ⊆ XC(P ) trivially, so by Theorem III.10.3[1] we have the inclusion (up to isomorphism) ∗ ∗ Hb (N(P ),M)(p) ⊆ Hb (C(P ),M)(p). Since H is a Sylow p-subgroup contained in C(P ), |W | and N(P ) N(P ) p are relatively prime and hence corC(P ) resC(P ) = |N(P ): C(P )| = |W | is an isomorphism on ∗ Hb (N(P ),M)(p), where the equality is due to Proposition III.9.5[1]. Thus the restriction map induces ∗ ∗ N(P ) a monomorphism Hb (N(P ),M)(p) ,→ Hb (C(P ),M). But as explained on pg84[1], if z = resC(P ) u ∗ N(P ) then z is N(P )-invariant; let the N(P )-invariants be denoted by Y ⊆ Hb (C(P ),M). Thus resC(P ) ∗ ∗ W maps Hb (N(P ),M)(p) monomorphically into Y . Since C(P ) /N(P ), Y = Hb (C(P ),M) as noted ∗ ∗ W on pg84[1]. Thus Hb (N(P ),M)(p) ⊆ Hb (C(P ),M) , and so combining the two inclusions we see that ∗ ∗ ∗ W ∗ W Hb (N(P ),M)(p) ⊆ Hb (C(P ),M)(p) ∩ Hb (C(P ),M) = Hb (C(P ),M)(p). For the other direction, if ∗ W N(P ) ∗ z ∈ Y(p) = Hb (C(P ),M)(p) then consider the element w = corC(P ) z. Since Hb (C(P ),M)(p) is annihi- ∗ lated by a power of p, w ∈ Hb (N(P ),M)(p). Regurgitating the proof of Theorem III.10.3[1] using the N(P ) 0 0 ∗ double-coset formula, we deduce that z = resC(P ) w where w = w/|W | ∈ Hb (N(P ),M)(p). This means W ∗ N(P ) ∗ that H(C(P ),M)(p) ⊆ Hb (N(P ),M)(p) because resC(P ) maps Hb (N(P ),M)(p) monomorphically into ∗ ∼ ∗ ∼ ∗ W the N(P )-invariants. Thus Hb (G, M)(p) = Hb (N(P ),M)(p) = Hb (C(P ),M)(p).

9.10: For any finite group G, the augmentation ideal I ⊂ ZG is a cyclic G-module if G is cyclic group by Exercise I.2.1(b); we could also just note that if G = hsi then I = ZG · (s − 1) because I consists of elements of the form sk − 1 [we then form the elements si − sj ∈ I via summation], and sk − 1 = N · (s − 1) where N = sk−1 + ··· + s + 1 ∈ ZG. Conversely, if I is cyclic as a G-module, so that I = ZG · x, then I claim that G admits a periodic resolution of period 2. Assuming this for the moment, G then has periodic cohomology (of period 2) by Theorem VI.9.1[1] and hence G is cyclic by Exercise VI.9.2. It suffices to prove the claim. The multiplication map G → G given by r 7→ rx has image I Zx Zε and kernel K, so we can form the exact sequence 0 → K → ZG → ZG → Z → 0. Under the category of abelian groups, the Rank-Nullity Theorem gives |G| = dimZZG = dimZI + dimZZ = dimZI + 1 for the augmentation map ε, and gives |G| = dim ZG = dim K +dim I for the x-multiplication map. The first Z Z Z ∼ equation implies dimZI = |G| − 1 and the second equation then implies dimZK = 1, i.e. K = Z as an |G| |G| abelian group. So with K = hki, G acts on K via gk0 = zk0 (for z ∈ Z). But then k0 = g k0 = z k0 x ε and hence z = 1, i.e. the G-action is trivial. Our exact sequence is now 0 → Z → ZG → ZG → Z → 0. Splicing together this sequence infinitely many times, we obtain the desired periodic resolution.

54 7 Chapter VII: Equivariant Homology and Spectral Sequences

0 00 2.1: Let 0 → C → C → C → 0 be a short exact sequence of chain complexes, and let {FpC} be 0 1 0 0 the filtration such that F0C = 0, F1C = C , and F2C = C. Then E1q = H1+q(C /0) = H1+q(C ) and 1 0 00 1 1 1 1 E2q = H2+q(C/C ) = H2+q(C ) and Epq = 0 for p ≤ 0 and p > 2. The differential d : Epq → Ep−1,q 00 0 1 gives maps ∂ : Hn(C ) → Hn−1(C ) with n = 2 + q and ∂ = d2q. 1 1 1 1 d3 1 ∂ 1 d1 1 2 1 Now E is given by · · · → E3q = 0 → E2q → E1q → E0q = 0 → · · · , and from E = H(E ) we see that 2 1 0 2 1 2 E1q = Kerd1/Im∂ = H(C )/Im∂ = Coker∂ and E2q = Ker∂/Imd3 = Ker∂ and Epq = 0 for p ≤ 0 and d2 d2 p > 2. In E2 we have Ker∂ →2q 0 and Coker∂ →1q 0 and trivial maps for p 6= 1 and p 6= 2, because the 2 2 2 2 2 differential d : Epq → Ep−2,q+1 is of bidegree (−2, 1). Thus H(E ) = E , and so the spectral sequence “collapses” at E2 (i.e. E2 =∼ E∞) because Er+1 =∼ H(Er) in general. ∞ ϕ Since Ep = Grp H(C) := FpH(C)/Fp−1H(C) where FpH(C) = Im{H(FpC) → H(C)}, we see that ∞ 2 ∼ ∞ 2 ∼ ∞ Ep = 0 for p ≤ 0 and p > 2 and Coker∂ = E = E1 = F1H(C) and Ker∂ = E = E2 = ∼ F2H(C)/F1H(C). Since F2H(C) = H(C) we have a surjection Hn(C) F2H(C)/F1H(C) = Ker∂, 0 ϕ and since F1H(C) = Im{H(C ) → H(C)} we have a commutative diagram H (C0) / H (C) n n O

ϕ % % ? F1Hn(C) ∞ hence an injection Coker∂ = F1Hn(C) ,→ Hn(C). We can now rewrite the E sequence 0 → F1Hn(C) → F2Hn(C)/F1Hn(C) → 0 as a short exact sequence 0 → Coker∂ ,→ Hn(C) Ker∂ → 0, and we then have a commutative diagram j H (C0) i / H (C) / H (C00) n 9 n : n

ϕ % % + $ $ , Coker∂ Ker∂ The top row is exact at Hn(C) because Kerj = Ker{Hn(C) Ker∂} = Coker∂ = Imϕ = Imi. Since Imj = Ker∂ and Keri = Kerϕ = Im∂, the top row of this commutative diagram extends to a long exact homology sequence. We have thus deduced the familiar long exact homology sequence from this spectral sequence, and the spectral sequence of a ﬁltered complex can be regarded as a generalization of the long exact sequence associated to a chain complex and a subcomplex.

3.1(a): Let C be a ﬁrst-quadrant double complex such that the associated spectral sequence to Fp(TC)n = L 1 1 1 Ci,n−i has Epq = 0 for q 6= 0, and let D be the chain complex E∗,0 with diﬀerential d . We i≤p L L L have the isomorphisms Hn(TC) = Ker[ p≤n Cp,n−p → p≤n−1 Cp,n−1−p]/Im[ p≤n+1 Cp,n+1−p → L L L 1 p≤n Cp,n−p] = [ p≤n−1 Hn−p(Cp,∗)]⊕X = [ p≤n−1 Ep,n−p]⊕X = 0⊕X, where X = Cn,0/(Im[Cn+1,0 → 1 Cn,0] ⊕ Im[Cn,1 → Cn,0]) = Hn(C∗,0/Im[C∗,1 → C∗,0]) = Hn(H0(C∗,∗)) = Hn(E∗,0) = Hn(D). Thus ∼ Hn(TC) = Hn(D).

3.1(b): Take τ : TC → D to be the canonical surjection, and note that this can be viewed as a map of double complexes C → D (where D is regarded as a double complex concentrated on the line 1 1 1 q = 0); this is obviously a ﬁltration-preserving chain map. Now Epq(D) = Hq(Ep,∗ ≡ Ep,0) which is 0 if 1 1 q 6= 0, and is Ker∂0 ≡ Ep,0 if q = 0 (since Ep,−1 = 0). This is precisely the spectral sequence associated to C, so the induced map on spectral sequences from τ is an isomorphism at the E1-level. Thus by Propo- ∼ sition VIII.2.6, τ induces an isomorphism H∗(TC) → H∗(TD) = H∗(D) and hence is a weak equivalence.

4.1: Suppose X is the union of subcomplexes Xα such that every non-empty intersection Xα0 ∩· · ·∩Xαp (p ≥ 0) is acyclic, and let K be the nerve of the covering {Xα}. Let C be the double complex L Cpq = σ∈K(p) Cq(Xσ), and let T be the [total] chain complex TC. As shown on pg167[1], we have 1 a spectral sequence with Epq equal to 0 if q 6= 0 and equal to Cp(X) if q = 0. Then Exercise 3.1(b)

55 implies that we have a weak equivalence T → C(X). Moreover, we have another spectral sequence with 1 1 Epq = Cp(K, Hq) where Hq ≡ {Hq(Xσ)} is a coeﬃcient system on K. Since each Xσ is acyclic, Epq is 0 for q 6= 0 and is Cp(X, Z) for q = 0. Then Exercise 3.1(b) implies that we have a weak equivalence ∼ T → C(K). Thus we have an isomorphism H∗(X) = H∗(K).

7.2: Let X be a G-complex such that for each cell σ of X, the isotropy group Gσ fixes σ pointwise; in this case the orbit space X/G inherits a CW-structure. Assume further that each Gσ is finite. We have a spectral sequence E1 = L H (G ,M ) ⇒ HG (X,M), where Σ is a set of representatives for X /G pq σ∈Σp q σ σ p+q p p (Xp is the set of p-cells of X) and Mσ = Zσ ⊗ M (Zσ is the Gσ-module additively isomorphic to Z, on which Gσ acts by the orientation character χσ : Gσ → {±1}). Since Gσ fixes σ pointwise, χσ(Gσ) = {1} and hence Zσ = Z and hence Qσ = Z ⊗ Q = Q (where we now take rational coefficients M = Q). Now for all q > 0, Hq(Gσ, Q) = Hq(Gσ) ⊗ Q = 0 where the first equality is proved in Exercise AE.6 and the 1 latter equality follows from the fact that Hq(Gσ) is finite (proved in Exercise AE.16). The E term is 2 1 then concentrated on the line q = 0, and the spectral sequence therefore collapses at E = H∗(E ) to ∗ yield HG(X, ) =∼ H (L H (G , )) = H (H (G, L IndG )) = H (H (G, C (X; ))) = ∗ Q ∗ σ∈Σp 0 σ Qσ ∗ 0 σ∈Σp Gσ Qσ ∗ 0 p Q H∗(Cp(X; Q)G) = H∗(X/G; Q), where the starred equality is the result of Shapiro’s lemma and the last equality is given by Proposition II.2.4[1]. If X is also contractible, then X is necessarily acyclic. Thus the above result and Proposition VII.7.3[1] ∼ G ∼ imply H∗(G, Q) = H∗ (X, Q) = H∗(X/G; Q). Note: The hypothesis that Gσ fixes σ pointwise is not very restrictive in practice. In the case of a simplicial action, it can always be achieved by passage to the barycentric subdivision X˜. Indeed, for 0 0 σ ⊂ X˜ which lies in σ ⊂ X, Gσ0 ⊆ Gσ. If Gσ0 did not fix σ pointwise then this would break continuity of the G-action on σ (consider two such simplices of X˜ which lie in σ and have a common face).

7.3: Let X be a G-complex and E a free contractible G-complex. There is a G-map X × E → X G G which is a homotopy equivalence, so Proposition VII.7.3 implies that H∗ (X × E) → H∗ (X) is an iso- G ∼ morphism. As G acts freely on X × E, H∗ (X × E) = H∗((X × E)/G). Thus we have the equivalence G ∼ G H∗ (X) = H∗(X × E).

7.5: Let X be a G-complex and let N be a normal subgroup of G which acts freely on X. Let Zσ be the orientation module and let Xp denote the set of p-cells of X. Let Σp be a set of representatives 0 for Xp/G and let Σp be a set of representatives for (X/N)p/(G/N); it is easy to see that both sets G ∼ G/N are in bijective correspondence. To prove that H∗ (X,M) = H∗ (X/N, M) with any G/N-module L ∼ L coefficient M, it suffices to show that Hq(Gσ,Mσ) = 0 0 Hq((G/N)σ0 ,Mσ0 ); this is because σ∈Σp σ ∈Σp we have a spectral sequence E1 = L H (G ,M ) ⇒ HG (X,M) and so the said isomorphism pq σ∈Σp q σ σ p+q will give isomorphic associated graded modules (since Er = H(Er−1) and E∞ is the associated graded G ∼ G/N module), and this will give H∗ (X,M) = H∗ (X/N, M) by Lemma VII.2.1[1]. In view of the bi- 0 0 ∼ jection Σp → Σp, σ 7→ σ , it suffices to show that Hq(Gσ,Mσ) = Hq((G/N)σ0 ,Mσ0 ). First note that Mσ ≡ Zσ ⊗ M = Zσ0 ⊗ M ≡ Mσ0 because the Gσ-action and the (G/N)σ0 -action on Z coincide (if g ∈ Gσ preserves the orientation of σ ∈ X, gσ = +σ, then gN preserves the orientation of σ0 = σ = nσ ∈ X/N because +σ0 = +σ = gσ = g · nσ = (gn)σ = (gn)σ0 ∈ X/N for all n ∈ N) and because the two actions coincide on M by definition (since M is a G/N-module). Thus it suffices ∼ to show (due to the Künnethformula) that Hq(Gσ) = Hq((G/N)σ0 ) where the homology is now us- ∼ 0 ing integer coefficients, and in turn it suffices to show that Gσ = (G/N)σ0 where σ is the image of σ under the quotient X → X/N. Consider the obvious monomorphism ϕ : Gσ → (G/N)σ0 given by g 7→ gN; it suffices to show that ϕ is surjective. But this is immediate, because if gN ∈ (G/N)σ0 0 0 −1 then gNσ = σ which implies gn1σ = n2σ for some n1, n2 ∈ N which implies n2 gn1 ∈ Gσ, and then −1 −1 −1 −1 −1 ϕ(n2 gn1) = n2 gn1N = n2 gN = n2 g(g n2g)N = gN where we note that N/G.

10.1: The proof of Theorem VII.10.5[1] used the assumption that |G| = p (p prime) to state that n ∼ dimZp Hb (G, Z)p = 1 for all n ∈ Z, because G = Zp has Tate cohomology group Zp in every dimension. The proof also used that assumption in order to apply Proposition VII.10.1[1] to the G-invariant G G subcomplex X of X; the isotropy group Gσ for every cell σ ∈ X − Y cannot equal G (because X is the largest subcomplex on which G acts trivially) and hence must be the trivial group (the only proper

56 ∼ subgroup of G = Zp).

10.2: The extended theorem holds for |G| = p1 by the original Theorem VII.10.5[1], so we proceed by induction, assuming the extended theorem holds for |G| = pr. Let |G| = pr+1 and remember the hypothesis that XH is a subcomplex for all H ⊆ G. Since G is a p-group we can choose a maximal normal subgroup N of index p, so that XG = (XN )G/N . Via induction (since |N| = pr) we can apply the extended theorem to X with XN [which is subcomplex by hypothesis], so every condition of the theorem on X is also satisfied on XN . Since |G/N| = p we can apply the original theorem to XN with (XN )G/N [which is a subcomplex since it is equal to XG], so every condition of the theorem on XN is also satisfied on (XN )G/N . Thus every condition of the extended theorem on X is also satisfied on (XN )G/N = XG, and the proof is complete.

∼ 2k 10.3: Let X be a finite-dimensional free G-complex (G finite) with H∗(X) = H∗(S ). Proposition G VII.10.1[1] (with Y = ∅) implies that Hb∗ (X,M) = 0, noting that Gσ is trivial for all σ since X is 2 G G-free. On the other hand, we have a spectral sequence Epq = Hbp(G, Hq(X,M)) ⇒ Hbp+q(X,M). Since the spectral sequence is concentrated on the horizontal lines q = 0 and q = 2k (the only 2k+1 nonzero homology terms of the 2k-sphere), it follows that the differential d : Hbp(G, H2k(X,M)) → Hbp−(2k+1)(G, H2k+(2k+1)−1(X,M)) = Hbp−2k−1(G, H4k(X,M)) = 0 is an isomorphism. To prove that every nontrivial element of G acts nontrivially on H2k(X) it suffices to show that every cyclic subgroup of G (generated by the elements of G) acts nontrivially on H2k(X), so we are immediately re- duced to the case where G is cyclic and nontrivial. Using the isomorphism d2k+1 with M = Z and p odd, we conclude that Hbodd(G, H2k(X)) = 0. This means G acts nontrivially on H2k(X), otherwise Hbodd(G, H2k(X)) = Hbodd(G, Z) = Z|G| which is not equal to 0. The proof is now complete. Note that ∼ a nontrivial G-action on H2k(X) = Z means that G = Z2, so having every nontrivial element of G act nontrivially on H2k(X) means that |G| ≤ 2 (the case |G| = 1 is satisfied vacuously since there are no nontrivial elements).

57 8 Chapter VIII: Finiteness Conditions

2.1: By deﬁnition, cd Γ = 0 iﬀ Z admits a projective resolution 0 → P → Z → 0 of length 0, i.e. Z =∼ P and Z is ZΓ-projective. But Exercise I.8.1 states that only the trivial group Γ = {1} makes Z a projective module. Thus the trivial group is the only group of cohomological dimension zero.

L 2.2: Take Γ = Z2. Then cd Γ = ∞ by Corollary 2.5. Now a free ZΓ-module F is a direct sum ZΓ where the Γ-action would be a parity-permutation on all or some of the summands, so Hn(Γ,F ) =∼ L n 2 L n [ H (Γ, (ZΓ) )] ⊕ [ H (Γ, ZΓ)] where the ﬁrst collection of summands has the nontrivial Γ-action. 2 Γ But then that module (ZΓ) is an induced module Ind{1}ZΓ, and so are the other modules ZΓ (trivially). So these modules are H∗-acyclic and hence Hn(Γ,F ) = 0 for all n > 0, i.e. sup {n | Hn(Γ,F ) 6= 0 for some F } = 0.

2.7(a): Induced Γ-modules ZΓ ⊗ A are cohomologically trivial (as noted on pg148[1]) and hence have projective dimension ≤ 1 by Theorem VI.8.12[1].

n+1 2.7(b): If proj dimRM ≤ n, then ExtR (M, −) = 0 by Lemma VII.2.1[1]. For any direct summand 0 n+1 0 n+1 M of M, we must then have ExtR (M , −) = 0 since ExtR (−, −) commutes with direct sums. Thus 0 proj dimRM ≤ n by Lemma VII.2.1[1].

2.7(c): Suppose Γ is finite and M is a Γ-module in which |Γ| is invertible. It suffices to show that M is a direct summand of an induced module Γ ⊗ A, for then proj dim ( Γ ⊗ A) ≤ 1 by part(a) and Z ZΓ Z hence proj dim M ≤ 1 by part(b). Take A = M , where M is the underlying abelian group of M. By ZΓ 0 0 −1 Corollary III.5.7[1] there is a Γ-module isomorphism ϕ : ZΓ⊗M → ZΓ⊗M0 given by g ⊗m 7→ g ⊗g m, P where ZΓ ⊗ M has the diagonal Γ-action. The inclusion i : M → ZΓ ⊗ M given by m 7→ g ⊗ m is P P P g∈Γ a Γ-module homomorphism because γ · i(m) = γ · ( g g ⊗ m) = g γg ⊗ γm = g g ⊗ γm = i(γ · m). 1 The Γ-module map π : ZΓ ⊗ M0 → M defined by g ⊗ m 7→ |Γ| gm is a Γ-splitting to ϕi [note: the P P −1 action on ZΓ ⊗ M0 is γ · (g ⊗ m) = γg ⊗ m]. Indeed, π[ϕi](m) = π[ϕ( g g ⊗ m)] = π[ g g ⊗ g m] = 1 P −1 1 |Γ| g gg m = |Γ| |Γ|m = m = idM (m). Thus the injection ϕi : M → ZΓ ⊗ M0 splits, and M is then (by definition of a splitting homomorphism) a direct summand of ZΓ ⊗ M0 as a Γ-module.

4.1: If Z is finitely presented as a ZΓ-module then Z is finitely generated and every surjection P Z (with P finitely generated and projective) has a finitely generated kernel (Proposition VIII.4.1[1]). In particular, the augmentation map ε : ZΓ → Z has kernel I which then must be finitely generated (obviously noting that ZΓ is free of rank 1). Exercise I.2.1(d) then implies Γ is a finitely generated group. Conversely, suppose Γ is a finitely generated abelian group, so that Γ = F (S)/R is a group presentation for Γ with |S| < ∞. Then there is an exact sequence (ZΓ)|S| → ZΓ → Z → 0 by Exercise IV.2.4(d) and hence Z is finitely presented as a ZΓ-module by Proposition VIII.4.1[1].

6.1: Let Γ be of type FL and cd Γ = n. Then Γ is of type FP and hence of type FP∞ by Propo- sition VIII.6.1, and so there is a partial resolution Fm → · · · → F0 → Z → 0 with each Fi free of finite rank by Proposition VIII.4.3 (for all m ≥ 0). Due to its cohomological dimension, we can make a finite projective resolution 0 → P → Fn−1 → · · · → F0 → Z → 0 with each Fi free and P projective. Then Proposition VIII.6.5 implies that P is stably free, and so there is some free module F of finite rank such that P ⊕ F is free. Take the free resolution 0 → F → F → 0 → · · · → 0 and consider its direct sum with the finite projective resolution. This gives us a finite free resolution for Z over ZΓ of length n.

6.3: Let Γ be of type FP and cd Γ = n. Then we have a finite projective resolution Pn → · · · → P0 → Z → 0, and taking the Hom-dual we obtain the resolution for cohomology which behaves as · · · → HomZΓ(Pn, ZΓ) → 0. Since Pn is a finitely generated projective module, so is its Hom-dual; thus Hn(Γ, ZΓ) is a finitely generated Γ-module.

58 9 Chapter IX: Euler Characteristics

1.1: For a ring A, suppose there is a Z-valued function r on ﬁnitely generated projective A-modules, satisfying r(P ⊕Q) = r(P )+r(Q) and r(A) = 1 and r(P ) > 0 if P 6= 0. I claim that A is indecomposable, i.e. that A cannot be decomposed as the direct sum of two non-zero left ideals. Indeed, a decomposition A = I ⊕ J yields the equation 1 = r(I) + r(J) because I and J are projective A-modules (direct summands of the free module A). Since both ideals are non-zero, r(I) ≥ 1 and r(J) ≥ 1, and this yields the desired contradiction 1 ≥ 1 + 1 = 2.

∗ 2.1: Let P be a ﬁnitely generated projective (left) A-module and let P = HomA(P,A) be its dual. ∗ ∗ ∼ Then P is a right A-module and we have P ⊗A P = HomA(P,P ) by Proposition I.8.3. Consider the diagram ∼ ∗ = P ⊗A P / HomA(P,P )

ev tr $ x T (A) where ev(u ⊗ x) = u(x) is the evaluation map. The isomorphism is given by u ⊗ x 7→ [p 7→ u(p) · x], and P P on basis elements ei ∈ P this image homomorphism is ei 7→ j u(ei)rj · ej, where x = j rjej. Thus P P P the composition is u⊗x 7→ tr[p 7→ u(p)·x] = i u(ei)ri = i ri · u(ei) = i u(riei) = u(x) = ev(u⊗x), and the diagram is commutative.

2.4: Let F be a ﬁnitely generated free module and e : F → F a projection operator of F onto a 2 direct summand isomorphic to P (this is idempotent since e = e). Then e = i ◦ idP ◦ π, where i and π are the inclusion and projection maps between F and P , so tr(e) = tr(i ◦ idP ◦ π) = tr(idP ) = R(P ). Thus R(P ) is equal to the trace of an idempotent matrix deﬁning P .

2.5: Let Γ be a group and ϕ : ZΓ → Z the augmentation map, and let P be a ﬁnitely generated projective Γ-module. Then Proposition 2.3 implies that trZ(Z⊗Γ idP ) = T (ϕ)(trZΓ(idP )) = T (ϕ)(RZΓ(P )). As this is an element of T (Z) = Z, it is immediate that trZ(Z ⊗Γ idP ) is precisely the Z-rank of PΓ = Z ⊗Γ P . Thus εΓ(P ) = T (ϕ)(RZΓ(P )).

2.6(a): Let Γ be a finite group. From the definition, trZΓ/Z : T (ZΓ) → T (Z) lifts to the map ZΓ → T (Z) = Z given by γ 7→ trZ(µγ ), where µγ : ZΓ → ZΓ is right-multiplication by γ. Taking µγ as a matrix over Z, it is the identity matrix for γ = 1 and is a matrix with zeros on the diagonal for ¯ 1 6= γ ∈ Γ(γ permutes the basis elements). Thus trZΓ/Z(1) = |Γ| and trZΓ/Z(¯γ) = 0 for 1 6= γ ∈ Γ. Consequently, there is a well-defined homomorphism τ : T (ZΓ) → Z such that τ(1)¯ = 1 and τ(¯γ) = 0 for 1 6= γ ∈ Γ, and one has trZΓ/Z = |Γ| · τ.

2.6(b): Applying Proposition 2.4 to α = idP and ϕ : Z ,→ ZΓ, and following the same method as in the proof of Exercise 2.5, and using the result of part(a), we have that rkZ(P ) = |Γ| · τ(RZΓ(P )) for any ﬁnitely generated projective Γ-module P . Thus ρΓ(P ) = τ(RZΓ(P )).

2.7: The previous exercise shows that ρΓ(P ) ∈ Z, for Γ a finite group. It is obvious from the definition of ρ that ρΓ(P ) > 0 if P 6= 0 (it must be greater than or equal to 1/|Γ|). Now for two finitely generated pro-

jective Γ-modules P and Q, they are necessarily free Z-modules and hence rkZ(P ⊕Q) = rkZ(P )+rkZ(Q). Furthermore, ρΓ(ZΓ) = rkZ(ZΓ)/|Γ| = |Γ|/|Γ| = 1. Thus ZΓ is indecomposable by Exercise 1.1.

4.1: The proof of Theorem IX.4.4[1] used the assumption that Γ was ﬁnite in order to apply Proposition IX.4.1[1] by replacing Γ by the cyclic subgroup Γ0 = hγi. The proposition requires Γ0 to be of ﬁnite index (for all γ ∈ Γ), and this will hold in general if |Γ| < ∞.

4.3(b): It is a fact from representation theory that a finitely generated kΓ-module is determined up to isomorphism by its character. By part(a), the character is in bijective correspondence with the Hattori-Stallings rank. Thus two finitely generated kΓ-modules are isomorphic iff they have the same

59 Hattori-Stallings rank.

4.3(c): Take k = Q. If Γ is finite and P is a finitely generated projective ZΓ-module then Q ⊗Z P is a finitely generated projective QΓ-module, hence a finitely generated projective ZΓ-module by Ex- ercise I.8.2. Then by Theorem 4.4 there is an integer r such that RΓ(Q ⊗Z P ) = r · [1]. But this r Hattori-Stallings rank is precisely that of (QΓ) , so Q ⊗Z P is a free QΓ-module by part(b).

4.4: Taking ρΓ(P ) = RΓ(P )(1) as a deﬁnition, the result is precisely Proposition 4.1 applied to γ = 1.

60 10 Additional Exercises

∼ G 1: Find a counterexample to the statement MG = M for a G-module M.

∼ ∼ ∼ For an arbitrary group G we have (ZG)G = Z ⊗ZG ZG = Z. Alternatively, (ZG)G = ZG/[I · ZG] = ZG/I =∼ Z (where the latter isomorphism follows from application of the 1st Isomorphism Theorem on the augmentation map). If G is finite, then the norm element N exists, and the integer multiples zN (z ∈ Z) are the only elements of ZG fixed by all g ∈ G (assuming left-multiplication action), so (ZG)G = Z · N [=∼ Z] is the ideal generated by N. But if G is infinite then there is no norm element, and (ZG)G = 0. P To prove this last statement, take any nonzero element x = i rigi of ZG (since it’s a finite sum we can assume 0 ≤ i ≤ n and all ri ∈ Z are nonzero) and consider the set S = {g0, . . . , gn}. To show that x∈ / (ZG)G it suffices to show that there is at least one nontrivial g with gS 6= S. We can assume 1 ∈/ S, 2 otherwise for any g∈ / S we have g · 1 = g∈ / S. Suppose that gS = S ∀ g ∈ G. Then g0 · g0 = g0 ∈ S, i n+1 n+2 and so through trivial induction we see that g0 ∈ S (1 ≤ i ≤ n + 1) and g0 · g0 = g0 ∈/ S (otherwise i |S| > n + 1), noting that g0 6= 1 since 1 ∈/ S. Thus we have arrived at a contradiction, and so the choices G = Z and M = ZG suffice. Another solution uses the choices G = Z2 = hxi and M = Z where G acts by x · n = −n. Then x · n = n ⇒ −n = n ⇒ 2n = 0, so the largest quotient on which G acts trivially is ZZ2 = Z2, and 2n = 0 only holds for n = 0 ∈ Z, hence ZZ2 = 0.

∗ ∗ 2: Let Z2 = hxi act on the additive complex numbers C by x · z = z , where z = x − iy is the Z2 ∼ ∼ 1 complex conjugate of z = x + iy.[Note: C = CZ2 = R]. Find H (Z2, C).

0 2 For a derivation f : Z2 → C we have f(x ) = f(x ) = f(1) = 0 and f(x) determines f. Now f(x2) = f(x) + x · f(x) = f(x) + f(x)∗ and so we must have f(x) = −f(x)∗ (i.e. a pure-imaginary number). Since iy = −(iy)∗ and 0 + iy = (iy)/2 + (iy)/2 = (iy)/2 − (−iy)/2 = (iy)/2 − (iy)∗/2, we have 1 1 ∗ 1 ∗ 1 1 ∗ 1 ∗ f(x) = 2 f(x) − 2 f(x) = − 2 f(x) − x · [ 2 f(x)] = x · [ 2 f(x) ] − [ 2 f(x) ]. Therefore, f is a principal 1 derivation and hence H (Z2, C) = 0, using the result of Exercise III.1.2 above.

k 1 3: Let the multiplicative cyclic group C2k = hxi act on Z by x · n = (−1) n. Calculate H (C2k, Z) under this action.

1 ∼ For k = 2m even, C2k acts trivially on Z, so H (C2k, Z) = Hom(C2k, Z) = 0 where this equation follows 1 from Exercise III.1.2 above. For k = 2m + 1 odd, we start by viewing the 1-cocycles as Z = Der(C2k, Z) 1 1 2i 2i−1 2i−1 and the 1-coboundaries as B = PDer(C2k, Z). For f ∈ Z , f(x ) = f(x)+x·f(x ) = f(x)−f(x ) and f(x2i) = f(x2i−1) + x2i−1 · f(x) = f(x2i−1) + (−1)(2i−1)(2m+1)f(x) = f(x2i−1) − f(x), so 2f(x) = 2f(x2i−1) ⇒ f(x) = f(x2i−1) and f(x2i) = 0. So Z1 consists of the derivations which are determined by f(x) and satisfy the derived properties, hence Z1 =∼ Z. For f ∈ B1, f(xi) = xi · n − n = (−1)in − n which satisﬁes f(xi) = 0 ∀ i = 2j and f(xi) = −2n ∀ i = 2j + 1, and so B1 =∼ 2Z. Thus 1 ∼ H (C2k, Z) = Z/2Z = Z2 when k is odd. 1 C2k Alternatively (for k odd), we know that H (C2k, Z) = KerN where the map N : ZC2k → Z is the norm map induced by multiplication on Z by the norm element N. Now Nm = 1 · m + x · m + x2 · 2k−1 m + ··· + x m = m − m + m − m + ··· + m − m = 0 and so NZ = 0 ⇒ KerN = ZC2k . Since i i x · n = (−1) n, the action is trivial for all g ∈ C2k if −n = n, hence the largest quotient on which C2k

acts trivially is ZC2k = Z2.

4: Noting that the only nontrivial map ϕ : Zi = hti → Zij = hsi is the canonical inclusion deﬁned j by t 7→ s , show that the induced map under H2n−1 is the same inclusion. This is the corestriction map Zij cor : H∗( i) → H∗( ij). Zi Z Z

Considering the two periodic free resolutions of Z, there exists an augmentation-preserving chain map f between them by Theorem I.7.5[1] and we look at the commutative diagram in low dimensions:

61 t−1 ε Z[Zi] / Z[Zi] / Z / 0

f f id 1 0 Z s−1 ε Z[Zij] / Z[Zij] / Z / 0 The right square yields f0(1) = 1, and the left square yields (s − 1)f1(1) = f0(t − 1) = t · f0(1) − f0(1) = j j−1 j−1 ϕ(t)1 − 1 = s − 1 = (s − 1)(1 + s + ··· + s ) ⇒ f1(1) = 1 + s + ··· + s . We assert that f is j−1 given by fn(1) = 1 for n even and fn(1) = 1 + s + ··· + s for n odd. Indeed, assuming inductively ij−1 that the chain map is valid up to n, it suffices to check commutativity (1 + s + ··· + s )fn+1(1) = i−1 fn(1 + t + ··· + t ) for n odd and (s − 1)fn+1(1) = fn(t − 1) for n even. For the latter case [n even] j−1 j j j we have (s − 1)(1 + s + ··· + s ) = (s − 1) and fn(t − 1) = (s − 1)fn(1) = s − 1, so commutativity i−1 j ij−j is satisfied. For the former case [n odd] we have fn(1 + t + ··· + t ) = (1 + s + ··· + s )fn(1) = j ij−j j−1 ij−1 ij−1 ij−1 (1+s +···+s )(1+s+···+s ) = 1+s+···+s and (1+s+···+s )fn+1(1) = (1+s+···+s ), so commutativity is satisfied. Using this chain map, and after moving to quotients, the cycle elements (for odd-dimensional homology) are mapped via ϕ∗(1) = 1 + 1 + ··· + 1 = j while the boundary elements are mapped via ϕ∗(1) = 1; thus the result follows.

n xn−1 5: Let d : G → A be a derivation. Prove the relation d(x ) = x−1 dx for x ∈ G.

0 1−1 For n = 0, d(1) = d(x ) = x−1 dx = 0, and we argue by induction on n. Assuming the relation at k+1 k k xk−1 x−1+xk+1−x xk+1−1 n = k holds, d(x ) = d(x · x ) = d(x) + x · d(x ) = [1 + x x−1 ]dx = [ x−1 ]dx = x−1 dx and we are ﬁnished. xn−1 n−2 n−1 Alternatively, we note that x−1 = 1 + x + ··· + x + x , so the relation immediately follows by successive calculations d(xn) = d(x) + x · d(xn−1) = d(x) + x[d(x) + x · d(xn−2)] = d(x) + x · d(x) + x2[d(x) + x · d(xn−3)] = [1 + x + x2]d(x) + x3 · d(xn−3), etc..

6: What information do we obtain about the homology of a group G by computing its homology with rational coeﬃcients?

Assuming Q is an abelian group with trivial G-action, we can apply the result of Exercise III.1.2 to obtain Z the short exact sequence 0 → Hn(G) ⊗ Q → Hn(G, Q) → Tor1 (Hn−1(G), Q) → 0. Since Q is torsion-free Z ∼ we have the equality Tor1 (Hn−1(G), Q) = 0 and hence the isomorphism Hn(G, Q) = Hn(G) ⊗ Q. Now |a|q q q A ⊗ Q = 0 for any torsion abelian group A because q ⊗ a = |a| ⊗ a = |a| ⊗ |a|a = |a| ⊗ 0 = 0. Therefore, dimQ(Hn(G, Q)) = rkZ(HnG). Moreover, if Hn(G, Q) is nontrivial then HnG is torsion-free. Ln 7: Let the multiplicative cyclic group Cn = hxi act on M = j=1 Z2 by x · ai = ai+1 where aj th 1 generates the j Z2-summand. Compute H (Cn,M).

1 n n−1 n−1 For f ∈ Z = Der(Cn,M) we have 0 = f(x ) = f(x) + x · f(x) ⇒ x · f(x) = −f(x) = f(x) [we can drop the negative sign because the maximum order for elements is 2]. So f is determined by f(x) = n−1 (z1, . . . , zn) and we must have (z1, . . . , zn) = x · (z1, . . . , zn) = (z2, . . . , zn, z1) ⇒ z1 = z2 = ··· = zn. 1 i i−1 Thus f(x) is 0 or (1, 1,..., 1), and Z = Z2. But f(x ) = f(x) + x · f(x) = f(x) + f(x) = 2f(x) = 0, and f(x) = (1, 1,..., 1) = x · n − n where n ∈ M is the element consisting of alternating 1’s and 0’s [the 1 1 case f(x) = 0 is trivial]. Thus f ∈ B = PDer(Cn,M), and so H (Cn,M) = 0.

n n 8: Let GLn(Z) act on Z by left matrix multiplication, where we consider Z as an n × 1 column n n vector with integral entries. Compute the induced map ψ : H∗(GLn(Z), Z ) → H∗(GLn(Z), Z ) under n the action of −[δij] on z ∈ Z .

The anti-identity matrix m = −[δij] is in the center Z(GLn(Z)) and so the conjugation action on −1 GLn(Z) by m is the identity. Thus we can rewrite the map of the action m · z as (g 7→ mgm = n g , z 7→ m · z = −z) ∈ (GLn(Z), Z ). By Proposition III.8.1[1] this map induces the identity on the respective homology with coeﬃcients, hence ψ = id∗.[But clearly ψ = −id∗ because it’s induced from n n z 7→ −z ∈ Z . Thus 2 · id∗ = 0 and H∗(GLn(Z), Z ) is all 2-torsion].

62 9: The cyclic group Cm is a normal subgroup of the dihedral group Dm = Cm o C2 (of symmetries −1 of the regular m-gon). There is a C2-action on Cm = hσi given by σ 7→ σ . Determine the action of −1 −1 C2 on the homology H2i−1(Cm, Z), noting that there is an element g ∈ Dm such that gσg = σ .

Letting c(g): Cm → Cm be conjugation by g, we can apply Corollary III.8.2[1] to obtain the in- ∼ duced action of Dm/Cm = C2 on H∗(Cm, Z) given by z 7→ c(g)∗z. It suffices to compute c(g)∗ on the chain level, using the periodic free resolution P of Cm, and using the trivial action on Z. Us- −1 ing the condition τ(hx) = [c(g)](h)τ(x) = h τ(x) on the chain map τ : P → P (for h ∈ Cm), we i m−i claim that τ2i−1(x) = τ2i(x) = (−1) σ x for i ∈ N and τ0(x) = x. Assuming this claim holds, i m−i the chain map P ⊗Cm Z → P ⊗Cm Z [in odd dimensions] is given by x ⊗ y 7→ (−1) σ x ⊗ gy = i m−i i i−m i (−1) σ x ⊗ y = (−1) x ⊗ σ y = (−1) x ⊗ y, and so c(g)∗ [hence the C2-action] is multiplication i by (−1) on H2i−1(Cm, Z). It suffices to prove the claim. Seeing that Nτ2i(1) = τ2i−1(N) = Nτ2i−1(1) where N is the norm element, we can restrict our attention to τ2i−1 and use induction on i since m−1 m−1 −1 (σ − 1)τ1(1) = (σ − 1)(−σ ) = σ − 1 = σ − 1 = τ0(σ − 1). This chain map must satisfy commutativity (σ − 1)τ2i−1(1) = τ2(i−1)(σ − 1), and this is indeed the case because (σ − 1)τ2i−1(1) = i m−i+1 m−i −1 i−1 m−i+1 i m−i+1 m−i (−1) (σ − σ ) and τ2(i−1)(σ − 1) = (σ − 1)(−1) σ = (−1) (σ − σ ). P 10: Assuming |G : H| < ∞, we define tr : MG → MH by tr(m) = g∈H\G gm, and we then define G the transfer map resH : H∗(G, −) → H∗(H, −) to be the unique extension of tr to a map of homological functors (using Theorem III.7.3[1]). Show that this agrees with the map defined by applying H∗(G, −) G ∼ G to the canonical injection M → CoindH M = IndH M and using Shapiro’s lemma. Assume that we know this latter map is compatible with ∂.

By Theorem III.7.3[1] it suﬃces to verify that the latter map equals tr in dimension zero. G ∼ G We ﬁrst want an explicit isomorphism for CoindH M = IndH M, asserting that it is given by ψ(f) = P −1 G g∈G/H g⊗f(g ). Following the proof of Proposition III.5.9[1], there is an H-map ϕ0 : M → CoindH M given by [ϕ0(m)](g) = {gm ∀ g ∈ H, 0 otherwise}, and by the universal property of induction G G this extends to a G-map ϕ : IndH M → CoindH M given by ϕ(g ⊗ m) = gϕ0(m). Now we have 0 0 P −1 0 00 00−1 0 00 00−1 0 0 ψϕ : g ⊗ m 7→ g [ϕ0(m)] 7→ g∈G/H g ⊗ [ϕ0(m)](g g ) = g ⊗ g g m = g g g ⊗ m = g ⊗ m, 00 00−1 0 −1 0 noting that there is one and only one g ∈ G/H such that g g ∈ H (otherwise g1 g = h1 and −1 0 g2 g = h2 gives g1h1 = g2h2 which contradicts their coset representations). In the other direction we P −1 P −1 0 00−1 0 00 0 00 00−1 have ϕψ : f 7→ g∈G/H g⊗f(g ) 7→ g∈G/H g[ϕ0(f(g ))](g ) = [ϕ0(f(g ))](g g ) = g g f(g ) = f(g0g00g00−1) = f(g0) = f, also noting that there is only one g00 ∈ G/H such that g0g00 ∈ H. Thus ϕ = ψ−1 and ψ is the desired isomorphism. G G Applying the zeroth homology functor H0(G, −) to the aforementioned map M → CoindH M → IndH M P −1 given by m 7→ (g 7→ gm) 7→ g∈G/H g ⊗ g m, and using the isomorphism from Shapiro’s lemma, P −1 P −1 ∼ we obtain the chain map x ⊗G m 7→ g∈G/H x ⊗G (g ⊗H g m) = g∈G/H xg ⊗G (1 ⊗H g m) = P −1 P −1 −1 P P g∈G/H xg ⊗H g m = g∈G/H g x ⊗H g m = g∈H\G gx ⊗H gm = g∈H\G g · (x ⊗H m). Using ∼ the natural isomorphism H0(G, −) = (−)G of Proposition III.6.1[1], this chain map yields the trace map tr described above.

∼ G 11: Prove that Hn(G, M) = Torn−1(I,M) where I is the augmentation ideal of ZG.

ε We have the short exact sequence of G-modules, 0 → I,→ ZG → Z → 0. By an analogue of The- G G orem 17.1.15[2] we have a long exact sequence of abelian groups · · · → Torn (ZG, M) → Torn (Z,M) → G G R Torn−1(I,M) → Torn−1(ZG, M) → · · · . It is a fact that if P is R-projective then Torn (P,B) = 0 for G any R-module B (n ≥ 1). Therefore, Torn (ZG, M) = 0 since ZG is a free [hence projective] ZG-module, G G G and so we obtain the isomorphisms Torn (Z,M) → Torn−1(I,M). Since Hn(G, −) = Torn (Z, −), the result follows.

∗ G G 12: Given |G : H| < ∞ and z ∈ H(G, M) where H(−, −) is either H∗ or H , show that corH resH z = |G : H|z.

63 G G Referring to Exercise AE.10 above, we have the map M → CoindH M → IndH M which yields the P G chain map x ⊗G m 7→ g∈H\G gx ⊗H gm, and this induces the transfer map resH in homology. Com- posing this with the corestriction map on the chain level x ⊗H m 7→ x ⊗G m, we obtain the chain map P P G G x ⊗G m 7→ g∈H\G gx ⊗G gm = g∈H\G x ⊗G m = |G : H|(x ⊗G m) which induces corH resH as multiplication by |G : H|. H G P On the other hand, we have the chain map Hom(F,M) → Hom(F,M) given by f 7→ g∈G/H [x 7→ −1 G gf(g x)] which induces the transfer map corH in cohomology; G acts diagonally. Composing this with P −1 the restriction map on the chain level, we obtain the chain map f 7→ f 7→ g∈G/H [x 7→ gf(g x)] = P g∈G/H [x 7→ f(x)] = |G : H|[x 7→ f(x)], noting that the domain element f is a G-invariant homomor- G G phism. This induces the aforementioned map corH resH .

Lm ∼ Lm ∗ 13: Give another proof that H(G, i Mi) = i H(G, Mi) where H(−, −) is either H∗ or H .

α β Applying Proposition III.6.1[1] to the short exact sequence of G-modules 0 → M1 → M1⊕M2 → M2 → 0, we obtain the standard long exact sequence in (co)homology. Since α is an injection onto a direct summand, the induced map α∗ under the covariant functor H∗(G, −) is an injection (refer to Exercise II.7.3); similarly, the induced map β∗ = H∗(G, β) is also an injection. Thus the long exact sequence breaks up into short exact sequences 0 → Hn(G, M1) → Hn(G, M1 ⊕ M2) → Hn(G, M2) → 0 (similar for coho- ∗ ∗ ∗ mology). These are split exact sequences because γ∗α∗ = id∗ and Γ β = id , where γ is the projection M1 ⊕ M2 → M1 and Γ is the inclusion M2 → M1 ⊕ M2. The result now follows by trivial induction on m.

n 14: Prove that if the G-module M has exponent p [prime] then H (G, M) and Hn(G, M) are Zp- vector spaces.

Showing that these groups are the speciﬁed modules is equivalent to showing that they are annihilated by p (i.e. have exponent dividing p). Considering homology, a chain z ∈ Cn(G, M) is a ﬁnite sum P P P of terms of the form m ⊗ [g1| · · · |gn], so pz = p(m ⊗ [g1| · · · |gn]) = (pm) ⊗ [g1| · · · |gn] = 0 = 0. 0 Thus pHn(G, M) = 0 ∀ n ≥ 0. Considering cohomology, a cochain f ∈ C (G, M) is an element of M and hence pf = 0 trivially. If n ≥ 1 then f ∈ Cn(G, M) is a function Gn → M, so pf is a function Gn → M → pM = 0 and hence pf = 0. Thus pHn(G, M) = 0 ∀ n ≥ 0.

15: If G is a ﬁnite group, show that Hn(G, M) is a Z|G|-module for n > 0. Thus we have the pri- L mary decomposition Hn(G, M) = p Hn(G, M)(p) where p ranges over the primes dividing |G|.

G G By Proposition III.9.5[1], cor{1}res{1}z = |G : {1}|z = |G|z where z ∈ H∗(G, M). But this composition G G factors through H∗({1},M) which is trivial in positive dimensions. Thus cor{1}res{1}z = 0 ⇒ |G|z = 0, and so Hn(G, M) is annihilated by |G| for n > 0.

16: Show that Hn(G, M) = 0 ∀ n > 0 if M is a finite abelian group and G is finite with relatively prime order, gcd(|G|, |M|) = 1. Furthermore, show that Hn(G, M) is a finite group (for n > 0) if G is finite and M is a finitely generated abelian group.

By Exercises AE.14+15 above, Hn(G, M) is annihilated by both |G| and |M|, and hence must be annihilated by a common factor. But gcd(|G|, |M|) = 1, so the common factor is 1 and Hn(G, M) = 0 for n > 0. By Exercise AE.15 above (assuming n > 0), |G|Hn(G, M) = 0 and so Hn(G, M) is all torsion. Since n |G| < ∞, Fn is a free G-module of finite rank r = |G| (F is the bar resolution of Z over ZG), where we ∼ Lr ∼ note that a G-basis for Fn is the (n+1)-tuples whose first element is 1. Thus Fn⊗GM = ( i ZG)⊗GM = Lr ∼ Lr i (ZG⊗G M) = i M is finitely generated since M is finitely generated, which implies that Hn(G, M) is also finitely generated. A finitely generated torsion group is finite, so the result follows.

G n n 17: Why does resH induce a monomorphism H (G, M)(p) ,→ H (H,M), where H is the Sylow p- subgroup of G?

64 n G Consider an element z ∈ H (G, M)(p) which lies in the kernel of resH . Composing this with the corestriction map gives you multiplication by |G : H| by Proposition III.9.5[1], and so |G : H|z = 0. But the order of z is |z| = pα and p does not divide |G : H|. Since pα and |G : H| both annihilate z, there is a common factor between them which also annihilates z, so z = 0 (the only common factor is 1). Thus G resH is a monomorphism when restricted to the p-primary component.

18: Compute the homology group H1(G) for any nonabelian simple group G.

The commutator group [G, G] is either 0 or G because the commutator group is a normal subgroup of G and a normal subgroup of a simple group must be either the trivial group or G itself. Since G is ∼ nonabelian, we cannot have [G, G] = 0, and so [G, G] = G (i.e. G is perfect). Thus H1(G) = G/[G, G] = G/G = 0. In particular, H1(An) = 0 ∀ n ≥ 5 where An is the alternating group on n letters (subgroup of Sn with index 2), by Theorem 4.6.24[2].

2 ∼ 19: Show that I/I = Gab where I is the augmentation ideal of ZG.

Apply Proposition III.6.1[1] to the short exact sequence 0 → I → ZG → Z → 0 to obtain the exact sequence H1(G, ZG) → H1(G) → IG → (ZG)G → Z → 0 in low dimensions. The latter map is an ∼ isomorphism (by the First Isomorphism Theorem) because the map is surjective and (ZG)G = Z. By ∼ 2 Proposition III.6.1[1], H1(G, ZG) = 0 because ZG is free [hence projective]. Thus H1(G) = IG = I/I ∼ by exactness, and we know that H1(G) = Gab, so the result follows. An explicit isomorphism is given by g[G, G] 7→ (g − 1) + I2.

1 20: Find a module M with trivial action such that H (D2n ,M) is nonzero, where D2n is the dihedral group with respect to a regular 2n−1-gon.

2 2 2n−1 A presentation for this group is D2n = hα, β | α = β = (αβ) = 1i. By the result of Exercise 1 ab III.1.2, H (D2n ,M) = Hom(D2n ,M). Now D2n quotiented by [D2n ,D2n ] forces the trivial relation 1 = 2n−1 2n−1 2n−1 ab 1 (αβ) = α β = 1 · 1 = 1, so D2n = Z2 ⊕ Z2. Thus H (D2n ,M) = Hom(Z2,M) ⊕ Hom(Z2,M), 1 ∼ which is nonzero if M = Z2, giving H (D2n , Z2) = Z2 ⊕ Z2.

21: Prove that a ﬁnitely generated abelian group G is cyclic if and only if H2(G, Z) = 0.

If G is a cyclic group Zn, then H2(Zn, Z) = 0 as explained on pg35[1]. If H2(G, Z) = 0 then Hopf’s formula (Theorem II.5.3[1]) gives R∩[F,F ] ⊆ [F,R], where F/R is a presentation for G. Since G is abelian we have 0 = [G, G] = [F/R, F/R] = [F,F ]R/R which implies [F,F ] ⊆ R. Alternatively, since G = F/R is abelian, [F,F ] ⊆ R by Proposition 5.4.7[2]. Thus [F,F ] = [F,R], and |F | < ∞ since G is finitely generated. If |F | = 1 then R = {xn} for some n ≥ 0, in which case G is obviously cyclic. It seems we’re stuck in terms of extracting any more information, but alternatively we can refer to V2 Theorem V.6.3[1] which says H2(G) is isomorphic to the second exterior power G for any abelian ∼ ∼ group. Thus it suffices to show that G is cyclic if G ⊗ G = h{g ⊗ g}i = G. Now if rkZ(G) = n then 2 2 ∼ rkZ(G ⊗ G) = n , so n = n gives n = 1 or n = 0. Also, Zi ⊗ Zj = Zgcd(i,j), so G must contain at most one Zm-summand (for each m) because the tensor product contains at least the squared-amount of those summands. Assume that G is not cyclic; then all m’s must be relatively prime (otherwise the tensor ∼ product will contain additional summands not in G due to greatest common divisors). Thus G = Z ⊕ Zn ∼ (for some n ≥ 2) since Zi ⊕ Zj = Zij iff gcd(i, j) = 1 [note that G must have the Z-summand in order to ∼ not be cyclic]. But (Z⊕Zn)⊗(Z⊕Zn) = Z⊕Zn ⊕Zn ⊕Zn which is not isomorphic to G, a contradiction (hence G is cyclic). Aside: A finitely generated abelian group G = F/R is cyclic iff [F,F ] = [F,R].

m+1 22: Let Zm = hti, Zm2 = hsi, and deﬁne a Zm-action on Zm2 by t · s = s . Compute the resulting Zm-module structure on Hj(Zm2 ), and then compute Hi(Zm,Hj(Zm2 )) where m is an odd prime.

65 First we must compute the change-of-rings map in integral homology for the map Za = hxi → Zb = hyi, x 7→ yr, where b|ra. Mimicking the solution to Exercise AE.4, the induced map in odd-dimensional homology is easily seen to be multiplication by r. Applying this result to the case where r = m + 1 and 2 p a = b = m , the Zm-action on Hj(Zm2 ) is multiplication by p(m+1) for t . If j is even, then the homolo- 2 2 2 Zm gies in question are trivial. By a result on pg58[1], Hodd(Zm, Zm ) = Coker(N :(Zm )Zm → (Zm ) ) 2 2 2 Zm and Heven(Zm, Zm ) = Ker(N :(Zm )Zm → (Zm ) ). Since m is an odd prime, the only nontrivial proper subgroup/quotient of Zm2 is Zm which is not ﬁxed by the Zm-action, so the norm map is 0 → 0 and thus Hi(Zm,Hj(Zm2 )) = 0 ∀ nonzero i, j.

23: Prove that if Q is injective then E(G, Q) is trivial.

Any short exact sequence 0 → Q → E → G → 0 splits for Q injective, so there is only one equivalence class of group extensions (namely, the split extension) and E(G, Q) = 0. Alternatively, Proposition III.6.1[1] and Theorem IV.3.12[1] imply E(G, Q) =∼ H2(G, Q) = 0 because Q is injective.

2 24: Prove that H (A5, Z2) 6= 0.

We have the central group extension 1 → Z2 → SL2(F5) → A5 → 1 as in Exercise I.5.7(b). It suf- ﬁces to show that E = Z2 o A5 is not isomorphic to SL2(F5), for then the above extension is nonsplit 2 ∼ and hence H (A5, Z2) = E(A5, Z2) 6= 0 by Theorem IV.3.12[1]. First note that Z2 must have the trivial A5-action, so E = Z2 × A5. Now it is a fact that SL2(F5) is perfect, while [E,E] ⊆ A5 by Proposition ∼ 5.4.7[2] because E/A5 = Z2 is abelian (hence E is not perfect). Thus H1(E) 6= 0 = H1(SL2(F5)) and so Z2 o A5 SL2(F5).

∼ 1 25: Let A = Z2×Z2 and let Aut(A) = S3 act on A in the natural fashion. Prove that H (S3, Z2×Z2) = 0.

In the semi-direct product E = AoS3 = Hol(A) [called the holomorph of A] we have a Sylow 3-subgroup ∼ 3 P = Z3 by Sylow’s Theorem, where |E| = 24 = 2 · 3. By Sylow’s Theorem, n3 | 8 and n3 ≡ 1 mod 3, where n3 is the number of Sylow 3-subgroups of E. Thus n3 is either 1 or 4, and this implies |NE(P )| is either 24 or 6 by the fact n3 = |E : NE(P )|. But we can exhibit at least two such subgroups [noting ∼ that S3 = Z3 o Z2], namely P1 = (0 × 0) o (Z3 o 0) and P2 = {(0, 0, 0, 0), (1, 0, 1, 0), (1, 0, 2, 0)}. Thus |NE(P )| = 6 which corresponds to n3 = 4, and S3 is then the normalizer of the Sylow 3-subgroup P = 0 o Z3 of E because P/S3 and |S3| = 6. Given a complement G to A in E [a group G is a complement to A in E if E = A o G], one Sylow 3-subgroup is Q/G and hence P = eQe−1 for some e ∈ E by −1 −1 −1 Sylow’s Theorem. Now S3 = NE(P ) = NE(eQe ) = eNE(Q)e = eGe , and so all complements are conjugate. Noting that the conjugacy classes of complements to A in E are the A-conjugacy classes of 1 splittings of E, we have H (S3, Z2 ×Z2) = 0 by Proposition IV.2.3[1] because there is only one conjugacy class.

26: In the proof of Theorem IV.3.12[1], all extensions were assumed to have normalized sections. Explain why this simpliﬁcation does not aﬀect the result of the theorem.

Given a factor set (2-cocycle) f : G × G → A, we assert that this lies in the same cohomology class as a normalized factor set (2-cocycle). Let δ1c be the coboundary of the constant function c : G → A defined by c(g) = f(1, 1). It suffices to show that F = f − δ1c is a normalized factor set, for then it belongs to the same cohomology class as the arbitrary f [it differs by a coboundary]. Now F (1, 1) = f(1, 1)−[δ1c](1, 1) = f(1, 1)−[c(1)+1·c(1)−c(1·1)] = f(1, 1)−f(1, 1) = 0 and so F : G×G → A is normalized. It is indeed a 2-cocycle (factor set) because gF (h, k) − F (gh, k) + F (g, hk) − F (g, h) = [gf(h, k) − f(gh, k) + f(g, hk) − f(g, h)] − {g[c(h) + hc(k) − c(hk)] − [c(gh) + ghc(k) − c(ghk)] + [c(g) + gc(hk) − c(ghk)] − [c(g) + gc(h) − c(gh)]} = 0 − gc(h) − ghc(k) + gc(hk) + c(gh) + ghc(k) − c(ghk) − c(g) − gc(hk) + c(ghk) + c(g) + gc(h) − c(gh) = [−gc(h) + gc(h)] + [−ghc(k) + ghc(k)] + [gc(hk) − gc(hk)] + [c(gh) − c(gh)] + [−c(ghk) + c(ghk)] + [−c(g) + c(g)] = 0 + 0 + 0 + 0 + 0 + 0 = 0. Since the bijection in the theorem is dependent on the cohomology class, and every class has a normalized factor set (hence normalized section), we are able to restrict our attention to those sections satisfying the normalization condition.

66 27: Show that H2(F,A) = 0 for F free by appealing to group extensions.

For any group extension 0 → A → E →π F → 1 deﬁne the set map S → E by s 7→ s˜, wheres ˜ ∈ π−1(s) is a lifting of s ∈ F = F (S). Then by the universal mapping property of free groups, this set map extends uniquely to a homomorphism ϕ : F → E which satisﬁes πϕ = idF by construction. Thus the extension splits, and by Theorem IV.3.12[1] we have H2(F,A) =∼ E(F,A) = 0. Note that this also follows from direct computation using the free resolution 0 → I → ZF → Z → 0, where the augmentation ideal I of ZF is free by Exercise IV.2.3(b). Indeed, the coboundary map is n δn : HomF (0,A) → HomF (0,A) for all n > 1 and hence H (F,A) = 0 ∀ n > 1.

2 28: Compute H (Q8, Z2), and determine the number of group extensions of Z2 by Q8. [These two problems are independent of each other].

4 2 The quaternion group Q8 is a non-abelian group of order 8 with presentation hx, y | x = 1, x = 2 2 ∼ y = (xy) i. Its center is C := Z(Q8) = Z2 which is also its commutator subgroup, and Inn(Q8) = ∼ ∼ ∼ Q8/Z(Q8) = Z2 × Z2. Also, Aut(Q8) = S4 and hence Out(Q8) = Aut(Q8)/Inn(Q8) = S3. As explained on pg102+104[1], Q8 is an S4-crossed module via the canonical map Q8 → Aut(Q8), and such an exten- ∼ sion gives rise to a homomorphism ψ : Z2 → Out(Q8) = S3. By Theorem IV.6.6[1] the set E(Z2,Q8, ψ) of equivalence classes of extensions giving rise to ψ is either empty or in bijective correspondence with 2 H (Z2,C), where C is a Z2-module via ψ [note: it is a fact that the center of a group is a characteristic subgroup, so Aut(Q8) acts naturally on C and hence Out(Q8) acts on C because any inner automorphism ∼ Z2 leaves C ﬁxed]. Now the only possible action on C = Z2 is the trivial one (giving CZ2 = C = C ), so 2 ∼ H (Z2,C) = CokerN = C/NC = C/0 = Z2. There are two possible choices of ψ, namely, the trivial map and the injection Z2 ,→ S3 = Z3 o Z2. If ψ is the trivial map then it automatically lifts to the trivial ho- ∼ momorphism Z2 → S4 and we have a direct product extension E = Q8 × Z2. For the injective ψ we have the semi-direct product Q8 oZ2 where Z2 switches the generators of Q8. Thus E(Z2,Q8, ψ) is nonempty, and Theorem IV.6.6[1] implies there are a total of 4 group extensions 1 → Q8 → E → Z2 → 1. To compute the second cohomology group of the group Q8 with coeﬃcients in Z2, embed Z2 in the ∗ H -acyclic module Hom(ZQ8, Z2) by z 7→ (q 7→ qz), and note that Coker(Z2 → Hom(ZQ8, Z2)) = Z2 because the evaluation map Hom(ZQ8, Z2) → Z2 given by f 7→ f(2) composes with the embedding to give the trivial map z 7→ (q 7→ qz) 7→ 2z = 0. The dimension-shifting argument then implies 2 ∼ 1 1 ∼ H (Q8, Z2) = H (Q8, Z2), and by Exercise III.1.2 we have H (Q8, Z2) = Hom(H1(Q8), Z2). Now ∼ ∼ ∼ H1(Q8) = Q8/[Q8,Q8] = Q8/C = Z2 × Z2 and so Hom(H1(Q8), Z2) = Hom(Z2, Z2) ⊕ Hom(Z2, Z2) = 2 ∼ 1 ∼ ∼ 2 Z2 × Z2. Therefore, H (Q8, Z2) = H (Q8, Z2) = H1(Q8) = Z2 ≡ Z2 × Z2.

29: Let p be a prime, let A be an abelian normal p-subgroup of a ﬁnite group G, and let P be a Sylow p-subgroup of G. Prove that G is a split extension of G/A by A iﬀ P is a split extension of P/A by A [Note: it is a fact that a normal p-subgroup is contained in every Sylow p-subgroup, so A ⊆ P ]. This result is known as Gaschu¨tz’s Theorem.

Suppose G splits over A, so that G =∼ A o G/A. Then the subgroup A o P/A is a Sylow p-subgroup of G and hence A o P/A =∼ P by Sylow’s Theorem, so P also splits over A. Conversely, suppose P splits ∼ over A (i.e. P = A o P/A). Note that P/A = Sylp(G/A) and multiplication by |G/A : P/A| = |G : P | is an automorphism of A [hence of H2(G/A, A)] since |A| is relatively prime to |G : P |. The composition H2(G/A, A) →res H2(P/A, A) cor→ H2(G/A, A) is then an isomorphism by Proposition III.9.5[1] because it is multiplication by |G : P |. In particular, the restriction homomorphism res : H2(G/A, A) → H2(P/A, A) is injective, so the only element of H2(G/A, A) which corresponds to the trivial element [split extension] of H2(P/A, A) is the trivial element [split extension]. Thus G splits over A, and the proof is complete.

∼ 2 ∗ 30: The Schur multiplier of a finite group G is defined as H2(G, Z) = H (G, C ) where the multiplicative group C∗ = C−{0} is a trivial G-module. Prove that the Schur multiplier (of a finite group) is finite.

Since |G| < ∞ and Z is ﬁnitely generated (as an abelian group), H2(G, Z) is a ﬁnite group by Ex- ercise AE.16. Alternatively, we shall show that every cohomology class contains a cocycle whose values lie in the nth

67 ∼ 2πi/n roots of unity hζi = Zn (where n = |G| and ζ = e ), for then there are only finitely many func- 2 2 ∗ n2 ∗ n tions/cocycles f : G → hζi and |H (G, C )| ≤ n < ∞. From the exact sequence 0 → Zn → C → ∗ 1 ∗ 1 ∗ 2 2 ∗ C → 0 we obtain a long exact sequence → H (G, C ) → H (G, C ) → H (G, Zn) → H (G, C ) → H2(G, C∗) → by Proposition III.6.1[1]. But the nth-power map on C∗ induces the n-multiplication map on H2(G, C∗) which is the zero map since n annihilates H2(G, C∗) by Corollary III.10.2[1], so the above 2 2 ∗ 2 long exact sequence gives us a surjection H (G, Zn) → H (G, C ) → 0. Now H (G, Zn) is finite by Exer- cise AE.16, so H2(G, C∗) is necessarily finite (by the 1st Isomorphism Theorem) and the proof is complete.

31: Show that ZG ⊗ ZG0 =∼ Z[G × G0] as (G × G0)-modules.

The map ZG×ZG0 → Z[G×G0] deﬁned by (zg, z0g0) 7→ zz0(g, g0) is obviously a Z-balanced map and hence gives a rise to a unique group homomorphism ϕ : ZG⊗ZG0 → Z[G×G0] by Theorem 10.4.10[2]. The obvious group homomorphism Z[G×G0] → ZG⊗ZG0 deﬁned by z(g, g0) 7→ z(g ⊗g0) = (zg ⊗g0) = (g ⊗zg0) is the inverse of ϕ because z(g, g0) 7→ (zg ⊗ g0) 7→ z1(g, g0) = z(g, g0) and (zg ⊗ z0g0) 7→ zz0(g, g0) 7→ zz0(g ⊗ g0) = (zg ⊗ z0g0). Thus ϕ is a group isomorphism, and it is a (G × G0)-module isomorphism because ϕ[(h, h0) · (zg ⊗ z0g0)] = ϕ(zhg ⊗ z0h0g0) = zz0(hg, h0g0) = zz0(h, h0)(g, g0) = (h, h0) · ϕ(zg ⊗ z0g0) where (h, h0) ∈ G × G0.

p q r s 32: Suppose u1 ∈ H G, u2 ∈ H G, v1 ∈ H G, and v2 ∈ H G. Prove that (u1 × v1) ` (u2 × v2) = qr p+q+r+s (−1) (u1 ` u2) × (v1 ` v2) in H (G, Z).

∗ 4 Note that u ` v := d (u × v) where d : G → G × G is the diagonal map. Let D : G × G → G be the 4 4 analogous diagonal map, and let P : G → G be the permutation (g1, g2, g3, g4) 7→ (g1, g3, g2, g4). Then ∗ ∗ ∗ D = P ◦(d×d), and we obtain (u1×v1) ` (u2×v2) = D (u1×v1×u2×v2) = (d×d) [P (u1×v1×u2×v2)] = qr ∗ qr ∗ ∗ qr (−1) (d × d) [u1 × u2 × v1 × v2] = (−1) d (u1 × u2) × d (v1 × v2) = (−1) (u1 ` u2) × (v1 ` v2). ∗ ∗ Another way is to perform the same calculation using u × v := p1(u) ` p2(v) and the fact that ∗ ∗ ∗ p (u ` v) = p (u) ` p (v), where p1 is the projection G × G → G × {1} = G and p2 is the projection G × G → {1} × G = G.

33: For the cap product, state the property of naturality with respect to group homomorphisms α : G → H. Also, provide the existence of an identity element for the cap product.

∗ Checking deﬁnitions, we have u a α∗z = α∗(α u a z) which is associated to the “commutative” diagram Hp(G, M) ⊗ H (G, N) a / H (G, M ⊗ N) D q q−p

α∗ α∗ α∗ Ô p a H (H,M) ⊗ Hq(H,N) / Hq−p(H,M ⊗ N) There is a left-identity element 1 ∈ H0(G, Z) = Z which is represented by the augmentation map ε, regarded as a 0-cocycle in HomG(F, Z). Let F be the standard resolution, let z ∈ H∗(G, M) with representation z = (g0, . . . , gi)⊗m, and take the diagonal approximation ∆ to be the Alexander-Whitney map. Pi degε·p P 0·i Then ε ⊗ z maps under a to p=0(−1) (g0, . . . , gp) ⊗ ε(gp, . . . , gi) ⊗ n = 0 + (−1) (g0, . . . , gi) ⊗ 1 ⊗ n = z, as explained on pg113[1]. Thus the element satisfies 1 a z = z for all z ∈ H∗(G, M), where we make the obvious identification Z ⊗ M = M of coefficient modules.

34: Prove that a ﬁnitely generated projective Z-module M is a ﬁnitely generated free Z-module; thus the two are actually equivalent (since free modules are projective).

∼ r By the Fundamental Theorem (Theorem 12.1.5[2]) M has the decomposition M = Z ⊕ Zn1 ⊕ · · · ⊕ Zni , where we note that Z is a Principal Ideal Domain. Since M is projective, all of its direct summands r must be projective. Now Z is free, hence projective. But Zn is not Z-projective because if it were then it would be a direct summand of a free Z-module F , and F would then have elements of ﬁnite order (a contradiction); alternatively we could note that applying the functor Hom(Zn, −) to the exact sequence n ∼ r 0 → Z → Z → Zn → 0 yields the sequence 0 → 0 → 0 → Zn → 0 which is not exact. Thus M = Z for

68 some r ∈ N ∪ {0} and hence M is Z-free of ﬁnite rank.

q 35: Let G = PSL2(Z) and let A be a G-module. Show that for every q ≥ 2 and for every x ∈ H (G, A), we have 6x = 0.

az+b The modular group is the group of M¨obius transformations T (z) = cz+d in the complex plane such a b that a, b, c, d ∈ Z with ad − bc = 1. It is isomorphic to G via the map T 7→ c d , and G is called the ∼ projective special linear group [the quotient of SL2(Z) by Z(SL2(Z)) = Z2]. Now it is a fact that G contains a free subgroup H of index |G : H| = 6 [this fact can be found in the paper The Number of Subgroups of Given Index in the Modular Group by W. Stothers]. By Exercise AE.27, Hq(H,A) = 0 for all q ≥ 2. Thus we can apply Proposition III.10.1[1] which states Hq(G, A) is annihilated by |G : H|, i.e. 6x = 0 for all x ∈ Hq(G, A) with q ≥ 2.

36: Let G be a finite cyclic group and let M be a G-module. The Herbrand quotient is defined to be h(M) = |H2(G, M)|/|H1(G, M)|, assuming both cohomology groups are finite. Show that h(Z) = |G| where Z has the trivial G-action, and show that h(M) = 1 for M finite.

2 ∼ G 1 We know that H (G, M) = M /NM and H (G, M) = NM/IM, where N is the norm element and I ≡ hσ − 1i is the augmentation ideal of G = hσi and NM := Ker(N : M → M). For M = Z with trivial 2 ∼ 1 action we have H (G, M) = Z|G| and H (G, M) = 0, so h(Z) = |ZG|/|{0}| = |G|/1 = |G|. If M is ﬁnite G G ∼ then h(M) = (|M | · |IM|)/(|NM| · |NM|) = |M | · |(σ − 1)M|/|M|, where we note that M/NM = NM. But the kernel K of the surjective map M σ→−1 (σ − 1)M is equal to M G because m ∈ M G maps to σm − m = m − m = 0 and if m ∈ K then σ|G|−1m = σ|G|−2m = ··· = σm = m, i.e. m ∈ M G. Thus |M|/|M G| = |(σ − 1)M| and so h(M) = |M|/|M| = 1.

37: Given the short exact sequence of G-modules 0 → A → B → C → 0 with G finite cyclic, show that h(B) = h(A)h(C) where we assume the cohomology groups for A and C [hence B] are finite. Here h is the Herbrand quotient (defined in the previous exercise).

ϕ φ Consider the long exact cohomology sequence H0(G, C) →δ0 H1(G, A) → H1(G, B) → · · · → H2(G, B) → H2(G, C) →δ2 H3(G, A). It is a fact (Exercise V.3.3(b)) that cupping this sequence with the generator of H2(G, Z) gives an isomorphism of that sequence onto itself (raising dimensions by 2), and hence ∼ gives the equality δ0 = δ2 which implies Kerϕ = Cokerφ. We can break the above sequence and ob- ϕ φ tain an exact sequence 0 → Kerϕ → H1(G, A) → · · · → H2(G, C) → Cokerφ → 0. I claim that Qm (−1)i if 0 → H1 → · · · → Hm → 0 is an exact sequence of finite abelian groups, then i=1 |Hi| = 1. Assuming this claim holds, and noting that the cohomology groups are finite abelian, we have |Cokerφ| |H1(G,A)| |H1(G,C)| |H2(G,B)| 1 1 1 = |Kerϕ| |H1(G,B)| |H2(G,A)| |H2(G,C)| = 1 · h(A) h(C) h(B) which implies h(B) = h(A)h(C) as desired. It suffices to prove the claim. The case m = 2 is trivial since the sequence yields the isomorphism ∼ H1 = H2 and hence |H1|/|H2| = 1, so we proceed by induction on m > 1. From the exact sequence φ ϕ 0 → H1 → · · · → Hm−1 → Hm → Hm+1 → 0 we obtain the exact sequence 0 → H1 → · · · Hm−2 → φ Qm−1 (−1)i (−1)m Hm−1 → Imφ → 0. Applying the inductive hypothesis we have i=1 |Hi| · |Imφ| = 1, and by exactness of the original sequence we see that |Imφ| = |Kerϕ|. But |Hm|/|Kerϕ| = |Imϕ| = |Hm+1|, so Qm−1 (−1)i (−1)m Qm+1 (−1)i |Kerϕ| = |Hm|/|Hm+1| and hence 1 = i=1 |Hi| ·(|Hm|/|Hm+1|) = i=1 |Hi| as desired.

38: If G and H are abelian groups with isomorphic group rings ZG =∼ ZH, show that G =∼ H.

The homology groups of G and H are independent of the choice of resolution up to canonical isomorphism, and the groups are deﬁned by HiG = Hi(FG) where F is a projective resolution of Z over ZG ∼ (similary for H). Since ZG = ZF , the G-modules Fi can be regarded as H-modules via restriction of scalars and hence the projective resolution F for the homology of G can also be used for the homology ∼ ∼ of H. We have FG = FH since Z ⊗ G F = Z ⊗ H F by the obvious map 1 ⊗ f 7→ 1 ⊗ f [using the Z Z ∼ isomorphism ϕ : ZG → ZH we have 1 ⊗ f = 1 ⊗ gf 7→ 1 ⊗ ϕ(g)f = 1 ⊗ f], and so Hi(G) = Hi(H) ∼ ∼ ∼ for all i. In particular, G/[G, G] = H1(G) = H1(H) = H/[H,H]. Since G and H are abelian groups,

69 [G, G] = 0 = [H,H] and hence G =∼ H.

39: Let F be a field and G a finite group of order |G| > 1. Show that the group algebra FG has zero divisors (hence is not a field) and show that the augmentation ideal I is a maximal ideal of FG.

ε Noting that F G/I =∼ F from the exact sequence 0 → I → FG → F → 0, I is a maximal ideal by Proposition 7.4.12[2] since F is a field [in a commutative ring R, an ideal M is maximal iff R/M is a field]. Now take g ∈ G such that gm = 1 with m > 1. Then (1−g)(1+g+···+gm−1) = 1−gm = 1−1 = 0 and hence 1 − g is a zero divisor.

40: Regarding Z2 as a module over the ring Z4, construct a resolution of Z2 by free modules over and use this to show that Extn ( , ) is nonzero for all n. Z4 Z4 Z2 Z2 2 The 2-multiplication map Z4 → Z4 has kernel {0, 2} and image {0, 2}, and the projection Z4 Z2 2 2 has kernel {0, 2} and image {0, 1}. Thus we have a free resolution · · · → Z4 → Z4 → Z4 Z2 → 0. Since Hom(Z4, Z2) = Z2, the 2-multiplication maps become trivial maps, and we obtain the sequence 0 0 · · · ← ← ← after applying Hom(−, ) to our free resolution. Thus Extn ( , ) =∼ 6= 0 Z2 Z2 Z2 Z2 Z4 Z2 Z2 Z2 for all n.

j 41: Let H be a subgroup of G of ﬁnite index, and let 0 → A0 →i A → A00 → 0 be an S.E.S. of G- modules. Show that the connecting homomorphisms are consistent with corestriction, i.e. the following diagram commutes. Hn(G, A00) δ / Hn+1(G, A0) O O cor cor

Hn(H,A00) δ / Hn+1(H,A0)

Let F be the resolution associated to G and let F 0 be the resolution associated to H, and let ∂∗ denote the [cochain complex] coboundary map. Consider the exact sequence of cochain complexes 0 0 iH 0 jH 0 00 0 → HomH (Fn,A ) → HomH (Fn,A) → HomH (Fn,A ) → 0, with similar notation for the sequence associated to G. The connecting map δ in cohomology is deﬁned as follows. For the cohomology-class 0 00 0 representative fb ∈ HomH (Fn,A ), there exists an f ∈ HomH (Fn,A) which maps onto fb via jH (by ∗ 0 0 surjectivity of jH ). Then it is realized that ∂ f = iH (f) for some f ∈ HomH (Fn,A ) which is unique by injectivity of iH , and f is the class representative of δ[fb] where [fb] is the cohomology class of fb. Thus in order for δ to commute with corestriction, it suﬃces to show that corestriction commutes with {∂∗, i∗, j∗} where i∗ and j∗ refer to both maps associated to H,G. Let’s show commutativity of the diagram j Hom (F ,A) G / Hom (F ,A00) G O n G O n cor cor

0 jH 0 00 HomH (Fn,A) / HomH (Fn,A ) P −1 P −1 Now cor[jH (f)] = cor[fb = j ◦ f] = g∈G/H gfb(g ) = g∈G/H gj[f(g )], and in the other di- P −1 P −1 P −1 rection we have jG[cor(f)] = jG[ g∈G/H gf(g )] = g∈G/H gf\(g ) = g∈G/H j[gf(g )] = P −1 g∈G/H gj[f(g )] where we note that j is a G-module homomorphism and hence commutes with the G-action. Thus the diagram is commutative, and similar calculations give commutativity with i∗ and ∂∗ since both are G-module homomorphisms.

42: Let p be a prime and let Sp be the symmetric group of degree p. Then each Sylow p-subgroup P is cyclic of order p, one such being the subgroup generated by the cycle (1 2 ··· p). Thus, H∗(P, Z) =∼ ∼ ∗ ∼ Z[ν]/(pν) where degν = 2. Show that NSp (P )/P = Zp and that it acts on P = Zp in the obvious way. ∗ ∼ p−1 Conclude that H (Sp, Z)(p) = Z[ν ]/(pν).

70 Since P is abelian, P ⊆ CSp (P ). Now |CSp (P )| = p(p − p)! = p as explained on pg127[2] where ∼ we note that P and its generator have the same centralizer, so we must have CSp (P ) = P . Corol- lary 4.4.15[2] states that NSp (P )/CSp (P ) is isomorphic to a subgroup of Aut(P ), and by Proposition ∼ ∗ 4.4.16[2] we know that Aut(P ) = Zp is cyclic of order ϕ(p) = p − 1 [ϕ is the Euler function]. Thus ∼ ∗ NSp (P )/P = H ⊆ Zp. The number of p-cycles in Sp is (p − 1)! as explained on pg127[2], and every conjugate of P contains exactly p − 1 p-cycles, so there are (p − 1)!/(p − 1) = (p − 2)! conjugates of P which is equal to the index |Sp : NSp (P )| by Proposition 4.3.6[2]; thus |NSp (P )| = p!/(p − 2)! = p(p − 1). ∼ ∗ ∼ ∗ This means |H| = |NSp (P )/P | = p − 1 and hence H = Zp ⇒ NSp (P )/P = Zp. This group acts by conjugation on P because it is a quotient of the normalizer, and this action is well-deﬁned because conjugation by elements of P is trivial, noting that P is abelian. ∗ ∼ Theorem III.10.3[1] along with a theorem of Swan (see Exercise III.10.1) states that H (Sp, Z)(p) = N (P ) H∗(P, Z) Sp . By Proposition II.6.2[1] the conjugation action by P induces the identity on H∗(P, Z), so ∗ ∗ ∼ the resulting action (Corollary II.6.3[1]) is the NSp (P )/P -action induced on H (P, Z). Thus H (Sp, Z)(p) = ∗ (Z[ν]/(pν))Zp . For a particular dimension j, the elements of the jth-cohomology group belong to Zp and ∗ hence the action on an element ν would be ν 7→ z∗ · ν = zν with 0 < z < p and z∗ ∈ Zp. Then in the i i i i i cohomology ring, the action is given by ν 7→ z∗ · ν = (z∗ · ν) ` ··· ` (z∗ · ν) = (zν) = z ν . So for ∗ i i i i an element to belong to the group of Zp-invariants, we must have ν = z ν ⇒ z ≡ 1 mod p for all z < p. This is only satisﬁed when i = p − 1 by Fermat’s Little Theorem (z is relatively prime to p), so P i(p−1) ∗ ∼ p−1 the invariant elements are of the form i ziν . Therefore, H (Sp, Z)(p) = Z[ν ]/(pν).

43: Prove that the Pontryagin product on H∗(G, Z) for G ﬁnite cyclic is the trivial map in positive dimensions (Z has trivial G-action). ∼ The map is given by HiG ⊗ HjG → Hi+jG for each i, j. Since HnG = G for n odd and HnG = 0 for n even, the domain of the map [hence the map] is trivial for i or j even. But if both i := 2c + 1 and j := 2c0 +1 are odd, then i+j = 2(c+c0 +1) is even, so the image of the map [hence the map] is trivial.

2r 44: Compute H (Zp ⊕ Zq) where p and q are not necessarily relatively prime.

2r ∼ L2r i 2r−i The cohomology K¨unnethformula of Exercise V.2.2 gives H (Zp ⊕Zq) = i=0 H (Zp)⊗H (Zq) ⊕ L2r+1 Z i 2r−i+1 ∼ r−1 ∼ r−1 i=0 Tor1 H (Zp),H (Zq) = (Z ⊗ Zq) ⊕ (Zp ⊗ Zq) ⊕ (Zp ⊗ Z) ⊕ [0] = Zp ⊕ Zq ⊕ Zgcd(p,q) for r ≥ 1. The Tor-parts were trivial because each summand became Tor(Heven,Hodd) = 0 and Tor(Hodd,Heven) = 0. Note that if p and q are relatively prime, then the result becomes Zp ⊕ Zq which agrees with the fact ∼ that Zp ⊕ Zq = Zpq is cyclic. n i n ∼ i n

For n = 1 we have H0(Z, Z) = H1(Z, Z) =∼ Z and Hi(Z, Z) = 0 for all i > 1 as proven on pg58[1]. This agrees with the proposed formula, so we argue by induction on n (assume the result holds up to and i n+1 i n ∼ Li p n including n). The Künnethformula of Exercise V.2.2 gives H (Z ) = H (Z ⊕ Z) = [ p=0 H (Z ) ⊗ i−p Li+1 Z p n i−p+1 ∼ Li p n i−p H (Z)] ⊕ [ p=0 Tor1 H (Z ),H (Z) ] = p=0 H (Z ) ⊗ H (Z), where the latter isomor- Z i−p+1 phism follows from the fact that Tor1 (−, Z) = 0 and H (Z) is either Z or 0 as stated above. Now n n i L p n i−p ∼ i n i−1 n ∼ i n i−1 n ∼ i i−1 p=0 H (Z ) ⊗ H (Z) = (H (Z ) ⊗ Z) ⊕ (H (Z ) ⊗ Z) = H (Z ) ⊕ H (Z ) = Z ⊕ Z , so n+1 n!(n−i+1) (n+1)! i n+1 ∼ i n n n! n! n!i H (Z , Z) = Z since i + i−1 = i!(n−i)! + (i−1)!(n−i+1)! = i!(n−i+1)! + i!(n−i+1)! = i!(n+1−k)! = n+1 i , and the inductive process is complete. The cohomological analog of the universal coefficient sequence of Exercise III.1.3 gives Hi(Zn, k) =∼ n n i n Z i+1 n ∼ i ∼ i Z r (H (Z ) ⊗ k) ⊕ Tor1 (H (Z ), k) = (Z ⊗Z k) ⊕ 0 = k , where we note that Tor1 (Z , −) = 0. Note: These integral cohomology groups are dual to the associated integral homology groups; see pg38[1]. n n This observation of Poincare´ duality is reflected in the relation i = n−i .

71 46: Let p be a prime, G a ﬁnite p-group, k a ﬁeld of characteristic p, and n ∈ N. If Hn(G, k) = 0, prove that Hn+1(G, Z) = 0.

The cohomological analog of the universal coeﬃcient sequence of Exercise III.1.3 gives Hn(G, k) =∼ n Z n+1 Z n+1 (H (G) ⊗ k) ⊕ Tor1 (H (G), k) = 0, which in particular implies Tor1 (H (G), k) = 0. Then since Z n+1 n+1 k contains the subring Zp, we must have Tor1 (H (G), Zp) = 0 and hence no element in H (G) has order p. But |G| = pα and so pαHn+1(G) = 0 by Corollary III.10.2[1], which means any element must have order a power of p. The only way this is satisﬁed with no element having order p is for every element to have order 1 (if an element h had order pa with a ≥ 2, then pa−1h would be an element of order p). Only the identity has order 1, so all elements are the same; thus Hn+1(G, Z) = 0.

47: Prove that the shuﬄe product is strictly anti-commutative.

The shuffle product on the bar resolution is given by P sgnσ [g1| · · · |gn] · [g1| · · · |gn] = σ(−1) [gσ−1(1)| · · · |gσ−1(n)|gσ−1(n+1)| · · · |gσ−1(2n)], s where σ ranges over the (n, n)-shuffles. But given any shuffled tuple (−1) [gi1 | · · · |g| · · · |g| · · · |gi2n ] we have a corresponding shuffle which simply swaps the two g’s, leaving the 2n-tuple fixed and altering the sign to (−1)s+1; the sign change arises from moving the left g an l amount of times and then moving the right g an l + 1 amount of times (the extra +1 is due to moving the right g around the left g) and this gives a total of 2l + 1 amount of moves with (−1)s+2l+1 = (−1)s+1. So if there are an odd number of shuffled tuples (due to n being even, i.e. the tuple is of even degree) then the sum will consist of paired tuples with different signs plus one extra tuple, giving a nontrivial sum. But if the degree is odd (i.e. an even number of shuffled tuples) then the sum will consist of only paired tuples with different signs, 2 giving a sum of 0. Thus x = 0 if degx is odd, where x = [g1| · · · |gn], and this is precisely the definition of strict anti-commutativity.

48: Let p be a prime and let Cp be the cyclic group of order p with trivial Fp-action. Explain how 2 ∼ the fact H (Cp, Fp) = Fp and the classiﬁcation of extensions of Cp by Fp matches up with the classiﬁca- tion theorem for groups of order p2.

If P is a group with |P | = p2 then it has nontrivial center, |Z(P )|= 6 1, by Theorem 4.3.8[2]. If |Z(P )| = p2 then P = Z(P ) and P is abelian. The only other scenario is |Z(P )| = p, in which case |P/Z(P )| = p; this implies P/Z(P ) is cyclic and hence it is a fact that P is abelian. By the Fundamental Theorem of Finitely Generated Abelian Groups, P must then be either the cyclic group Cp2 or the elementary abelian group Cp × Cp; another proof of this is given in Corollary 4.3.9[2]. This means there are only two extension groups of Cp by Fp, but this does not necessarily mean that there are only two classes of extensions (pos- ∼ sible short exact sequences). Theorem IV.3.12[1] states that E(Cp, Fp) = Fp and hence there are a total of p classes of group extensions. There is the canonical split extension 0 → Fp → Cp × Cp → Cp → 1, and so the other extensions must ﬁt into p − 1 classes and arise from projections Cp2 Cp (this is th because the only injection Cp ,→ Cp2 is the canonical inclusion, i.e. the p -power map). Switching to βi p multiplicative notation, these extensions are 1 → Cp ≡ hai → Cp2 ≡ hbi → Cp → 1 with a 7→ b and i βi(b) = a for 1 ≤ i ≤ p − 1.

49: Let G be a finite group, let H ⊆ G, and let K be any group. Let F be a field which acts triv- ∼ ∗ ∗ = ∗ ially on G and K, and consider the Künnethisomorphism κ : H (G, F) ⊗F H (K, F) → H (G × K, F). G×K G G×K G Show that resH×K ◦ κ = κ ◦ (resH ⊗ id) and trH×K ◦ κ = κ ◦ (trH ⊗ id).

Both equations are straightforward, so we will only prove the latter (concerning the transfer map). For ∗ ∗ ∗ ∗ uH ∈ H (H, F) and v ∈ H (K, F), κ is deﬁned on H (H, F)⊗F H (K, F) by κ(uH ⊗v) = huH ×v, x⊗yi = degv·degx huH , xi · hv, yi, where we hide the factor (−1) for convenience. Then (tr ◦ k)(uH ⊗ v) = P −1 P trhuH × v, x ⊗ yi = (g,k)∈(G×K)/(H×K)h(g, k) · uH × v, (g, k) x ⊗ yi = (g,1)∈(G/H)×{1}hguH × −1 P −1 P −1 1v, g x⊗1yi = g∈G/H hguH , g xi·hv, yi. But [κ◦(tr⊗id)](uH ⊗v) = κ( g∈G/H guH (g x)⊗v(y)) = P −1 g∈G/H hguH , g xi · hv, yi, so the two compositions are in fact equal, as desired.

50: For an abelian group G whose order is divisible by the prime p, Theorem V.6.6[1] states that

72 V ∼= the isomorphism ρ : (Gp) ⊗ Γ (pG) → H∗(G, p) is natural if p 6= 2. Why? Zp Zp Zp Z

Referring to pg126[1], where pG = Tor(G, Zp) and Gp = G ⊗ Zp = G/pG, we have a split-exact uni- V2 versal coeﬃcient sequence 0 → (Gp) → H2(G, Zp) → pG → 0. Now ρ(x ⊗ y) = ψ(x)ϕ(y), where V ψ : (Gp) → H∗(G, Zp) is the natural map of Theorem V.6.4[1] and ϕ : Γ(pG) → H∗(G, Zp) is the Zp-algebra homomorphism extended from a splitting φ : pG → H2(G, Zp) of the above sequence. Since ψ is natural and ϕ is an extension of the splitting φ, the question of naturality of ρ reduces to the question of naturality of φ (in dimension 2). The splitting is made by choice, and if p is odd, we may use the V2 canonical splitting H2(G, Zp) → (Gp) given in Exercise V.6.4(b) since 2 is invertible in Zp for p 6= 2 (i.e. prime p odd); remember that a splitting can be made on either side of the sequence. Thus the isomorphism ρ is natural if p 6= 2. But if p = 2 then we do not have a known canonical splitting, and we cannot prove naturality in this case since we do not have an explicit map.

51: Referring to the proof of Proposition VI.2.6[1], if η : M → Q0 is an admissible injection (i.e. a split injection of H-modules) and M is projective as a ZH-module, then why is Cokerη projective as a ZH-module?

η Since η is H-split, we have a split-exact sequence 0 → M → Q0 → Cokerη → 0, and so Q0 =∼ M ⊕Cokerη. But Q0 is ZG-projective by Corollary VI.2.2.[1], hence ZH-projective by Exercise I.8.2. Now a direct sum is projective iﬀ each direct summand is projective, so Cokerη must be ZH-projective.

52: Let G be a goup with order r. Show that for each q there exists G-modules C with Hb q(G, C) cyclic of order r.

0 For q = 0 we can take C = Z since Hb (G, Z) = Z/|G|Z = Zr. Then using the dimension-shifting tech- 0 ∼ 1 0 ∼ −1 nique, we can ﬁnd G-modules C1 and C2 such that Hb (G, Z) = Hb (G, C1) and Hb (G, Z) = Hb (G, C2); see property 5.4 on pg136[1]. Therefore, through repeatable applications of dimension-shifting, we can q ∼ range over all q to obtain Hb (G, C) = Zr.

53: Fill in the details to the proof of Proposition VI.7.1[1] which states that the evaluation pairing i 0 0 ρ : H (G, M ) ⊗ Hi(G, M) → Q/Z is a duality pairing, where M = Hom(M, Q/Z).

Let F be a projective resolution of Z over ZG. The evaluation pairing is obtained by composing the 0 0 0 pairing h·, ·i : HomG(F,M ) ⊗ (F ⊗G M) → M ⊗G M with the evaluation map M ⊗G M → Q/Z. The pairing hu, z = x ⊗ mi is given by u ⊗ (x ⊗ m) = u(x) ⊗ m, and composing this with the evalu- i 0 0 ation map gives ρ(u ⊗ z) = [u(x)](m). Note that ρ gives rise to the mapρ ¯ : H (G, M ) → Hi(G, M) 0 ∼ deﬁned by [¯ρ(u)](z) = ρ(u ⊗ z) = [u(x)](m). We have HomG(F,M ) = HomG(F, Hom(M, Q/Z)) = (∗) 0 HomG(F ⊗M, Q/Z) = Hom(F ⊗GM, Q/Z) = (F ⊗GM) . The equality (∗) arises from Exercise III.1.3, as 0 ∗ 0 ∼ 0 Q/Z has trivial G-action. Since ( ) is exact, we can pass to homology to obtain H (G, M ) = H∗(G, M) . ∼ This isomorphism resulted from HomG(F, Hom(M, Q/Z)) = HomG(F ⊗M, Q/Z) which is given by Theo- rem 10.5.43[2], and this is precisely the mapρ ¯. Therefore, sinceρ ¯ is an isomorphism, ρ is a duality pairing.

54: Let G = Z2 = hgi and let A = Z8, written additively. Make G act on A by x 7→ 3x, and let G act trivially on B = Z2. Show that A is cohomologically trivial, but A ⊗ B is not.

First note that Hb ∗({1},M) = 0 for any M; it is clearly trivial in all dimensions not equal to −1 and 0, and in those two dimensions it is the kernel and cokernel of the norm map M → M which is the identity (so the kernel and cokernel are trivial). Thus for A to be cohomologically trivial it suﬃces to show that Hb ∗(G, A) = 0. The complete resolution from Exercise VI.3.1 implies that Hb n(G, A) = CokerN for n even n ∼ and Hb (G, A) = KerN for n odd (see pg58[1]). The action on A gives AG = Z8/Z4 = {0+Z4, 1+Z4} = Z2 G since 1 7→ 3 ≡ 1 mod 2, and A = Z2 = {0, 4} since 4 7→ 12 ≡ 4 mod 8. Thus the norm map N : Z2 → Z2 is given by 1 7→ N1 = 1 · 1 + g · 1 = 1 + 3 = 4, which is the identity. Thus KerN = CokerN = 0 and Hb ∗(G, A) = 0. However, repeating the above with coeﬃcient module A ⊗ B we see that the norm map is the trivial map, 1 ⊗ 1 7→ 1 ⊗ 1 + g · (1 ⊗ 1) = 1 ⊗ 1 + 3 ⊗ 1 = 4 ⊗ 1 = 2 ⊗ 2 = 0. Thus

73 2i+1 ∼ Hb (G, A) = KerN = Z2 ⊗ Z2 = Z2 and A ⊗ B is not cohomologically trivial.

55: Let H/G and let M be a G-module. Then M H is naturally a G/H-module, and the pair (ρ : G → G/H, α : M H → M) is compatible in the sense that α[ρ(g) · m] = g · α(m). Thus we have a homomorphism inf : Hn(G/H, M H ) → Hn(G, M) called the inﬂation map. Using this, show that for the semi-direct product G = H o K and module M with trivial G-action, the group Hn(K,M) is isomorphic to a subgroup Hn(G, M).

From the inclusion i : K,→ G and the surjection ρ : G → G/H = K we can pass to cohomology to obtain the restriction res : Hn(G, M) → Hn(K,M) and the inﬂation inf : Hn(K,M) → Hn(G, M). Since n n ρ ◦ i = idK , the composite inf ◦ res is also the identity. Thus inf is injective, so H (K,M) ⊆ H (G, M) up to isomorphism.

55: In Exercise AE.34 it was shown that a module is finitely generated Z-projective iff it is finitely generated Z-free. Weakening the hypothesis, show that Z-projective = Z-free.

Free modules are projective, so it suﬃces to show that Z-projective implies Z-free. If P is Z-projective then it is a submodule of a Z-free module. But any submodule of a Z-free module is free by Theorem I.7.3[5], so P is Z-free.

57: Prove that the symmetric group S3 has periodic cohomology, and ﬁnd its period.

∼ Since S3 = Z3 o Z2 and |S3| = 3! = 6, its Sylow subgroups are Z2 and Z3 which are both cyclic. Then by Theorem VI.9.5[1], S3 has periodic cohomology. Alternatively, any proper subgroup must have order 1 or 2 or 3 and hence must be cyclic, so Theorem VI.9.5[1] implies that S3 has periodic cohomology. 4 ∼ Alternatively, H (S3, Z) = Z6 by Exercise III.10.1 and so S3 has periodic cohomology by Theorem VI.9.1[1]. n ∼ n+4 We can see this periodicity via Exercise III.10.1, because H (S3) = H (S3) for all n > 0. Thus the period is 4.

58: Let N : Z → ZG denote the G-module homomorphism z 7→ Nz, where N ∈ ZG is the norm element, and let ε : ZG → Z be the augmentation map. Prove that if α : Z → ZG is a G-module homomorphism then α = aN for some a ∈ Z, and if β : ZG → Z is a G-module homomorphism then β = bε for some b ∈ Z.

First note that α is determined by where it sends the identity, α(1) = x. Then for α to be compatible with the G-action we must have g · x = g · α(1) = α(g · 1) = α(1) = x, so x ∈ (ZG)G = Z · N where the equality is shown in Exercise AE.1; thus α = aN for some a ∈ Z. Now β is also determined by β(1) = z, so we must have z = g · z = g · β(1) = β(g) = z0 and hence β maps G onto a single integer b; thus β = bε for some b ∈ Z (where we note that ε(G) = 1).

59: If G1 and G2 are perfect groups with universal central extensions E1 and E2, respectively, prove that E1 × E2 is a universal central extension of G1 × G2.

We have universal central extensions 0 → H2(Gi) → Ei → Gi → 1 by hypothesis (see Exercise IV.3.7). Since the direct sum of two extensions is an extension, we have a central extension 0 → H2(G1)⊕H2(G2) → E1 ×E2 → G1 ×G2 → 1. I claim that this extension is a universal central extension. Indeed, H2(G1)⊕H2(G2) is isomorphic to H2(G1×G2) by the K¨unnethformula, since H1(Gi) = 0 by per- fectness of Gi. Now, the universal central extension of G1 ×G2 is 0 → H2(G1 ×G2) → E → G1 ×G2 → 1 ∼ by deﬁnition, so E1 × E2 = E by the Five-Lemma (applied to the two sequences of G1 × G2) and hence E1 × E2 is the universal central extension of G1 × G2.

60: In the proof of Proposition VIII.2.4[1], with Γ0 ⊂ Γ, where did we use the hypothesis that |Γ:Γ0| < ∞?

74 We considered a free Γ-module F , and noted that if F 0 is a free Γ0-module of the same rank then ∼ Γ 0 n 0 0 ∼ n F = IndΓ0 F . We then applied Shapiro’s lemma to yield H (Γ ,F ) = H (Γ,F ). But Shapiro’s n 0 0 ∼ n Γ 0 Lemma is actually given by H (Γ ,F ) = H (Γ, CoindΓ0 F ). We therefore used the isomorphism Γ 0 ∼ Γ 0 0 CoindΓ0 F = IndΓ0 F of Proposition III.5.9[1], which holds if |Γ:Γ | < ∞.

75 11 References

[1] Kenneth Brown, Cohomology of Groups, Springer GTM 87 (1982).

[2] David Dummit and Richard Foote, Abstract Algebra (3rd Ed.), John Wiley & Sons (2004).

[3] Allen Hatcher, Algebraic Topology, Cambridge University Press (2006).

[4] James Munkres, Elements of Algebraic Topology, Perseus Books Publishing (1984).

[5] Serge Lang, Algebra (Revised 3rd Ed.), Springer GTM 211 (2002).

[6] James Munkres, Topology (2nd Ed.), Prentice Hall (2000).

Other Useful Literature Alejandro Adem & James Milgram, Cohomology of Finite Groups (2nd Ed.), Springer (2004). David Benson, Representations and Cohomology I+II, Cambridge University Press (1995). Jon Carlson & others, Cohomology Rings of Finite Groups (with Appendix), Springer (2003). Henri Cartan & Samuel Eilenberg, Homological Algebra, Princeton University Press (1956). Leonard Evens, The Cohomology of Groups, Oxford University Press (1991). Ross Geoghegan, Topological Methods in Group Theory, Springer GTM 243 (2008). Karl Gruenberg, Cohomological Topics in Group Theory, Springer (1970). Serge Lang, Topics in Cohomology of Groups, Springer (1996). Saunders Mac Lane, Homology, Springer (1995). Jean-Pierre Serre, Local Fields, Springer GTM 67 (1979). Jean-Pierre Serre, Galois Cohomology, Springer (2002). Charles Thomas, Characteristic Classes and Cohomology of Finite Groups, Cambridge University Press (1986). Edwin Weiss, Cohomology of Groups, Academic Press (1969).