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 g2C=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 = fg − 1 j g 2 G; g 6= 1g 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 8g 2 A. To show that I = span(A), let αgg 2 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 = (fs − 1 j s 2 Sg) over ZG. P P P P A ⊆ I since "(A) = " ( rgg)(s − 1) = "( rgg)"(s − 1) = y · 0 = 0. P s Pg Ps Pg If x 2 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 2 A for any g 2 G so that x 2 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 j I = (fs − 1 j s 2 Sg). Then every element of I is a sum of elements of 0 0 ±1 Pn−1 0 the form g − g j g = g s . For g 2 G, g − 1 2 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, 9 i0 j 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 j 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 − 1ig2G 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 2 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 jGj = n and let t be the image of T in R = Z[T ]=(T n − 1) =∼ ZG. The el- ement 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 irre- ducible 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 2 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 2 Z[G=H] has the expression x = zi(giH), and there exists an element fixed by H, namely, x0 = g0H = H where g0 2 H. H is annihilated by I = Ker" = (fs − 1g) since (s − 1)H = sH − H = H − H = 0 8s 2 S, and so Ix0 = 0. We have g − 1 2 I since "(g − 1) = "(g) − "(1) = 1 − 1 = 0. Hence (g − 1)x0 = 0 ) gx0 = x0 8g 2 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 2 Cn(X), g 2 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 2 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 2 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 ' : X ! X=G be the quotient map, which sends the disjoint collection of giW 's to '(W ). Since −1 ` ' '(W ) = i giW , giW ! '(W ) is a bijective map (restriction of ') and thus it is a homeomorphism (' continuous and open). This covering space is regular because G acts transitively on '−1(Gx) by def- inition. Elements of G are obviously deck transformations since Ggx = Gx, hence G ⊆ Aut(X). Given Γ 2 Aut(X) with Γ(a) = b, those two points are mapped to the same orbit in X=G (since ' ◦ Γ = '), and so 9 g 2 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 ' is the universal cover of X=G, so X=G is a K(G; 1) with universal cover X.