Indecomposable Modules for a Group with a Normal Cyclic Sylow P-Subgroup Adam Wood July 31, 2018

Indecomposable Modules for a Group with a Normal Cyclic Sylow p-subgroup Adam Wood July 31, 2018

Let G be a finite group and let k be a field of characteristic p. The goal is to describe the indecomposable modules of kG when G has normal cyclic Sylow p-subgroup. We first review definitions and basic theorems concerning indecomposable modules in general. We then investigate the indecomposable modules of kG when G is cyclic of order n. The primary reference is Local Representation Theory, by J.L. Alperin [1]. Most proofs can be found in [1]. This note is primarily expository and focuses on summarizing and understanding the arguments for describing the indecomposable kG-modules.

Background and Preliminary Lemmas

Let A be a k-algebra. Recall that an A-module M is said to be indecomposable if it cannot be decomposable into a direct sum of nontrivial nonzero submodules. A simple module is indecomposable, but an indecomposable module does not necessarily have to be simple. The indecomposable modules have a characterization in terms of their endormorphism ring. We introduce a local algebra, prove a lemma, and then proceed with the characterization of indecomposable modules. A k-algebra A is local if

A/rad A =∼ k.

Lemma. A k-algebra A is local if and only if every element of A is nilpotent or invertible.

Proof. Suppose that A is local and that a ∈ A. If a ∈ rad A, then a is nilpotent since the radical is a nilpotent ideal. Otherwise a + rad A is a nonzero element of A/rad A and since A/rad A =∼ k, we can write a = λ + r, where λ ∈ k and r ∈ rad A. We can explicitly construct an inverse for a. Note that a = λ + r = λ−1(1 − (−r/λ)). Using the geometric series we have that λ−1(1 − λ−1r + λ−1r2 − · · · ) is an inverse for a. This sum stops eventually since r is nilpotent. Thus, a is invertible. Conversely, suppose that every element of A is nilpotent or invertible. Then, the same statement holds true for the quotient A/rad A. Since A/rad A is semisimple, we can decompose this quotient into a direct sum of matrix algebras over k. If there was more than one matrix algebra in this decomposition, then (1, 0,..., 0) would be an element of A/rad A that is not nilpotent and not invertible. So, A/rad A is isomorphic to a matrix algebra over k. Now, a matrix of dimension greater than one has elements that are not nilpotent or not invertible, which shows that this matrix algebra must consist of 1 × 1 matrices over k. Therefore, A/rad A =∼ k. Thus, A is local. Lemma. A matrix T is invertible if and only if zero is not an eigenvalue. Proof. A matrix T is invertible if and only if det(T ) 6= 0 if and only if zero is not an eigenvalue. Note that the determinant is the product of the eigenvalues. Lemma. A matrix T is nilpotent if and only if zero is the only eigenvalue.

Proof. If T is nilpotent, then T n = 0. Let λ be an eigenvalue of T with eigenvector v. Then, T v = λv and T n = 0 imply that λ = 0. If all of the eigenvalues of T are zero, then the characteristic polynomial of T , T n − λI = 0, becomes T n = 0 so that T is nilpotent. The characterization of indecomposable modules is as follows.

Proposition. Let M be an A-module. Then, M is indecomposable if and only if End(M) is local. Proof. If M is ﬁnitely generated, there is a nice proof that uses Fitting’s Lemma. Suppose that End(M) is not local and let ρ ∈ End(M) be an element that is not invertible and not nilpotent. The generalized eigenspaces of the matrix corresponding to ρ form a direct sum decomposition for M. Since ρ is not invertible, zero is

1 an eigenvalue of ρ. Since ρ is not nilpotent, there exists a nonzero eigenvalue. Therefore, the generalized eigenspaces form a nontrivial direct sum decomposition. So, M is not indecomposable. Conversely, suppose that M is decomposable and write M = M1 ⊕M2 for submodules M1 and M2. The projections onto M1 and onto M2 are elements of End(M) that are not invertible and not nilpotent. Thus, End(M) is not local. The Krull-Schmidt Theorem gives a uniqueness statement concerning direct sum decompositions of modules into indecomposable modules.

Theorem. Let M is an A-module and let M = N1 ⊕ · · · ⊕ Nr and N = L1 ⊕ · · · ⊕ Ls be two direct sum decompositions of M into indecomposable submodules. Then, r = s and, up to permutation of indices, ∼ Ui = Vi for all i. A corollary gives the property of cancelation of direct sums.

Corollary. If M, U, and V are A-modules satisfying M ⊕ U =∼ M ⊕ V , then U =∼ V . Proof. Decompose M, U, and V as a direct sum of indecomposable modules. Then, M ⊕ U =∼ M ⊕ V have the same decompositions in the sense of the Krull-Schmidt Theorem. That is, the isomorphism classes of indecomposable and the multiplicity of each isomorphism class are same. Subtract the multiplicities of the indecomposable modules appearing in the decomposition of M from M ⊕ U and M ⊕ V . The remaining decompositions are therefore the same. Thus, U =∼ V . Cyclic Group of Order n

Let G = hgi be a cyclic group of order n. We have shown in another document that in the characteristic zero case, n−1 n−1 M M kG =∼ k[x]/(xn − 1) =∼ k[x]/(x − ζi) =∼ k =∼ kn i=0 i=0 where ζ is a primitive nth root of unity. In the characteristic zero case, we know that kG is semisimple and decomposes into a direct sum of its irreducible representations, and hence, into a direct sum of its simple modules. So, when k has characteristic zero, there are n distinct irreducible representations of G and we can exhibit their character tables explicitly. If k has characteristic p and p - |G|, we have the same decomposition of kG and hence, the same result. However, if p | |G|, then kG is not semisimple and we do not have the same result. Note that xn − 1 does not factor into distinct linear factors and we do not have the same decomposition of kG. We are interested in ﬁnding the indecomposable kG modules.

Let V be a kG-module. We will show that V decomposes into Jordan blocks and that these Jordan blocks are the indecomposable kG-modules. Consider the action of G on V by multiplication by g, the generator of G. This action deﬁnes a kG-automorphism of V ; let T be the matrix of this automorphism of V . Now, kG =∼ k[x]/(xn − 1) implies that kG is a PID. Looking at multiplication by g is “essentially” the same as looking at multiplication by x is the polynomial ring since g generates G and gn = 1. So, talking about the Jordan canonical form of the matrix T makes sense. Note that gnv = v since g has order n, so T n = I. If λ is an eigenvalue of T , then T nv = λnv implies that λn = 1 so that any eigenvalue is an nth root of unity. The theory of Jordan canonical form tells us that T decomposes into its Jordan blocks. Since V is a ﬁnitely generated module over the PID kG, V decomposes into a direct sum of modules, which correspond to the Jordan blocks of T . That is, we can write

V = V1 ⊕ · · · ⊕ Vs where Vi has a basis v1, . . . , vr satisfying T (vi) = λvi + vi+1 for all 1 ≤ i ≤ r − 1 and T (vr) = λvr and λ is an eigenvalue of T . Then, each Vi corresponds to a single eigenvalue. By the uniqueness of the Jordan canonical form, each Vi is indecomposable. If not, then there would be more Jordan blocks. We have shown that every Jordan block is an indecomposable kG-module. Also, any kG-module decomposes into a direct

2 sum of Jordan blocks corresponding to an nth root of unity. Since every kG-module decomposes uniquely into indecomposable modules, this argument shows that the indecomposable kG-modules are Jordan blocks th Jr(λ) of dimension r for an n roots of unity λ. We next show that there are n indecomposable kG-modules. Fix a Jordan block V corresponding to λ of dimension r. Suppose that n = pam, where p - m. We will show that λ is actually an mth root of unity and r ≤ pa. Since k has characteristic p, the function x 7→ xp a is injective as it is a ﬁeld homomorphism. So, λn = (λm)p = 1 implies that λm = 1. Therefore, λ is an mth root of unity. We can factor xm − 1 into distinct factors,

m m Y x − 1 = (x − λi). i=1

For simplicity, assume that λ = λ1. Then, λi for 2 ≤ i ≤ m is not an eigenvalue for T on V since V corresponds to the eigenvalue λ. That is, for v ∈ V ,(T − λiI)v 6= 0 for all 2 ≤ i ≤ m. Since k has a a a characteristic p,(xm − 1)p = (xm)p − 1p = xn − 1. Therefore,

n m pa pa pa pa 0 = T − I = (T − 1) = (T − λI) (T − λ2I) ··· (T − λmI) .

pa Since the terms involving λi for 2 ≤ i ≤ m are nonsingular on V , the above shows that (T −λI) = 0. Recall that a basis for V is given by v1, . . . , vr, where T vi = λvi + vi+1 for 1 ≤ i ≤ r − 1 and T vr = λvr. That is, k r a (T − λI) vk 6= 0 for k < r and (T − λI) vr = 0. then, we must have that r ≤ p . Consider the Jordan block V corresponding to an nth root of unity λ of dimension r. We have shown that λ is an mth of unity and that r ≤ pa. Therefore, we have m · pa choices for possible Jordan blocks. That is, the Jordan blocks have one of n possible structures. It remains to show that these are distinct. Each Jordan block for T is a kG-module. Since λ and r characterize the Jordan block up to isomorphism, there are n distinct Jordan blocks. Thus, the are n indecomposable kG-modules. We have found the indecomposable modules, but what about the simple modules? Let V be a Jordan block of dimension r corresponding to eigenvalue λ. If r = 1, then V is simple. If r > 1, then there is a basis v1, . . . , vr as above. The submodule generated by v2, . . . , vr is a proper submodule of V ; it is also a Jordan block corresponding to λ but it has dimension r − 1. So, the simple kG-modules are the Jordan blocks of dimension one. There are exactly m of these since there are m Jordan blocks of dimension r for every r ≤ pa. We next look at the radical and the socle series of V for a Jordan block for λ of dimension r. We will show that the quotients of both of these series are simple and hence, that V is uniserial.

Let W be the subspace of V spanned by v2, . . . , vr. We claim that (g − λ1)V = W . By our choice of basis, (g − λ1).vi = vi+1 for all 1 ≤ i ≤ r − 1 and (g − λ1).vr = 0. So, (g − λ1)V = W . Now,

m g − 1 = (g − λ1)(g − λ21) ··· (g − λm1) and (g − λ21) ··· (g − λm1)V = V imply that (gm − 1)V = (g − λ1)V . Also,

a a a (gm − 1)p = gp m − 1p = gn − 1 = 0

shows that (gm −1) is a nilpotent element of kG. Since G is commutative, kG is commutative, and therefore, (gm − 1) ∈ rad kG. So, W ⊆ (rad kG)V = rad(V ). But, W is a subspace of V of codimension one and is therefore equal to rad(V ). Since W is a Jordan block for λ with dimension one less than that of V , rad(V ) is a Jordan block for λ of dimension one less than that of V . Applying the above argument to rad(V ), we get that the radical series descends one dimension at a time. Also, V/W is a simple kG-module, with multiplication by g given by λ. So, the quotients of the radical series are all simple. We can also consider the socle series for V . The socle of V is semisimple and is therefore a direct sum of simple submodules. We have shown that the simple kG-modules are Jordan blocks of dimension one. But, V only contains one Jordan block of dimension one, the space spanned by vr. So, the socle of V is simple. Consider V/soc(V ),

3 which has basis {v1 + soc(V ), . . . , vr−1 + soc(V )}. Therefore, V/soc(V ) is a Jordan block for λ and simi- 2 larly, has a simple socle, in this case, the space spanned by vr−1 + soc(V ). Recall that soc (V ) is deﬁned to be the submodule so that soc2(V )/soc(V ) is the socle of V/soc(V ). Therefore, soc2(V ) is spanned by vr and vr−1. Continuing this argument inductively, we can see that each quotient of the socle series is simple.

We can now fully describe the structure of kG as a kG-module. We have shown that there are m simple kG-modules, which are Jordan blocks of dimension one. The indecomposable kG-modules are the n uniserial kG-modules corresponding to a Jordan block of dimension k for an mth root of unity, where 1 ≤ k ≤ pa. Recall that each uniserial kG-module has simple composition factors; if a uniserial kG-module corresponds th to an m root of unity λ, then g acts by λ on the simple composition factors. Call these factors Sλ. Let P be an indecomposable projective kG-module of dimension pa corresponding to the mth root of unity λ. This module is projective by a future corollary below. Also, since every projective kG-module has dimension over k divisible by pa, the indecomposable kG-modules of dimension pa are the only indecomposable projective ∼ kG-modules. Then, P/rad(P ) = Sλ and we can see that P is the unique projective indecomposable module th corresponding to Sλ. We then get a projective indecomposable module for each m root of unity λ. Let Pλ denote the projective indecomposable module corresponding to Sλ. Then, ∼ kG = Pλ1 ⊕ · · · ⊕ Pλm ,

th a a where λi are distinct m roots of unity. Since kG has dimension p m and each Pλi has dimension p , the dimensions above agree. I think it is common to visualize this decomposition in the following picture, where the vertical lines are the composition series and the diﬀerent columns correspond to the diﬀerent projective indecomposable kG-modules.

Pλ1 Pλm

◦ ◦

. . . ⊕ ··· ⊕ .

◦ ◦

We next look at the decomposition of kG in terms of factoring the polynomial xn − 1. As above, suppose that n = pam, where p - m.Then, since k has characteristic p,

a a xn − 1 = xp m − 1 = (xm − 1)p

and we can see that m n Y pa x − 1 = (x − λi) , i=1

4 th where λi are distinct m roots of unity. Therefore,

∼ n ∼ pa kG = k[x]/(x − 1) = k[x]/(x − λi)

which agrees with the decomposition above.

Sylow p-subgroups

Theorem. Let P be a projective kG-module and let H be a subgroup of G. Then, PH , the restriction of P to kH, is a projective kH-module. Proof. Since a projective module is a direct summand of a free module, it suﬃces to show that kG is free as a kH-module. Let x1, . . . , xh be a complete set of representatives of distinct left cosets of H in G. Then

h [ G = xiH. i=1

Also, the set kxiH is a kH-module since it is invariant under H. Deﬁne

φi : kH → kxiH

λh → λxih

Then, φi is a well deﬁned kH-module isomorphism. So,

h ∼ M ∼ M ∼ h (kG)H = kxiH = kH = (kH) i=1 i=1 so that kG is a free kH-module. Corollary. Let P be a Sylow p-subgroup of G of order pa. Then, every projective kG-module has dimension over k divisible by pa. Proof. Since P is a p-group, kP has a unique simple module of dimension one over k. Consider kP and decompose kP into a direct sum of indecomposable projective kP -modules, which is possible since kP is a free kP -module. Recall that the indecomposable projective modules are in bijection with the simple modules. Since there is only one simple module of dimension one, there is only one indecomposable projective modules in the direct sum decomposition of kP . So, kP is indecomposable as a kP -module. Therefore, any projective kP -module is free. Since kP has dimension pa, any free kP -module must have dimension divisible by pa. We have shown that any projective kP -module has dimension divisible by pa. Now, let Q be a projective kG-module. By the previous theorem QP is a projective kP -module and hence has dimension divisible by pa. Thus, Q has dimension divisible by pa.

Lemma. Let R be a normal Sylow p-subgroup of G and let M be a kG-module. Then, rad(M) = rad(MR). If R is cyclic with generator x, then rad(M) = (1 − x)M. Proof. Since R is normal in G, g ∈ G induces an automorphism of kR via conjugation. Therefore, −1 grad(kR)g = rad(kR). By deﬁnition, grad(MR) = grad(kR) · MR = grad(kR) · M. By the above −1 comment, grad(kR) · M = grad(kR)g · gM = rad(kR)M = rad(MR). So, rad(MR) is a kG-submodule of M. We want to show that M/rad(MR) is a semisimple kG-module. We can form the quotient M/rad(MR) in kR and see that this quotient is a semisimple kR-module. Therefore, it is a direct sum of simple modules. Since R is a p-group, the only simple kR-module is the trivial module k. So, any element of R acts as the identity on M/rad(MR) and hence, M/rad(MR) is a k[G/R]-module. Since k does not divide the order of G/R, we have that k[G/R] is semisimple. Thus, M/rad(MR) is semisimple and rad(M) ⊆ rad(MR).

5 Conversely, by Cliﬀord’s Theorem, since M/rad(M) is a semisimple kG-module, it is also a semisimple kR- module. So, rad(MR) ⊆ rad(M). Thus, rad(M) = rad(MR). Now, suppose that R is cyclic with generator x. Recall that the only simple kR-module is the trivial module k. Also, x acts on the quotient M/(1 − x)M as the identity since m + (1 − x)M 7→ xm + (1 − x)M and m − xm ∈ (1 − x)M. Then, each element of R induces the identity linear transformation on M/(1 − x)M and therefore, M/(1 − x)M is semisimple since it is a sum of copies of the trivial module. These comments show that rad(MR) ⊆ (1 − x)M. Since 1 − x annihilates the trivial module k (as x acts as 1 on k) and k is the only simple module, 1 − x ∈ rad(kR). Then, (1 − x)M ⊆ rad(kR)M = rad(MR). Thus, rad(M) = (1 − x)M. Group with Cyclic Normal Sylow p-Subgroup

Let G be a group and suppose that R is a cyclic normal Sylow p-subgroup of G of order pa with generator x. Previously, we have described the indecomposable kG-modules when G is cyclic of order n. Now, we just have that G has a cyclic normal Sylow p-subgroup that is cyclic of order pa. In the cyclic case, we have shown that the indecomposable kG-modules are Jordan blocks which are uniserial modules and that the projective indecomposable modules are the top dimension Jordan blocks. We now describe the indecomposable projective kG-modules. Recall that each indecomposable projective module P corresponds ∼ to a simple module S via P/rad(P ) = S. If S has dimension d, then PR is isomorphic to d copies of kR, P has radical length pa, and each quotient of the radical series is d dimensional. Lemma. Let k be a ﬁeld of characteristic p and let R be a p-group. Then, kR is a local ring. Proof. Consider the augmentation map

ε : kR → k X X arr 7→ ar r∈R r∈R with kernel I, the augmentation ideal of kR. Since kR/I =∼ k, I is a maximal ideal of kR. Therefore, if P J(kR) denotes the Jacobson radical of kR, J(kR) ⊆ I. Conversely, suppose that r∈R arr ∈ I. Then, P pn r∈R ar = 0. Also, since R is a p-group, r = 1 for all r ∈ R. Then, !pn !pn X X pn X pn X arr = (arr) = ar = ar = 0 r∈R r∈R r∈R r∈R

so that I ⊆ J(kR). These arguments show that I is the unique maximal ideal of kR. Thus, kR is a local ring. If R is cyclic, then we have that

n n kR =∼ k[x]/(xp − 1) =∼ k[x]/(x − 1)p .

Then, via these isomorphisms, (x − 1) is the unique maximal ideal of kR. Proposition. If P is an indecomposable projective kG-module corresponding to the simple module S of dimension d, then ∼ PR = kR ⊕ · · · ⊕ kR, d times.

Proof. By a previous lemma, PR is a projective kR-module. Since kR is a local ring, any projective is free and hence, PR is a free module. So, PR is a direct sum of copies of kR. Recall that the radical splits across direct sums and that rad(P ) = rad(PR). Since P/rad(P ) is d dimensional, also PR/rad(PR) is d dimensional. Then, ∼ ∼ ∼ PR/rad(PR) = kR⊕· · ·⊕kR/rad(kR⊕· · ·⊕kR) = kR⊕· · ·⊕kR/rad(kR)⊕· · ·⊕rad(kR) = kR/rad(kR)⊕· · ·⊕kR/rad(kR).

6 Since kR is local with maximal ideal rad(kR), by the lemma above, kR/rad(kR) =∼ k. So, kR/rad(kR) is one dimensional. Note that the dimensions of both sides in the isomorphism must agree. Thus, PR is a direct sum of d copies of kR. Proposition. If P is an indecomposable projective kG-module corresponding to the simple module S of dimension d, then P has radical length pa.

a Proof. Follows since kR is has composition length p and rad(P ) = rad(PR). Proposition. If P is an indecomposable projective kG-module corresponding to the simple module S of dimension d, then each quotient of the radical series of P is d dimensional.

Proof. Follows since kR is uniserial and rad(P ) = rad(PR). We next look at the structure of the indecomposable projective kG-modules. We will show that they are uniserial and explicitly find their radical series. We first establish some notation. Let Pk denote the projective ∼ indecomposable kG-module that is the projective cover of the trivial kG-module k. Then, Pk/rad(Pk) = k. In the case when G is a p-group, Pk = kG; however, in this case, we do not have that result. We claim that 2 2 there is a well defined map from Pk/rad (Pk) to Pk/rad(Pk). Indeed, since rad (Pk) ⊆ rad(Pk), this map is well defined. Also note that this map is surjective. Then, we have an exact sequence

2 2 0 → rad(Pk)/rad (Pk) → Pk/rad (Pk) → Pk/rad(Pk) → 0.

2 2 For notational purposes, let V = rad(Pk)/rad (Pk) and let U = Pk/rad (Pk). We can simplify the short exact sequence above using this notation by

0 → V → U → k → 0.

Since the radical quotients for the uniserial module Pk all have dimension one, V is a simple kG-module of dimension one. Let S be a simple kG-module; in particular, note that P/rad(P ) =∼ S for some indecomposable projective kG-module P . We verify a few isomorphisms and equalities that will useful in the classiﬁcation of the projective indecomposable kG-modules.

Proposition. With the notation as above, U is uniserial of composition length two.

a Proof. Since Pk is uniserial of composition length p , we have a unique composition series for Pk given by

pa pa−1 2 0 = rad (Pk) ⊆ rad (Pk) ⊆ · · · ⊆ rad (Pk) ⊆ rad(Pk) ⊆ Pk.

2 Since Pk is uniserial, also Pk/rad (Pk) is uniserial. In particular, this quotient has a unique composition 2 series given by its radical series. Reducing the composition series for Pk by rad (Pk), we get a composition series 2 2 2 2 0 = rad (Pk)/rad (Pk) ⊆ rad(P )/rad (P ) ⊆ Pk/rad (Pk). Since rad2(P ) 6= rad(P ), this series gives the composition series of length two for U. Therefore, rad(U) = V , U/V =∼ k is simple, and V is simple. ∼ Proposition. With the notation as above, (S ⊗k U)/(S ⊗k V ) = S. ∼ ∼ ∼ Proof. Since S is free over k,(S ⊗k U)/(S ⊗k V ) = S ⊗k (U/V ). Then, S ⊗k (U/V ) = S ⊗k k = S using the short exact sequence above.

Proposition. With the notation as above, S ⊗k V is simple.

7 Proof. We have shown earlier that a simple kG-module tensored over k with a one dimensional kG-module is simple. A more instructive proof goes as follows. Since V is one dimensional, suppose that V =∼ kv for some v ∈ V . If g ∈ G, suppose that g.v = λv; this gives the action of kG on V . Deﬁne V 0 to be a one 0 ∼ 0 −1 0 ∼ dimensional simple kG-module with V = kv and g.v = λ v. We will show that V ⊗k V = k. Since V 0 0 0 and V are both one dimensional, also V ⊗k V is one dimensional. The action of g ∈ G on V ⊗k V can be 0 0 given on the generator, v ⊗ v of V ⊗k V . We have that g.(v ⊗ v0) = g.v ⊗ g.v0 = λv ⊗ λ−1v0 = λλ−1v ⊗ v0 = v ⊗ v0

−1 0 since λ ∈ k. We have shown that V ⊗k V is a one dimensional kG-module with a trivial action of G; 0 ∼ 0 ∼ ∼ hence, V ⊗k V = k, the trivial kG-module. Now, note that S ⊗k V ⊗k V = S ⊗k k = S is simple. Suppose, 0 for the sake of contradiction, that S ⊗k V has a non trivial proper submodule, M. Then, M ⊗k V is a non 0 trivial proper submodule of S ⊗k V ⊗k V , which is a contradiction since the latter module is simple. Thus, S ⊗k V is simple.

Proposition. With the notation as above, S ⊗k U is not semisimple.

Proof. We will show that rad(S ⊗k U) 6= 0. Since rad(U) 6= 0, by a lemma above, if x is the generator of the cyclic Sylow p-subgroup of G, then (1 − x)U 6= 0. Choose u ∈ U with u 6= 0 so that (1 − x)u 6= 0. Note that, since S is simple, SR is semisimple and is therefore a sum of simple kR-modules. However, since R is a p-group and k has characteristic p, the only simple kR-module is the trivial module k. Therefore, S has a trivial action by R. Let s ∈ S with s 6= 0. Then, by the above discussion, xs = s. So, (1 − x)(s ⊗ u) = s ⊗ u − xs ⊗ xu s ⊗ u − s ⊗ xu = s ⊗ (u − xu) = s ⊗ (1 − x)u 6= 0

since s 6= 0 and (1 − x)u 6= 0. Thus, (1 − x)(S ⊗k U) 6= 0, which shows that rad(S ⊗k U) 6= 0 by the lemma above.

Proposition. With the notation as above, S ⊗k U is uniserial of composition length two.

Proof. Since S ⊗k U is not semisimple, rad(S ⊗k U) 6= 0. Since (S ⊗k U)/(S ⊗k V ) is simple, rad(S ⊗k U) ⊆ ∼ S ⊗k V . Since S ⊗k V is simple, rad(S ⊗k U) = S ⊗k V . Then, (S ⊗k U)/rad(S ⊗k U) = S is simple. Also, 2 since S ⊗k V is simple, rad (S ⊗k U) = 0 and S ⊗k U is uniserial of composition length two. Let P be the projective indecomposable module corresponding to the simple module S and suppose that S has dimension d.

Show that the same or similar arguments work for an arbitrary kG-module M with M/rad(M) =∼ S. 2 ∼ Proposition. rad(P )/rad (P ) = S ⊗k V ∼ ∼ Proof. We have shown that (S ⊗k U)/rad(S ⊗k U) = S. By assumption, P/rad(P ) = S. So, we have an isomorphism P/rad(P ) → (S ⊗k U)/rad(S ⊗k U).

Since P is projective, this isomorphism lifts to a homomorphism P → S ⊗k U. Since the original map is surjective, the lifted map is surjective. One can use Nakayama’s Lemma to prove this fact. So, P surjects 2 2 onto S ⊗k U. Therefore, rad(P )/rad (P ) surjects onto rad(S ⊗k U)/rad (S ⊗k U). But, 2 ∼ rad(S ⊗k U)/rad (S ⊗k U) = S ⊗k V. 2 By the classiﬁcation of indecomposable projective modules, rad(P )/rad (P ) has dimension d. Also, S ⊗k V has dimension d. Thus, the two modules are isomorphic. In particular, rad(P )/rad2(P ) is simple.

8 a k−1 k ∼ Proposition. For all 1 ≤ k ≤ p , rad (P )/rad (P ) = S ⊗k V ⊗k · · · ⊗k V . | {z } (k−1) times Proof. We have shown the claim for k = 2. Assume that the claim holds for k. We must show that k k+1 ∼ k−1 k ∼ rad (P )/rad (P ) = S ⊗k V ⊗k · · · ⊗k V . By induction, we know that rad (P )/rad (P ) = S ⊗k | {z } k times 0 V ⊗k · · · ⊗k V is simple and therefore, the projective cover of this simple module, say P , surjects onto | {z } (k−1) times k−1 0 0 ∼ rad (P ). By the above proposition, we know that since P /rad(P ) = S ⊗k V ⊗k · · · ⊗k V , is simple, also | {z } (k−1) times 0 2 0 ∼ 0 2 0 rad(P )/rad (P ) = S ⊗k V ⊗k · · · ⊗k V . Then, rad(P )/rad (P ) surjects onto | {z } k times

rad(radk−1(P ))/rad2(radk−1(P )) = radk(P )/radk+1(P ).

Since the dimensions agree, we have that

k k+1 ∼ rad (P )/rad (P ) = S ⊗k V ⊗k · · · ⊗k V . | {z } k times By induction, we have proven the claim. In summary, the projective indecomposable kG-modules are as follows. There is one for each simple kG- module S. Suppose that P corresponds to the simple module S of dimension d. Then, P is uniserial of composition length pa with k−1 k ∼ rad (P )/rad (P ) = S ⊗k V ⊗k · · · ⊗k V | {z } (k−1) times

a 2 for all 1 ≤ k ≤ p , where, if Pk is the projective cover of the trivial module k, then V = rad(Pk)/rad (Pk).

We next consider an arbitrary indecomposable kG-module. We will show that any indecomposable kG- module is uniserial; each indecomposable module will be uniquely determined by its simple quotient by its radical and its composition length. We will use the classification of the projective indecomposable kG- modules to find the indecomposable kG-modules. The general ideal of the proof is as follows. Let M be an indecomposable kG-module. Then, M is the homomorphic image of a projective indecomposable kG-module. This will show that M is uniserial. We then explain why M being uniserial gives us a characterization of the indecomposable kG-modules. To prove that M is the homomorphic image of a projective indecomposable kG-module, we take two different approaches. The first is in [1] and is slightly more roundabout and the second is from [2] and is slightly more direct. The proof in [2] is more general and applies to any module over a finite dimensional Nakayama algebra over a field. A Nakayama algebra is an algebra in which the indecomposable projective and injective modules are all uniserial. In particular, we have shown that the projective indecomposable kG-modules are uniserial. Since the projective and injective kG-modules are the same, kG is a Nakayama algebra. Proposition. Let M be an indecomposable kG-module. Then, M is the homomorphic image of a projective indecomposable kG-module. Proof. We will show that M decomposes into a direct sum of the image of a projective indecomposable kG-module and a possibly zero submodule; the indecomposability of M will imply that M is equal to the image of the projective indecomposable kG-module. Let V ⊆ U ⊆ M be submodules so that U/V is uniserial and, among all uniserial quotients of submodules of M, U/V has the maximum composition length. Since rad(U/V ) is a submodule of U/V , there exists a submodule W ⊆ U containing V so that W/V = rad(U/V ). By the third isomorphism theorem, U/W =∼ S for some simple module since U/V is

9 uniserial. Choose P , an indecomposable projective module, corresponding to S, that is, P/rad(P ) =∼ S. Consider the map ψ : P → U/W given by the composition P → P/rad(P ) → U/W since the latter modules are both isomorphic to S and note that ψ is surjective. Also consider the natural projection homomorphism π : U → U/W and note that π is surjective. Since P is projective, there exists a map φ : P → U so that ψ = π ◦ φ. U φ π ψ P U/W

Let U0 = φ(P ) and note that U0 is non zero. Since ψ = π ◦ φ and ψ is surjective, U0/W = U/W . Therefore, U = U0 + W . Note that U/V is uniserial and therefore has unique maximal submodule rad(U/V ) = W/V . Also, (U0 + V )/V ⊆ U/V and (U0 + V )/V + W/V = U/V . This shows that (U0 + V )/V is not contained in the maximal module of U/V and must therefore be equal to U/V . So, U0 + V = U. To see that this ∼ sum is direct, we must show that U0 ∩ V = 0. By the third isomorphism theory, U0/(U0 ∩ V ) = U/V . If U0 ∩ V 6= 0, then U0 has greater composition length than U/V , which is a contradiction. Write `(M) for the composition length of a module. Since ` is additive in short exact sequences, `(M) = `(N) + `(M/N). So, `(U0/(U0 ∩ V )) = `(U0) − `(U0 ∩ V ) < `(U0) since U0 ∩ V 6= 0 and therefore has nonzero composition length. But, `(U0/(U0 ∩ V )) = `(U/V ), and we have the contradiction. Thus, U = U0 ⊕ V . We next decompose M into a direct sum of U0 and another submodule. Since the quotient of a uniserial module is uniserial, also the image of a uniserial module is uniserial. So, U0 is uniserial and therefore has a simple socle, soc(U0). Since every simple module is isomorphic to the socle of an indecomposable injective module, let Q be the ∼ indecomposable injective module with soc(U0) = soc(Q). Let ψ : soc(U0) → Q be the composition of the isomorphism and the inclusion soc(U0) → soc(Q) → Q and let ι : soc(U0) → M denote the inclusion map. Since Q is injective, there exists a map φ : M → Q so that φ ◦ ι = ψ.

ψ soc(U0) Q φ ι M

Let V0 = Ker(φ). If U0 ∩ V0 6= 0, then U0 ∩ V0 would contain the unique simple submodule of U0, which is soc(U0). Then, φ(soc(U0)) = 0, which contradicts the fact that φ ◦ ι = ψ and ψ is nonzero. So, U0 ∩ V0 = 0. ∼ By the third isomorphism theorem, (U0 + V0)/V0 = U0/(U0 ∩ V0) = U0 is uniserial. Also, by the ﬁrst isomorphism theorem, M/V0 is isomorphic to a submodule of Q, which is uniserial, and is therefore uniserial. We have that (U0 +V0)/V0 and M/V0 are both uniserial quotient of submodules of M. Also, using the above isomorphisms, `((U0 + V0)/V0) = `(U0) and `(U0) = `(U/V ). By the maximality of the composition length of U/V ,(U0 + V0)/V0 = M/V0. Therefore, M = U0 + V0. Since we have shown that U0 ∩ V0 = 0, we have that M = U0 ⊕ V0. But, since M is indecomposable and U0 is non zero, M = U0. Proof. Since M is a ﬁnitely generated kG-module we have a surjective homomorphism φ :(kG)n → M. Pk Decompose kG into k indecomposable projectives, Pi. Then, M = i=1 φ(Pi), where this sum might not be direct. Now,

k k k X X X rad(M) = rad(kG)M = rad(kG) φ(Pi) ⊆ rad(A)φ(Pi) = rad(φ(Pi)) i=1 i=1 i=1

j j so that if rad (φ(Pi)) = 0 for some ﬁxed j for all i, then rad (M) = 0. If there did not exist i so that j j rad (φ(Pi)) = 0, then rad (M) 6= 0. This shows that there exists i so that φ(Pi) has composition length equal to that of M. Let P be the indecomposable projective submodule of (kG)n with image having equal composition length to that of M and let U = φ(P ). Since P is indecomposable and projective, P is uniserial and hence, U is uniserial. Since U is uniserial, U has a simple socle. Consider the injective hull I of U with

10 essential monomorphism ε : U → I. We will show that soc(I) is simple. We claim that soc(U) is an essential submodule of U. Indeed, since U is uniserial, if N ⊆ U is any submodule, then either N = 0 or soc(U) ⊆ N. So, if N ∩ soc(U) = 0, then N = 0 and soc(U) is an essential submodule of U. Since the injective hull of an essential submodule is the injective hull of the module, we have that I is the injective hull of soc(U). But, since soc(U) is simple, it is equal to the socle of its injective hull. Hence, soc(U) = soc(I). In particular, soc(I) is simple, which shows that I is indecomposable. Since I is injective, there exists a map ψ : M → I so that ψ ◦ ι = ε. U ε I ψ ι M Note that `(Im(M)) = `(M/Ker(ε)) = `(M) − `(Ker(ε)) so that `(Im(M)) ≤ `(M). But, `(U) = `(M) shows that `(M) ≤ `(Im(M)) ≤ `(M) so that `(Im(M)) = `(M). Since I is indecomposable and injective, I is uniserial, and hence has a unique submodule of each composition length; therefore, ψ(U) = ψ(M). We have a map ι : U → M. We will show that this map splits. Since ψ ◦ ι = ε, ψ(U) = ε(U). Then, U =∼ ε(U) = ψ(U) = ψ(M) since ε is injective; let ε denote this isomorphism from U to ε(U). We then have a map ε−1 ◦ ψ : M → U with (ε−1 ◦ ψ) ◦ ι = ε−1 ◦ ε = id. So, U → M splits and U is a direct summand of M. Since M is indecomposable, M = U.

Proposition. Every indecomposable kG-module is uniserial. Proof. Every indecomposable kG-module is the image of a projective indecomposable, and hence uniserial, kG-module. Since the image of a uniserial module is uniserial, every indecomposable kG-module is uniserial.

We now comment on the ramiﬁcations of this result. Let M be an indecomposable kG-module. Then, M is the homomorphic image of a projective indecomposable kG-module or equivalently, M is uniserial. So, M is completely characterized by its simple quotient, M/rad(M), and its composition length. Indeed, since M is the homomorphic image of some indecomposable projective kG-module P , it is isomorphic to a quotient of P . But P is uniserial and hence has a unique quotient of each composition length so that M is uniquely determined by its quotient M/rad(M) =∼ P/rad(P ) and its composition length. In particular, suppose that there are s simple kG-modules and hence, s projective indecomposable kG-modules. Recall that each projective indecomposable kG-module has composition length pa. Then, each projective indecomposable kG-module P has pa corresponding indecomposable kG-modules, one for each integer k with 1 ≤ k ≤ pa, each satisfying M/rad(M) =∼ S, where S is the simple module so that P/rad(P ) =∼ S. Therefore, there are spa indecomposable kG-modules. One can make a similar picture to the picture above of the indecomposable kG-modules.

References

[1] J.L. Alperin. Local Representation Theory, volume 11 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, 1986. [2] Peter Webb. A Course in Finite Group Representation Theory, volume 161 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, 2016.