# Finitely Generated Modules over Group Rings of a Direct Product of Two Cyclic Groups

Authors: Ahmed Najim; Mohammed Elhassani Charkani
Journal: Algebra
 (2014)

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## Abstract

LetK be a commutative field of characteristic p>0 and let G=G1×G2, where G1 and G2 are two finite cyclic groups. We give some structure results of finitely generated K[G]-modules in the case where the order of G is divisible by p. Extensions of modules are also investigated. Based on these extensions and in the same previous case, we show that K[G]-modules satisfying some conditions have a fairly simple form.


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## Body

## 1. Introduction

LetK be a field of characteristic p>0 and let G be a finite group. The study of K[G]-modules in the case where the order of G is divisible by p is a very difficult task. When G is a finite abelian p-group, we find in [1] the following statement: a complete classification of finitely generated K[G]-modules is available only when G is cyclic or equal to C2×C2, where C2 is the cyclic group of order 2. In [2] we find this classification in these two cases. Still more, in the case where the Sylow p-subgroup P of G is not cyclic, the groups G such that p=2 and P is dihedral, semidihedral, or generalized quaternion are the only groups for which we can (in principle) classify the indecomposable K[G]-modules (see [2]). These reasons just cited show the importance of the study of K[G]-modules when G is of order divisible by p and equal to a direct product of two cyclic groups.Now, letK be a commutative field of characteristic p>0 and let G=G1×G2, where G1 and G2 are two finite cyclic groups. Let M be a finitely generated K[G]-module. When M is considered as a module over a subalgebra K[H] of K[G] for a subgroup H of the group G, we write M↓H.In Section2, we show that if G1 is a cyclic p-group and the characteristic of K does not divide the order of G2, then we can have a complete system of indecomposable pairwise nonisomorphic K[G]-modules. In the rest, we assume that G1=σ1 and G2=σ2 are cyclic p-groups. Under conditions that M↓G1 is a free K[G1]-module and that M/(σ1-1)M is a free K[G2]-module, we show that M is a free K[G]-module. We also show that if σ2 is of order pn, n≠0, and H2 is the subgroup of G2 generated by σ2pn-r with 0<r≤n, then under certain conditions M is a free K[G1×H2]-module. The fact that M↓G1 must be a free K[G1]-module is one of these conditions, and exactly in the end of this section we give a result that shows when this condition is satisfied. In Section 3 and always in the case where G1 and G2 are cyclic p-groups, we show that under some conditions K[G]-modules have a fairly simple form. But in case p=2, G1 and G2 are two cyclic groups of respective orders 2 and 2n, n≠0; these modules have this simple form without any other assumptions other than that they must be finitely generated over K[G].


## 2. FreeK[Cpm×Cpn]-Modules of Finite Rank

Throughout this paper, rings are assumed to be commutative with unity. We begin this section by giving a weak version of Nakayama’s lemma with an elementary proof.Lemma 1 (Nakayama).
LetG be a p-group with p odd, R a ring of characteristic pk where k is a natural number, M an R[G]-module (not necessarily finitely generated), and N a submodule of M and σ∈G. Then, one has the following:(1)
if(σ-1)M=M, then M=0;(2)
ifM=(σ-1)M+N, then M=N;(3)
ifxi, i∈I, are representatives in M of a generating family of M/(σ-1)M, then (xi)i∈I generate M.Proof.
(1) Letpr be the order of G. We have
(1)σ-1pr=∑i=0pr‍Cpri-1iσpr-i=σpr+-1pr+∑i=1pr-1‍Cpri-1iσpr-i=∑i=1pr-1‍Cpri-1iσpr-i=p∑i=1pr-1‍1pCpri(-1)iσpr-i.
For 1≤i≤pr-1, p∣Cpri, so (1/p)Cpri is a natural number. So
(2)σ-1kpr=σ-1prk=p∑i=1pr-1‍1pCpri-1iσpr-ik=pk∑i=1pr-1‍1pCpri-1iσpr-ik=0,
since R has characteristic pk. Now M=(σ-1)M=(σ-1)2M=⋯=(σ-1)kprM=0.
(2) IfM=(σ-1)M+N, then M/N=((σ-1)M+N)/N=(σ-1)(M/N), and then, by (1), M/N=0 and then M=N.
(3) IfN is the submodule generated by xi, then M=(σ-1)M+N, and then by (2) we have M=N.Remark 2.
Lemma1 remains true if p=2 and R is of characteristic 2.For a ringR of prime characteristic p and for a cyclic group G of order pr generated by an element σ, we have the following lemma.Lemma 3.
Letk=pr′ with 0<r′≤r and let H be a subgroup of G generated by σpr/k. Then, one has R[G]/(σ-1)kR[G]≅R[H] (as R-algebras).Proof.
Define(3)ψ:RGσ-1kRG⟶RH,σ¯⟼σpr/k,
where ψ is a well-defined R-algebra homomorphism. It is easy to see that ψ is surjective. As R[G]/(σ-1)kR[G] and R[H] are finite free modules of the same rank k over R, ψ is an isomorphism.Remark 4.
With the notation of Lemma3, H is simply the subgroup of G generated by σpr-r′.LetK be a commutative field of characteristic p>0 and let G be a direct product of two finite groups G1 and G2. We have K[G]=K[G1×G2]≅R[G2], where R=K[G1].Assume thatG1 is a cyclic group of order pm generated by σ1 and p does not divide the order of G2. R=K[G1] is a principal Artinian local ring. Indeed K[G1]≅K[X]/(X-1)pm; this isomorphism is induced by the homomorphism Ψ:K[X]→K[G] defined by Ψ(X)=σ1. K[X]/(X-1)pm is a principal Artinian local ring with residue field K (up to isomorphism) whose maximal ideal is generated by X-1¯. So R is a principal Artinian local ring with residue field K (up to isomorphism) whose maximal ideal is generated by σ1-1. We have K[G]≅R[G2], where R is a principal Artinian local ring of residue field K. The characteristic of K does not divide the order of G2. Under these conditions, we can apply [3, Theorem 3.6] to have a complete system of indecomposable pairwise nonisomorphic K[G]-modules.In the remainder of this section, we assume thatG1=Cpm and G2=Cpn are two cyclic groups of respective orders pm and pn and are generated, respectively, by σ1 and σ2. We have K[G]≅R[G2]. As R is a commutative ring and local and G2 is a p-group, by [4, Proposition 10, page 239], R[G2] is a local ring. Therefore K[G] is a local ring. As K is commutative ring and local and G is a p-group, K[G] is a local ring by [4, Proposition 10, page 239]. So the K[G]-projective modules are free K[G]-modules.Lemma 5.
LetM be a K[G]-module. Then, M/(σ1-1)M is a K[G2]-module (also M/(σ2-1)M is a K[G1]-module).Proof.
This lemma is a particular case of a more general result (see [5, page 386]). But for this particular case, we can give the following direct proof: M/(σ1-1)M is a (K[G1]/(σ1-1)K[G1])[G2]-module, and we have already seen that (σ1-1)K[G1] is the unique maximal ideal of K[G1] and K[G1]/(σ1-1)K[G1]≅K. So M/(σ1-1)M is a K[G2]-module.
Similarly we show thatM/(σ2-1)M is a K[G1]-module.Proposition 6.
LetM be a free K[G]-module of rank l. Then, M/(σ1-1)M is a free K[G2]-module and M/(σ2-1)M is a free K[G1]-module of the same rank l.Proof.
AsK[G] is a local ring, K[G]-projective modules are free K[G]-modules, and therefore this proposition is only a particular case of a more general result (see [5, Lemma 2.2.]). But for this particular case, we can give the following specific proof: we have M≅(K[G])l=(K[G1×G2])l. So M≅(K[G1][G2])l. Then, we have
(4)σ1-1M≅σ1-1KG1][G2l≅σ1-1KG1][G2l.
Hence,
(5)Mσ1-1M≅KG1][G2lσ1-1KG1][G2l≅KG1][G2σ1-1KG1][G2l≅KG1σ1-1KG1G2l.
As K[G1]/(σ1-1)K[G1]≅K (as we have already seen), M/(σ1-1)M≅(K[G2])l. So M/(σ1-1)M is a free K[G2]-module of rank l.
Similarly we show thatM/(σ2-1)M is a free K[G1]-module of rank l.Proposition 7.
LetM be a K[G]-module. If M↓G1 is a free K[G1]-module and M/(σ1-1)M is a free K[G2]-module, then M is a free K[G]-module.Proof.
R
=
K
[
G
1
] is a principal Artinian local ring with residue field K and σ1-1 is a generator of its maximal ideal. M↓G1 is a free R-module and Q=M/(σ1-1)M is a projective K[G2]-submodule of M/(σ1-1)M. Then, M=P⊕M′, where P is a projective R[G2]-module and P/(σ1-1)P≅Q=M/(σ1-1)M (according to [3, Proposition 4.13]). We have
(6)Mσ1-1M≅P⊕M′σ1-1P⊕M′≅P⊕M′σ1-1P⊕σ1-1M′≅Pσ1-1P⊕M′σ1-1M′≅Mσ1-1M⊕M′σ1-1M′.
So M′/σ1-1M′=0. By Nakayama’s lemma and the remark following it, M′=0. Therefore, M=P which is projective R[G2]-module. As R[G2]≅K[G] is a local ring, M is a free K[G]-module.LetJi be the Jacobson radical of K[Gi] for i∈{1,2}. Note that if K is of characteristic p (as here) and G′ is a cyclic p-group, then the Jacobson radical of K[G′] is none other than (σ-1)K[G′], where σ is a generator of G′ (see [5, page 122]).LetM be a finitely generated K[G]-module and k a natural number such that 1≤k≤pn. As K[G]≅R[G2], M is a R[G2]-module. So M/J1M is a K[X]/(X-1)pn-module. M is called of type k if M/J1M is a free K[X]/(X-1)k-module (terminology of [6]).Lemma 8.
IfM is a K[G]-module of type k with k=pr and 0<r≤n and H2 is the subgroup of G2 generated by σ2pn-r, then M/J1M is a free K[H2]-module.Proof.
AsM is of type k, M/J1M is a free K[X]/(X-1)k-module. Define
(7)ψ:KXX-1k⟶KH2,X¯⟼σ2pn-r,
where ψ is a well-defined K-algebra homomorphism. It is not difficult to show that ψ is an isomorphism (using an argument similar to that done in the proof of Lemma 3). So M/J1M is a free K[H2]-module.Theorem 9.
LetM be a K[G]-module of type k, with J2kM=0, and let H2 be the subgroup of G2 generated by σ2pn-r with 0<r≤n. If M↓G1 is R-free and k=pr, then M is a free K[G1×H2]-module.Proof.
M is an R[G2]-module R-free. We have J2kM=0, so (σ2-1)kM=0, and therefore ((σ2-1)kR[G2])M=0. So M is an R[G2]/(σ2-1)kR[G2]-module R-free. By Lemma 3, R[G2]/(σ2-1)kR[G2]≅R[H2]; then M is an R[H2]-module R-free. M/J1M is a free K[X]/(X-1)k-module, so by Lemma 8 this is a free K[H2]-module. In conclusion M is a K[G1×H2]-module such that
(8)M↓G1isafreekG1-module,MJ1MisafreeKH2-module.
So by Proposition 7  M is a free K[G1×H2]-module.In Theorem9 we assumed that the k[G]-module M satisfies the following condition: M↓G1 is R-free. So it is useful to know when this condition is satisfied. This is the subject of the following result.Theorem 10.
LetM be a k[G]-module and σ an element of G of order pr. The following conditions are equivalent:(1)
M↓〈σ〉 is free;(2)
dim⁡k(M)=(pr/(pr-1))dim⁡k((σ-1)M);(3)
dim⁡k(M)=dim⁡k((σ-1)M)+dim⁡k((σ-1)pr-1M).Proof.
(
1
)
⇒
(
2
) Assume that M↓〈σ〉 is free. There exists a nonzero natural number n such that M↓〈σ〉≅(k[〈σ〉])n. The endomorphism φ of M defined by φ(m)=(σ-1)m for all m∈M is nilpotent of nilpotency index pr, and k[〈σ〉] is an indecomposable k[〈σ〉]-module. Therefore M has a basis in which the matrix of φ is a Jordan matrix. This matrix is formed of n blocks of order pr all equal to(9)01⋱⋱⋱10.
So dim⁡k(M)=pr×n. We can easily see that dim⁡k((σ-1)M)=n×(pr-1). Therefore dim⁡k(M)=(pr/(pr-1))dim⁡k((σ-1)M).
(
2
)
⇒
(
3
) Now, assume that dim⁡k(M)=(pr/(pr-1))dim⁡k((σ-1)M). So dim⁡k(M)=pr×(dim⁡k(M)-dim⁡k((σ-1)M)). As dim⁡k(M)-dim⁡k((σ-1)M) is the number of blocks of the Jordan matrix of φ, the order of each block is less than or equal to pr, and dim⁡k(M) is equal to the sum of the orders of these blocks, then the order of each block is pr. Therefore dim⁡k((σ-1)pr-1M) is equal to the number of Jordan blocks of φ. So dim⁡k(M)-dim⁡k((σ-1)M)=dim⁡k((σ-1)pr-1M); that is, dim⁡k(M)=dim⁡k((σ-1)M)+dim⁡k((σ-1)pr-1M).
(
3
)
⇒
(
1
) Assume that dim⁡k(M)=dim⁡k((σ-1)M)+dim⁡k((σ-1)pr-1M). So dim⁡k((σ-1)pr-1M) is equal to the number of Jordan blocks of φ. Therefore the order of each Jordan block is equal to pr. So the modules contained in a decomposition of M↓〈σ〉 as a direct sum of indecomposable modules are of the form k[σ]; that is, M↓σ is free.


## 3. Classification of Finitely GeneratedK[Cpm×Cpn]-Modules: Use of Module Extensions

LetG be a finite group and let R be a ring. Let M1 and M2 be two R[G]-modules. We put T=HomR(M2,M1). T has a natural structure as a (R[G],R[G])-bimodule centralized by R (see [7, section 25]). Explicitly, we have
(10)λfm2=λ·fm2,fλm2=fλm2hh∀λ∈RG,m2∈M2,f∈T.
A derivation F:R[G]→T is an R-homomorphism satisfying
(11)Fλ′λ=λ′Fλ+Fλ′λ∀λ,λ′∈RG.
Derivations from RG into T form an R-module Der(R[G],T). For F∈Der(R[G],T) we equip M1×M2 with an R[G]-module structure by
(12)λm1,m2=λm1+Fλm2,λm2∀m1,m2∈M1×M2,λ∈RG.
This R[G]-module is denoted by M1×FM1 as in [8].An extension ofM2 by M1 is an R[G]-exact sequence O→M1→X→M2→O. Let O→M1→uX→vM2→O and O→M1→u′X′→v′M2→O be a pair of extensions of M2 by M1. These two extensions are equivalent if there exists an isomorphism of R[G]-modules Φ:X→X′ such that Φou=u′ and v′oΦ=v. These equivalence classes of extensions form an R-module Ext⁡R[G]1(M2,M1). The R[G]-modules sequence O→M1→iM1×FM2→jM2→O, where i and j denote, respectively, the canonical injection from M1 to M1×FM2 and the second projection from M1×FM2 to M2, is exact. The equivalence class of this sequence is denoted by [M1×FM2].Remark 11.
With the previous notations, derivationsF and modules M1×FM2 play the same role as the cocycles α and modules M1×αM2 defined in [8].From Proposition 25.10 of [7] we have the following result.Proposition 12.
The correspondenceθ:Der(RG,T)→ExtR[G]1(M2,M1) defined by θ(F)=[M1×FM2] is surjective whenever M2 is finitely generated and projective as R-module.From Theorems 5.2 and 5.3 of [9] we have the following result.Proposition 13.
LetG be a cyclic group of order pr generated by an element σ, K a field of characteristic p, and M an indecomposable K[G]-module. Then, M is isomorphic to (σ-1)sK[G], where s is a natural number strictly less than pr.Lemma 14.
LetR be a ring and G=G1×G2 a direct product of two finite groups. Let M be an R[G]-module such that the action of G1 on M is trivial and let M′ be an R[G2]-module. If M is isomorphic to M′ as R[G2]-modules and if we extend the action of G2 on M′ to G by σ·m′=m′, ∀(σ,m′)∈G1×M′, then M is isomorphic to M′ as R[G]-modules.Proof.
Letψ:M→M′ be an isomorphism of R[G2]-modules. We extend the action of G2 on M′ to G by σ·m′=m′, ∀(σ,m′)∈G1×M′. We easily see that the application ψ:M→M′ is an isomorphism of R[G]-modules.LetK be a commutative field of characteristic p>0. Let G=G1×G2, where G1=Cpm and G2=Cpn are two cyclic groups of respective orders pm and pn and are generated, respectively, by σ1 and σ2, and let J1 be the Jacobson radical of K[G1].Proposition 15.
LetM be a finitely generated K[G]-module. If J1M=0, then there exists a nonzero natural number n′ such that M≅⊕i=1n′(σ2-1)kiK[G2], 0≤ki<pn, as K[G]-modules, where the action of G1 on ⊕i=1n′(σ2-1)kiK[G2] is trivial.Proof.
IfJ1M=0, then the action of G1 on M is trivial since J1=(σ1-1)K[G1]. By Proposition 13, there exists a nonzero natural number n′ such that M≅⊕i=1n′(σ2-1)kiK[G2], 0≤ki<pn, as K[G2]-modules. Then, Lemma 14 allows concluding the following.Theorem 16.
LetM be a finitely generated K[G]-module. If J12M=0, then there exist two nonzero natural numbers n′ and n′′ and two K[G]-modules M1=⊕i=1n′(σ2-1)kiKG2, 0≤ki<pn, and M2=⊕i=1n′′(σ2-1)ki′K[G2], 0≤ki′<pn, where the action of G1 on M1 and M2 is trivial, and there is a derivation F from K[G] in HomK(M2,M1) such that M≅M1×FM2.Proof.
We have the exact sequence ofK[G]-modules O→J1M↪M→M/J1M→O. As J12M=0, J1(J1M)=0. So by Proposition 15 there exists a nonzero natural number n′ such that J1M≅⊕i=1n′(σ2-1)kiK[G2], 0≤ki<pn, as K[G]-modules, where the action of G1 on ⊕i=1n′(σ2-1)kiK[G2] is trivial. We set M1=⊕i=1n′(σ2-1)kiK[G2]. We have J1(M/J1M)=0. So by Proposition 15 there exists a nonzero natural number n′′ such that M/J1M≅⊕i=1n′′(σ2-1)ki′K[G2], 0≤ki′<pn, as K[G]-modules, where the action of G1 on ⊕i=1n′′(σ2-1)ki′K[G2] is trivial. We set M2=⊕i=1n′′(σ2-1)ki′K[G2]. Then, Proposition 12 shows that M≅M1×FM2 for a derivation F from K[G] in HomK(M2,M1).Ifp=2, G1=C2, and G2=C2n, then we have the following corollary.Corollary 17.
For all finitely generatedK[G]-modules there exist two nonzero natural numbers n′ and n′′ and two K[G]-modules M1=⊕i=1n′(σ2-1)kiKG2, 0≤ki<2n, and M2=⊕i=1n′′(σ2-1)ki′K[G2], 0≤ki′<2n, where the action of G1 on M1 and M2 is trivial, and there is a derivation F from K[G] in HomK(M2,M1) such that M≅M1×FM2.Proof.
We haveJ1=(σ1-1)K[G1] and as (σ1-1)2=0 since the field K is of characteristic p=2, J12=0. So J12M=0 for K[G]-module of finite type M. The rest is a simple application of Theorem 16.Now we return to casesG1=Cpm and G2=Cpn, where G2 is generated by an element σ2, and let J2 be the Jacobson radical of K[G2]. For an integer k=pr with 0<r≤n and for the subgroup H2 of G2 generated by σ2pn-r, we have the following result.Theorem 18.
LetM be a finitely generated K[G]-module, with J2k+1M=0. If M/J2kM↓G1is R-free and of type k, then there exist two nonzero natural numbers n′ and n′′ and two K[G]-modules M1=⊕i=1n′(σ1-1)kiKG1, 0≤ki<pn, and M2=(K[G1×H2])n′′, where the action of G2 on M1 is trivial, and there is a derivation F from K[G] in HomK(M2,M1) such that M≅M1×FM2.Proof.
We have the following exact sequence:(13)O⟶J2kM↪M⟶MJ2kM⟶O.
As J2(J2kM)=0, by Proposition 15, there exists a nonzero natural number n′ such that J2kM≃⊕i=1n′(σ1-1)kiKG1, 0≤ki<pn, as K[G]-modules, where the action of G2 on ⊕i=1n′(σ1-1)kiKG1 is trivial. M/J2kM is a K[G]-module of type k with J2k(M/J2kM)=0, more M/J2kM↓G1 is R-free, and k=pr with 0<r≤n. Then, Theorem 9 shows that M/J2kM is a free K[G1×H2]-module. Therefore there exists a nonzero natural number n′′ such that M/J2kM≃(K[G1×H2])n′′. The rest is a simple application of Proposition 12.



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