On Modules over a G-set

0.17 MB·

by  Mehmet Uc

On Modules over a G–set
7 Mehmet Uc, Mustafa Alkan
1
0 January 24, 2017
2

n
a Abstract
J
0 LetRbeacommutativeringwithunity,M amoduleoverRandletS beaG–setfora
2 finitegroupG. WedefineasetMS tobethesetofelementsexpressedastheformalfinite

] sumoftheform P msswherems ∈M. ThesetMS isamoduleoverthegroupringRG
A s∈S
under the addition and the scalar multiplication similar to the RG–module MG defined
R
byKosan,LeeandZhouin[9]. Withthisnotion,wenotonlygeneralizebutalsounifythe
.
h theories of both of the group algebra and the group module, and we also establish some
at significant properties of (MS)RG. In particular, we describe a method for decomposing
m a given RG–module MS as a direct sum of RG–submodules. Furthermore, we prove the
[ semisimplicity problem of (MS)RG with regard to theproperties of MR, S and G.

1
v
1 Introduction
4
4
4
Throughout this paper, G is a finite group with identity element e, R is a commutative ring
6
0 withunity1, M isanR–module,RGisthe groupring,H ≤GdenotesthatH isasubgroupof
.
1 GandS isa G–setwitha groupactionofGonS. IfN is anR–submodule ofM,itis denoted
0
by N ≤M .
7 R R
1 MS denote the set of all formal expressionof the form mss where ms ∈M and ms =0
: s∈S
v X
i for almost every s. For elements µ = mss, η = nss ∈ MS, by writing µ = η we mean
X
s∈S s∈S
X X
r ms =ns for all s∈S.
a
We define the sum in MS componentwise
µ+η = (m +n )s
s s
s∈S
X
It is clear that MS is an R–module with the sum defined above and the scalar product of
m s by r ∈R that is (rm )s.
s s
s∈S s∈S
X X
For ρ= r g ∈RG, the scalar product of m s by ρ is
g s
g∈G s∈S
X X
ρµ = r m (sg), sg =s′ ∈S,
g s
s∈S
X
= ms′s′ ∈MS
s′∈S
X
ItiseasytocheckthatMS isaleftmoduleoverRG,andalsoasanR–module,itisdenoted
by (MS) and (MS) , respectively. The RG–module MS is called G–set module of S by
RG R
1
2
M over RG. It is clear that MS is also a G–set. If S is a G–set and H is a subgroup of
G, then S is also an H–set and MS is an RH–module. In addition, if S is a G–set and a
group, and M = R, then it is easy to verify that RS is a group algebra. On the other hand,
if a group acts on itself by multiplication then naturally we have (MS) = (MG) . Since
RG RG
there is a bijective correspondence between the set of actions of G on a set S and the set of
homomorphisms from G to Σ (Σ is the group of permutations on S), the G–set modules
S S
is a large class of RG–modules and we would say that (MG) introduced in [9] considering
RG
the group acting itself by multiplication is a first example of the G–set modules. That is why
the notion of the RG–module MS presents a generalization of the structure and discussions
of RG–module MG and some principal module-theoretic questions arise out of the structure
of (MS) . Therefore, this new concept generalizes not only the group algebra but also the
RG
group module, and also unifies the theory of these two concepts.
The purpose of this paper is to introduce the concept of the RG–module MS, and show
the close connection between the properties of (MS) , M , S and G. The semisimplicity
RG R
of (MS) with regard to the properties of M , S and G and the decomposition of (MS)
RG R RG
into RG–submodules will occupy a significant portion of this paper. In Section 1, we present
some examples and some properties of (MS) to show that an R–module can be extended
RG
to RG–modules in various ways via the change of the G–set and the group ring. In Section 2,
we give our first major result about the decomposition of a given RG–module MS as a direct
sum of RG–submodules. In Section 3, in order to go further into the structure of (MS) , we
RG
firstrequireε thatis anextensionof the usualaugmentationmap ε andthe kernelof ε
MS R MS
denoted by △ (MS). Then we give the condition for when △ (MS) is an RG–submodule of
G G
(MS) . Finally,weareinterestedinthesemisimplicityof(MS) accordingtotheproperties
RG RG
of M , S and G.
R
We start to set out the idea of G–set modules in more detail by considering some examples
of G–set modules and establishing some properties of (MS) . The following examples for
RG
(MS) show how useful the notion of G–set module for extension of an R–module M to an
RG
RG–module. They also point the relations among G–set S, RG–module MS, G and H where
H ≤ G. Example 1.1 shows that for different group actions on different G–sets of the same
finite group we get different extensions of an R–module M to an RG–module. Moreover, we
see that these are also RH–modules unsurprisingly in Example 1.2.
Example 1.1 Let M be an R–module, G = D = a,b:a3 =b2 =e,b−1ab=a−1 and r =
6
r g =r e+r a+r a2+r b+r ba+r ba2 ∈RD .
g 1 2 3 4 5 6 (cid:10) 6 (cid:11)
gX∈D6
1. Let S = G and let the group act itself by multiplication. Then MS = MG is an RG–
module.
2. Let S = {D ,C ,C ,Id} and let G act on its set of subgroups C = a:a3 =e ≤ D ,
6 3 2 3 6
C = b:b2 =e ≤ D , Id = {e} ≤ D by g∗H = gHg−1 for H ≤ G, g ∈ G. Then
2 6 6 (cid:10) (cid:11)
(cid:10) (cid:11)
3
MS ={ m s=m Id+m C +m C +m D |m ∈M} and we get
s Id C2 2 C3 3 D6 6 s
s∈S
X
rµ = (r m +r m +r m +r m +r m +r m )Id
1 1 2 1 3 1 4 1 5 1 6 1
+(r m +r m +r m +r m +r m +r m )C
1 C2 2 C2 3 C2 4 C2 5 C2 6 C2 2
+(r m +r m +r m +r m +r m +r m )C
1 C3 2 C3 3 C3 4 C3 5 C3 6 C3 3
+(r m +r m +r m +r m +r m +r m )D .
1 D6 2 D6 3 D6 4 D6 5 D6 6 D6 6
3. Let S = K ={e,b},K ={a,ba},K = a2,ba2 that is the set of right cosets of a
1 2 3
fixed subgroup H = C = b:b2 =e ≤D and let G act on S by g∗(Hx) = H(gx) for
(cid:8) 2 6(cid:8) (cid:9)(cid:9)
x,g ∈ G. Then MS = { m s = m K +m K +m K | m ∈ M} and we have
(cid:10) s (cid:11)K1 1 K2 2 K3 3 s
s∈S
X
the following relations such that
K 1=K K 1=K K 1=K
1 1 2 2 3 3
K a=K K a=K K a=K
1 2 2 1 3 1
K a2 =K K a2 =K K a2 =K
1 3 2 3 3 2
K b=K K b=K K b=K
1 1 2 3 3 2
K ba=K K ba=K K ba=K
1 2 2 1 3 3
K ba2 =K K ba2 =K K ba2 =K .
1 3 2 2 3 1
So, we get
rµ = (r m +r m +r m +r m +r m +r m )K
1 K1 4 K1 3 K2 5 K2 2 K3 6 K3 1
+(r m +r m +r m +r m +r m +r m )K
2 K1 5 K1 1 K2 6 K2 3 K3 4 K3 2
+(r m +r m +r m +r m +r m +r m )K .
3 K1 6 K1 2 K2 4 K2 1 K3 5 K3 3
Example 1.2 Let M be an R–module, G=D = a,b:a3 =b2 =e,b−1ab=a−1 , H =C =
6 3
a:a3 =e ≤D and k = k g =k e+k a+k a2 ∈RC .
6 g 1 2 (cid:10)3 3 (cid:11)
(cid:10) (cid:11) gX∈D6
1. Let S = G and let the group act itself by multiplication. Then MS = MG is an RH–
module.
2. Let S = {D ,C ,C ,Id} with the group action defined in Example 1.1 (2). For µ =
6 3 2
m s=m Id+m C +m C +m D ∈MS, we get
s Id C2 2 C3 3 D6 6
s∈S
X
kµ = (k m +k m +k m )Id+(k m +k m +k m )C
1 1 2 1 3 1 1 C2 2 C2 3 C2 2
+(k m +k m +k m )C +(k m +k m +k m )D .
1 C3 2 C3 3 C3 3 1 D6 2 D6 3 D6 6
3. Let S = K ={e,b},K ={a,ba},K = a2,ba2 with the group action defined in
1 2 3
Example 1.1 (3). For µ= m s=m K +m K +m K ∈MS, we get
(cid:8) s K1 1(cid:8) K2(cid:9)(cid:9)2 K3 3
s∈S
X
kµ = (k m +k m +k m )K +(k m +k m +k m )K
1 K1 3 K2 2 K3 1 2 K1 1 K2 3 K3 2
+(k m +k m +k m )K
3 K1 2 K2 1 K3 3
Now, we make a point of some relations between the R–submodules of M and the RG–
submodules of MS by the following results.
4
Lemma 1.3 Let N , N be R–submodules of M. Then N S + N S = MS if and only if
1 2 1 2
N +N =M.
1 2
Proof Let N S +N S = NS. Take m ∈ M and so ms ∈ MS for any s ∈ S. We write
1 2
ms = n s + n s for n s ∈ N S and n s ∈ N S where n ∈ N ,
si i sj j si i 1 sj j 2 si 1
sXi∈S sXj∈S sXi∈S sXj∈S
n ∈N S. So, there exists i,j such that m=m +m .
sj 2 si sj
Let N +N =M and µ= m s∈MS. For all s∈S, we canwrite m =n +n′ where
1 2 s s s s
s∈S
X
n ∈N , n′ ∈N . Hence, µ= n s+ n′s, and so N S+N S =NS. (cid:3)
s 1 s 2 s s 1 2
s∈S s∈S
X X
Lemma 1.4 LetN , N beR–submodulesof M. Then N S∩N S =0if andonlyif N ∩N =
1 2 1 2 1 2
0.
Proof Let N S +N S = 0. Take n ∈ N ∩N , and so ns ∈ N S ∩N S. So, n = 0 since
1 2 1 2 1 2
ns=0.
Conversely, let N ∩N =0. Take η = n s∈N S∩N S. So n ∈N ∩N and n =0
1 2 s 1 2 s 1 2 s
s∈S
for all s∈S. Hence, N S∩N S =0. X (cid:3)
1 2
From [2] we recall that if G is a finite group, S and T are G–sets, then ϕ: S −→T is said
to be a G–set homomorphismif ϕ(gs)=gϕ(s) for any g ∈G, s∈S. If ϕ is bijective, then ϕ is
a G–set isomorphism. Then we say that S and T are isomorphic G–sets, and we write S ≃T.
Fors∈S,Gs={gs:g ∈G}istheorbitofs. ItiseasytoseethatGsisalsoaG–setunder
the action induced from that on S. In addition, a subset S′ of S is a G–set under the action
induced from S if and only if S′ is a union of orbits.
Proposition 1.5 Let M be an R–module, N an R–submodule of M, G a finite group, S a
G–set. Then MS ≃(M)S.
NS N
Proof We know that NS is an RG–submodule of MS. Define a map θ such that
θ : MS −→ (M)S , µ= mss 7−→ θ(µ) = (ms+N)s
N
s∈S s∈S
X X
θ(gµ) = θ(g m s)
s
s∈S
X
= gθ(µ)
So, θ is a G–set homomorphism. It is clear that θ is a G–set epimomorphism. Furthermore, θ
is an RG–epimorphism and we get kerθ =NS. (cid:3)
Lemma 1.6 Anyproper subset of an orbit Gs of s∈S is not a G–set under theaction induced
from S.
Proof Suppose that a proper subset T of an orbit Gs of s ∈ S is a G–set. Then there exist
sg ∈G, sg ∈/ T for some g ∈G. Take an element sh in T, h∈G, and so
(gh−1)(hs) = g(h−1(hs))
= gs∈/ T.
5
Hence,wecalltheorbitGsofs∈S theminimalG–set. Moreover,S = Gs whereI denotes
i
i∈I
the index of disjoint orbits of S. Hence, we have S
MS =M( Gs ).
i
i∈I
[
(cid:3)
Lemma 1.7 Let N be an R–submodule of an R–module M, S a G–set. Let I denote the index
of disjoint orbits of S, J a subset of I and S′ = Gs and let Gs be an orbit Gs of s ∈ S
j i i
j∈J
for i∈I. Then we have the following results: S
1. NGs is an RG–submodule of MS for s ∈ S. Moreover, NGs is a minimal RG–
i i i
submodule of MS containg N under the action induced from that on S.
2. NS′ =N( Gs )= (NGs ).
j j
j∈J j∈J
S S
3. NS′ is an RG–submodule of MS.
Proof
1. It is clear that NGs ⊆ MS. Let η = n gs ∈ NGs , r ∈ R, h ∈ G. Then we
i g i i
g∈G
have rη ∈ NGs and hη = h( n gs ) =P n hgs = n g′s ∈ NGs . Hence,
i g i g i g i i
g∈G g∈G hg=g′∈G
NGs is an RG–submodule ofPMS. AssumePthat there is aPn RG–submodule N of MS
i 1
such that N ≤ (N ) ≤ (NGs ) . Take an element n ∈ N, and so nhs ∈ N for
R 1 RG i RG i 1
some h ∈ G since (N ) ≤ (NGs ) . Then h−1(nhs ) = (nes ) = ns ∈ N and
1 RG i RG i i i 1
g(ns )=ngs ∈N for all g ∈G. This means that N =NGs .
i i 1 1 i
2,3. Clear by the definition of MS.
(cid:3)
Lemma 1.8 Let L be an RG–submodule of MS, a fixed s∈S. Then,
1. L ={x∈M | there is y ∈L such that y =xs+k, k ∈MS} is an R–submodule of M.
s
2. S ={s∈S |there is x∈M, and also k ∈L such that y =xs+k ∈L } is a G–set in S
L
under the action induced from that on S.
Proof
1. It is obvious that Ls is in M. Let x1, x2 ∈Ls′ and r ∈R. Then, there is y1 =x1s+k1,
y = x s + k ∈ L and y + y = (x + x )s + k + k ∈ L where x + x ∈ MS.
2 2 2 1 2 1 2 1 2 1 2
Furthermore, ry =rx s+rk ∈L, and so rx ∈L .
1 1 1 1 s
2. Let s∈S′ and g, h∈G. Then ∃x∈M,∃k ∈L such that y =xs+k ∈L and
xs+k =y =ey =e(xs+k)=xes+ek =xes+k
6
So, s=es. Since s is also an element of S, we have
(hg)y = (hg)(xs+k)
= (hg)xs+(hg)k.
Hence, we get (hg)s=h(gs).
(cid:3)
Lemma 1.9 Let M be an R–module and S a G–set. Let I denote the index of disjoint orbits
of S such that S = Gs and let Gs be an orbit of s ∈ S for i ∈ I. If NGs is a simple
i i i i
i∈I
RG–submodule of MSS, then N is a simple R–submodule of M and G is a finite group whose
order is invertible in End (M) (|G|−1 ∈End (M)).
R R
Proof Assume that there is an R–submodule L of M such that L ≤ N ≤ M. Then
(LGs ) ≤ (NGs ) , and by Lemma 1.6 this is a contradiction. So, N is a simple R–
i RG i RG
submodule of M. (cid:3)
Theorem 1.10 Let L be a simple RG–submodule of MS. Then there is a unique simple R–
submodule N of M and a unique orbit Gs such that L=NGs.
Proof For some s ∈ S, by Lemma 1.8 L is a non-zero R–module. And so, L Gs 6= 0 is an
s s
RG–submodule of L. Since L is simple RG–submodule, we have L Gs =L. Then, by Lemma
s
1.9 L is a simple R–submodule of M.
s
Takeanelements′ ∈S suchthatLs′ is non-zeroR–submoduleofM. Hence, Ls′Gs′ =L=
LsGs. Take an element x∈Ls′Gs′. And so, we write
n n
x= l g s′ = k g s
i i i i
i=1 i=1
X X
where li ∈ Ls′, ki ∈ Ls, gi ∈ G and n = |G|. Then, there exists gj ∈ G such that g1s = gjs′,
and s=g−1g s′. So, we get Gs=Gs′. That is why we can write
1 j
Gs=S ={s∈S |there is x∈M, and also k ∈L such that y =xs+k ∈L}.
L
Moreover, N =Ls =Ls′ is unique by the definition of MS. (cid:3)
On the other hand, the following example shows that the converse of the theorem does not
hold.
Example 1.11 Let R =Z , M = Z , G =C = a:a2 =e and RG= Z C . If S = G and
3 3 2 3 2
Gactson itselfbygroupmultiplication thenMS =Z C whereZ C issemisimple RG–module
(cid:10) 3 2 (cid:11) 3 2
since |G|≤∞ and characteristic of R does not divide |G| by Maschke’s Theorem. Since Z C
3 2
is semisimple there is a unique decomposition of Z C by Artin-Weddernburn Theorem. Then,
3 2
Z C ≃ Z ⊕Z as R–module since |C | = 2. Here, Z is a simple R–submodule of Z C .
3 2 3 3 2 3 3 2
Moreover, by [11] we have Z C ≃ Z C (1+a)⊕Z C (1−a) as RG–module where Z C (1+a)
3 2 3 2 2 3 2 2 3 2 2
and Z C (1−a) are simple RG–submodules of Z C . Let N =Z that is a simple R–submodule
3 2 2 3 2 3
of M. Hovewer, NGs=Z C is not simple RG–module.
3 2
7
Lemma 1.12 Let {M : i ∈ I} be a family of right R−modules, G a finite group and S a
i
G−set. Then
M S = M S
i i
! ! !
Mi∈I RG Mi∈I RG
Proof Consider the following map
(i) (i)
(...,m ,...)s 7−→ (...,m s,...)
M S −→ M S , s s
i i
! s∈S s∈S
i∈I i∈I X X
M M
that is an isomorphism. (cid:3)
Theorem 1.13 An R–module M is projective if and only if (MS) is projective.
R RG
Proof Assume that M is projective. Then for an index I, (R)(I) ≃ M ⊕A where A is a
R
right R–module. So, by Lemma 1.12
((RS)(I)) ≃ ((R)(I)S)
RG RG
≃ ((M ⊕A)S)
RG
≃ (MS) ⊕(AS)
RG RG
So, (MS) is projective.
RG
Now, assume that (MS) is projective. Then ((RS)(I)) ≃ (MS) ⊕ B where B
RG RG RG
is a right RG–module for some set I. All this concerning modules are also R–modules and
((RS)(I)) ≃(MS) ⊕B . ((RS)(I)) is a free module because (RS) is free. Since (MS)
R R R R R R
is direct summand of a free module, it is projective. So, M is projective. (cid:3)
R
2 The Decomposition of (MS)
RG
The theme of this section is the examination of a G–set module (MS) throughthe study of
RG
a decomposition of it. The decompositions of RG and (MG) obtained from the idempotent
RG
defined as e = Hˆ , where |H| is the order of H and Hˆ = h, explained in [11] and
H |H|
h∈H
[15], respectively. A similar method give a criterion for the decomPposition of a G–set module
(MS) . In addition, End MS denotes all the RG–endomorphisms of MS.
RG RG
Lemma 2.1 Let M be an R-module and H a normal subgroup of finite group G. If |H|, the
order of H, is invertible in R then e = Hˆ is an idempotent in End (MS). Moreover, e
H |H| RG H
is central in End (MS).
RG
e e
Proof Firstly, we will show that e is an RG–homomorphism. We start with proving that
H
Hˆg = gHˆfor g ∈G. Since for all h ∈H, there is h ∈ H such that h g = gh , we have that
i ig i ig
Hˆg = h g = gh =gHˆ. Teherefore, Hˆ rg =rg Hˆ and we have e (rgm)=rge (m)
i ig |H| |H| H H
hi∈H hi∈H
form∈PMS,r ∈RPandg ∈G. Itisalsoclearthate (m+n)=e (m)+e (n)form,n∈MS,
H H H
e e
g ∈G.
e e e
8
Secondly, by using the fact that Hˆ.Hˆ =|H|.Hˆ, we get
Hˆ
e (e (m)) = e ( m)
H H H
|H|
= e (m)
e e eH
So, eH is an idempotent. e
Finally, we prove that e is a central idempotent in End (MS). We will show that e
H RG H
comemutes with every element of End (MS). Let f be in End (MS) and so Hˆf(m) =
RG RG
f(Hˆm) for m∈MS. Thuse, we have e
Hˆ
e f(m) = f(m)
H
|H|
Hˆ
e = f( m)=fe (m).
H
|H|
e (cid:3)
For µ= m g ∈MG and s ∈S, we write
g i
g∈G
P
µs = m (gs )
i g i
g∈G
X
= m (gs )∈MS
gsi i
gXsi∈S
Then for i ∈ I and α ∈ M(Gs ), we write α = m gs . Moreover, we write β =
i gsi i
gsi∈Gsi
m gs for β = m s∈MS since MS =PM( Gs ).
gsi i s i
i∈Igsi∈Gsi s∈S i∈I
P LePt H be a normal subgProupof G. It is well known thaSt on G/H we have the group action
g(tH)=gtH for g,t∈G. Consider g( m (sH))=( m (gsH)) for m ∈M.
s s s
s∈S s∈S
Let S′ ⊂ S be a G/H–set. Then PS′ = G/Hs′Pwhere J denotes the index of disjoint
j
j∈J
orbits of S′ and MS′ = M( G/Hs′j). ThSen for η = ms′s′ ∈ MS, we can write η =
j∈J s′∈S′
ms′s′. S P
j∈Js′∈G/Hs′
j
PHenPce, we have the following result.
Lemma 2.2 Let M be an R–module, G a finite group, H a normal subgroup of G, S a
G–set and S′ ⊂ S a G/H–set. Then MS′ is an RG–module with action defined as gη =
g( ms′s′)= g( ms′(tHs′j)= ms′(gtHs′j) where
j∈Js′∈G/Hs′ j∈Js′∈G/Hs′ j∈Js′∈G/Hs′
j j j
Pη = P ms′s′ ∈PMS′Pand s′ =tHs′j for Pt∈G.P
j∈Js′∈G/Hs′
j
P P
Theorem 2.3 Let H be a normal subgroup of G, |H| invertible in R and e , defined above,
H
then we have MS = e .MS ⊕(1−e ).MS and there exists a G/H–set S′ ⊂ S such that
H H
e .MS ≃MS′. More precisely, e
H
e e
e
e .MS =e M( Gs ) ≃M( e Gs )
H H i H i
!
i∈I i∈I
[ [
e e e
9
Proof Firstly, we know that MG = e .MG⊕(1−e ).MG and e .MG ≃ M(G/H) by
H H H
the theorem in [15]. Since e is a central idempotent by Lemma 2.1, we get MS =e .MS⊕
H H
(1−e ).MS. Now, consider θ : G−→Ge.e where g 7→ege . This isea group homomorphism
H H H
since θ(gh) = ghe = ghe2e= ge he = θ(g)θ(h). It is clear that θ is a group epimeorphism.
H H H H
We heave kerθ ={g ∈G|ge =e }={ge∈G|(g−1)e e=0}=H since (g−1) Hˆ =0 and
H H H |H|
gHˆ =Hˆ for g ∈He. Moreeover,wee gete G = G ≃ Imθ =Ge . So,
erθ H H
e e e
e
e .MS =e M( Gs ) =M( Ge s )≃M( (G/H)s )
H H i H i i
!
i∈I i∈I i∈I
[ [ [
e e e
SincegHs =gHs fors ,s ∈S,i,l∈I,wegetaG/H–setS′⊂Swhere (G/H)s =S′ ⊆S.
i l i l j
j∈J
Hence S
e .MS ≃M( (G/H)s )=M( (G/H)s )=MS′
H i j
i∈I j∈J
[ [
So, e .MS ≃MS′. e (cid:3)
H
e
Theorem 2.4 Let M be an R–module and G a finite group. For a G–set S = Gs (I
i
i∈I
denotes the index of disjoint orbits of S), MS ≃ MG\kerθ where θ : MG−→MSGs are
i i i
i∈I
RG–epimorphisms. L
Proof Since MGs ∩MGs = ∅ for i 6= j ∈ I where S = Gs and I denotes the index of
i j i
i∈I
disjoint orbits of S , we have MS =M( Gs )= MGs .S
i i
i∈I i∈I
Consider S L
θ : MG −→ MGs , mgg 7−→ mggsi
i i
g∈G g∈G
P P
For µ= m g ∈MG, r ∈R, h∈G, we have
g
g∈G
P
θ (rµ) = θ (r m g)=θ ( rm g)= rm gs
i i g i g g i
g∈G g∈G g∈G
X X X
= r m gs =rθ ( m g)=rθ (µ).
g i i g i
g∈G g∈G
X X
θ (hµ) = θ (h m g)=θ ( m hg)= m hgs
i i g i g g i
g∈G g∈G g∈G
X X X
= h( m gs )=hθ ( m g)=hθ (µ).
g i i g i
g∈G g∈G
X X
Hence,θ isanRG–homomorphism. Itisclearthatθ isanepimorphism. Moreover,MG\kerθ ≃
i i i
Imθ =MGs . Then,
i i
MS =M( Gs )= MGs ≃ MG\kerθ .
i i i
i∈I i∈I i∈I
[ M M
(cid:3)
10
3 Augmentation Map on MS and Semisimple G–set Mod-
ules
In the theory of the groupring, the augmentation ideal denoted by △(RG) is the kernelof the
usual augmentation map ε such that
R
ε : RG −→ R , rgg 7−→ rg .
R
g∈G g∈G
P P
The augmentation ideal is always the nontrivial two-sided ideal of the group ring and we have
△(RG)= r (g−1):r ∈R,g ∈G . Theaugmentationideal△(RG)isofuseforstudy-
g g
(g∈G )
ing not onlyPthe relationship between the subgroups of G and the ideals of RG but also the
decomposition of RG as direct sum of subrings.
In [9], ε is extended to the following homomorphism of R–modules
R
ε : MG −→ M , mgg 7−→ mg .
M
g∈G g∈G
P P
The kernel of ε is denoted by △(MG) and
M
△(MG)= m (g−1):m ∈M,g ∈G .
 g g 
gX∈G 
We devotethis sectionto εMS thatis anextensionof εM, andto thekernelof εMS denotedby
△ (MS).
G
Definition 3.1 The map
m s 7−→ m
ε : MS −→ M , s s
MS
s∈S s∈S
X X
is called augmentation map on MS.
In addition, ε (m s )=ε (m s )=m for m s , m s ∈MS where m ∈M, s ,s ∈
MS s 1 MS s 2 s s 1 s 2 s 1 2
S, however m s 6=m s . Hence, ε is not one-to-one.
s 1 s 2 MS
Lemma 3.2 Let M be an R–module, G a group and S a G–set. Then ε (rµ) = ε(r) ε (µ)
MS MS
for µ= m s∈MS, r = r g ∈RG. In particular, ε is an R–homomorphism.
s g MS
s∈S g∈G
X X
Proof Let µ= m s∈MS, r = r g ∈RG, then
s g
s∈S g∈G
X X
ε (rµ) = ε (r m )(gs)
MS MS g s
 
gs∈S
X
 
= εMS ms′s′ , ms′ =rgms,gs=s′ ∈S,
s′∈S !
X
= r m
g s
  !
g∈G s∈S
X X
= ε(r)ε (µ).
MS