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A note on the semisimplicity of rings

1. (Semi)group rings Let k be any unitary (not necessarily commutative), and let S be a semigroup, define

kS := {a: S → k | #Supp(a) ∈ N} (where Supp(a) := {s ∈ S | a(s) 6= 0k}) to be the set of all functions from S to k taking only a finite number of non zero values, this is naturally a left k- and M kS = ks,b where sb: S −→ k s.t. sb(x) = δs,x (i.e. sb(x) = 0k if x 6= s, and sb(x) = 1k if x = s ); s∈S with respect to this decomposition one can write X X a = (a(s))s∈S = a(s)sb = assb (where the sum is actually finite) for any a ∈ kS. s∈S s∈S The k-module kS has also a natural ring structure via the product (it has also the pointwise product, but this is not the one we want), given a, b ∈ kS: X ab = a ∗ b is the function defined as ab(s) := a(u)b(v), (u,v)∈S×S|uv=s P P P P P that is ab = ( u∈S auub)( v∈S avvb) = (u,v)∈S×S aubvuvc = s∈S( (u,v)∈S×S|uv=s aubv)sb. From this it is not difficult to check that (a, b) 7→ ab is associative and distributive.

1.1. Remarks. — Obviously kS is a non zero if and only if k is a non zero commutative ring and S is a non empty commutative semigroup. Assume from now on that S is a monoid, i.e. it has a neutral element e ∈ S.

1.1.1. . — The element 1keb ∈ kS is the unit of the ring kS. Moreover, by definition, k = keb, and S = 1kSb commute each other. 1.1.2. Units. — If G = U(S) is the group (submonoid) of invertible elements in S then U(k)U[(S) = U(k)Gb ⊆ U(kS), where U(·) denotes the set of invertible elements of a unitary ring or monoid.

1.1.3. Decomposable coefficients. — Let k = k1 × k2 be a direct product of two (non trivial) unitary rings. Then 1k = x + y with x, y ∈ c(k) two non trivial central and mutually orthogonal idempotents. Then, for any monoid S we get ∼ kS = (k1 × k2)S = k1S × k2S, because 1kS = 1kecS = (x + y)ecS = xecS + yecS with xecS, yecS central and mutually orthogonal idempotents. 1.1.4. Decomposable semigroups. — Let S = S1 × S2 be a direct product of two semigroups, and let k be a unitary ring. Then one can check that ∼ kS = k(S1 × S2) = (kS1)S2. 1.1.5. Group rings. — If S = G is a group, and a, b ∈ kG then   ! X X X X ab =  aubv sb = aubu−1s s,b s∈S (u,v)∈S×S|uv=s s∈S u∈S which is the usual form of the convolution product used in Fourier theory.

1 2

1.1.6. Functoriality. — The construction kS is clearly covariantly functorial both in k and in S. Any monoid ϕ: S1 −→ S2 induces a (unitary) ring homomorphism kϕ: kS1 −→ kS2 which is covariantly functorial. And any unitary ring homomorphism f : k1 −→ k2 induces a (unitary) ring homomorphism fS : k1S −→ k2S which is covariantly functorial. It is not difficult to check that the group ring construction is the left adjoint to the “units” functor from unitary rings to groups; this means that to give a ring homomorphism

ϕ: k1G −→ k2 where k1 and k2 are rings, and G is a group it is equivalent to give ( a ϕ|G : G −→ U(k2), and

a unitary ring homomorphism ϕ|k1 : k1 −→ k2 such that ϕ(k1) ⊆ ck2 (ϕ(G)). 1.1.7. Augmentation (or Norm) and Trace. — Let S be a monoid and k be a unitary ring, we then always have the surjective unitary ring homomorphism, called augmentation, or norm, X X α: kS −→ k, assb 7→ as which is a retraction of the injective unitary ring homomorphism

k −→ kS, a 7→ aecS. Hence k is a direct summand of kS, indeed kS = k ⊕ Ker(α) for x = α(x) + (x − α(x)) ∈ k ⊕ Ker(α). Obviously, for any s ∈ S the element sb− eb is inside the two-sided Ker(α), hence Ker(α) contains the two-sided ideal ({sb− eb | s ∈ S}), and actually Ker(α) = ({sb− eb | s ∈ S}) being X X X X X x = assb = assb− 0 = assb− ( as)eb = as(sb− eb) P for any x = assb ∈ Ker(α). If S = G is a group, and A ⊆ G generates G as a group then ({gb − eb | g ∈ G}) = ({gb − eb | −1 −1 g ∈ A}) =: IA in kG, indeed if g ∈ A then gd − eb = −gd(gb − eb) ∈ IA, and if g1, g2 ∈ A then −1 d−1 −1 gb2 − gd1 = (gb2 − eb) − (g1 − eb) ∈ IA, and gd1g2 − eb = gb1(gb2 − gd1 ) ∈ IA, and from this it is not difficult to conclude ({gb − eb | g ∈ G}) = IA . Besides the norm we also have a k-linear trace X T : kS −→ k, a = a s 7→ a(e ) = a , sb S eS s∈S such that T (ab) = T (ba) for each a, b ∈ ck(k)S. 1.1.8. Involution. — If S = G is a group, then the ring kG has an involution:

∗ X X −1 : kG −→ kG, aggb 7→ aggd g∈G g∈G which is k-linear and satisfies (x∗)∗ = x, (xy)∗ = y∗x∗. Hence any left kG-module M is actually also a right kG-module via mx := x∗m for any m ∈ M and x ∈ kG. 1.1.9. The c(kG). — Let us consider the center of the ring kG

c(kG) = ckG(kG) = {a ∈ kG | ab = ba for any b ∈ kG} = ckG(kecG ∪ Gb) P which clearly contains the commutative unitary ring ck(k). Let a = g∈G aggb ∈ ckG(kG) then P P a(xecG) = g∈G agxgb = (xecG)a = g∈G xaggb for each x ∈ k implies ag ∈ ck(k) for each g ∈ G; moreover, for each h ∈ G we must have X X X X abh = agghc = agh−1 gb = bha = aghgc = ah−1ggb g∈G g∈G g∈G g∈G 3 that is (using the functional notation) (abh)(g) = (bha)(g) for each g, h ∈ G ⇔ a(gh−1) = a(h−1g) for each g, h ∈ G this tourns out to be equivalent to the following condition a(h−1gh) = a(g) for each g, h ∈ G.

Hence a ∈ ckG(kG) if and only if a is a function from G to k with finite support such that: (1) a takes values in ck(k); (2) a(h−1gh) = a(g) for each g, h ∈ G, this means that a has to be constant on each conjugacy class of G, that is a is a . P Let C ⊆ G be a finite coniugacy class of G, then σC := g∈C gb ∈ ck(k)G ⊆ kG satisfies both the above requirements and therefore σC ∈ c(kG). On the other hand it follows easily that c(kG) is a free ck(k)-module with bases {σC | C is a finite coniugacy class of G}. 1.2. Examples. — n h ∼ 1.2.1. kCn. — Let Cn = {z ∈ C | z = 1} = {ζ | h = 0, . . . , n − 1} =< ζ >= Z/nZ, where 2π i n−1 ζ = e n ∈ C, be the of n. Then we have the group ring ZCn = Z⊕Zζb⊕· · ·⊕Zζ[, and the unitary ring homomorphism

Z[t] −→ ZCn, p(t) 7→ p(ζb) (1) ∼ n ∼ ∼ n clearly induces an isomorphism ZCn = Z[t]/(t −1). More generally kCn = k ⊗Z ZCn = k[t]/(t −1) for any unitary ring k. 2 3 4 Note that, for n = 5 we have ab = 1 in ZC5 with a = 1 − ζb − ζb and b = 1 − ζb− ζb , hence U(ZC5) is strictly bigger than U(Z)Cc5 = ±Cc5. ∼ 1.2.2. kC∞. — Let C∞ =< ζ >= Z be an infinite cyclic group. Then we have the group ring C = L ζn, and the unitary ring homomorphism Z ∞ n∈Z Zc Z[t] −→ ZC∞, p(t) 7→ p(ζb) ∼ −1 ∼ ∼ −1 clearly induces an isomorphism ZCn = Z[t]t = Z[t, t ]. More generally kC∞ = k ⊗Z ZC∞ = k[t, t ] for any unitary ring k.

1.2.3. kΣ3. — Let Σ3 be the permutation group of order 6. It can be shown that, for any zero field K, we have ∼ KΣ3 = K × K × Mat2(K). 1.2.4. kQ. — Let Q be the group of order 8. It can be shown that ∼ ∼ RQ = R × R × R × R × H, while CQ = C × C × C × C × Mat2(C). 1.3. Linear representations. — Let G be a group. We know that the of k-linear repre- sentions is equivalent to the category of left kG-modules. Actually, to give a linear representation of G on a k-left module M ρ: G −→ Autk(M) is the same as giving a unitary (k-linear) ring homomorphism

ϕ: kG −→ Endk(M).

Hence ring theoretic relations holding in kG are also valid in Endk(M).

(1) By definition, if G is finite, ZG is always a free of rank |G|. If we consider the smallest of C 2π i 2π i 2 containing ζ we find Z[ζ] = {p(ζ) ∈ C | p(t) ∈ Z[t]}. For example C ⊇ Z[e 3 ] = Z ⊕ Ze 3 =∼ Z[t]/(t + t + 1), in which 2π i 2 4π i π i 2π i 2π i 2π i 2π i 4π i (e 3 ) = e 3 = −e 3 = −e 3 − 1. The ring Z[e 3 ] = Z ⊕ Ze 3 is not isomorphic to ZC3 = Z ⊕ Zed3 ⊕ Zed3 . 4

Let us collect some standard examples. 1.3.1. kG. — The left kG-module kG is the . If G is finite it always contains P P the left ideal kG · g∈G gb = k g∈G gb (which is a free k-submodule of rank 1). This submodule corresponds to the one dimensional trivial representation. In the general case, one can recover the trivial representation as the quotient module k =∼ kG/ Ker(α), where α is the augmentation map. 1.3.2. M G. — Let M be a left kG-module, then there always exists a maximal left kG-submodule of M on which G acts trivially, that is G \ M := {m ∈ M | gm = m for each g ∈ G} = Ker(g − IdM ), g∈G This is the G-invariant part of M. It can very well happen that M G = 0.

1.3.3. Homk(M,N), with k commutative. — The rings of the form kG, with k commutative, have some special features which distinguish them among general (non commutative) rings R. For example if M,N are left R-modules, the abelian groups HomR(M,N) ⊆ HomZ(M,N) are no longer R-modules in any natural way, and moreover we do not even have a good notion of tensor product. Without enter into details let us just note that the k-module Homk(M,N), with M and N left kG- modules, is a left kG-modules with g ∗ T : M −→ N, (g ∗ T )(m) := gT (g−1m), where T ∈ Homk(M,N), m ∈ M, g ∈ G. This defines the structure of left kG-module on (2) Homk(M,N) , and moreover G Homk(M,N) = {T ∈ Homk(M,N) | g ∗ T = T for each g ∈ G} −1 = {T ∈ Homk(M,N) | gT (g m) = T (m) for each m ∈ M, g ∈ G} −1 −1 = {T ∈ Homk(M,N) | T (g m) = g T (m) for each m ∈ M, g ∈ G}

= {T ∈ Homk(M,N) | T (gm) = gT (m) for each m ∈ M, g ∈ G}

= HomkG(M,N)

Hence the “natural” structure of left kG-module on HomkG(M,N) is the trivial one. Moreover, the k-module M ⊗k N is a left kG-module via the diagonal action

g · (m ⊗k n) := gm ⊗k gn, for each m ∈ M, n ∈ N, g ∈ G; and under this definition one has isomorphisms of left kG-modules: ∼ ∼ ∼ N ⊗k k = k ⊗k N = N, Homk(M,N) = Homk(M, k) ⊗k N, where k is the trivial kG-module, and M is finitely generated.

1.4. Reynolds operator. — Let G be a finite group and let k be a commutative ring such that |G| is a unit in k (in particular char(k) - |G|). Then, given any k-linear representation V of G (equivalently, a kG-module V ) the following construction makes sense 1 X R : V −→ V G, v 7→ gv. V |G| g∈G G Obviously RV is a well defined k-linear map, moreover it is a retraction of the inclusion V ,−→ V , indeed 1 X 1 X |G| v ∈ V G ⇒ R (v) = gv = v = v = v. V |G| |G| |G| g∈G g∈G

(2) The left kG-module Homk(M,N) is the “internal hom” of the category of left kG-modules. 5

G G G That is RV : V −→ V gives a projection of V onto V , so that V = V ⊕ Ker(RV ) as k-modules. G But more than this: RV : V −→ V is also G-invariant (which, in this special case, is the same as kG-linear) 1 X 1 X R (gv) = h(gv) = gv = R (v)[= g R (v), for G acts trivially on V G]. V |G| |G| V V h∈G g∈G G Hence Ker(RV ) is also a left kG-module, and hence V = V ⊕ Ker(RV ) also as kG-modules. Moreover, from a categorical point of view R is a natural transformation from the identity functor of the category of k-linear represntations to the “invariant part” functor: for any G-equivariant map of G representations f : V −→ W (i.e. f is kG-linear) we have f ◦ RV = RW ◦f. Up to terminology, this means that finite groups are linearly reductive over fields whose characteristic does not divide their orders.

1.4.1. RHomk(V,W ). — Let us note some special features of the Reynolds operator of the representations G HomkG(V,W ) = Homk(V,W ) ⊆ Homk(V,W ).

a) Obviously RHomkG(V,W ) is the identity, because G acts trivially on HomkG(V,W ) = G Homk(V,W ) ;

b) R := RHomk(V,W ) : Homk(V,W ) −→ HomkG(V,W ), therefore to any k-linear map T : V −→ W it attaches a kG-linear map R(T ): V −→ W . In particular Ker(R(T )) and Im(R(T )) are always kG-submodules. c) Let S ∈ Homk(U, V ) and T ∈ Homk(V,W ) then 1) if S is G-linear then RU,W (T ◦ S) = RV,W (T ) ◦ S; 2) if T is G-linear then RU,W (T ◦ S) = T ◦ RU,V (S). Let us prove the first assertion (the second is similar); given u ∈ U we have:   1 X 1 X R(T ◦ S)(u) = g ∗ (T ◦ S) (u) = (g ∗ (T ◦ S)(u)) |G|  |G| g∈G g∈G 1 X = gT (S(g−1u)) |G| g∈G 1 X = gT (g−1S(u)) |G| g∈G 1 X = g ∗ T (S(u)) |G| g∈G   1 X = g ∗ T (S(u)) |G|  g∈G = (R(T ) ◦ S)(u).

1.5. First version of Maschke’s Theorem. — Let G be a finite group and let k be a field such that |G| is a unit in k (that is char(k) - |G|), and let V be a k-linear representation of G. Then any subrepresentation W of V is a direct summand of V (as representations). Proof. — We have a G-linear inclusion S : W,−→ V , we have to show that there exists a subrep- resentation W 0 ⊆ V such that V = W ⊕ W 0. As V and W are vector spaces over k ther exists T ∈ Homk(V,W ) such that T ◦ S = IdW (and in particular V = W ⊕ Ker(T )). Then, applying the

Reynolds operator RHomk(W,W ) we have

IdW = R(IdW ) = R(T ◦ S) = RV,W (T ) ◦ S, 6

G therefore R(T ) ∈ Homk(V,W ) = HomkG(V,W ) gives a G-linear retraction of S. Hence V = W ⊕ Ker(R(T )) as representations.

2. Semisimplicity We just recall some basic definitions and known facts (without proofs). Let R be a unitary ring (not necessarily commutative).

2.1. Simple (irreducible), semisimple (completely reducible), and indecomposable mod- ules. — 2.1.1. A simple (or irreducible) R-module is a non zero module M such that if N is a submodule of M then N = 0 or N = M. 2.1.2. A direct summand of an R-module M is a submodule N of M such that there exists an R- submodule N 0 of M with M = N ⊕ N 0 (internal ). 2.1.3. A semisimple (or completely reducible) module is an R-module M satisfying the following equivalent conditions a) each R-submodule of M is a direct summand; b) M is the (internal) direct sum of a family of simple R-submodules; c) M is the (internal) sum of a family of simple R-submodules. 2.1.4. An indecomposable R-module is a non zero module M such that if N is a direct summand of M then N = 0 or N = M, that is M is not of the form N ⊕ N 0 with N,N 0 proper non zero submoules.

2.2. Remarks and examples. — Trivially: M is simple ⇔ M is semisimple and indecomposable. 2.2.1. Stability properties. — The notion of semisimpleness is stable by taking direct sums, sub- modules, quotient modules. 2.2.2. Simple modules. — If M is a simple (left) R-module then for each m ∈ M \{0} we have ∼ that M = Rm = R/AnnR(m) and AnnR(m) = {r ∈ R | rm = 0} is a maximal left ideal of R. 2.2.3. Left vector spaces. — If R = D is a , left D-modules V are just left vector spaces. In this case V is simple if and only if V =∼ D, and any V is semisimple by the existence of bases. 2.2.4. Abelian groups. — Abelian groups are R-modules with R = Z, the simple Z-modules are therefore the finite cyclic groups of prime orders. Note that Z/4Z is not semisimple but it is indecomposable.

2.2.5. rings over division rings. — Let D be any division ring, and let R = Matn(D), then M = Dn is a left R-module (in the obvious way) which is simple.

2.3. Simple and semisimple rings. — 2.3.1. A simple ring(3) is a ring R such that if I is a two-sided ideal of R then I = (0) or I = R. 2.3.2. A semisimple ring is a ring R which is semisimple as a left R-module.

2.4. Remarks and examples. —

(3)Lang defines R to be simple if R is semisimple with only one isomorphism class of simple left ideals. This is a more restricted class of rings. It can be shown that a ring R is simple in the sense of Lang ⇔ R is a left (or equivalently, right) ∼ artinian simple ring in our sense ⇔ R = Mn(D) for a uniquely determined n ∈ N and a uniquely determined division ring D. 7

2.4.1. It can be shown that a ring R is semisimple as a left module over itself ⇔ R is semisimple as a right module over itself. Hence the notion of semisimple ring is two-sided. 2.4.2. If R is a semisimple ring then any R-module M is such, indeed M is a quotient of a (possibly infinite) direct sum of copies of R.

2.4.3. Simple rings need not be semisimple: R = A1(Q) = Q < x, y > /(xy − yx − 1), the first Weyl over Q, is a non commutative which is a simple but not semisimple ring. 2.4.4. The commutative simple rings are exactly the fields.

2.4.5. Any division ring D is simple. For any division ring D and any n ≥ 2, the Mn(D) is both a simple ring and a (right and left) semisimple ring, but it is is not a division ring. 2.4.6. Any direct product of finitely many semisimple rings is such. 2.4.7. Wedderbund-Artin Theorem.— Let R be any (left) semisimple ring. Then R =∼

Matn1 (D1) × · · · × Matnr (Dr) for suitable division rings D1,...,Dr and positive n1, . . . , nr. The number r is uniquely determined, as are the pairs (n1,D1),..., (nr,Dr) (up to a permutation). There are exactly r mutually nonisomorphic left simple modules over R, say M1,...,Mr; ni is the ∼ Lr ⊕ni number of times Mi occurs as a composition factor in the left R-module R (so that R = i=1 Mi ); and Di is the ring of Mi, defined as a ring of right operators on Mi. 2.4.8. The commutative semisimple rings are exactly the finite products of fields, in particular they are all reduced rings, that is they don’t have non zero elements. 2.4.9. The center of a semisimple ring is a direct product of finitely many fields, in particular they are all reduced rings, that is they don’t have non zero nilpotent elements. 2.4.10. If R is a semisimple ring then such is any of its ring retracts, that is any (unitary) subring S ⊆ R such that there exists a ring homomorphism π : R −→ S such that π(s) = s for each s ∈ S (this implies that S is a direct summand of the S-module R, indeed R = S ⊕ Ker(π), and moreover Ker(π) is a two-sided ideal of R). 2.4.11. It’s not too difficult to prove what follow. Let R be a semisimple ring: (a) any ideal of R is a sum of simple components(4) of R; (b) any quotient ring of R is semisimple.

2.5. Non semisimple group rings. — Let p be a prime number, and let R := FpCp be the group ring of the cyclic group of order p, Cp, over the field with p elements Fp. Then choosing a generator of Cp and using Frobenius we get ∼ p ∼ p FpCp = Fp[t]/(t − 1) = Fp[t]/((t − 1) ), a non reduced , which is not a semisimple ring having the non zero nilpotent t − 1. Note p−1 also that the trivial one dimensional subrepresentation V := Fp(1 + t + ··· + t ) ⊆ R is not a direct summand of R; indeed R, being local, has no non zero elements.

3. Semisimplicity of group rings 3.1. Generalized Maschke’s Theorem. — Let k be any unitary ring and let G be a group. The group ring kG is semisimple if and only if the following three conditions are satisfied a) k is semisimple; b) G is finite; × c) |G| ∈ k (that is |G|1k is a unit in k).

(4)I.e. the sum of all minimal left ideals of R in the same isomorphism class as left R-modules. 8

Proof. — Assume that R = kG is semisimple. a) The ring k is semisimple because k =∼ R/ Ker(α) with α: R −→ k the augmentation map, and any quotient ring of a semisimple one is still such. b) The group G is finite. The kernel of the augmenation map is a two-sided ideal of R = kG, which is semisimple, hence there exists a suitable left ideal I ⊆ R such that M R = Ker(α) I. Hence there are mutually ortogonal idempotents x, y ∈ kG such that Ker(α) = Rx, I = Ry, 1 = x + y.

Being (0) ( Ker(α) ( R, it is x 6= 0 and y 6= 0. On the other hand Ker(α)y = Rxy = 0 ⇒ (g − 1)y = 0, that is gy = y, for each g ∈ G. L P Now y ∈ R = kG = g∈G kgb hence y = g∈G y(g)gb with the coefficients y(g) ∈ K uniquely determined. Being that hy = y for each h ∈ G we must have that the coefficient hy(g) = y(h−1g) of g ∈ G in hy is the same as the coefficient y(g) of g ∈ G in y, that is to say y(g) = y(h−1g) for each g, h ∈ G.

As y 6= 0 we have that y is a non zero multiple of the sum of all gb with g ∈ G, hence G must be finite by the definition of R: the support of the function y : G −→ k must be finite. c) The ring k is assumed to be semisimple, hence by Wedderbund-Artin’s Theorem, k = k1 ×· · ·×kr ∼ with ki = Matni (Di) for uniquely determined ni ∈ N and division rings Di. Now kG = k1G × · · · × krG semisimple ⇒ each kiG is still such. On the other hand

U(kG) = U(k1G × · · · × krG) = U(k1G) × · · · × U(krG).

Hence it’s enough to prove the theorem in the case k = Matn(D), that is when k is a semisimple indecomposable ring. Note that by hypothesis ∼ kG = Matn(D)G = Matm1 (K1) × · · · × Matmt (Kt)

for some mi ∈ N and some division rings Ki. The center of k = Matn(D) is a field, call it

K = ck(k) = cMatn(D)(Matn(D)) = cD(D); while the center of kG is a product of fields, in which K sits diagonally. Obviously we have |G| ∈ ck(k) ⊆ ckG(kG) (note that k commutes with G inside kG, hence ck(k) ⊆ ckG(kG), but k itself may be not contained in ckG(kG)). P In the group algebra kG, with G finite, we always have the non zero element y = g∈G gb. This is a non zero G-invariant central element, indeed: X X bhy = bh gb = hgc = y = ybh for each h ∈ G, g∈G g∈G P and more generally xy = yx = α(x)y for any x = g∈G aggb ∈ kG, where α: kG −→ k is the augmentation map. Hence y ∈ c(kG). Moreover

2 X X X y = y gb = ygb = y = |G|y holds in c(kG). g∈G g∈G g∈G But kG semisimple ⇒ c(kG) is a product of fields, hence y 6= 0 in c(kG) ⇒ y2 6= 0 in c(kG) and therefore |G|1K cannot be 0 in K. Hence |G| is a unit in kG. Let us now assume the conditions a), b) and c). The converse follows as in the first version of Maschke’s Theorem by means of the Reynolds operator. Let us rewrite the argument in the general case, with k any unitary ring. We start with a kG-linear inclusion S : W,−→ V , and we have to show 9 that there exists a kG-submodule W 0 ⊆ V such that V = W ⊕ W 0. As V and W k-modules with k-semisimple, there exists T ∈ Homk(V,W ) such that T ◦ S = IdW . Then we define the k-linear map 1 X R(T ): V −→ V, v 7→ gT (g−1v). |G| g∈G 1 P −1 1 P Since |G| g∈G gT (g v) ∈ |G| g∈G gW ⊆ W we have that actually R(T ) ∈ Homk(V,W ). Now we note that 1 X 1 X |G| w ∈ W ⇒ R(T )(w) = gT (g−1w) = gg−1w = w = w ⇒ R(T ) ◦ S = Id , |G| |G| |G| W g∈G g∈G and hence V = W ⊕ Ker(R(T )) as k-modules. Finally we check that R(T ) is actually kG-linear: X X R(T )(( aggb)v) = R(T )( ag(gv)) g∈G g∈G X = ag R(T )(gv) g∈G ! X 1 X = a hT (h−1gv) g |G| g∈G h∈G ! X 1 X = a hT ((g−1h)−1v) g |G| g∈G h∈G ! X 1 X = a gg−1hT ((g−1h)−1v) g |G| g∈G h∈G ! X 1 X = a g hT (h−1v) g |G| g∈G h∈G X = agg R(T )(v) g∈G X = ( aggb) R(T )(v) g∈G

Hence R(T ) ∈ HomkG(V,W ) gives a kG-linear retraction of S. Hence V = W ⊕ Ker(R(T )) as kG-modules.

Alessio Del Padrone, Dipartimento di Matematica, Università di Genova, Via Dodecaneso 35, 16146 Genova, Italy, Tel: +390103536923, Fax: +390103536752 • E-mail : [email protected] Url : http://www.dima.unige.it/~delpadro/