Table Of ContentSOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER
WEBB
CI˙HAN BAHRAN
I changed the notation in some of the questions.
Chapter 8
1. Prove that if G is any finite group then the only idempotents in the integral
group ring ZG are 0 and 1.
[If e is idempotent consider the rank of the free abelian group ZGe and also its
image under the homomorphism ZG → F G for each prime p dividing |G|, which
p
is a projective F G-module. Show that rank ZGe is divisible by |G|. Deduce from
p Z
this that if e (cid:54)= 0 then e = 1.]
Let e be an idempotent in ZG and fix a prime p. Let ϕ : ZG → F G be the suggested
p
ring homomorphism. Write f = ϕ(e), now as e and f are both idempotents, we have
ZG = ZGe⊕ZG(1−e) and F G = F Gf ⊕F G(1−f). Note that ϕ maps ZGe onto
p p p
F Gf. Therefore if we pick a Z-basis {a ,...,a } for ZGe, then {ϕ(a ),...,ϕ(a )}
p 1 n 1 n
generate F Gf as a Z-module; hence as an F -vector space. Thus
p p
dim F Gf ≤ n = rank ZGe.
Fp p Z
Similarly,
dim F G(1−f) ≤ rank ZG(1−e).
Fp p Z
On the other hand, we have
|G| = rank ZG = rank ZGe+rank ZG(1−e)
Z Z Z
|G| = dim F G = dim F Gf +dim F G(1−f).
Fp p Fp p Fp p
Thus the above inequalities can’t be strict. Denoting the p-part of |G| by |G| , by
p
Corollary 8.3, |G| divides dim F Gf = rank ZGe. Since this happens for every
p Fp p Z
prime p, we deduce that |G| divides rank ZGe ≤ |G|. Therefore either rank ZGe = 0
Z Z
or rank ZGe = |G|. The former yields e = 0 and the latter yields e = 1.
Z
2. (a) Let H = C ×C and let k be a field of characteristic 2. Show that (IH)2 is
2 2
a one-dimensional space spanned by H = (cid:80) h.
h∈H
Write H = (cid:104)a,b | a2 = b2 = 1,ab = ba(cid:105). So IH is generated by a−1 and b−1 as a left
kH-module. Hence pairwise products of these generators generate (IH)2. Note that
1
SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 2
since chark = 2,
(a−1)2 = a2 −1 = 0
(b−1)2 = b2 −1 = 0
(a−1)(b−1) = ab−a−b+1 = ab+a+b+1 = H
(b−1)(a−1) = ba−b−a+1 = ab+a+b+1 = H.
Thus (IH)2 is generated by H as a left kH-module, but also as a k-vector space by
Exercise 6.2(a).
(b) Let G = A = (C × C ) (cid:111) C and let F be the field with four elements.
4 2 2 3 4
Compute the radical series of each of the three indecomposable projectives for
F A and identify each of the quotients
4 4
RadnP /Radn+1P .
S S
Now do the same for the socle series. Hence determine the Cartan matrix of F A .
4 4
[Start by observing that F A has 3 simple modules, all of dimension 1, which one
4 4
might denote by 1, ω and ω2. This exercise may be done by applying the kind of
calculation which led to Proposition 8.9]
By Proposition 8.8(a), the simple F G-modules are the simple F C -modules. Write
4 4 3
C = (cid:104)t | t3 = 1(cid:105) and F = {0,1,ω,ω2}. Since F∗ ∼= C , there are three group
3 4 4 3
homomorphisms from C to F∗. They send t to 1, ω and ω2, respectively. Each of these
3 4
homomorphisms yield a one-dimensional F C -module, which we also denote by 1, ω
4 3
and ω2. Clearly, here 1 is the trivial representation and the other two are nontrivial.
Also ω (cid:29) ω2, as the element ωt − ω2 ∈ F C annihilates the module ω but not ω2.
4 3
Thus 1, ω and ω2 are non-isomorphic. Since dimF C = 3, they form a complete list
4 3
of simple F C -modules (because every simple occurs at least once in the composition
4 3
series of the regular representation).
Write H = C × C as in (a). Note that a − 1 and b − 1 both annihilate H, so
2 2
(IH)3 = 0. By Proposition 8.8, the radical series of P (P is the projective cover of
1 1
the trivial module 1, it is P in the notation of the proposition) is
k
0 ⊆ (IH)2 = (cid:104)H(cid:105) ⊆ IH ⊆ F H = P .
4 1
TheactionofC ontheabovemodulesisbyconjugation,wemaychoosetheconjugation
3
by t on H acting as a (cid:55)→ b (cid:55)→ ab (cid:55)→ a.
To determine the radical layers of P , note that C and H both permute the elements
1 3
of H, so they fix H. Hence the bottom layer is trivial, that is, isomorphic to 1. By the
virtue of being a projective cover of 1, the top layer is also 1. Now we investigate the
middle layer IH/(cid:104)H(cid:105). Note that 1+ωa+ω2b ∈ IH −(cid:104)H(cid:105) and we have
a·(1+ωa+ω2b) = a+ω +ω2ab
≡ ω +a+ω2(a+b+1) (mod H)
= ω +a+ω2a+ω2b+ω2
= 1+ωa+ω2b
SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 3
b·(1+ωa+ω2b) = (b+ωab+ω2)
≡ ω2 +b+ω(a+b+1) (mod H)
= ω2 +b+ωa+ωb+ω
= 1+ωa+ω2b
t·(1+ωa+ω2b) = (1+ωb+ω2ab)
≡ 1+ωb+ω2(a+b+1) (mod H)
= 1+ωb+ω2a+ω2b+ω2
= ω +ω2a+b
= ω(1+ωa+ω2b).
Thus the one-dimensional F -vector space spanned by the coset (1+ωa+ω2b)+(cid:104)H(cid:105) is
4
an F A -submodule of IH/(cid:104)H(cid:105)which is isomorphic to ω (Note that checking the action
4 4
of t would have been enough above because as the radical layers are semisimple, H
must act trivially on them).
Now consider the element 1+ω2a+ωb. We have
t·(1+ω2a+ωb) = 1+ω2b+ωab
≡ 1+ω2b+ω(a+b+1) (mod H)
= 1+ω2b+ωa+ωb+ω
= ω2 +ωa+b
= ω2(1+ω2a+ωb).
Thus the one-dimensional F -vector space spanned by the coset (1+ω2a+ωb)+(cid:104)H(cid:105)
4
is an F A -submodule of IH/(cid:104)H(cid:105)which is isomorphic to ω2.
4 4
Ä ä
Since dimIH = dimF H −1 = 3, we have dim IH/(cid:104)H(cid:105) = 2. Thus we conclude that
4
IH/(cid:104)H(cid:105) ∼= ω ⊕ω2.
UsingProposition8.8again,wealsoconcludethattheradicallayersofP areω,ω2⊕1,ω
ω
and the radical layers of P are ω2,1 ⊕ ω,ω2. Thus the Cartan matrix of F A is
ω2 4 4
2 1 1
1 2 1 (with respect to any ordering of 1,ω,ω2). The socle series of these indecom-
1 1 2
posable projectives coincide with their radical series because of the following:
Proposition 1. Let k be a field and G a finite group. If S is a simple kG-module
such that the Loewy length of P is at most 3, then the socle and radical series of P
S S
coincide.
Proof. We give a proof when the Loewy length is 3, other cases are similar or trivial.
We have 0 (cid:54)= Rad2P ⊆ SocP . Since SocP ∼= S is simple, we get Rad2P = SocP .
S S S S S
We also have RadP ⊆ Soc2P (cid:54)= P . As P /RadP ∼= S is simple, we get RadP =
S S S S S S
Soc2P . (cid:3)
S
SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 4
(c) Now consider F A where F is the field with two elements. Prove that the 2-
2 4 2 ñ ô
0 1
dimensional F -vector space on which a generator of C acts via is a simple
2 3 1 1
F C -module. Calculate the radical and socle series for each of the two indecom-
2 3
posable projective modules for F A and hence determine the Cartan matrix of
2 4
F A .
2 4
®ñ ô ´
d
We are asked to show that S = : d,e ∈ F is a simple F C -module where the
e 2 2 3
ñ ô ñ ô
d e
generator t acts via t · = . Suppose V is a proper nonzero submodule
e d+e
of S. Since dim S = 2, we have dim V = 1. Because there is only one group
F F
2 2
homomorphism from C to (F )×, V must be a trivial module. So there exists 0 (cid:54)=
ñ ô ñ ô3 ñ2ô ñ ô
d d d e
∈ V such that = t · = , which yields d = e = 0; a contradiction.
e e e d+e
So we have at least two simple F C -modules: the trivial module k and S. Since
2 3
dimk +dimS = 3 = dimF C , there are no other simples. These are also the simple
2 3
modules of F A .
2 4
Similar to part (b), the radical series of P is 0 ⊆ (cid:104)H(cid:105) ⊆ IH ⊆ P and the top
k k
and bottom radical layers are k. We give two ways to show that the middle layer is
isomorphic to S: First, if it wasn’t S, it would have to be k, making all composition
factors of P isomorphic to k. This is impossible by Theorem 8.10 because A does not
k 4
have a normal 2-complement.
Second approach is by direct calculation. Consider the elements a + 1 and b + 1 in
IH. We claim that the cosets corresponding to these elements in IH/(cid:104)H(cid:105) are F -
2
independent. So assuming there exists c,d,e ∈ F such that
2
c(a+1)+d(b+1) = eH,
we show that c and d must be zero. Indeed,
ca+c+db+d = ea+eb+eab+e
(c+d+e)+(c+e)a+(d+e)b+eab = 0.
Hence c+e = d+e = e = 0, so c = d = 0. So these cosets form a basis for IH/(cid:104)H(cid:105).
Moreover, the action of t ∈ C on these basis elements is given by
3
t·(a+1) = b+1
t·(b+1) = ab+1 ≡ a+b ≡ (a+1)+(b+1) (mod H).
ñ ô
0 1
This is the same as the action of , therefore IH/(cid:104)H(cid:105) ∼= S as F A -modules.
1 1 2 4
By Proposition 8.8, the radical layers of P are S, S⊗S and S from bottom to top. We
S
give two ways two decompose the semisimple module S⊗S into a direct sum of simples.
First way is computational, dealing with basis. Note that S ⊗S is a 4-dimensional F -
2
0 0 0 1
0 0 1 1
vector space where t acts via the matrix A = . By calculation, the Smith
0 1 0 1
1 1 1 1
SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 5
X 0 0 1 1 0 0 0
0 X 1 1 0 1 0 0
normal form of XI − A = XI + A = is .
0 1 X 1 0 0 X +1 0
1 1 1 X +1 0 0 0 X3 +1
Thus as an F [X]-module where X acts as A, S ⊗S is isomorphic to
2
F [X] F [X] F [X] F [X] F [X]
2 2 ∼ 2 2 2
⊕ = ⊕ ⊕
(X +1) (X3 +1) (X +1) (X +1) (X2 +X +1)
1 0 0 0
0 1 0 0
Hence A is similar to the matrix . From here we conclude that as an
0 0 0 1
0 0 1 1
F C -module (hence as an F A -module), S ⊗S ∼= k ⊕k ⊕S.
2 3 2 4
The second way to see is more conceptual. As a 2-dimensional simple, the dual module
S∗ must be isomorphic to S. And we have
∼
HomF2C3(S ⊗F2 S,k) = HomF2C3(S,HomF2(S,k))
= Hom (S,S∗)
F2C3
= Hom (S,S).
F2C3
Hence
dimHom (S ⊗S,k) = dimEnd (S) = 2
F2C3 F2C3
because the multiplicity of S in F C is 1. Thus we conclude that the multiplicity of k
2 3
∼
in S ⊗S is 2 and hence S ⊗S = k ⊕k ⊕S.
In summary, the radical layers of P are k,S,k and those of P are S, k⊕k⊕S, S. Thus
k S ñ ô
2 2
theCartan matrix ofF A , with theorderk,S ofthe simples, is . Thesocle series
2 4 1 3
of P and P coincide with their radical series by Proposition 1. We see that the Cartan
k S
matrix is not symmetric here, and the reason is because the field F is not “big enough”.
2
3. Let G = H (cid:111)K where H is a p-group, K is a p(cid:48)-group, and let k be a field of
characteristic p. Regard kH as a kG-module via its isomorphism with P , so H
k
acts as usual and K acts by conjugation.
(a)Showthatforeachn, (IH)n isakG-submoduleofkH, andthat(IH)n/(IH)n+1
is a kG-module on which H acts trivially.
(IH)n is evidently preserved under the action of H, so it suffices to show that K ·
(IH)n ⊆ (IH)n. Employ induction on n. Now an arbitrary element of IH is of the
(cid:88) (cid:88)
form a h where a = 0. So for g ∈ K, we have
h h
h∈H h∈H
g · (cid:88) a h = (cid:88) a ghg−1
h h
h∈H h∈H
(cid:88)
= a h.
g−1hg
h∈H
(cid:88) (cid:88)
Since h (cid:55)→ g−1hg permutes H, we have a = a = 0. This finishes the basis
g−1hg h
h∈H h∈H
step.
SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 6
Note that the action of K on H respects the group multiplication, that is g ·(hh(cid:48)) =
(g ·h)(g ·h(cid:48)) for g ∈ K. Thus using the induction hypothesis, we have
K ·(IH)n+1 = K ·((IH)(IH)n) = (K ·IH)(K ·(IH)n) ⊆ (IH)(IH)n = (IH)n+1.
Since H is a p-group, as a kH-module, (IH)n/(IH)n+1 = RadnkH/Radn+1kH is
semisimple, hence has trivial H-action by Proposition 6.3.
(b) Show that
P = kH ⊇ IH ⊇ (IH2) ⊇ (IH)3 ⊇ ···
k
is the radical series of P as a kG-module.
k
This follows from Proposition 8.8(e).
(c) Show that there is a map
IH/(IH)2 ⊗ (IH)n/(IH)n+1 → (IH)n+1/(IH)n+2
k
x+(IH)2 ⊗y +(IH)n+1 (cid:55)→ xy +(IH)n+2
which is a map of kG-modules. Deduce that (IH)n/(IH)n+1 is a homomorphic
image of (IH/(IH)2)⊗n.
It suffices to show that
IH/(IH)2 ×(IH)n/(IH)n+1 → (IH)n+1/(IH)n+2
(x+(IH)2,y +(IH)n+1) (cid:55)→ xy +(IH)n+2
is a well-defined k-bilinear map. The multiplication map IH × (IHn) → (IH)n+1 is
clearly k-bilinear, so we only need to show well-definition. So suppose x,x(cid:48) ∈ IH and
y,y(cid:48) ∈ (IH)n such that x−x(cid:48) ∈ (IH)2 and y −y(cid:48) ∈ (IH)n+1. Then
xy −x(cid:48)y(cid:48) = xy −xy(cid:48) +xy(cid:48) −x(cid:48)y(cid:48) = x(y −y(cid:48))+(x−x(cid:48))y(cid:48) ∈ (IH)n+2.
For the last claim, induct on n. For n = 1, the claim is trivial via the identity map.
Suppose that there is a surjective map
(IH/(IH)2)⊗n → (IH)n/(IH)n+1.
Tensoring with the identity map on IH/(IH)2, we get a surjective map
(IH/(IH)2)⊗n+1 = (IH/(IH)2)⊗n ⊗IH/(IH)2 → (IH)n/(IH)n+1 ⊗IH/(IH)2.
Finally post composing this with the (clearly surjective) map defined in the first map
we get a surjective map
(IH/(IH)2)⊗(n+1) → (IH)n+1/(IH)n+2.
(d) Show that the abelianization H/H(cid:48) becomes a ZG-module under the action
g ·xH(cid:48) = gxg−1H(cid:48). Show that the isomorphism IH/(IH)2 → k⊗ H/H(cid:48) specified
Z
by (x − 1) + (IH)2 (cid:55)→ 1 ⊗ xH(cid:48) of Chapter 6 Exercise 17 is an isomorphism of
kG-modules.
Suppose xH(cid:48) = yH(cid:48) where x,y ∈ H. So x−1y ∈ H(cid:48) and for g ∈ G we have
(gxg−1)−1(gyg−1) = gx−1g−1gyg−1 = gx−1yg−1 ∈ gH(cid:48)g−1 = H(cid:48)
SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 7
because H(cid:48) is characteristic in H and H is normal in G, which implies that H(cid:48) is
normal in G. So the action is well-defined. And since conjugation by g is a group
homomorphism on H, the induced action on the abelian group H/H(cid:48) is Z-linear.
Denote the specified map by ϕ. It suffices to show that ϕ preserves both the H-action
and the K-action. The H-action on IH/(IH)2 is trivial by part (a) and the H acts
trivialy on H/H(cid:48) too (H(cid:48) absorbs conjugation by H). And for a ∈ K,
Ä Ä ää Ä ä
ϕ a· (x−1)+(IH)2 = ϕ (axa−1 −1)+(IH)2
= 1⊗axa−1H(cid:48)
= 1⊗a·(xH(cid:48))
= a·(1⊗xH(cid:48))
ÄÄ ää
= a· ϕ (x−1)+(IH)2 .
4. The group SL(2,3) is isomorphic to the semi direct product Q (cid:111) C where
8 3
the cyclic group C acts on Q = {±1,±i,±j,±k} by cycling the three generators
3 8
i, j and k. Assuming this structure, compute the radical series of each three
indecomposable projectives for F SL(2,3) and identifyy the radical quotients
4
RadnP /Radn+1P .
S S
[Use Chapter 6 Exercise 15. ]
Will come back to this after I fully understand the exercise in Ch. 6.
5. Let G = P (cid:111)S be a group which is the semi direct product of a 2-group P and
3
the symmetric group of degree 3. (Examples of such groups are S = V (cid:111)S where
4 3
V = (cid:104)(1,2)(3,4),(1,3)(2,4)(cid:105) and GL(2,3) = Q (cid:111)S where Q is the quatertnion
8 3 8
group of order 8.)
(a)Letk beafieldofcharacteristic2. ShowthatkGhastwonon-isomorphicsimple
modules.
ByCorollary6.4, thenormal2-subgroupP ofGactstriviallyonthesimplekG-modules
and consequently the simple kG-modules coincide with the simple kS -modules via the
3
projection G → S .
3
Since the abelianization of S is C and there is no element of order 2 in k×, there
3 2
is no group homomorphism from S to k× other than the trivial one. Hence the only
3
one-dimensional (simple) kS -module is the trivial module k.
3
There is a group homomorphism
S → GL (k)
3 2
ñ ô
0 1
(1,2) (cid:55)→
1 0
ñ ô
0 1
(1,2,3) (cid:55)→
1 1
(see page 2 noting that chark = 2) which yields a two-dimensional kS -module V. It
3
is straightforward to check that V has no nonzero elements fixed by G, hence does not
SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 8
contain a copy of k. Since k is the only one-dimensional kS -module, we deduce that
3
V is simple.
Let r the multiplicity of V in the radical quotient of kS . Then by Theorem 7.14, r =
3
dim (V) where D = End (V). Observe that by our description of V, D is precisely
D kS3 ñ ô ñ ô ñ ôV
0 1 0 1 a b
the centralizer of the matrices and in M (k). Let A = ∈ D. First,
1 0 1 1 2 c d
we have
Çñ ô ñ ôå Çñ ô ñ ôå
0 1 0 1 0 1 0 1
A − = − A
1 1 1 0 1 1 1 0
ñ ô ñ ô
0 0 0 0
A = A
0 1 0 1
ñ ô ñ ô
0 b 0 0
=
0 d c d
ñ ô
a 0
so b = c = 0 and A = . Second, we have
0 d
ñ ô ñ ô
0 1 0 1
A = A
1 0 1 0
ñ ô ñ ô
0 a 0 d
= ,
d 0 a 0
ñ ô
a 0
so a = d and A = . Therefore D consists of scalar matrices, hence is isomorphic
0 a
to k. Consequently r = dim (V) = 2. It follows that the radical quotient kS /RadkG
k 3
contains a copy of k ⊕V ⊕V. But this already covers 5 dimensions and since kS is
3
not semisimple, the radical quotient can’t have dimension more than 5. Thus
∼
kS /RadkS = k ⊕V ⊕V
3 3
and so k,V are the only simple kS -modules.
3
(b) Let e , e , e be the orthogonal idempotents which appeared in Example 7.5.
1 2 3
Show that each e is primitive in F G and that dimF Ge = 2|P| for all i.
i 4 4 i
[Use the fact that F Ge are projective modules.]
4 i
Consider the (natural) surjective ring homomorphism ϕ : F G → F [G/P] = F S .
4 4 4 3
Note that ϕ fixes e ’s. Using Exercise 10 and 3 in Chapter 6, we have that
i
kerϕ = F G·IP = F G·RadF P ⊆ RadF G
4 4 4 4
is nilpotent. Since e ’s were shown to be primitive in F S in Example 7.5, they are
i 4 3
also primitive in F G by Theorem 7.10.
4
Since each F Ge is a projective F G-module, by Corollary 8.3 |G| = 2|P| divides
4 i 4 2
dimF Ge . In particular, as e is nonzero we have 2|P| ≤ dimF Ge . Because e(cid:48)s are
4 i i 4 i i
orthogonal, we get
6|P| = |G| = dimF G ≥ dim(F Ge +F Ge +F Ge )
4 4 1 4 2 4 3
= dimF Ge +dimF Ge +dimF Ge
4 1 4 2 4 3
≥ 6|P|.
SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 9
Thus the above inequalities can’t be strict; so F G = F Ge ⊕ F Ge ⊕ F Ge and
4 4 1 4 2 4 3
dimF Ge = 2|P|.
4 i
(c) Show that if e = ()+(1,2,3)+(1,3,2) then F S e is the projective cover of
1 4 4 1
the trivial module and that F S e and F S e are isomorphic, being copies of the
4 4 2 4 4 3
projective cover of a 2-dimensional module.
Since S = V (cid:111) S , the largest semisimple quotients of F S and F S are the same,
4 3 4 4 4 3
which is k⊕S ⊕S by Example 7.5. Thus by Theorem 7.14, the multiplicity of P and
k
P in F S are 1 and 2, respectively.
S 4 4
Now let n = (cid:80) g ∈ F S . Since ne = 3n = n, the indecomposable projective
F S e containsga∈St4rivial su4b4module. Th1us by Theorem 8.15, F S e ∼= P . And by
4 4 1 4 4 1 k
Theorem 7.14, F S e ∼= P ∼= F S e .
4 4 2 S 4 4 3
(d) Show that F Ge ∼= F (cid:104)(1,2,3)(cid:105)e ↑G for each i.
4 i 4 i (cid:104)(1,2,3)(cid:105)
Write H = (cid:104)(1,2,3)(cid:105). Note that each e lies in F H. Since F H is a semisimple ring,
i 4 4
F He are projective F H-modules, hence F He ↑G are projective F G-modules. By
4 i 4 4 i H 4
orthogonality of the e and the fact that dimF H = 3, we deduce that dimF He = 1.
i 4 4 i
Thus
|G| |S ||P|
dimF He ↑G= |G : H| = = 3 = 2|P|,
4 i H 3 3
whichyields,by(c),thatF He ↑G areindecomposable. Andagainby(c),weonlyneed
4 i H
todistinguishwhichoneistheprojectivecoverofthetrivialmodule. Now,F He isgen-
4 1
eratedby(cid:80) h,henceisacopyofthetrivialF H-module. ThusF He ↑G∼= F [G/H]
h∈H 4 4 1 H 4
is a permutation F G-module. Since permutation modules surject on the trivial mod-
4
ule, it must be isomorphic to P .
k
6. Let A be a finite-dimensional algebra over a field k, and let A be the right
A
regular representation of A. The vector space dual (A )∗ = Hom (A ,k) becomes
A k A
a left A-module via the action (af)(b) = f(ba) where a ∈ A, b ∈ A and f ∈ (A )∗.
A A
Prove that the following two statements are equivalent:
(a) (A )∗ ∼= A as left A-modules.
A A
(b) There is a non-degenerate associative bilinear pairing A×A → k.
An algebra satisfying these conditions is called a Frobenius algebra. Prove that, for
a Frobenius algebra, projective and injective modules are the same thing.
By basic set theory, there is a bijection
Φ : {functions from A×A to k} → {functions from A to kA}
defined by Φ(f)(a)(b) = f(a,b). Now we investigate step by step how Φ preserves the
various structures. Let f : A×A → k.
• Φ(f) is k-linear if and only if f is linear in the first argument. For Φ(f) is
k-linear if and only if
Φ(f)(λa+a(cid:48)) = λΦ(f)(a)+Φ(f)(a(cid:48))
SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 10
for every a ∈ A and λ ∈ k. And that happens if and only if
Φ(f)(λa+a(cid:48))(b) = (λΦ(f)(a)+Φ(f)(a(cid:48)))(b)
Φ(f)(λa+a(cid:48))(b) = λΦ(f)(a)(b)+Φ(f)(a(cid:48))(b)
f(λa+a(cid:48),b) = λf(a,b)+f(a(cid:48),b)
for every b ∈ A.
• Φ(f) maps A to A∗ if and only if f is linear in the second argument. Indeed,
for a ∈ A, Φ(f)(a) ∈ A∗ if and only if
Φ(f)(a)(λb+b(cid:48)) = λΦ(f)(a)(b)+Φ(f)(a)(b(cid:48))
f(a,λb+b(cid:48)) = λf(a,b)+f(a,b(cid:48))
for every λ ∈ k and b ∈ A.
Hence Φ restricts to a bijection
Φ : {f : A×A → k| f is a k-bilinear form} → Hom (A,A∗).
k
Now, we observe the following:
• AssumeΦ(f)isanisomorphism. First,supposethereexistsasuchthatf(a,b) =
0foreveryb ∈ A. SoΦ(f)(a)(b) = 0foreveryb ∈ AandthismeansΦ(f)(a) = 0.
Thus a = 0 since Φ(f) is injective. Second, assume that there exists b such that
f(a,b) = 0 for every a ∈ A. Suppose b is not zero. Then there exists δ ∈ A∗
such that δ(b) = 1 and since Φ(f) is surjective, δ = Φ(f)(a ) for some a . Hence
0 0
0 = f(a ,b) = Φ(f)(a )(b) = δ(b) = 1, a contradiction. We conclude that f is
0 0
non-degenerate.
• Conversely, assume f is non-degenerate. We show that kerΦ(f) = 0, which is
equivalent to showing Φ(f) is an isomorphism because A and A∗ have the same
dimension: Suppose Φ(f)(a) = 0 for some a. This means 0 = Φ(f)(a)(b) =
f(a,b) for all b ∈ A; hence by non-degeneracy a = 0.
Thus, Φ further restricts to a bijection
Φ : {non-degenerate k-bilinear forms on A×A} → Iso (A,A∗).
k
Finally, we observe that Φ(f) : A → ( A)∗ is a right A-module homomorphism ( A
A A A
is a right A-module via (g ·c)(d) = g(cd)) if and only if
Φ(f)(ac) = Φ(f)(a)·c
for every a,c ∈ A. And that happens if and only if
Φ(f)(ac)(b) = (Φ(f)(a)·c)(b)
Φ(f)(ac)(b) = Φ(f)(a)(cb)
f(ac,b) = f(a,cb)
for every b ∈ A. That is, Φ(f) is a right A-module homomorphism if and only if f
is associative. Note that in this case the dual map of Φ(f) yields a left A-module
homomorphism from A to (A )∗. That is, Φ establishes the equivalence of (a) and
A A
(b) up to a dualization. Alternatively, defining
Ψ : {functions from A×A to k} → {functions from A to kA}
by Ψ(f)(a)(b) = f(b,a) would directly lead to the desired isomorphism in a similar
fashion with Φ.
