For the moment, let k be a field of any arbitrary field. Suppose U is a kG-module, H a subgroup of G and V is a kH-module. In the following, let V ↑GH denote the induced representation of V from H to G. Let U ↓GH denote the restriction of U to kH-module.
If U′ is a direct summand of U , then we write U′ | U.
We first state different definitions of H-projectiveness. These three definitions are actually equivalent by [43] Corollary 11.3.4.
Definition 1. A kG-module U is said to be H-projective if U is a direct summand of
T ↑GH for some kH-module T .
Definition 2. We say a kG-module U H-projective if for a given exact sequence 0 −→f E1 −→ Eg 2 → U → 0 of kG-modules, it splits if and only if it splits as kH-modules.
Definition 3. We say a kG-module U H-projective if U|U ↓GH↑GH. Example 4. Every kG-module U is G-projective.
Example 5. Since every projective kG-module is a direct summand of (kG)n, a kG-module P is projective if and only if it is 1G-projective.
Proposition 6. Suppose U is a Q-projective kG-module with Q a minimal subgroup satisfying this condition. Then this minimal subgroup Q is unique up to conjugacy.
Proof. Suppose there exists a minimal subgroup Q′ of G such that U is Q′-projective.
Since U is both Q- and Q′-projective, we have U|U ↓GQ↑GQ↓GQ′↑GQ′. From Mackay’s formula
(ref. [36] Proposition 22),
U ↓GQ↑GQ↓GQ′↑GQ′= (U ↓GQ↑GQ↓GQ′)↑GQ′ = ( ⊕
s∈Q′\G/Q
((U ↓GQ)s ↓ssQQ∩Q′)↑QsQ′∩Q′)↑GQ′
= ⊕
s∈Q′\G/Q
((U ↓GQ)s ↓ssQQ∩Q′)↑GsQ∩Q′, where sQ = sQs−1
Since U is indecomposable, U must be a direct summand of ((U ↓GQ)s ↓ssQQ∩Q′)↑GsQ∩Q′ for some s. Since Q′ is minimal, we have sQ∩ Q′ = Q′. This means that Q′ is conjugate to a subgroup of Q. Similarly, Q is conjugate to a subgroup of Q′. Hence Q is conjugate to Q′
Definition 7. Let U be a kG-module, and Q a subgroup of G as in Proposition 6. We
say Q is a vertex of U .
Proposition 8. Suppose H is a subgroup of G such that |G/H| is invertible in k. Then every kG-module M is H-projective.
Proof. See [43] Proposition 11.3.5.
Corollary 9. Suppose U is an indecomposable kG-module, chark = p, with a vertex Q.
This Q must be a p-subgroup of G.
Proof. Let H be a Sylow p-subgroup P of G, then by Proposition8, M is P -projective.
Hence a vertex of M is a p-group.
Proposition 10. Let U be an indecomposable kG-module. Suppose U has a vertex Q, and
one has U|T ↑GQ for some kQ-module T . If we choose such kQ-module T indecomposable, then T is unique up to conjugacy by an element belongs to NG(Q).
Proof. At first, we choose our T specifically. Observe that since U|U ↓GQ↑GQ and U is indecomposable, we can choose some indecomposable kQ-module T with T|U ↓GQ such
that U|T ↑GQ. Now suppose there exists an indecomposable kQ-module T′ such that U|T′ ↑GQ. Since U|T ↑GQ, we have
T|U ↓GQ|T′ ↑GQ↓GQ, and T′ ↑GQ↓GQ= ⊕
s∈Q\G/Q
(T′ ↓QsQ∩Q)s↑QsQ∩Q .
Hence T|(T′ ↓Qs′Q∩Q)s′ ↑Qs′Q∩Q) for some s′. Since Q is a vertex, it is minimal. Therefore,
s′Q = Q. We must have s′ ∈ NG(Q).
Definition 11. Let Q, T, U be as in Proposition 10, we say T is a source of U .
The following proposition is not directly related to the content of this section. How-ever, this property looks similar to Definition 3, and will be used later.
Proposition 12. For any kH-module V , we have V|V ↑GH↓GH. Proof. From Mackay’s formula, we have
V ↑GH↓GH= ⊕
s∈H\G/H
(V ↓HsH∩H)s↑HsH∩H .
When s∈ [e], we have V = (V ↓HsH∩H)s ↑HsH∩H. Therefore, V|V ↑GH↓GH.
Theorem 13 (Krull-Schmidt-Azumaya). Let R be a complete discrete valuation ring
or a field. If Λ is an R-algebra and finitely generated as R-module, then every finitely generated Λ-module M can be written as ⊕n
i=1Mi with Mi indecomposable R-modules, and the set counting with the multiplicity {Mi} is uniquely determined by M.
Proof. See [10] Theorem 6.12.
Theorem 14 (Green’s Correspondence). Let k be a field of characteristic p. Let Q be a
p-subgroup of G and L a subgroup of G containing the normalizer of Q in G.
(1). Suppose U is an indecomposable kG-module with vertex Q, then there exists a unique indecomposable kL-module f (U ) with vertex Q, such that f (U )|U ↓GL. Also if X|U ↓GL and X ̸∼= f (U ), then X is H-projective, for some H =xQ∩ L and x ∈ G ∖ L.
(2). Suppose V is an indecomposable kL-module with vertex Q, then there exists a unique indecomposable kG-module g(V ) with vertex Q, such that g(V )|V ↑GL.And if Y|V ↑GL andY ̸∼= g(V ), then Y is H-projective for some H =yQ∩ Q, and y ∈ G ∖ L.
(3). Moreover, we have gf (U ) ∼= U, and f g(V ) ∼= V .
Proof. Step 0 : At first, suppose H is the subgroup in (2), then |H| is strictly smaller than |Q|. This is because if |H| is equals to |Q|, we have H = yQ∩ Q = Q for some y ∈ G ∖ L. Therefore, we have y ∈ NG(Q) ⊂ L, which can not happen. Also, the subgroup H in (1) can not be conjugate to Q under L. If so, we have yQ ∩ L = xQ for some x ∈ L. Consequently, yx−1Q equals to Q, hence yx−1 ∈ NG(Q). Therefore, y belongs to xNG(Q)⊂ L, and we get a contradiction.
Now we prove (2) first.
Step 1 : Suppose V has source T , then we write T ↑LQ= V ⊕ Z. By Proposition 12, we have
V ↑GL↓GL= V ⊕ V′, Z ↑GL↓GL= Z⊕ Z′.
Now, we look at T more carefully to understand V . We have
T ↑GQ↓GL= ⊕
s∈L\G/Q
Ts ↓sLQ∩sQ↑LsL∩Q= V ⊕ V′ ⊕ Z ⊕ Z′.
Since V has a vertex Q, we have
V ⊕ Z = T ↑LQ= Ts↓sLQ∩sQ↑LsL∩Q, s∈ L.
For indecomposable summands of V′, Z′, they must be sL∩ Q-projective, s ̸∈ L. How-ever, L∩sQ is not conjugate to Q under L by the first paragraph. Hence V ↑GL↓GL has the unique summand V with vertex Q.
Step 2 : (Existence) Now consider V ↑GL, write V ↑GL as direct sum of indecomposable
kG-modules. We can pick an indecomposable one, say U , such that U ↓GL contains V and such U must have a vertex Q. We have that U is Q-projective. Suppose it has a vertex Q′ proper subgroup of Q and a source T′. V|U ↓GL |T′ ↑GQ′↓GL, similar to step 1, V must has a vertex strictly smaller than Q, contradiction to the minimality of the vertex Q.
Step 3 : (Uniqueness) Suppose U′|V ↑GL, and U′ has a vertex Q′ < Q < L. Since U′|U′ ↓GL↑GL, one of indecomposable summands of U′ ↓GL must contain a source of U′, when we restrict this summand to Q′. We denote this indecomposable summand by X. This X must has vertex a Q′, otherwise it will contradict with U′ has a vertex Q′. Also X|U′ ↓GL |V ↑GL↓GL, by step 1, we have that X is L∩sQ-projective. But since X is an indecomposable kL-module, we have that Q′ must be L-conjugate to a subgroup of L∩sQ for some s ̸∈ L. Also lQ′ < L∩sQ implies Q′ < L∩sl−1Q. Since Q′ < Q, we have Q′ < Q∩s′Q for s′ ̸∈ L, since s ̸∈ L, and we have Q′ < Q. By step 0, we have that
|Q′| ̸= |Q|, hence Q′ is a proper subgroup of Q.
For proof of (1): Now suppose U is an indecomposable kG-module with a vertex Q.
Since U is also L-projective, U|U ↓GL↑GL. Hence U ↓GL contains an indecomposable kL-module V such that U|V ↑GL. And such a kL-module V must have a vertex Q and source T since U has. Now suppose V′ ̸∼= V is another indecomposable summand of U ↓GL | T ↑GQ↓GL. By Step 1 of proof (2), V′ must be xQ∩ L-projective for some x ∈ L − Q. By Step 0, we know thatxQ∩L is not L-conjugate to Q, hence V′ is not an indecomposable kL-module with a vertex Q.
For proof of (3): This follows from
U|U ↓GL↑GL, V|V ↑GL↓GL,
and the uniqueness of (1), (2).
Theorem 15 (Schur–Zassenhaus). Suppose G is a finite group, and N is a normal
subgroup of G such that the order of G/N is coprime to the order of N . Then G is a semidirect product of N and G/N .
Proof. See [11] Theorem 17.4.39.
Definition 16. We let Sk(G) denote the set of all non-isomorphic simple kG-modules.
Proposition 17. Suppose G is a finite group with the (unique) normal Sylow p-subgroup
P cyclic of order pn. Hence G is a semi-direct product of P and a subgroup K with |K|
coprime to p. Then the number of isomorphism classes of indecomposable kG-modules is pn· |Sk(K)|.
Proof. See [43], Corollary 11.2.2.
Corollary 18. Suppose G has a Sylow p-group P of order p, then there are
(p− 1) · |Sk(NG(P ))| + |Sk(G)|
indecomposable kG-modules.
Proof. By the Schur-Zassenhaus Theorem, we have NG(P ) = P ⋊ K. And we have |Sk(NG(P ))| = |Sk(K))|(cf. [43], Corollary 6.2.2). By Proposition 17, there are p·
|Sk(NG(P ))| non-isomorphic classes of kNG(P )-modules. Since there are|Sk(NG(P ))| in-decomposable projective kG-modules (up to isomorphism), i.e. 1-projective, p·|Sk(NG(P ))|−
|Sk(NG(P ))| of them must be P -projective. By Green’s correspondence, there are p|Sk(NG(P ))|−
|Sk(NG(P ))| indecomposable and not projective kG-modules with a vertex P , and |Sk(G)| indecomposable projective kG-modules.
Definition 19. A ring R is said to be of finite representation type, if there are only
finitely many isomorphism classes of indecomposable R-modules.
Proposition 20. Let k be a field of characteristic p and P be a Sylow p-subgroup of G.
Then kG is of finite representation type if and only if kP is of finite representation type.
Proof. (⇒): For an indecomposable kP -module V , we have V |V ↑GP↓GP. Since V is kP -indecomposable, V| U ↓GP for some indecomposable kG-summand U of V ↑GP. Since kG is of finite representation type and by Theorem13for any indecomposable kG-module U , we have that U ↓GP has only finitely many indecomposable kP -summands. Our kP can only be of finite representation type.
(⇐): By Proposition 8, we know that every indecomposable kG-module U must be P -projective, and U|V ↑GP for some indecomposable kP -module V . Again, by Theorem 13and finiteness of kP , kG can only be of finite representation type.
If we assume the following Example in Section 2.2 and using the proposition above, we get a criterion of G that kG is of finite representation type.
Proposition 21. Let P be a cyclic p-group of order q = pn, k be a field of characteristic p, then kP is of finite representation type.
Proof. Suppose M is an indecomposable kP -module with the representation ρ : P → GL(M ), and P is generated by g. Clearly, ρ(g)q = id. Since characteristic of k = p, (ρ(g)− id)q = 0. Since M is indecomposable, there are only one Jordan block of ρ(g), and the size of Jordan block is not bigger than q. Particularly, dimkM ≤ q. Hence there are only finitely many indecomposable kP -modules up to isomorphism.
Theorem 22. Let k be a field of characteristic p. Then kG is of finite representation type if and only if any of its Sylow p-subgroup is cyclic.
Proof. By Proposition 20, it is equivalent to prove that kP has finite representation type if and only if it is cyclic. By Proposition 21, we know that if P is cyclic then kP is of finite representation type. Now suppose P is not cyclic. Considering the Frattini subgroup Φ(P ) of P (cf. [11] p. 199), we have that P /Φ(P ) ∼= Cpd. Since P is not cyclic, we have d ≥ 2. Therefore, Cp × Cp is a homomorphic image of P . In Section 2.2 we will show that kCp × Cp is not of finite representation type. Since any indecomposable
kCp×Cp-module can be also viewed as an indecomposable kP -module, kP is not of finite
Here 1 above is the identity element (1, 1) of G.
u1 u2 un
Since n is arbitrary, once we prove that each M2n+1 is indecomposable, we have infinitely many indecomposable kG-modules. We prove this in two ways.
The first one is to prove that the endomorphism ring E = EndkG(M2n+1) is a local ring, then by [10] Proposition 6.10, M2n+1 is indecomposable kG-module. We can use the