2.2
Behaviour Under Induction and Restriction
Given a finite group G and a p-subgroupQ≤G, the most direct path to calculating
S(G, Q) would be as follows:
(i) find an indecomposable decomposition of k[G/Q];
(ii) determine which direct summand in this indecomposable decomposition contains a submodule isomorphic to kG.
Modern day algebraic programming languages are well-equipped to handle both of these steps, but the first involves increasingly lengthy computation times as |G:Q| grows larger. Our first result in this section provides a means of avoiding this inefficiency when calculatingS(G, Q).
Proposition 2.2.1. Suppose thatGis a finite group andQ≤H ≤G. ThenS(G, Q) is
a direct summand of S(H, Q)↑G, and there is precisely one direct summand isomorphic
to S(G, Q) in any indecomposable decomposition of S(H, Q)↑G.
Proof. Since S(H, Q) is a direct summand of k[H/Q], it follows from Lemmas 1.1.4
and 1.1.19 that S(H, Q)↑G is a direct summand of k[G/Q]. Furthermore, S(H, Q)↑G
contains a submodule isomorphic tokG, sinceS(H, Q) contains a submodule isomorphic
to kH. Thus, by the Krull-Schmidt Theorem and the Scott-Alperin Theorem, we know
that in any indecomposable decomposition of S(H, Q)↑G there is precisely one direct
summand which is isomorphic to S(G, Q), as required.
Remark 2.2.2. This provides a more efficient means of calculating S(G, Q), through
repeated induction. To be more precise, suppose that we have a chain of subgroups:
P =H0 < H1 < H2 <· · ·< Hr−1 < Hr =G.
Then the above shows that S(Hi+1, Q) is a direct summand of S(Hi, Q)↑HHi+1i for
0≤i≤r−1. Using this approach to calculating S(Hi, Q), we keep the dimension of
the modules we are decomposing much smaller than a more direct approach.
A partial converse to the above statement is given by the following, which was proved in [6].
38 Scott Modules Proposition 2.2.3. Suppose thatGis a finite group andQ≤H ≤G. Assume thatV is an indecomposable direct summand ofk[H/L] for some subgroupL≤H. If S(G, Q)
is a direct summand of V↑G, then Q∈Syl
p(L) andV ∼=S(H, Q).
Proof. By Lemma 1.1.4, we know that S(G, Q) is a direct summand ofk[G/L]. Thus,
S(G, Q) is a direct summand of k[G/L] containing kG as a submodule; by (iii) of the
Scott-Alperin theorem, we therefore know that Q∈Sylp(L). Thus S(H, L)∼=S(H, Q),
by Proposition2.1.5. Furthermore, by the hypothesis and Proposition 2.2.1, we know that S(G, Q) is a direct summand of both S(H, Q)↑G and V↑G. If V ̸∼=S(H, Q), then
we have
k[H/L]∼=V ⊕ S(H, Q)⊕ X for some direct summand X and hence
k[G/L]∼=V↑G⊕ S(H, Q)↑G⊕ X ↑G.
Thus k[G/L] contains two distinct direct summands in an indecomposable decomposi-
tion which are isomorphic to kG, a contradiction. SoV ∼=S(H, Q), as required.
We can rephrase the previous two results slightly differently in terms of the concept of multiplicities of Scott modules. Given a kG-module M, we say that the multi- plicity of S(G, Q) inMis the number of distinct direct summands appearing in an indecomposable decomposition of M which are isomorphic toS(G, Q). Proposition
2.2.1 therefore states that the multiplicity of S(G, Q) in S(H, Q)↑G is one; on the
other hand, Proposition 2.2.3 tells us that if the multiplicity of S(G, Q) in V↑G is
nonzero and V is an indecomposable p-permutation kH-module, then V ∼=S(H, Q).
The multiplicity of S(G, Q) in a generalkG-moduleMis given by the following result, due to Green (see [15, 4.10]). Before stating the result, we set up some notation: given a kG-module M, we setIH(M) to be the H-fixed points of M, i.e.,
2.2 Behaviour Under Induction and Restriction 39 moreover, we let IH,G(M) = trGH(IH(M)), where trGH denotes the relative trace map,
i.e., the map trG
H :IH(M)→ M given by trG H(m) = X t∈[G/H] tm for all m ∈IH(M).
Theorem 2.2.4. Suppose that Mis akG-module. The multiplicity of S(G, Q) inM is equal to dim(IQ,G(M)/J), where
J ={x∈IQ,G(M) :α(x) = 0 for all α ∈IQ(M∗)}.
We now turn our attention towards what happens when we restrict S(G, Q) to a
subgroup H. The following serves as a complement to Proposition 2.2.1.
Proposition 2.2.5. Suppose that Gis a finite group and Q≤H ≤G. Then S(H, Q)
is a direct summand of S(G, Q)↓H.
Proof. SinceS(G, Q) is a relativelyQ-projective cover ofkG, it follows that there exists
a surjective kG-homomorphism δ:S(G, Q)→kG which is Q-split. In particular, δH :
S(G, Q)↓H →kH is a surjective kH-homomorphism which isQ-split andS(G, Q)↓H is
relatively Q-projective, since S(G, Q) has vertex Q. It follows from Proposition 1.1.12
that S(H, Q) is a direct summand of S(G, Q)↓H, as required.
We note one key difference between Propositions2.2.1 and 2.2.5;S(H, Q)↑G has a single direct summand in any given indecomposable decomposition which is isomor-
phic to S(G, Q), whereasS(G, Q)↓H may contain multiple distinct direct summands
isomorphic to S(H, Q) in an indecomposable decomposition, by Mackey’s Theorem.
The following more general result covers the case where Q is not necessarily a
subgroup of H.
Proposition 2.2.6. [22, Theorem 1.7] Let G be a finite group, H ≤ G and Q be
a p-subgroup of G. Suppose that R is an element which is maximal in the poset
{Qg∩H :g ∈G}. ThenS(H, R) is a direct summand of S(G, Q)↓H.
Proof. By assumption, R = Qx ∩H for some x ∈ G. We shall prove the result by
40 Scott Modules
R ≤H, so the result follows from Proposition 2.2.5. We may therefore assume that
|Q|>|R|. Set N =NG(R) and let S be a maximal element in the poset
Ω = {Qg∩N :R < Qg∩N, g ∈G}.
We note that Ω is nonempty, since R < Qx and hence N
Qx(R) > R, so Qx∩N =
NQx(R)> R. ThusS exists and S =Qy ∩N for some y∈G. Note that R < S and
hence |Q|/|S|<|Q|/|R|, so by the inductive hypothesis, it follows that S(G, Q)↓N has
a direct summandU ∼=S(N, S). In particular, U ↓NH(R) has a direct summandU which
contains a submodule isomorphic to kNH(R); moreover, by Mackey’s Theorem, we have
(k[N/S])↓NH(R)
∼
= M
n∈[NH(R)\N/S]
k[NH(R)/NH(R)∩nSn−1].
Note that for all n∈N we have
R≤NH(R)∩nSn−1 ≤H∩(Qyn −1 ∩N)≤H∩Qyn−1, so R=NH(R)∩nSn−1. Thus (k[N/S])↓NH(R) ∼= M n∈[NH(R)\N/S] k[NH(R)/R]
and it follows that U ∼= S(NH(R), R). In particular, there exists a direct summand
V of S(G, Q)↓H such that V↓NH(R) contains U as a direct summand. By the Green
correspondence and [1, Lemma 11.3], it follows that vx(V) = vx(U) = R and hence
V ∼= S(H, R), since V is the unique direct summand of U ↑H with vertex R. Thus,
V ∼=S(H, R) and we are done.