## Divisibility theorems for group representations II October 14, 2009

Posted by Akhil Mathew in algebra, representation theory.
Tags: , , ,

So last time we proved that the dimensions of an irreducible representation divide the index of the center. Now to generalize this to an arbitrary abelian normal subgroup.

There are first a few basic background results that I need to talk about.

Induction

Given a group ${G}$ and a subgroup ${H}$ (in fact, this can be generalized to a non-monomorphic map ${H \rightarrow G}$), a representation of ${G}$ yields by restriction a representation of ${H}$. One obtains a functor ${\mathrm{Res}^G_H: Rep(G) \rightarrow Rep(H)}$. This functor has an adjoint, denoted by ${\mathrm{Ind}_H^G: Rep(H) \rightarrow Rep(G)}$.

It can be described explicitly using tensor products in a simple manner. Given an inclusion ${H \rightarrow G}$, there is a map ${\mathbb{C}[H] \rightarrow \mathbb{C}[G]}$ from which one gets the restriction functor ${Rep(G) \simeq \mathbf{Mod}(\mathbb{C}[G]) \rightarrow \mathbf{Mod}(\mathbb{C}[H]) \simeq Rep(H)}$. As is well-known for rings ${A \rightarrow B}$, the adjoint to this functor is given by the tensor product:

$\displaystyle \mathrm{Ind}_H^G(M) := \mathbb{C}[G] \otimes_{\mathbb{C}[H]} M.$

It follows that if ${N = \mathrm{Ind}_H^G(M)}$, then ${M}$ is a ${H}$-submodule of ${N}$. Also, if ${G = \bigcup_i g_i H}$ is a coset decomposition, then

$\displaystyle M = \bigoplus g_i N$

as vector spaces. In particular, ${\dim N = (G:H ) \dim N}$. Conversely, if the above decomposition holds then ${M = \mathrm{Ind}_H^G(N)}$.

The adjointness relation is usually called Frobenius reciprocity, and is written as

$\displaystyle \hom_G(\mathrm{Ind}_H^G(X), Y) \simeq \hom_H(X, \mathrm{Res}^G_H(Y)), \quad \forall X \in Rep(H), Y \in Rep(G).$

Restriction

First, we discuss the restriction to abelian normal subgroups via a lemma, which will enable us to induct on ${|G|}$

Lemma 1 Let ${G}$ be a group, ${A}$ an abelian normal subgroup. Let ${M}$ be a simple ${G}$-representation. Then there are two possibilities:

1. ${A}$ acts by scalar matrices on ${M}$.
2. There is a subgroup ${H}$ with ${A \subset H \subsetneq G}$ and

$\displaystyle M = \mathrm{Ind}_H^G(M')$

for some ${M' \subset M}$

A more general version of this is in Serre or Curtis-Reiner, in the latter attributed to Clifford.  The more general result drops the abelian hypothesis and replaces 1 above by “the restriction to A is isotypic, i.e. a direct sum of isomorphic simple objects.”

So, by restriction ${M}$ is an ${A}$-module. By the previous post, we can write as ${A}$-modules

$\displaystyle M = \bigoplus_{\chi \in \hom(A, \mathbb{C}^*)} M_{\chi},$

where

$\displaystyle M_{\chi} := \{ m \in M: am = \chi(a)m, \forall a \in A \}.$

Perhaps ${M_{\chi}}$ ought to be thought of as a weight space, as in the theory of semisimple Lie algebras, when ${A}$ is replaced by a Cartan subalgebra. It turns out that in this case, the elements of ${G}$ permute the spaces ${M_{\chi}}$: if ${m \in M_{\chi}}$, then ${gm \in M_{\chi_g}}$ where ${\chi_g(x) = \chi(g^{-1}xg)}$ because of the “fundamental calculation” (Fulton and Harris’ phrase)

$\displaystyle xgm = g (g^{-1} x g )m = g \chi_g(x)m, \quad x \in H.$

Now pick ${\chi}$ with ${M_{\chi} \neq 0}$. If ${M_{\chi}=M}$ then we are in the first case. Otherwise, take ${H}$ to be the stabilizer of ${M_{\chi}}$ and the result follows.

Finally, the theorem

Theorem 2 Let ${G}$ be a finite group and ${A}$ an abelian normal subgroup. Then each simple representation of ${G}$ has dimension at most ${|G|/|A|}$

Induction on ${G}$. Assume the theorem proved for smaller groups.

First of all, let’s make some reductions. Assume that ${V}$ is a simple representation of ${G}$ which is faithful, because otherwise we could replace ${G}$ by the image in ${Aut(V)}$ to get a representation of a quotient ${G'}$ of ${G}$. The image ${A'}$ of ${A}$ in that quotient satisfies ${(G':A') \mid (G:A)}$, so we reduce to the faithful case.

Next, I claim that we may assume that ${A}$ does not act by scalars on ${V}$. If it did, then ${A}$ would be contained in the center of ${G}$ (by faithfulness), in which case the theorem is already proved.

By the lemma, this means there is a subgroup ${H \subsetneq G}$ containing ${A}$ with ${V = \mathrm{Ind}_H^G(W)}$ for some (necessarily simple) ${W \in Rep(H)}$. Then by the inductive hypothesis

$\displaystyle \dim V = (G:H) \dim W \mid (G:H)(H:A),$

proving the theorem.