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. 

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