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

*Posted by Akhil Mathew in algebra, representation theory.*

Tags: Clifford's theorem, Frobenius reciprocity, induction, restriction

trackback

Tags: Clifford's theorem, Frobenius reciprocity, induction, restriction

trackback

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 and a subgroup (in fact, this can be generalized to a non-monomorphic map ), a representation of yields by **restriction** a representation of . One obtains a functor . This functor has an adjoint, denoted by .

It can be described explicitly using tensor products in a simple manner. Given an inclusion , there is a map from which one gets the restriction functor . As is well-known for rings , the adjoint to this functor is given by the tensor product:

It follows that if , then is a -submodule of . Also, if is a coset decomposition, then

as vector spaces. In particular, . Conversely, if the above decomposition holds then .

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

**Restriction **

First, we discuss the restriction to abelian normal subgroups via a lemma, which will enable us to induct on :

Lemma 1Let be a group, an abelian normal subgroup. Let be a simple -representation. Then there are two possibilities:

- acts by scalar matrices on .
- There is a subgroup with and
for some .

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 is an -module. By the previous post, we can write as -modules

where

Perhaps ought to be thought of as a weight space, as in the theory of semisimple Lie algebras, when is replaced by a Cartan subalgebra. It turns out that in this case, the elements of permute the spaces : if , then where because of the “fundamental calculation” (Fulton and Harris’ phrase)

Now pick with . If then we are in the first case. Otherwise, take to be the stabilizer of and the result follows.

**Finally, the theorem **

Theorem 2Let be a finite group and an abelian normal subgroup. Then each simple representation of has dimension at most .

Induction on . Assume the theorem proved for smaller groups.

First of all, let’s make some reductions. Assume that is a simple representation of which is faithful, because otherwise we could replace by the image in to get a representation of a quotient of . The image of in that quotient satisfies , so we reduce to the faithful case.

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

By the lemma, this means there is a subgroup containing with for some (necessarily simple) . Then by the inductive hypothesis

proving the theorem.

Hi Akhil; please excuse my contacting you this way. We really appreciate the additions you’ve made to the nLab, and we hope that you’ll also set up a user account at the nForum. That way you’ll be able to log any changes or new entries you’ve created, and people who have subscribed to the RSS feed will get an immediate heads-up and check out what you’ve done (without having to hit ‘Recently Revised’, which had given us problems in the past).

It’s pretty straightforward; just go here and hit Account to get started. You might also consider subscribing to the RSS feed.

To any others reading this: the nLab is a wiki originally created by hosts and patrons of the n-Category Cafe blog to record working notes on all aspects of mathematics. The home page describes pretty well what we’re about. We always welcome new contributors; please do not be intimidated by the content. Contributors come from a wide variety of backgrounds and with varying levels of expertise.

@Todd: Thanks, I just set up the nForum account and logged the changes I made yesterday.