There are many elegant results on the dimensions of the simple representations of a finite group , of which I would like to discuss a few today.

The final, ultimate goal is:

Theorem 1Let be a finite group and an abelian normal subgroup. Then each simple representation of has dimension dividing .

To prove it, we need to talk about quite a few topics.

**Simple representations of abelian groups **

This is quite simple: **They are all of degree 1**. Indeed, if is a simple -module for abelian, then multiplication by is a -morphism , hence a scalar by Schur. (Recall that for now we are working over an algebraically closed field whose characteristic is prime to .) Thus any one-dimensional subspace is -stable, so by simplicity .

The character of a simple representation is just a homomorphism for the ground field.

**The dimension divides the order (not necessarily abelian ) **

The main goal here is somewhat more interesting:

Theorem 2For arbitrary, the dimensions of the simple representations divide .

This is less straightforward, and relies on the notion of algebraic integer.

To prove it, first recall for a **class function** (i.e. with , the element is central and so acts on a simple representation by a scalar (by Schur’s lemma).

Lemma 3The scalar is

Indeed, the central element acts with trace by the definition of , and also with trace .

I claim now that is necessarily an algebraic integer if is so for all —this is the crux of the proof. By linearity of the map , it is sufficient to do this when , by taking combinations of the characteristic functions of the conjugacy classes.

Let be the conjugacy classes and be the corresponding central elements. The span the center of , which consists of the class functions. So we have a commutative ring

this follows because for has integral coefficients, so is closed under multiplication.

There is a ring-homomorphism by the action on , and by Schur again. Since is integral over —by Cayley-Hamilton, as it is a finitely generated abelian group— must consist of algebraic integers. This proves the claim.

Now is always a sum of roots of unity, so an algebraic integer; thus by the lemma and the orthogonality relations

is an algebraic integer, hence an integer, proving the theorem.

**The dimension divides the index of the center **

There are various refinements of the previous result, of which here is one which uses the center.

Theorem 4The dimensions of the simple representations of divide the index for the center.

The proof (due to Tate) is an amusing application of the tensor power trick; one fixes an irreducible , and considers the representation of the direct product , which is simple. The subgroup consisting of

acts trivially on , because by Schur each acts by a scalar on . In particular, is an irreducible representation of . So , where . Whence .

*A recent (published) paper had near the beginning the passage `The object of this paper is to prove (something very important).’ It transpired with great difficulty, and not till near the end, that the `object’ was an unachieved one.*

—Littlewood’s Miscellany

Well, this sure isn’t a paper, but we haven’t achieved our objective yet—we have to replace by an arbitrary abelian subgroup. Fortunately this is not too difficult, but to keep this post from becoming too long I’ll stop here for now.

October 11, 2009 at 10:31 pm

In my mind, the results you’ve proved here in this post are more interesting and deeper than your finite ultimate Theorem 1. If V is a simple representation of G and is an irreducible consitutent of V considered as a representation of A, then by Frobenius reciprocity V is a direct summand of the induced representation . Now consider the dimensions.

October 11, 2009 at 10:43 pm

Perhaps do you mean to say that each simple representation has dimension dividing ?

October 14, 2009 at 12:29 am

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