## Divisibility theorems for group representations October 11, 2009

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

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

The final, ultimate goal is:

Theorem 1 Let ${G}$ be a finite group and ${A}$ an abelian normal subgroup. Then each simple representation of ${G}$ has dimension dividing ${|G|/|A|}$.

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 ${X}$ is a simple ${G}$-module for ${G}$ abelian, then multiplication by ${\sigma \in G}$ is a ${G}$-morphism ${X \rightarrow X}$, hence a scalar by Schur. (Recall that for now we are working over an algebraically closed field whose characteristic is prime to ${|G|}$.) Thus any one-dimensional subspace is ${G}$-stable, so by simplicity ${\dim \ X = 1}$.

The character of a simple representation is just a homomorphism ${G \rightarrow k^*}$ for ${k}$ the ground field.

The dimension divides the order (not necessarily abelian ${G}$

The main goal here is somewhat more interesting:

Theorem 2 For ${G}$ arbitrary, the dimensions of the simple representations divide ${|G|}$

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

To prove it, first recall for a class function ${f: G \rightarrow k}$ (i.e. with ${f(tst^{-1}) = f(s), \forall s,t \in G}$, the element ${\sum_s f(s) s}$ is central and so acts on a simple representation ${V}$ by a scalar ${c}$ (by Schur’s lemma).

Lemma 3 The scalar ${c}$ is $\displaystyle \frac{1}{\dim V} \sum_s f(s) \chi_V(s).$

Indeed, the central element ${\sum_s f(s) s}$ acts with trace ${\sum_s f(s) \chi_V(s)}$ by the definition of ${\chi_V}$, and also with trace ${(\dim V) c}$.

I claim now that ${c}$ is necessarily an algebraic integer if ${f(s)}$ is so for all ${s \in G}$—this is the crux of the proof. By linearity of the map ${f \rightarrow c}$, it is sufficient to do this when ${f(G) \subset \mathbb{Z}}$, by taking combinations of the characteristic functions of the conjugacy classes.

Let ${c_1, \dots, c_t \subset G}$ be the conjugacy classes and ${d_i := \sum_{c_i} s}$ be the corresponding central elements. The ${d_i}$ span the center of ${\mathbb{C}[G]}$, which consists of the class functions. So we have a commutative ring $\displaystyle R = \bigoplus \mathbb{Z} d_i ;$

this follows because ${d_i d_j}$ for ${i,j \in [1,t]}$ has integral coefficients, so ${R}$ is closed under multiplication.

There is a ring-homomorphism ${\phi: R \rightarrow \mathbb{C} \subset End(V)}$ by the action on ${V}$, and by Schur again. Since ${R}$ is integral over ${\mathbb{Z}}$—by Cayley-Hamilton, as it is a finitely generated abelian group— ${\phi(R)}$ must consist of algebraic integers. This proves the claim.

Now ${\chi(s) }$ is always a sum of roots of unity, so an algebraic integer; thus by the lemma and the orthogonality relations $\displaystyle c = \frac{1}{\dim V} \sum \chi(s)\overline{\chi(s)} = \frac{g}{\dim V}$

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 4 The dimensions of the simple representations of ${G}$ divide the index ${(G:Z)}$ for ${Z \subset G}$ the center.

The proof (due to Tate) is an amusing application of the tensor power trick; one fixes an irreducible ${V}$, and considers the representation ${V^{\otimes N}}$ of the direct product ${G^N}$, which is simple. The subgroup ${T_N \subset G^N}$ consisting of $\displaystyle (x_1, \dots, x_N): \ \forall x_i \in Z , \quad x_1 \dots x_N = 1$

acts trivially on ${V^{\otimes N}}$, because by Schur each ${(1, 1, \dots, x_i, 1, 1, \dots)}$ acts by a scalar on ${V^{\otimes N}}$. In particular, ${V^{\otimes N}}$ is an irreducible representation of ${G^N/T_N}$. So ${\dim V^{\otimes N} \mid \mathrm{card}(G^N/T_N) = g^N/z^{N-1}}$, where ${z = \mathrm{card} Z, g = \mathrm{card} G}$. Whence ${\dim V \mid g/z}$.

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 ${Z}$ 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. 1. Peter McNamara - October 11, 2009

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 $\chi$ 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 $Ind_A^G \chi$. Now consider the dimensions. 2. Peter McNamara - October 11, 2009

Perhaps do you mean to say that each simple representation has dimension dividing $|G|/|A|$? Akhil Mathew - October 12, 2009

Ack, yes, thanks for pointing that out. (Corrected.) The version I previously had stated, the bound for the dimension, is much more trivial. It doesn’t even need normality, by the argument you just gave (or, if one wishes to avoid using the notion of induced representation, by taking an irreducible $A$-module $W$ and considering the sum of the $g_i W$ for $g_i$ left coset representatives for $A$ in $G$.)

3. Divisibility theorems II « Delta Epsilons - October 14, 2009

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