# Irreducible Degrees and the Order of a Finite Group

This is a question of aesthetics.

For a finite group of order $n$, the proof that the degree $d$ of a complex irreducible representation divides $n$ goes by showing that the rational number $n/d$ is an algebraic integer. As an application of the fact that $\mathbf Z$ is an integrally closed domain, this proof is really spectacular. But I feel that this is an indirect proof, not providing an insight into what is actually going on.

Being a statement of very basic nature perhaps there are other 'natural' or alternative ways of seeing why this happens. I would be grateful if experts here can point out other proofs or explain what is happening in the traditional proof.

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Let me mention: mathoverflow.net/questions/108406/… – Alexander Chervov Apr 9 '13 at 12:45

$\def\ZZ{\mathbb{Z}}$I answered a similar question over on math.SE. Some general thoughts:

All proofs I know rely on constructing an element of the group ring which acts by $|G|/\dim V$, or some closely related quantity, on $V$. Since $G$ acts on the regular rep by matrices in $GL_n(\ZZ)$, and $V$ is a summand of the regular rep, this shows that $|G|/\dim V$ is an integer. (Exactly how easy that argument is depends how much algebra your audience has.)

Most books seem to use $\sum_{\chi} \sum_{g \in G} \chi(g) g$, where the sum is over the irreducible characters of $\chi$. This acts by $|G|/\dim V$ on $V$, and clearly lies in $\ZZ^{alg}[G]$, where $\ZZ^{alg}$ is the ring of algebraic integers.

In fact, setting $a_g = \sum_{\chi} \chi(g)$, the $a_g$ are integers, so this element is in $\ZZ[G]$. I spent some time a few months ago trying to find a combinatorial proof that the $a_g$ are integers, without developing the theory of algebraic integers, but failed. For a while I believed that $a_g \geq 0$, but that turned out to be false. Let $X \subset \mathbb{F}_2^5$ be the subgroup $\{ (x_1, x_2, x_3, x_4, x_5) : x_1+x_2+x_3+x_4+x_5=0 \}$ and let $x = A_5 \ltimes X$. If I recall correctly (my notes are at home), I got that $a_g$ is negative at $((12)(34),\ (1,0,0,0,1) )$.

An alternative hack is to use the element $\sum_{g, h \in G} ghg^{-1} h^{-1}$. This is manifestly in $\ZZ[G]$ and it acts by $(|G|/\dim V)^2$ on $V$. That let's you avoid mentioning algebraic integers, but it still involves character theory computations that I find unenlightening.

I'd be interested in hearing other proofs!

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@David Speyer: Thanks for the detailed answer here and at SE. I am teaching rep theory without using group algebras and fumbling. In Lagrange's theorem we see cosets are of same cardinality and provide a partition of $G$. About degree of intermediate fields in finite extensions also we have a transparent proof. I am looking for such a simple underlying idea. As odd ordered cyclic groups have 2-dimensional irreps as symmetries of regular polygons we need to bring the dependence on complex numbers (via Schur's lemma). Not that easy perhaps. :-( – P Vanchinathan Apr 8 '13 at 23:38
Correction to my earlier comment: I meant real two-dimensional representation for odd cyclic groups. – P Vanchinathan Apr 9 '13 at 2:56
Bump and Ginzburg remark on page 4 of their paper "Generalized Frobenius-Schur Numbers" that the Mathieu group $M_{11}$ provide another example where some of the $a_g$ are negative (and that the observation goes back to Solomon and Thompson). @David, your example is smaller (order 1920 versus 7920), so maybe it is not well-known. – Gene S. Kopp Apr 10 '13 at 23:24
Another interesting point about the Bump and Ginzburg paper is that it provides a combinatorial interpretation for the $a_g$ assuming the existence of an involution on $G$ with certain properties. – Gene S. Kopp Apr 10 '13 at 23:27
Thanks for the reference to the Bump and Ginzburg paper! That's very nice. – David Speyer Apr 11 '13 at 2:58

Here is an exposition of the alternative hack'' to which David Speyer alluded. For concreteness, I have fixed a group; the argument will go through in general, of course.

Let $g = (12) \in S_3$, a transposition in the symmetric group on three elements. The conjugacy class of $g$ consists of all three transpositions. Form $a = \frac{1}{3}\left[(12) + (13) + (23) \right] \in \mathbb{C}S_3$, the average of $g$'s conjugacy class. It is clear that for any $x \in G$, $$xax^{-1} = a$$ since conjugating $a$ will simply rearrange the terms in the sum. Equivalently, $$xa = ax,$$ and $a$ lies in the center of $\mathbb{C}S_3$. It follows that if $\rho: S_3 \longrightarrow \mbox{Aut}(V)$ is an irreducible representation, the matrix $$\rho(a) = \frac{1}{3}\left[\rho(12) + \rho(13) + \rho(23) \right]$$ commutes with any $\rho(x)$. Schur's lemma tells us that $\rho(a)$ is a scalar matrix, that is, $\rho(a) = \lambda I$ for some $\lambda \in \mathbb{C}$.

Summarizing, for each irreducible representation $\rho$, we may define a class function $\psi_{\rho}$ which associates to any $g \in S_3$ the scalar $\lambda$ by which $a=\frac{1}{|S_3|}\sum_{x \in S_3} xgx^{-1}$ acts on $V$.

As it happens, $\psi_{\rho}(g)$ can be computed easily in terms of the character $\chi^{\rho}$: after all, every element conjugate to $g$ has the same trace; by linearity, $\mbox{Tr}(\rho(a)) = \chi^{\rho}(a) = \chi^{\rho}(g)$. It follows that $$\psi_{\rho}(g) = \frac{\chi^{\rho}(g)}{\mbox{dim}V}.$$ A natural next step is to eliminate reference to a particular $g$ using an inner product: $$\langle \psi_{\rho},\chi^{\rho} \rangle = \frac{1}{\mbox{dim}V}.$$ Expanding the definition of the inner product on class functions, $$\mbox{Tr}\left[\frac{1}{|S_3|^2}\sum_{g \in S_3} \left(\sum_{x \in S_3} \rho(xgx^{-1}) \right) \rho(g^{-1}) \right] = \frac{1}{\mbox{dim}V},$$ and $$\frac{1}{|S_3|^2}\sum_{g \in S_3} \sum_{x \in S_3} \chi^{\rho}(xgx^{-1}g^{-1}) = \frac{1}{\mbox{dim}V}.$$

We are led to consider the formal sum $d=\sum_{g,h \in S_3} ghg^{-1}h^{-1}$ mentioned in David Speyer's post. This sum is invariant under any automorphism of $S_3$ (in particular inner automorphisms) and so lies in the center of the group algebra $\mathbb{C}S_3$. Schur's lemma tells us that the image of $d$ under any irreducible representation $\rho$ is a scalar. By the above we get $$\sum_{g \in S_3} \sum_{h \in S_3} \rho(ghg^{-1}h^{-1}) = \left[\frac{|S_3|}{\mbox{dim}V}\right]^2I.$$

Considering now the regular representation $\mathbb{C}S_3$, we see that the element $d$ acts with rational number eigenvalues. In other words, its characteristic polynomial splits completely over $\mathbb{Q}$. But we can also see by inspection that $d$ acts by an integer matrix. The characteristic polynomial is a monic integer polynomial and splits into linear factors, so its roots are integers. Since $\mathbb{C}S_3$ contains every irreducible representation as a summand at least once, we see that each $$\frac{|S_3|}{\mbox{dim}V} \in \mathbb{Z}.$$ In particular, $1$, $2$, and $1$ all divide $6$.

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Thanks, John for a clear writing. – P Vanchinathan Apr 11 '13 at 9:10