TITLE: Proofs that the degree of an irrep divides the order of a group
QUESTION [33 upvotes]: It is a theorem in basic representation theory that the degree of an irreducible representation on $G$ over $\mathbb{C}$ divides the order of $G$.  The usual proof of this fact involves algebraic integers (see for example Fulton & Harris' Representation Theory, Serre's Linear representations of finite groups, or Simon's Representations of finite and compact groups).  However I find this proof somewhat unsatisfying precisely because it uses algebraic integers, which don't show up much elsewhere in basic representation theory, and it is not at all evident why algebraic integers should be used.  I feel that there has to be another proof of this theorem that uses techniques of group theory and representation theory, but the only other proof I know is one by Kopp and Wiltshire-Gordon, but that proof seems to use even more complicated ideas if not machinery!
What are some other proofs of this theorem?

REPLY [3 votes]: Here's my attempt at an "elementary" proof, which is basicially a geodesic path through the proof that Serre gives in Linear representations of finite groups.
Theorem.
Let $\rho$ be an irreducible representation of $G$.  Then $d=\dim
  \rho$ divides $n=\lvert G\rvert$.
For any $x\in G$, consider the operator
$$\def\Cl{\operatorname{Cl}}T_x := \sum_{g\in \Cl(x)} \rho_g.$$
Note that this only depends on the conjugacy class of $x$.
We have $\rho_a T_x\rho_{a^{-1}} = T_x$ for any $a\in G$, so by
Schur's lemma, $T_x=\lambda_x I$ for some $\def\C{\mathbb{C}}\lambda_x\in \C$.
We can actually compute $\lambda_x$ by taking traces:
$$\lambda_x = \frac{\lvert\Cl(x)\rvert}{d} \chi(x)$$
where $\chi$ is the character of $\rho$.
In particular, $\lambda_e=1$.
By taking the trace of $\sum_{g\in G} \chi(g^{-1})\rho_g = \sum_{x_i}
\chi(x_i^{-1}) T_{x_i}$  (sum over a set $\{x_i\}$ of representatives
of conjugacy classes), and using the orthogonality relation
$\langle\chi,\,\chi\rangle=1$, we get an identity
$$\frac{n}{d} =
\sum_{x_i}\chi(x_i^{-1})\lambda_{x_i}.$$
Let $R=$ the abelian subgroup of $\C$ generated under addition by the finite set of elements
$$\zeta^k \lambda_x, \qquad 0\leq k<n,\quad x\in G,$$
where $\zeta=e^{2\pi i/n}$.  Since $\chi(g)$ is a sum of $n$-th roots
of unity, we have from the above identity that $n/d\in R$.
Claim.  For any $g,h\in G$, there is a formula
$$T_gT_h = \sum_{x_i} m_{x_i} T_{x_i}\qquad \text{and therefore}\qquad
\lambda_g\lambda_h = \sum_{x_i} m_{x_i} \lambda_{x_i},$$
where the sum is over $\{x_i\}$ a set of representatives of conjugacy
classes,
and $m_x=$ size of $\{(u,v)\in \Cl(g)\times \Cl(h)\;|\;uv=x\}$, and
thus $\def\Z{\mathbb{Z}}m_x\in \Z$.  Proof. Straightforward.
By the above claim, $R$ is in fact a subring of $\C$, so
multiplication by any element of $R$ such as $n/d$ defines an
endomorphism $\phi\colon R\to R$ of the underlying abelian group.
As an abelian group $R$ is certainly
finitely generated, and since $R\subseteq \C$ it is torsion free.  So by
the classification of
finitely generated abelian group it is isomorphic to $\Z^N$ for some
$N\geq 1$.  Choose a
basis for $R$ over $\Z$, and let $A$ be the integer matrix representing $\phi$ in this basis.
Since $\def\Q{\mathbb{Q}}n/d\in \Q$, for any $r\in R$ we must have $d\phi(r)=nr$, so $dA=nI$, i.e., $A=(n/d)I$.  But $A$ is an $N\times N$ integer matrix with $N\geq1$, so $n/d\in  \Z$ as desired.