proof that dimension of complex irreducible representation divides order of group

Theorem Let $G$ be a finite group and $V$ an irreduciblecomplex representation of finite dimension $d$. Then $d$ divides$|G|$.

Proof: Given any $\alpha$ in the group ring of $G$ (denoted$\mathbb{Z}G$) we may define a sequence of submodules of$\mathbb{Z}G$ (regarded as a module over $\mathbb{Z}$)by $A_i$ equals the $\mathbb{Z}$ linear span of$\{1,\alpha ,{\alpha }^2,\mathrm{\cdots },{\alpha }^i\}$.

$\mathbb{Z}G$ is Noetherian as a module over $\mathbb{Z}$ so wemust have $A_i=A_{i-1}$ for some $i$. Hence ${\alpha }^i$ may beexpressed as a $\mathbb{Z}$ linear combination of lower powers of$\alpha$. In other $\alpha$ solves a monic polynomial ofdegree $i$ with coefficients in $\mathbb{Z}$.

Given a conjugacy class $C$ in $G$, we may set ${\varphi }_C={\sum }_{g\in C}g$. Then ${\varphi }_C$ is central in $\mathbb{Z}G$, as given $h\in G$, we have:

$${\varphi }_{C}h=h\sum _{g\in C}{h}^{-1}gh=h\sum _{g\in C}g=h{\varphi }_C$$

Hence applying ${\varphi }_C$ to $V$ induces a $ℂG$ linear map$V\to V$. By Schur’s lemma this must bemultiplication by somecomplex number ${\lambda }_C$. Then ${\lambda }_C$ is an algebraicinteger as it solves the same monic polynomial as ${\varphi }_C$.

Also any $g\in G$ has finite order so the map it induces on $V$must have eigenvalues which are roots of unity and hence algebraicintegers. Hence the sum of the eigenvalues, ${\chi }_V(g)$, mustalso be an algebraic integer.

Now $V$ is irreducible so,

$$|G|=\sum _{g\in G}{\chi }_V(g){\chi }_V{(g)}^{*}=\sum _{C\subset G}\mathrm{tr}({\varphi }_C){\chi }_V{(C)}^{*}=d\sum _{C\subset G}{\lambda }_C{\chi }_V{(C)}^{*}$$

Therefore $|G|/d$ is both rational and an algebraic integer. Henceit is an integer and $d$ divides $|G|$.