# proof that dimension of complex irreducible representation divides order of group

Theorem Let $G$ be a finite group and $V$ an irreducible complex 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 $\{1,\alpha,\alpha^{2},\cdots,\alpha^{i}\}$.

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

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

 $\phi_{C}h=h\sum_{g\in C}h^{-1}gh=h\sum_{g\in C}g=h\phi_{C}$

Hence applying $\phi_{C}$ to $V$ induces a $\mathbb{C}G$ linear map $V\to V$. By Schur’s lemma this must be multiplication by some complex number $\lambda_{C}$. Then $\lambda_{C}$ is an algebraic integer as it solves the same monic polynomial as $\phi_{C}$.

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 algebraic integers. Hence the sum of the eigenvalues, $\chi_{V}(g)$, must also be an algebraic integer.

Now $V$ is irreducible so,

 $|G|=\sum_{g\in G}\chi_{V}(g){\chi_{V}(g)}^{*}=\sum_{C\subset G}{\rm tr}(\phi_{% C}){\chi_{V}(C)}^{*}=d\sum_{C\subset G}\lambda_{C}{\chi_{V}(C)}^{*}$

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

Title proof that dimension of complex irreducible representation divides order of group ProofThatDimensionOfComplexIrreducibleRepresentationDividesOrderOfGroup 2013-03-22 17:09:04 2013-03-22 17:09:04 whm22 (2009) whm22 (2009) 8 whm22 (2009) Proof msc 20C99