# Abstract Nonsense

## A ‘Lemma’ (Pt. I)

Point of post: In this post we will prove a lemma which shall prove to be absolutely necessary in proving Burnside’s lemma.

Motivation

The basic idea of this post is to use a variant of the left $\ast$-representation of $G$ generated by an irrep $\rho$ to create a mapping $\omega:\text{irr}(G)\to\mathbb{C}$. We shall then use this to show that the number $\displaystyle \frac{\chi(g)}{\chi(e)}\#\left(\mathcal{C}_g\right)$ is algebraic. This will prove to be surprising important in our proof of Burnside’s theorem.

The Preamble

Let $G$ be a finite group and $\chi$ be an irreducible character of $G$. Let then $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ be any irrep which admits $\chi$ as its character. Recall that we may then define the induced $\ast$-representation $\rho^\ast:\mathcal{A}(G)\to\mathcal\mathcal{U}\left(\mathscr{V}\right)$ by $\displaystyle \rho^\ast(f)=\sum_{g\in G}f(g)\rho(g)$. Our first claim is that $\rho^\ast$ is irreducible. Indeed:

Theorem: Let $\rho^\ast$ be defined as above. Then, $\rho^\ast$ is irreducible.

Proof: Suppose that$\mathscr{W}\leqslant\mathscr{V}$ is $\rho^\ast$-invariant then in particular we must have that $\left(\rho^\ast(g_0)\right)\left(\mathscr{W}\right)\subseteq\mathscr{W}$ but $\rho^\ast(g_0)=\rho(g_0)$ so that $\left(\rho(g_0)\right)\left(\mathscr{W}\right)\subseteq\mathscr{W}$. Since this was true for every $g_o\in G$ it follows that $\mathscr{W}$ is $\rho$-invariant and thus $\mathscr{W}$ is either $\{\bold{0}\}$ or $\mathscr{V}$. Since $\mathscr{W}$ was abritrary the conclusion follows.$\blacksquare$

Note though from basic properties of $\ast$-representations we have that for any $f\in\text{Cl}(G)$ (i.e. a class function) one has that $\rho^\ast(f)\rho^\ast(a)=\rho^\ast(f\ast a)=\rho^\ast(a\ast f)=\rho^\ast(a)\rho^\ast(f)$ for any $a\in\mathcal{A}(G)$.  Thus, by Schur’s lemma we may conclude that $\rho^\ast(f)=z_f\mathbf{1}$ for some $z_f\in\mathbb{C}$. Since this mapping is conjugation invariant (i.e. $\psi(f)=z_f\mathbf{1}$ for every $\psi=t\rho t^{-1}$ for some isomorphism $t$) and every irrep which admits $\chi$ as its character is conjugate it follows that $\psi^\ast(\chi^{(\alpha)})=z_\alpha\mathbf{1}$ for every $\psi$ which admits $\chi$ as its character. Thus, we can define the function $\omega_\chi:\text{Cl}(G)\to\mathbb{C}$ by $\omega_\chi(f)\mathbf{1}=\rho^\ast(f)$ for any $\rho^\ast$ where $\rho$ is an irrep which admits $\chi$ as its character. Our first theorem says that $\omega_\chi$ is a homomorphism Indeed:

Remark: We now begin using $\text{irr}(G)$ to denote $\left\{\chi^{(\alpha)}:\alpha\in\widehat{G}\right\}$.

Theorem: Let $\chi\in\text{irr}(G)$, and let $\omega_\chi$ be defined as above. Then $\chi$ is a $\mathbb{C}$-homomorphism (i.e. it’s an associative algebra homomorphism)

Proof: Let $f,f'\in\text{Cl}(G)$ and $z,z'\in\mathbb{C}$. Then, for any representation $\rho$ admitting $\chi$ as it’s character we have that

\begin{aligned}\rho^\ast(zf+z'f') &=z\rho^\ast(f)+z'\rho^\ast(f')\\ &=z\omega_\chi(f)\mathbf{1}+z'\omega_\chi(f)\mathbf{1}\\ &=(z\omega_\chi(f)+z'\omega_\chi(f'))\mathbf{1}\end{aligned}

and so by definition $\omega_\chi(zf+z'f')=z\omega_\chi(f)+z'\omega_\chi(f')$. Moreover, let $f,f'\in\text{Cl}(G)$ then one has that

\begin{aligned}\rho^\ast(f\ast f') &=\rho^\ast(f)\rho^\ast(f')\\ &=\omega_\chi(f)\mathbf{1}\omega_\chi(f')\mathbf{1}\\ &=(\omega_\chi(f)\omega_\chi(f'))\mathbf{1}\end{aligned}

Furthermore, let $f,f'\in\text{Cl}(G)$ then one has that

\displaystyle \begin{aligned}\rho^\ast(f\ast f') &= \rho^\ast(f)\rho^\ast(f')\\ &= \omega_\chi(f)\mathbf{1}\omega_\chi(f')\mathbf{1}\\&=(\omega_\chi(f)\omega_\chi(f'))\mathbf{1}\end{aligned}

and so $\omega_\chi(f\ast f')=\omega_\chi(f)\omega_\chi(f')$ from where the conclusion follows. $\blacksquare$

In particular this tells us that $\omega_\chi$ is entirely determined on the canonical basis $\left\{\mathbf{1}_{\mathcal{C}_j}\right\}_{j=1}^k$ for $\text{Cl}(G)$ where $\mathcal{C}_1,\cdots,\mathcal{C}_k$ are the conjugacy classes of $G$.

Note though that there is a natural way to relate $\omega_\chi\left(\bold{1}_{\mathcal{C}_j}\right)$ and any $g\in\mathcal{C}_j$. Indeed:

Theorem: Let $G$ be a finite group with conjugacy classes $\mathcal{C}_1,\cdots,\mathcal{C}_k$ then for any $j\in[k]$ and any $g_0\in\mathcal{C}_j$ one has that

$\displaystyle \omega_\chi\left(\mathbf{1}_{\mathcal{C}_j}\right)=\frac{\chi(g)\#\left(\mathcal{C}_j\right)}{\chi(e)}$

Proof: We note that by definition one has for any character $\rho$ which admits $\chi$ as its character

\displaystyle \begin{aligned}\chi(e)\omega_\chi\left(\bold{1}_{\mathcal{C}_j}\right) &= \text{tr}\left(\omega_\chi\left(\mathbf{1}_{\mathcal{C}_j}\right)\mathbf{1}\right)\\ &= \text{tr}\left(\rho^\ast\left(\mathbf{1}_{\mathcal{C}_j}\right)\right)\\ &= \text{tr}\left(\sum_{g\in G}\mathbf{1}_{\mathcal{C}_j}(g)\rho(g)\right)\\ &= \sum_{g\in G}\mathbf{1}_{\mathcal{C}_j}(g)\chi(g)\\ &= \sum_{g\in\mathcal{C}_j}\chi(g)\\ &= \#\left(\mathcal{C}_j\right)\chi(g_0)\end{aligned}

from where the conclusion follows. $\blacksquare$

References:

1. Isaacs, I. Martin. Character Theory of Finite Groups. New York: Academic, 1976. Print.