# Abstract Nonsense

## A ‘Lemma’ (pt. II)

Point of post: This post is a continuation of this one.

The ‘Lemma’

We next study as a tool to prove our still looming lemma the multiplicative relation between the elements of the canonical basis of $\text{Cl}(G)$. Namely, we see that:

Theorem: Let $G$ be a finite group with conjugacy classes $\mathcal{C}_1,\cdots,\mathcal{C}_k$. Then, for any $i,j\in[k]$ there exists $c_{i,j,r}\in\mathbb{N}\cup\{0\}$ for every $r\in[k]$ such that $\displaystyle \mathbf{1}_{\mathcal{C}_j}*\mathbf{1}_{\mathcal{C}_k}=\sum_{r=1}^{k}c_{i,j,r}\mathbf{1}_{\mathcal{C}_r}$.

Proof: Recall that $\text{Cl}(G)$ is real the center of the group algebra $\mathcal{A}(G)$. But, recall that the center of an algebra is a subalgebra. In particular since $\text{Cl}(G)$ is a subalgebra of $\mathcal{A}(G)$ we have that  $\text{Cl}(G)$ is in particular closed under multiplication–convolution in this case. Thus, since $\mathbf{1}_{\mathcal{C}_i}$ and $\mathbf{1}_{C_j}$ are class functions then so is $\mathbf{1}_{\mathcal{C}_i}\ast\mathbf{1}_{\mathcal{C}_j}$ . Thus, since $\left\{\mathbf{1}_{\mathcal{C}_r}\right\}_{r=1}^k$ is a basis for $\text{Cl}(G)$ we know that there does exist constants $c_{i,j,r}\in\mathbb{C}$ for $r=1,\cdots,k$ such that $\displaystyle \mathbf{1}_{\mathcal{C}_i}\ast\mathbf{1}_{\mathcal{C}_j}=\sum_{r=1}^{k}c_{i,j,r}\mathbf{1}_{\mathcal{C}_r}$ and in fact (as has already been proven in general) $c_{i,j,r}=\left(\mathbf{1}_{\mathcal{C}_i}\ast\mathbf{1}_{\mathcal{C}_j}\right)(g_r)$ for any $g_r\in\mathcal{C}_r$. But, by definition one has that

\displaystyle \begin{aligned}\left(\mathbf{1}_{\mathcal{C}_i}\ast\mathbf{1}_{\mathcal{C}_j}\right)(g_r) &= \sum_{g\in G}\mathbf{1}_{\mathcal{C}_i}\left(xg^{-1}\right)\mathbf{1}_{\mathcal{C}_j}(g)\\ &= \#\left\{(a,b)\in \mathcal{C}_i\times\mathcal{C}_j:ab=x\right\}\end{aligned}

which is surely a nonnegative integer. The conclusion follows. $\blacksquare$

From this obtain the following:

Theorem : Let $G$ be a finite group with conjugacy classes $\mathcal{C}_1,\cdots,\mathcal{C}_k$ and $\chi\in\text{irr}\left(G\right)$. Then, if $\omega_\chi$ is defined as before then $\omega_\chi\left(\mathbf{1}_{\mathcal{C}_r}\right)\in\mathbb{A}$ where $\mathbb{A}$ are, as usual, the algebraic integers.

Proof: We first claim that $\omega_\chi\left(\mathbf{1}_{\mathcal{C}_i}\right)\omega_\chi\left(\mathbf{1}_{\mathcal{C}_j}\right)$ is a $\mathbb{Z}$-linear combination of $\left\{\omega_\chi\left(\mathbf{1}_{\mathcal{C}_1}\right),\cdots,\omega_{\mathcal{C}_k}\right\}$. Indeed, one has that there exists $c_{i,j,r}\in\mathbb{Z}$ for $r=1,\cdots,k$ such that $\displaystyle \mathbf{1}_{\mathcal{C}_i}\ast\mathbf{1}_{\mathcal{C}_j}=\sum_{r=1}^{k}c_{i,j,r}\mathbf{1}_{\mathcal{C}_r}$. Thus, applying $\omega_\chi$ to both sides and recalling that it’s an algebra homomorphism we may conclude that $\displaystyle \omega_\chi\left(\mathbf{1}_{\mathcal{C}_i}\right)\omega_\chi\left(\mathbf{1}_{\mathcal{C}_j}\right)=\sum_{r=1}^{k}c_{i,j,r}\omega_\chi\left(\mathbf{1}_{\mathcal{C}_r}\right)$ Fix then $\ell\in[k]$. Rephrasing the above,we know that there exists $k^2$ elements of $\mathbb{Z}$, call them $c_{r,s}$ for $r,s\in[k]$ such that $\displaystyle \omega_{\chi}\left(\mathbf{1}_{\mathcal{C}_i}\right)\omega_\chi\left(\mathbf{1}_{\mathcal{C}_r}\right)=\sum_{s=1}^{k}c_{r,s}\omega_{\chi}\left(\mathbf{1}_{\mathcal{C}_s}\right)$. Let $M=[c_{r,s}]$ and $v=\left(\omega_\chi\left(\mathbf{1}_{\mathcal{C}_1}\right),\cdots,\omega_\chi\left(\mathbf{1}_{\mathcal{C}_k}\right)\right)^{\top}$.  A quick check then shows that $Mv=\omega_\chi\left(\mathbf{1}_{\mathcal{C}_i}\right)v$  and so $\omega_\chi\left(\mathbf{1}_{\mathcal{C}_i}\right)$ is in the spectrum of some matrix in $\text{Mat}\left(\mathbb{Z}\right)$ and so by a prior characterization we know that $\omega_\chi\left(\mathbf{1}_{\mathcal{C}_i}\right)\in\mathbb{A}$. Since $i\in[k]$ was arbitrary the conclusion follows. $\blacksquare$

From this we get the follwing:

Theorem (The ‘Lemma’): Let $G$ be a finite group and $\chi\in\text{irr}\left(G\right)$. Then, for every $g\in G$ one has that $\displaystyle \frac{\chi(g)}{\chi(e)}\#\left(\mathcal{C}_g\right)$ where $\mathcal{C}_g$ is the conjugacy class of $G$ containing $g$.

Proof: This follows immediately from the previous theorem and the already proven fact that $\displaystyle \frac{\chi(g)}{\chi(e)}\#\left(\mathcal{C}_g\right)=\omega_\chi\left(\mathbf{1}_{\mathcal{C}_g}\right)$. $\blacksquare$

Corollary: Let $G$ be a finite group and $\chi\in\text{irr}(G)$.Then, either $\chi(g)$ is irrational or $\chi(e)\mid \chi(g)\#\left(\mathcal{C}_g\right)$.

So our ‘lemma’ turned out to be a theatrical production in and of itself.

References:

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