Abstract Nonsense

The Index of the Center of a Character

Point of post: In this post we explore the index of the center of a character within a finite group.

Motivation

In our last post we saw how every character $\chi$ on a finite group $G$ naturally creates a host of subgroups of $G$, namely the center $\bold{Z}(\chi)$ of the character. We also saw how a lot of information pertaining to the index of $\bold{Z}(\chi)$ in $G$.

Index of the Center of a Character

We discussed bef0re the notion of taking a representation $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ and restricting it to a subgroup $H\leqslant G$ to create a representatino $\rho:H\to\mathcal{U}\left(\mathscr{V}\right)$. Moreover, we discussed how the same metholody can take a character $\chi$ on $G$ and produce a character $\chi$ on $H$ by restricting $\chi$ to  $H$. Of course in general the representation and thus the the character that results from this restriction doesn’t have to be irreducible even if the original was (take any representation and $H=\{e\}$). But, there is a way in which one can bound the inner product of any character restricted to $H$ with itself by the value of the inner product of the original character with itself. Specifically:

Theorem: Let $G$ be a finite group and $H\leqslant G$. Moreover, let $\chi$ be a character on $G$ and $\chi_{H}$ the restriction of $\chi$ to $H$. Then

$\displaystyle \left\langle \chi_H,\chi_H\right\rangle_H\leqslant \left(G:H\right)\left\langle \chi,\chi\right\rangle_G$

where $\langle\cdot,\cdot\rangle_H$ is meant to recall that we’re taking the inner product considering $\chi_H$ an element of the group algebra $\mathcal{A}\left(H\right)$ and $\langle\cdot,\cdot\rangle$ the inner product on the group algebra $\mathcal{A}(G)$. Moreover, equality holds if and only if $\chi$ vanishes on $G-H$.

Proof: One merely notes that

$\displaystyle |H|\left\langle\chi_H,\chi_H\right\rangle_H=\sum_{h\in H}\left|\chi(h)\right|^2\leqslant\sum_{g\in G}|\chi(g)|^2=|G|\left\langle \chi,\chi\right\rangle_G$

Moreover, the fact that equality holds if and only if $\chi(g)$ vanishes on $G-H$ is also obvious from this. $\blacksquare$

From this e get the interesting corollary:

Corollary: Let $G$ be a finite group and $\chi$ an irreducible character on $G$. Then, $\chi(e)^2\leqslant \left(G:\bold{Z}(\chi)\right)$.

Proof: Recall from previous posts that $\chi_{\bold{Z}(\chi)}=\chi(e)\lambda$ for some degree-one irrep $\lambda$ of $\bold{Z}(\chi)$. Thus, by first principles we know that $\left\langle \lambda,\lambda\right\rangle_{\bold{Z}(\chi)}=1$ and so by our previous theorem we get that

\displaystyle \begin{aligned}\chi(e)^2 &=\chi(e)^2\frac{1}{|\bold{Z}(G)|}\sum_{g\in\bold{Z}(G)}|\lambda(g)|^2\\ &= \frac{1}{|\bold{Z}(G)|}\sum_{g\in\bold{Z}(G)}\left|\chi(e)\lambda(g)\right|^2\\ &=\left\langle\chi_{\bold{Z}(G)},\chi_{\bold{Z}(G)}\right\rangle_{\bold{Z}(G)}\\ &\leqslant\left(G:\bold{Z}(G)\right)\langle\chi,\chi\rangle\\ &=\left(G:\bold{Z}(G)\right)\end{aligned}

And this equality holds if and only if $\chi$ vanishes on $G-\bold{Z}(G)$. $\blacksquare$

In fact, we can say even more about $\chi(e)^2$ when $G/\bold{Z}(\chi)$ is abelian. Indeed:

Theorem: Let $G$ be a finite group and $\chi$ a character on $G$ for which $G/\bold{Z}(\chi)$ is abelian. Then $\chi(e)^2=\left(G:\bold{Z}(\chi)\right)$.

Proof: By our previous theorem it suffices to show that $\chi$ vanishes on $G-\bold{Z}(\chi)$. So, let $g\in G-\bold{Z}(G)$. Now, we clearly must have that $g^{-1}h^{-1}gh\notin\ker\chi$ for some $h\in G$. Indeed, if this were true then we’d have that $g\in\mathcal{Z}\left(G/\ker\chi\right)$ but since $\chi$ is an irrep we know from previous theorem that $\mathcal{Z}\left(G/\ker\chi\right)=\bold{Z}(\chi)/\ker\chi$ and so $g\in\bold{Z}(\chi)$ contrary to assumption. But, since $G/\bold{Z}(\chi)$ is abelian we know that $\underbrace{g^{-1}h^{-1}gh}_{z}\in\bold{Z}(\chi)$ and so by the fundamental characterization of $\bold{Z}(\chi)$ we have that $\rho(z)=\lambda(z)\mathbf{1}$ (where $\rho$ is any representation which induces $\chi$). But, since $g\notin\ker\chi=\ker\rho$ we have that $\lambda(g)\ne 1$. Note though that $\rho(gz)=\rho(g)\rho(z)=\lambda(z)\rho(g)$ and so $\chi(gz)=\lambda(z)\chi(g)$. But $gz=h^{-1}gh$ and since $\chi$ is a class function this implies that $\chi(gz)=\chi(g)$. Thus, $\chi(g)=\chi(gz)=\lambda(g)\chi(g)$ and since $\lambda(g)\ne 1$ this implies that $\chi(g)=0$. The conclusion follows. $\blacksquare$

References:

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