# Abstract Nonsense

## The Center of a Character (Pt. I)

Point of post: In this post we discuss the notion the ‘center of a character’ and its relations to other concepts we’ve previously learned.

Motivation

In previous posts we’ve seen how we can take concepts from plain group theory and attempt to make an analogy for characters. In this post we extend this further by defining the ‘center’ of a character, which, as we shall see are just those $g\in G$ for which $|\chi(g)|=\chi(e)$. In other words, its just those $g\in G$ for which the modulus of $\chi(g)$ is maximized. This shall prove useful in the future, in particular it shall help us devolop enough machinery to prove Burnside’s Theorem–the classic use of representation theory.

Center of a Character

Let $G$ be a finite group and $\chi$ some character of $G$. Then, we define the center of $\chi$, denoted $\bold{Z}(\chi)$, to be

$\bold{Z}(\chi)=\left\{g\in G:\left|\chi(g)\right|=\chi(e)\right\}$

Our first claim is that, despite what is not immediately apparent, $\bold{Z}(\chi)\leqslant G$ for every $\chi$. Indeed, this follows immediately from:

Theorem: Let $G$ be a finite group and $\chi$ the character of the representation $\rho$. Then,

$\bold{Z}\left(\chi\right)=\left\{g\in G:\rho(g)=z\mathbf{1}\;\; z\in\mathbb{C}\right\}$

Proof: It’s clear by the unitarity of $\rho(g)$ for every $g\in G$ that if $\rho(g)=zI$ then $|z|=1$. Thus, $\chi(g)=\text{tr}\left(zI\right)=z\chi(e)$ and so $\left|\chi(g)\right|=\chi(e)$ as desired.

Conversely, suppose that $g\in\bold{Z}\left(\chi\right)$. Recall from the spectral theorem that we have that $\rho(g)$ is similar to $\text{diag}(\lambda_1,\cdots,\lambda_n)$ where $n=\deg\rho$ and $\lambda_1,\cdots,\lambda_n$ are the eigenvalues of $\rho(g)$ with multiplicity. But, by assumption we have that

$\displaystyle \left|\sum_{j=1}^{n}\lambda_j\right|=n$

from where it follows from the fact that $|\lambda_j|=1$ one must have that $\lambda_1=\cdots=\lambda_j$ and so $\rho(g)$ is similar to $\lambda_1 \mathbf{1}$ and thus $\rho(g)=\lambda_1\mathbf{1}$. The conclusion follows. $\blacksquare$

Corollary:Let $G$ be a finite group and $\chi$ a character for $G$. Then, $\bold{Z}(G)\leqslant G$.

Proof: This clearly follows since if $g,g'\in \bold{Z}(G)$ then $\rho(g)=z\mathbf{1}$ and $\rho(g')=z'\mathbf{1}$ and so evidently $\rho(gg'^{-1})=\rho(g)\rho(g')^{-1}=z\mathbf{1}z'^{-1}\mathbf{1}=(zz'^{-1})\mathbf{1}$ and so $gg'^{-1}\in\bold{Z}(G)$. The conclusion follows. $\blacksquare$

If we have a representation $\rho$ of a finite group $G$ one can easily verify that for any subgroup $H$ of $G$ the restriction of $\rho$ to $H$ is a representation of $H$. Moreover, the character of $\rho$ restricted to $H$ is also a character on $H$. Also, it’s apparent that if $\rho_{\mid H}$ is irreducible then $\rho$ is irreducible. Thus, if $\chi_{\mid H}\cong \chi^{(\alpha)}$ for some $\alpha\in\widehat{H}$ then $\chi\cong \chi^{(\beta)}$ for some $\beta\in\widehat{G}$. With this in mind, we make the following claim:

Theorem: Let $G$ be a finite group and $\chi$ a character of $G$. Then, $\chi_{\mid \bold{Z}(\chi)}=\chi(e)\chi_\lambda$ for some degree-one irrep $\lambda$.

Proof: We know by definition that for every $g\in \bold{Z}(G)$ there exists some $z_g\in\mathbb{C}$ for which $\rho(g)=z_g\mathbf{1}$. Define $\lambda:\bold{Z}(G)\to\mathbb{C}^{\times}$ by $\lambda(g)=z_g$. This is clearly a degree-one irrep and moreover it’s clear that $\rho_{\bold{Z}(G)}=\lambda(g)\mathbf{1}$. Thus,

$\chi_{\rho_{\mid\bold{Z}(G)}}(g)=\text{tr}\left(\rho_{\bold{Z}(G)}(g)\right)=\text{tr}\left(\lambda(g)\mathbf{1}\right)=\chi(e)\lambda(g)$

from where the conclusion follows. $\blacksquare$

References:

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

March 8, 2011 -

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