# Abstract Nonsense

## The Center of a Character (Pt. II)

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

We now show some connections between the center of a character and its kernel. Namely:

Theorem: Let $G$ be a finite group and $\chi$ a character of $G$. Then: a) $\bold{Z}(\chi)/\ker\chi$ is cyclic, b) $\bold{Z}(\chi)/\ker\chi\subseteq\mathcal{Z}\left(G/\ker\chi\right)$ and c) if $\chi$ is an irreducible character then $\bold{Z}(\chi)/\ker\chi=\mathcal{Z}\left(G/\ker\chi\right)$.

Proof:

a) We first claim that $\ker\chi=\ker\lambda$. To see this we merely note that $g\in\ker\chi$ if and only if $\rho(g)=\mathbf{1}$ but $\rho(g)=\lambda(g)\mathbf{1}$ and so $\rho(g)=\mathbf{1}$ if and only if $\lambda(g)=1$ from where the conclusion follows. It follows then from the first isomorphism theorem that $\bold{Z}(\chi)/\ker\chi\cong\lambda(G)$. But, $\lambda(G)$ is a finite multiplicative subgroup of $\mathbb{C}$ and thus clearly cyclic.

b) Evidently one has that $\rho\left(\bold{Z}\left(\chi\right)\right)\subseteq\mathcal{Z}\left(\rho\left(G\right)\right)$ (this is immediate since each $\rho(g)$ for $g\in\bold{Z}(\chi)$ is a scalar matrix). Thus, one has that for any $k\in\bold{Z}(\chi)$ (since $\ker\chi=\ker\rho$) and $g\in G$ one has that $\rho(gk)=\rho(g)\rho(k)=\rho(k)\rho(g)=\rho(kg)$ so that $\rho(gkg^{-1}k^{-1})\in\ker\rho$ and so $\ker\rho=gkg^{-1}k^{-1}\ker\rho$ and thus $gk\ker\rho=kg\ker\rho$ so that $k\ker\rho\in\mathcal{Z}\left(G/\ker\chi\right)$. Since $k\ker\chi\in\bold{Z}\left(\chi\right)$ was arbitary the conclusion follows.

c) If $\rho$ is irreducible and $g\ker\chi\in \mathcal{Z}\left(G/\ker\chi\right)$ it clearly follows that $\rho(g)\in\mathcal{Z}\left(\rho(G)\right)$ and so by Schur’s Lemma it follows that $\rho(g)=z\mathbf{1}$ and so $g\ker\chi\in\bold{Z}/\ker\chi$. $\blacksquare$

The real interesting thing now comes from the next theorem:

Theorem: Let $G$ be a finite group. Then,

$\displaystyle \mathcal{Z}(G)=\bigcap_{\alpha\in\widehat{G}}\bold{Z}\left(\chi^{(\alpha)}\right)$

Proof: Clearly if $g\in\mathcal{Z}(G)$ then we have clearly from Schur’s lemma that $\rho(g)=z\mathbf{1}$ with $|z|=1$ and so $g\in\bold{Z}(\chi^{(\alpha)})$ for every $\alpha\in\widehat{G}$.

Conversely, suppose that $g\in\bold{Z}(\chi^{(\alpha)})$ for every $\alpha\in\widehat{G}$. It follows then from the comments of the previous theorem that $g\ker\chi^{(\alpha)}\in\mathcal{Z}\left(G/\ker\chi^{(\alpha)}\right)$ for every $\alpha\in\widehat{G}$. Thus, for any $h\in G$ one has that $ghg^{-1}h^{-1}\ker\chi^{(\alpha)}=\ker\chi^{(\alpha)}$ or that $ghg^{-1}h^{-1}\in\ker\chi^{(\alpha)}$. It follows then that $\displaystyle ghg^{-1}h^{-1}\in\bigcap_{\alpha\in\widehat{G}}\ker\chi^{(\alpha)}$. But, recalling that every normal subgroup is the realization of some intersection of the $\ker\chi^{(\alpha)}$‘s, and so in particular $\{e\}$ is the intersection of some of the $\ker\chi^{(\alpha)}$‘s we may conclude that $\displaystyle \bigcap_{\alpha\in\widehat{G}}\ker\chi^{(\alpha)}=\{e\}$ and so $ghg^{-1}h^{-1}=e$ and so $g$ and $h$ commute. Since $h\in G$ was arbitrary it follows that $g\in\mathcal{Z}\left(G\right)$ and so the conclusion follows. $\blacksquare$

References:

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

March 9, 2011 -

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