Abstract Nonsense

Crushing one theorem at a time

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).


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



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


March 9, 2011 - Posted by | Algebra, Representation Theory | , , , ,


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