Abstract Nonsense

Crushing one theorem at a time

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.

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

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