# Abstract Nonsense

## Irreducible Characters (Pt. II)

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

Irreducible Characters Span Class Functions

We now show that $\displaystyle \text{span}\left\{\chi^{(\alpha)}:\alpha\in\widehat{G}\right\}=\text{Cl}(G)$. Indeed:

Theorem: Let $G$ be a finite group then $\displaystyle \text{span}\left\{\chi^{(\alpha)}:\alpha\in\widehat{G}\right\}=\text{Cl}(G)$.

Proof: Let $f\in\text{Cl}(G)$. Since in particular $f\in\mathcal{A}(G)$ and $\displaystyle \text{span }\mathcal{D}=\mathcal{A}(G)$ as was already proven. Thus, there exists constants $c^{(\alpha)}_{i,j}$ for every $\alpha\in\widehat{G}$ and $i,j\in[d_\alpha]$ such that

$\displaystyle f=\sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}c^{(\alpha)}_{i,j}D^{(\alpha)}_{i,j}$

It follows then that for any $x\in G$ one has that

\begin{aligned}f(x) &= \frac{1}{|G|}\sum_{g\in G}f\left(gxg^{-1}\right)\\ &= \frac{1}{|G|}\sum_{g\in G}\sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}c^{(\alpha)}_{i,j}D^{(\alpha)}_{i,j}\left(gxg^{-1}\right)\\ &=\frac{1}{|G|}\sum_{g\in G}\sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}c^{(\alpha)}_{i,j}\sum_{p,q=1}^{d_\alpha}D^{(\alpha)}_{i,p}D^{(\alpha)}_{p,q}(x)\overline{D^{(\alpha)}_{j,q}(g)}\\ &= \sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}c^{(\alpha)}_{i,j}\sum_{p,q=1}^{d_\alpha}D^{(\alpha)}_{p,q}(x)\frac{1}{|G|}\sum_{g\in G}D^{(\alpha)}_{i,p}(g)\overline{D^{(\alpha)}_{j,q}(g)}\end{aligned}

which (since the inner most sum is the conjugate of the inner product) is equal to

\displaystyle \begin{aligned}\sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}c^{(\alpha)}_{i,j}\sum_{p,q=1}^{d_\alpha}D^{(\alpha)}_{p,q}(x)\frac{1}{d_\alpha}\delta_{p,q}\delta_{i,j} &= \sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}\frac{1}{d_\alpha}c^{(\alpha)}_{i,j}\delta_{i,j}\sum_{p,q=1}^{d_\alpha}D^{(\alpha)}_{p,q}(x)\delta_{p,q}\\ &=\sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}\frac{1}{d_\alpha}c^{(\alpha)}_{i,j}\delta_{i,j}\sum_{p=1}^{d_\alpha}D^{(\alpha)}_{p,p}(x)\\ &=\sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}\frac{1}{d_\alpha}c^{(\alpha)}_{i,j}\delta_{i,j}\chi^{(\alpha)}(x)\\ &=\sum_{\alpha\in\widehat{G}}\left(\frac{1}{d_\alpha}\sum_{i=1}^{d_\alpha}c^{(\alpha)}_{i,i}\right)\chi^{(\alpha)}(x)\end{aligned}

and since $x\in G$ was arbitrary it follows that

$\displaystyle f=\sum_{\alpha\in\widehat{G}}\left(\frac{1}{d_\alpha}\sum_{i=1}^{d_\alpha}c^{(\alpha)}_{i,i}\right)\chi^{(\alpha)}$

And, since $f\in\text{Cl}(G)$ was arbitrary the conclusion follows. $\blacksquare$

From this we get the obvious corollary that $\#\left(\widehat{G}\right)=k$ where $k$ is the number of classes of $G$. From this we can now prove a theorem which, in the future, will help us prove a few interesting structural results about groups. Namely:

Theorem: Let $G$ be a finite group. Then, $G$ is abelian if and only if every one of it’s irreps has degree one.

Proof: The necessity of this theorem was a previous theorem and so it suffices to show sufficiency. To do this we merely note that if each irrep has degree one then

$\displaystyle k=\#\left(\widehat{G}\right)=\sum_{\alpha\in\widehat{G}}1^2=\sum_{\alpha\in\widehat{G}}d_\alpha^2=|G|$

and thus the number of conjugacy classes of $G$ is equal to its order, which is true if and only if $G$ is abelian. The conclusion follows. $\blacksquare$

References:

1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print

February 25, 2011 -

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