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We let $G$ be a finite group.

If $\chi$ is a complex character of $G$, we define $\overline{\chi}:G \to \mathbb{C}$ by $\overline{\chi}(g)=\overline{\chi(g)}$ for all $g \in G$. We write

$\nu(\chi):= \frac{1}{|G|}\displaystyle\sum_{g \in G}\chi(g^2)$

for the Frobenius Schur Indicator.

We let Irr($G$) denote the set of irreducible complex characters of $G$. We want to show that:

$\displaystyle\sum_{\chi\in Irr(G)}\nu(\chi)\chi(1)=|\{h \in G:h^2=1\}|$

There is a hint: Define $\alpha:G \to \mathbb{C}$ by $\alpha(g)=|\{h \in G: h^2 = g\}|$. Prove that $\alpha$ is a class function and use that Irr$(G)$ is an orthonormal basis of the vector space $R(G)$ of class function of $G$.

So we first try to show that $\alpha$ is a class function, i.e. we want to show that $|\{h \in G:h^2=g\}|=|\{h \in G:h^2=xgx^{-1}\}|$, for all $x,g \in G$, but I really cannot see how this is true.

As for the second part, assuming that $\alpha$ is indeed a class function, we can write $\alpha$ (second part of the hint) as $\alpha=\displaystyle \sum_{\chi \in Irr(g)}\langle\alpha,\chi\rangle\chi = \displaystyle \sum_{\chi \in Irr(g)}\frac{1}{|G|}\displaystyle \sum_{g \in G}\langle\alpha(g),\overline{\chi(g)}\rangle \chi$

but I am not sure at all how to proceed from here.

This is all in relation with this question

Any help is much appreciated.

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  • $\begingroup$ Conjugation by $x$ is an automorphism of $G$. $\endgroup$ – Lord Shark the Unknown Apr 10 at 14:02
  • $\begingroup$ Thank you. I'm not sure how the result ($\alpha$ being a class function) follows from this. $\endgroup$ – amator2357 Apr 10 at 14:16
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Let us first define the set $A(g)=\{h \in G: h^2=g\}$ and $\alpha(g)=|A(g)|$, its cardinality. First observe that $\alpha$ is a class function, that is, it is constant on conjugacy classes: fix for the moment an $x \in G$ and define a map from $A(g) \rightarrow A(x^{-1}gx)$ by $h \mapsto x^{-1}hx$. This map is well-defined: $(x^{-1}hx)^2=x^{-1}h^2x=x^{-1}gx$, so $x^{-1}hx \in A(x^{-1}gx)$. The map is also injective: if $x^{-1}hx=x^{-1}kx$, then obviously $h=k$. And it is surjective: if $k \in A(x^{-1}gx)$ then $xkx^{-1} \in A(g)$ and $xkx^{-1}$ maps to $k$. Hence $\alpha(g)=\alpha(x^{-1}gx)$ for every $x \in G$.

Now $\alpha$ is a class function, and it takes non-negative integer values. This does not make it into a character, but since the irreducible characters of $G$ form an orthonormal basis for the class functions we can write $$\alpha=\sum_{\chi \in Irr(G)}\nu(\chi)\chi $$, with $\nu(\chi) \in \mathbb{Z}_{\geq 0}$. Now we need to show that in fact $$\nu(\chi)=\frac{1}{|G|}\sum_{g \in G}\chi(g^2)$$ From the formula for $\alpha$ it follows that $\nu(\chi)=[\chi,\alpha]=\frac{1}{|G|}\sum_{g \in G}\chi(g)\overline{\alpha(g)}=\frac{1}{|G|}\sum_{g \in G}\chi(g)\alpha(g)$. Note that $\chi(g)\alpha(g)=\sum_{\{h \in G: h^2=g\}}\chi(h^2)$, we get the formula for $\nu(\chi)$. Finally, observe that $\alpha(1)=|\{h \in G: h^2=1\}|$. So $$\alpha(1)=\sum_{\chi \in Irr(G)}\nu(\chi)\chi(1)= |\{h \in G: h^2=1\}|$$ as wanted.

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