# Question on irreducible complex characters

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.

• Conjugation by $x$ is an automorphism of $G$. – Angina Seng Apr 10 '19 at 14:02
• Thank you. I'm not sure how the result ($\alpha$ being a class function) follows from this. – amator2357 Apr 10 '19 at 14:16

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.