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:
$\displaystyle\sum_{\chi\in Irr(G)}\nu(\chi)\chi(1)=|\{h \in G:h^2=1\}| (1)$
We define $\alpha:G \to \mathbb{C}$ by $\alpha(g)=|\{h \in G: h^2 = g\}|$, which is a class function. Thus we can write $\alpha$ as
$\alpha=\displaystyle \sum_{\chi \in Irr(g)}\langle\alpha,\chi\rangle\chi $, which follows from the fact that $\alpha$ is a class function and $Irr(G)$ is an orthonormal basis of the vector space $R(G)$ of class functions of $G$. We can see that equivalently what we want to show is that
$\displaystyle\sum_{\chi\in Irr(G)}\nu(\chi)\chi(1)=\alpha(1)$.
We have
\begin{split} \alpha(g) &= \displaystyle \sum_{\chi \in Irr(G)} \langle \alpha(g),\chi(g) \rangle \chi(g)\\ &= \displaystyle \sum_{\chi \in Irr(G)} \frac{1}{|G|} \displaystyle \sum_{g \in G} \alpha(g) \overline{\chi(g)}\chi(g) \end{split}
Also
\begin{split} \displaystyle\sum_{\chi\in Irr(G)}\nu(\chi)\chi(1) &= \displaystyle \sum_{\chi \in Irr(G)} \frac{1}{|G|} \displaystyle \sum_{g \in G} \chi(g^2)\chi(1)\\ &= \displaystyle \sum_{\chi \in Irr(G)} \langle \chi(g^2),\chi(1) \rangle \end{split} but I am really not sure where I should go from there to show that $(1)$ is true. I'd appreciate some hints.