Let G be a finite group of odd order n. Suppose that g is in a real conjugacy class C (a real conjugacy class C is a conjugacy class so that $C$ = $C^{-1}$ ). So $h^{−1}gh = g^{−1}$ for some h $\in G$.
Why can we follow from that: $h^{−2}gh^2 = g$ ?
This means $h^2 \in C_G(g)$. Since n is odd, the order of h is odd, say 2k + 1. It follows that h = ($h^2)^{k+1}$
Why can we follow from that: h $\in C_G(g)$ ?
Thankfull for any help.
On
Rewrite the equality as $h^{-1}g^{-1}h=g$ and now(here’s the trick!) replace the g in $h^{-1}g^{-1}h$ with $h^{-1}g^{-1}h$ (because g and $h^{-1}g^{-1}h$ are equal). So we get that $g=h^{-1}g^{-1}h=h^{-1}(h^{-1}g^{-1}h)^{-1}h=h^{-2}gh^2$ keep replacing g inside $h^{-2}gh^2$ with $h^{-1}g^{-1}h$ to eventually get that $h^{-2k}gh^{2k}=g$, but $h^{2k}=h^{-1}$ so by the last equality $h$ and $g$ commute
The order of $|G|$ is odd, say $G=2m-1$, for some positive integer $m$. Hence by Lagrange $h^{|G|}=1=(h^2)^m \cdot h^{-1}$. It follows that $h=(h^2)^m$, so if $h^2 \in C_G(g)$, then $h \in C_G(g)$.