Let $G$ be a finite group. If $H\le G$ is a $p$ - subgroup then $[G:H]\equiv_p [N_G(H):H]$. The proof is fairly straightforward (using a previous lemma on finite $p$ - group actions)
The lemma used is: $H$ finite $p$ - group acting on a finite set $S$. If $S_0=\{x\in S:hx=x\quad\forall h\in H\}$ then $|S|\equiv_p|S_0|$
Can we generalize like this:
Let $G$ be a group and $H\le G$ a $p$ - subgroup. If $[G:H]$ is finite then so is $[N_G(H):H]$ and $[G:H]\equiv_p [N_G(H):H]$
" Proof " Let $S=\{gH:g\in G\}$ and $H$ acting on $S$ by $h\cdot(gH)=hgH$. Then $$S_0=\{gH\in S: hgH=gH\quad\forall h\in H\}=$$ $$=\{gH\in S: g^{-1}Hg\subseteq H\}$$
When $G$ was finite we could make $g^{-1}Hg\subseteq H$ into $g^{-1}Hg=H$ and get $|S_0|=[N_G(H):H]$ but now that $G$ may be infinite is there a way to fix it?
I know of a group-subgroup $H\le G$ such that $xHx^{-1}\subseteq H$ but $xHx^{-1}\ne H$, but $[G:H]$ is not finite so it does not work. Is there any simple counterexample?