TITLE: How to prove that $[G:xHx^{-1}] = [G:H]$ given $H \le G$?
QUESTION [6 upvotes]: The problem is as follows:

Let $G$ be a group and $H$ a subgroup of $G$ (i.e., $H \le G$); let $x$ be any element of $G$ (i.e., $x \in G$). To prove that $[G:xHx^{-1}] = [G:H]$.

I am able to justify that the $xHx^{-1}$ is indeed a subgroup of $G$ (i.e., $xHx^{-1} \le G$) and that $|xHx^{-1}| = |H|$. 
To prove that $[G:xHx^{-1}] = [G:H]$, it is necessary to provide a bijective function $F$ between the set of (different, left) cosets of $H$ (denoted $S = \{ aH : a \in G \}$) and that of $xHx^{-1}$ (denoted $T = \{ a (xHx^{-1}) : a \in G\} $). 
I have tried the function $F(aH) = a(xHx^{-1})$, but failed to show it is bijective. Could someone offer a hint of the feasible function?

REPLY [6 votes]: Try the map $aH \mapsto x(aH)x^{-1}$. Note that $x(aH)x^{-1} = (xax^{-1})xHx^{-1}$.
More generally, if $\phi$ is an automorphism of $G$ and $H \leq G$, then $[G:H] = [G : \phi(H)]$ since $aH \mapsto \phi(aH)$ is a bijection between left cosets of $H$ and $\phi(H)$. Your problem is the case where $\phi(g) = xgx^{-1}$.