How to prove that $[G:xHx^{-1}] = [G:H]$ given $H \le G$?

abstract-algebra group-theory

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 $\phi(g) = xgx^{-1}$.

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