Prove that if $H \leq G, M \trianglelefteq G, M \leq H,$ and $xHx^{-1}/M =H/M$, then $xM \in N_G (H)/M$.
This question arises from trying to prove that $\overline{N_G(H)} = N_\overline{G}(\overline{H})$. I proved the $\subset $ direction and now I am trying to prove the $\supset $ direction by solving the above lemma. Now it suffices to find an element $y \in N_G (H)$ such that $y^{-1}x \in M $ but I am not sure how to find such an element $y $. If we have that $M \leq xHx^{-1}$ then by the lattice isomorphism theorem we would have that $xHx^{-1} = H $ so that we may take $y=x $ but I am not sure with the case when $M \not\leq xHx^{-1}$.