# $\overline{N_P(H)} = N_\overline{P}(\overline{H})$?

Let $P$ be a $p$-group. For a subgroup $K$ of P containing $Z(P)$, by $\overline{K}$ we denote the quotient $K/Z(P)$. Let $H$ be a proper subgroup of $P$ containing $Z(P)$.

The proof of the first theorem in section 6.1 from the book Abstract Algebra by Dummit and Foote uses the following equality:

$\overline{N_P(H)} = N_\overline{P}(\overline{H})$

They say it follows directly from Lattice Isomorphism theorem. I can't see why. Any thoughts? I am looking for a formal argument using Lattice Isomorphism theorem.

In general, if $H$ is a subgroup of $G$, and $N$ a normal subgroup, with $N \subseteq H$, then $N_\overline{G}( \overline H)=\overline{N_G(H)}$, where $\overline \cdot$ is modding out by $N$.