D+F 3.1.24

Show that the intersection of a subgroup $H$ and a normal subgroup $N$ is normal in $H$. i.e.
$H \leq G$ and $N \unlhd G \Longrightarrow H \cap N \unlhd H$.

Note: this is one of the results of the second/diamond isomorphism theorem.

$\textit{Proof:}$ Let $x\in{H \cap N}$. Since $H \leq G$, it is closed, so $\forall h\in{H}$, $hxh^{-1} \in{H}$. $N$ is normal, so it is invariant under conjugation. So when $x\in{H \cap N}$, $hxh^{-1} \in{N}$. We have $hxh^{-1}\in{H}$ and $hxh^{-1}\in{N}$, so $hxh^{-1} \in{H \cap N}$ $\therefore H \cap N \unlhd H \quad \blacksquare$