## The normalizer of a subgroup is the join of all subgroups in which it is normal

Let $G$ be a group and $H \leq G$. Prove that if $N$ is a normal subgroup of $H$ then $H \leq N_G(N)$. Deduce that $N_G(N)$ is the largest subgroup of $G$ in which $N$ is normal (i.e. the join of all subgroups $H$ for which $N \leq H$ is normal).

Note that if $h \in H$, then $hNh^{-1} = N$ since $N$ is normal in $H$. Hence $H \subseteq N_G(N)$, so that $H \leq N_G(N)$ by a lemma to a previous exercise.

Now if we consider the set $\mathcal{A} = \{ H \leq G \ |\ N \vartriangleleft H \}$, note that $N_G(N) \in \mathcal{A}$ and if $H \in \mathcal{A}$ then $H \leq N_G(N)$. Hence $N_G(N)$ is $\subseteq$-greatest among the subgroups of $G$ in which $N$ is normal.