If H is a subgroup of G, show that H is normal to the normalizer $\displaystyle (N(H))$
$\displaystyle H$ is clearly a subgroup of $\displaystyle N_G(H)$.
To see that $\displaystyle H \lhd N_G(H)$, take any $\displaystyle x \in N_G(H)$. By definition, $\displaystyle xHx^{-1}=H$. Since this is true for all $\displaystyle x \in N_G(H)$, the statement follows.