Math Help - One more normalizer q

1. One more normalizer q

If H is a subgroup of G, show that H is normal to the normalizer $(N(H))$

2. $H$ is clearly a subgroup of $N_G(H)$.

To see that $H \lhd N_G(H)$, take any $x \in N_G(H)$. By definition, $xHx^{-1}=H$. Since this is true for all $x \in N_G(H)$, the statement follows.