Is my proof of $C_G(H) \le N_G(H)$ correct? Let $x\in C_G(H)$. This means $xh = hx$ for all $h \in H$. Then $xH = Hx$ (This is that part I'm not so sure about). Hence, $x \in N_G(H)$, so that we have $C_G(H) \le N_G(H)$.

An other way to see this is as follows. $N_G(H)$ acts on $H$ By conjugation as automorphisms. The kernel of this action is precisely $C_G(H)$. So even $C_G(H) \unlhd N_G(H)$, and the quotient, $N_G(H)/C_G(H)$ embeds homomorphically in (so can be viewed as a subgroup of) Aut$(H)$.
