Normal Subgroups in the Center of a Group

Question

This is from Dummit and Foote Abstract Algebra, page 134.

If $H$ is a subgroup of a group $G$, and $H \cong \mathbb{Z}_2$, then we can deduce from the automorphism group that $N_G(H) = C_G(H)$.

I understood this, since $H$ has an element of order 1 and 2, and they have to map to elements of the same order. So the Aut$(H) = 1$. From this, we know $1 = \frac{N_G(H)}{C_G(H)}$, so $N_G(H) = C_G(H)$.

If in addition, $H$ is a normal subgroup of $G$, then $H \subset Z(G)$

How do we know this?

Answer 1

If $N$ is normal, $N_G(H) = G$. Then, since $N_G(H)= C_G(H)$, it follows that $C_G(H)=G$. This,by definition, means that $H \subset Z(G)$.

Answer 2

I'll turn my comment into a "hint answer":

Let $x \in H$ be the unique nontrivial element of the normal subgroup $H \cong \mathbb{Z}_2$. Let $g \in G$ be any element. What can you say about $g x g^{-1}$?