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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?

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  • $\begingroup$ A key observation, I suppose, is that $H=N_G(H)$ if $H$ is normal. $\endgroup$ – Shaun Sep 28 at 15:31
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    $\begingroup$ 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 $g x g^{-1}$? $\endgroup$ – Lee Mosher Sep 28 at 15:43
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    $\begingroup$ @Shaun if $H$ is normal, isn't $N_G(H) = G$? $\endgroup$ – Jess Sep 28 at 15:59
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    $\begingroup$ Yes indeed, @Jess; thank you for correcting me! $\endgroup$ – Shaun Sep 28 at 16:04
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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 $g x g^{-1}$?

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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 $H \subset Z(G)$.

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