# Normal Subgroups in the Center of a Group

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?

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

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 $$g x g^{-1}$$?
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 $$H \subset Z(G)$$.