Exercise 2.2.10 (Normalizer and centralizer of a subgroup of order $2$)

Proof. We know that $C_G(H)\subseteq N_G(H)$ is always true (see p. 50). Suppose that $|H|=2$, so that $H=\{1,a\}$ where $a\in G,a\ne 1$.

If $g\in N_G(H)$, then $\mathit{gH}{g}^{-1}\subseteq H$, therefore $\mathit{ga}{g}^{-1}\in \{1,a\}$. If $\mathit{ga}{g}^{-1}=1$ then $a=1$. This is impossible, so $\mathit{ga}{g}^{-1}=a$. Then the equalities $\mathit{ga}=\mathit{ag}$ and $\mathit{ge}=\mathit{eg}$ show that $g\in C_G(H)$. This shows $N_G(H)\subseteq C_G(H)$, so

If $N_G(H)=G$ (that is if $H$ is normal in $G$), then $G=C_G(H)$, i.e. every element of $G$ commutes with every element of $H$. Hence $H\le Z(G)$.

(For instance $H=\{1,-1\}\le Q_8=G$ satisfies $N_G(H)=C_G(H)=G$ and $H\le Z(G)$.) □