## Every normalizer contains the group center

Let $G$ be a group. Prove that $Z(G) \leq N_G(A)$ for any subset $A \subseteq G$.

If $A = \emptyset$, the statement is vacuously true since $N_G(A) = G$. If $A$ is not empty, let $x \in Z(G)$. Then $xax^{-1} = a$ for all $a \in A$, so that $xAx^{-1} = A$. Hence $x \in N_G(A)$.