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).

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