Normal subgroups of order 2 are central

Let H be a subgroup of order 2 in G. Show that N_G(H) = C_G(H). Deduce that if N_G(H) = G, then H \leq Z(G).

Say H = \{ 1, h \}.

We already know that C_G(H) \subseteq N_G(H). Now suppose x \in N_G(H); then \{ x1x^{-1}, xhx^{-1} \} = \{ 1, h \}. Clearly, then, we have xhx^{-1} = h. Thus x \in C_G(H). Hence N_G(H) = C_G(H).

If N_G(H) = G, we have C_G(H) = G. Then ghg^{-1} = h for all h \in H, so that gh = hg for all h \in H, and thus H \leq Z(G).

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