Let G be a finite group and let n be a divisor of |G|. Show that if H is the only subgroup of G of order n, then H must be normal in G.
I need to show gh^g-1 is in H.
We have |G|=nb since n is a divisor.
Merely recall that for any $\displaystyle g_0\in G$ the inner automorphism $\displaystyle i_{g_0}:G\to G:g\mapsto g_0gg_0^{-1}$ is...well..an automorphism. Thus, since $\displaystyle H\leqslant G$ and $\displaystyle |H|=n$ then $\displaystyle {g_0}H g_0^{-1}=i_{g_0}(H)\leqslant G$ and $\displaystyle |g_0 Hg_0^{-1}|=n$................so?