# Thread: finite groups, normal subgroups, divisors

1. ## finite groups, normal subgroups, divisors

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.

2. Originally Posted by kathrynmath
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?

3. I'm confused. I've never heard the word automorphism

4. Originally Posted by kathrynmath
I'm confused. I've never heard the word automorphism
Show that $\displaystyle g_0 Hg_0^{-1}\leqslant G$ and has the same order as $\displaystyle H$.

5. Originally Posted by Drexel28
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?
Then H is normal