# finite groups, normal subgroups, divisors

• Dec 2nd 2010, 09:08 AM
kathrynmath
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.
• Dec 2nd 2010, 10:48 AM
Drexel28
Quote:

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 $g_0\in G$ the inner automorphism $i_{g_0}:G\to G:g\mapsto g_0gg_0^{-1}$ is...well..an automorphism. Thus, since $H\leqslant G$ and $|H|=n$ then ${g_0}H g_0^{-1}=i_{g_0}(H)\leqslant G$ and $|g_0 Hg_0^{-1}|=n$................so?
• Dec 2nd 2010, 03:02 PM
kathrynmath
I'm confused. I've never heard the word automorphism
• Dec 2nd 2010, 05:22 PM
Drexel28
Quote:

Originally Posted by kathrynmath
I'm confused. I've never heard the word automorphism

Show that $g_0 Hg_0^{-1}\leqslant G$ and has the same order as $H$.
• Dec 2nd 2010, 05:58 PM
kathrynmath
Quote:

Originally Posted by Drexel28
Merely recall that for any $g_0\in G$ the inner automorphism $i_{g_0}:G\to G:g\mapsto g_0gg_0^{-1}$ is...well..an automorphism. Thus, since $H\leqslant G$ and $|H|=n$ then ${g_0}H g_0^{-1}=i_{g_0}(H)\leqslant G$ and $|g_0 Hg_0^{-1}|=n$................so?

Then H is normal