# [SOLVED] Prove a finite subgroup is normal, given order conditions.

• Jan 20th 2010, 11:54 AM
hatsoff
[SOLVED] Prove a finite subgroup is normal, given order conditions.
Let $\displaystyle G$ be a finite group, and let $\displaystyle n$ be a divisor of $\displaystyle |G|$. Show that if $\displaystyle H$ is the only subgroup of $\displaystyle G$ of order $\displaystyle n$, then $\displaystyle H$ must be normal in $\displaystyle G$.

I'm not really sure what to do with this. It's been a while since I worked with cyclic groups or Lagrange's theorem, and so I may be forgetting some essential tricks.

Any help would be much appreciated!
• Jan 20th 2010, 12:24 PM
Drexel28
Quote:

Originally Posted by hatsoff
Let $\displaystyle G$ be a finite group, and let $\displaystyle n$ be a divisor of $\displaystyle |G|$. Show that if $\displaystyle H$ is the only subgroup of $\displaystyle G$ of order $\displaystyle n$, then $\displaystyle H$ must be normal in $\displaystyle G$.

I'm not really sure what to do with this. It's been a while since I worked with cyclic groups or Lagrange's theorem, and so I may be forgetting some essential tricks.

Any help would be much appreciated!

Let $\displaystyle i_g\left[H\right]$ denote the inner automorphism of $\displaystyle H$. In other words, $\displaystyle i_g\left[H\right]=gHg^{-1}$. First prove that $\displaystyle i_g\left[H\right]$ is a subgroup of $\displaystyle G$ and then observe that the mapping $\displaystyle \phi:H\mapsto i_g\left[H\right]$ is a bijection. In other words we have that $\displaystyle gHg^{-1}$ is a subgroup with the same order as $\displaystyle H$. Thus, by hypothesis we have that $\displaystyle gHg^{-1}=H$ and the conclusion follows.
• Jan 20th 2010, 12:52 PM
hatsoff
Thanks man! I guess I just have to better familiarize myself with this stuff.