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

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

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'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!

2. Originally Posted by hatsoff
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'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 $i_g\left[H\right]$ denote the inner automorphism of $H$. In other words, $i_g\left[H\right]=gHg^{-1}$. First prove that $i_g\left[H\right]$ is a subgroup of $G$ and then observe that the mapping $\phi:H\mapsto i_g\left[H\right]$ is a bijection. In other words we have that $gHg^{-1}$ is a subgroup with the same order as $H$. Thus, by hypothesis we have that $gHg^{-1}=H$ and the conclusion follows.

3. Thanks man! I guess I just have to better familiarize myself with this stuff.