# Finite subgroups of unique orders are normal

1. Nov 15, 2009

### redone632

1. The problem statement, all variables and given/known data
Let G be a finite group and H a subgroup of G having order m. Show that if H is the only subgroup of order m in G, then H is normal in G.

2. Relevant equations
A subgroup $H$ of $G$ is normal in $G$ if and only if $xHx^{-1} \subseteq H \forall x \in G$

3. The attempt at a solution
Suppose that H is the only subgroup of order m. Then elements in G\H cannot have order m.

If $x \in H$ then clearly, $xHx^{-1} \subseteq H$

If $x \notin H$ that is, $x \in G \backslash H$ so $|x| \neq m$ then ...

This is where I'm not seeing anything. Any help to point me in the right direction would be greatly appreciated!

2. Nov 15, 2009

### Dick

Is xHx^(-1) a subgroup? How many elements does it have?

3. Nov 15, 2009

### redone632

Since H is a subgroup of order m, xHx^(-1) will be a subgroup of order m. Since H is the only subgroup of order m there will be no other subgroup such that xKx^(-1) is of order m. Therefore, xHx^(-1) = H.

I think that's it, but I can't connect the dots to the last statement.

4. Nov 15, 2009

### Dick

I don't see the problem. If xHx^(-1) is a subgroup of order m, and H is the ONLY subgroup of order m, then xHx^(-1) must be the same subgroup as H. What's K got to do with it?

5. Nov 15, 2009

### redone632

Oh wow, I swear I miss the most obvious things. Thanks for your help again!