A problem which I've found quite difficult it is the following:
Let $G$ be a group with order $2^n k$ with $k$ odd, and suppose there is an element of order $2^n$ (i.e. the Sylow-$2$ subgroup is cyclic). Prove that the elements of odd order in $G$ form a normal subgroup of order $k$.
One major difficulty is in $G$ not being abelian. I've seen one proof of it, which used induction on $n$ and the group action of $G$ on itself by multiplication, noting that a permutation $\lambda_g$ is odd iff $2^n$ divides the order of $g$.
However, I've not been able to find any other methods to prove it, even after learning more group theory than I had learned at the time. I thought about using the Sylow theorems, but that didn't result in anything.
Can anyone think of other ways to prove this result? Full proofs aren't necessary; any directions which are or could be fruitful are appreciated.