A group of order $2p^n$ has a unique Sylow p-subgroup.

Question

I want to prove that if $G$ is a group of order $2p^n$ with $p$ a prime number$\neq 2$, then $G$ has a a normal $p$ subgroup.

By the first theorem of Sylow there exist a Sylow p-subgroup of order $p^n$, now if I prove that this subgroup is unique, by the second theorem of Sylow, it will be normal.

I tried proving it by assuming that there are $2$ p-subgroups and getting to a contradiction using the third theorem without succeeding.

Answer 1

Let $S$ be a Sylow $p$-subgroup of $G$. Since $|G:S| = 2$, $S$ is normal in $G$. Consequently, $S$ is fixed by conjugation with elements of $G$. Since (second Sylow) $G$ acts transitively by conjugation on the Sylow $p$-subgroups, $S$ is the only Sylow $p$-subgroup.

Answer 2

The cardinal of the set $S$ of Sylow $p$-group divides $2$. If it $2$, $G$ acts transitively by conjugation on $S$, the cardinal of the kernel of this action is $p^n$. So this kernel is a normal Sylow group and it is unique since the Sylow subgroups are conjugated.