[Math] Showing that the intersection of all subgroups of order $n$ is normal

Question

I have been thinking about this problem:

"Suppose that a group $G$ has a subgroup of order $n$. Prove that the intersection of all subgroups of $G$ of order $n$ is a normal subgroup of $G$."

Unfortunately, I don't have much of a start. I know that the intersection will be a subgroup of each of the subgroups of order $n$, and thus will have order dividing $n$. I also know that conjugation preserves the order of elements, which suggests that $xNx^{-1}$ could be a subset of $N$ ($N$ being the intersection of all subgroups of order $n$), but I don't see why conjugation necessarily takes elements of $N$ to other elements of $N$.

I'd really appreciate a hint on how to go about showing this. Thanks.

Answer 1

Hint: Conjugation permutes the set of subgroups of order $n$.