TITLE: Showing that the intersection of all subgroups of order $n$ is normal
QUESTION [11 upvotes]: 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.

REPLY [2 votes]: All group structure-related property is preserved in automophisms. So inner automophism just permutes conjugacy class. and intersection is preserved. so conjugation of elements of intersection subgroup is simply mapped onto intersection group.