# Showing that the intersection of all subgroups of order $n$ is normal

"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.

• Fix an $x$, then the map $a \rightarrow xax^{-1}$ is an isomorphism. What happens to a subgroup of order $n$ under an isomorphism? – RghtHndSd Aug 21 '13 at 3:13
• Note that you have a Characteristic subgroup - which is a stronger property, because every automorphism (including conjugation) permutes the subgroups of order $n$. – Mark Bennet Aug 21 '13 at 4:00
• Could someone take a look at my comment under the answer? I'd like a response. – Alex Petzke Aug 24 '13 at 17:45
• A nice answer to this was given in February 2010 here: answers.yahoo.com/question/index?qid=20100219223444AAfdd2N – a student Mar 4 '16 at 6:40

Hint: Conjugation permutes the set of subgroups of order $n$.
• Alright, I see that. I'm still not quite seeing it though. Not all subgroups of order $n$ have to be isomorphic, so I guess taking the intersection kind of accounts for that. So then restricting conjugation to the intersection results in an automorphism so that $xNx^{-1} \subseteq N$? – Alex Petzke Aug 23 '13 at 13:25