Prove that if $g^{n} \in H$, n divides |g|

Let H be a subgroup of a finite group G. Suppose that g belongs to G and n is the smallest positive integer such that $g^{n} \in H$. Prove that n divides |g|.

I couldn't get anywhere with this problem due to the condition "n is the smallest positive integer such that $g^{n} \in H$." I tried letting |g|=m, then I could show $a^{m-n}\in H$. I also realized $a^{kn} \in H\ (k\in N^{+})$ as well, since $H$ contains the identity element. But how do these things lead to anywhere?

• If you showed $a^{m-n}\in H$, you can show $a^{m-2n}\in H$, and $a^{m-3n}\in H$, etc. How far can this go? – Prometheus Sep 19 '14 at 7:10
• It can go up to $a^{m-(m-1)n}=a^{n} \in H$ since both are inverse elements of $a^{-n}$. But how does this imply $m-(m-1)n$ is divisible by $n$, or m is divisible by n? – user177196 Sep 19 '14 at 7:25
• Consider only positive powers of $a$ that belong to $H$. You know $a^m$ does, and $n$ is the smallest positive power. You can show $a^{m-qn}$ belongs to $H$ for any integer $q$. How small a positive integer can you make $m-qn$ be? – Prometheus Sep 19 '14 at 7:36
• Thank you for your great hint! m-qn = n, which implies n divides m. Is that right? – user177196 Sep 19 '14 at 7:57
• Many thanks for your help! – user177196 Sep 19 '14 at 8:12

If $n$ didn't divide $|g|$ then $|g| = nq + r$ for some $q \in \mathbb{N}$ and $r \in (0,n)$. Since $e=g^{|g|} = g^{nq}g^{r}$, derive a contradiction with the minimality of $n$.