Originally Posted by

**Tachyon** This is what I have so far:

Let $\displaystyle Na \in G/N$ have order *n* for some $\displaystyle a \in G$. Then $\displaystyle (Na)^n = Na^n = e$, so $\displaystyle a^n \in N$. Since N is finite, let $\displaystyle ord(N) = m$. Then $\displaystyle (a^n)^m = e$, so $\displaystyle ord(a)|nm$.

When I pursued this line of reasoning, I was trying to show that $\displaystyle n|ord(a)$, and thus $\displaystyle n = ord(a)$. However, I'm stuck here. Have I made a mistake somewhere, or am I missing something?

Thanks for any insights.