Simple proof that the order of an element of a group divides order of the group itself

The following theorem have been given to us:

"Let $$G$$ be a group and $$a \in G$$ be an element of it. Let $$k \in \mathbb Z^+$$ be a positive integer. Suppose that $$|a|=n$$. Then $$=$$ and $$|a^k| = \frac{n}{gcd(n,k)}$$."

I would like to know how to show from this theorem as a corollary, that for any finite cyclic group $$G = $$ generated by arbitrary element $$a \in G$$, the order of any element $$x \in G$$ divides the order of the group $$G$$. Most proofs I've seen online are based around Lagrange's Theorem, but in this case, I don't want to be using Lagrange's Theorem.

Is there a very simple and straightforward way to prove this?

• If $G$ is generated by $a$, then $\vert G \vert = \vert a \vert$. – Robert Shore Jul 16 at 0:07
• @RobertShore I made a typo there. I meant to say any element of the group, and not just $a$. – Tim Jul 16 at 0:09
• The $<x>$ is a subgroup of $<a>$ and $<a>$ is a subgroup of $G$. By Lagrange theorem $|x|$ divides $|<a>|$ and $|<a>|$ divides $|G|$. So $|x|$ divides $|G|$. QED. Now tell me why you don't want to use Lagrange theory. And tell me why you want to prove something that Lagrange theory makes utterly obvious. – fleablood Jul 16 at 0:19
This is immediate from the theorem that you're given. If $$a$$ generates $$G$$ and $$\vert a \vert = n$$, then $$\vert G \vert =n$$ and the Theorem tells you $$\vert a^k \vert = \frac{n}{\gcd (n, k)}$$, so $$\vert a^k \vert \gcd(n, k) = n$$ and $$\vert a^k \vert$$ divides $$n$$.