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 $<a^k>=<a^{gcd(n,k)}>$ 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 $G = <a>$ 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?
Thanks in advance.
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 $n$.