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.

  • $\begingroup$ If $G$ is generated by $a$, then $\vert G \vert = \vert a \vert$. $\endgroup$ Jul 16, 2019 at 0:07
  • $\begingroup$ @RobertShore I made a typo there. I meant to say any element of the group, and not just $a$. $\endgroup$
    – Tim
    Jul 16, 2019 at 0:09
  • $\begingroup$ Why do you object to using Lagrange's theorem? $\endgroup$
    – Rob Arthan
    Jul 16, 2019 at 0:15
  • $\begingroup$ 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. $\endgroup$
    – fleablood
    Jul 16, 2019 at 0:19
  • $\begingroup$ @fleablood Because what I'm reading hasn't covered Lagrange's Theorem yet, and we need to prove it directly from the previous theorem, hence it being a corollary. $\endgroup$
    – Tim
    Jul 16, 2019 at 0:23

1 Answer 1


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 $n$.


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