# Let $a$ be an element of a finite group $G$. Prove that the order of the cyclic subgroup divides the order of the group

Let $a$ be an element of a finite group $G$. Prove the order of the cyclic subgroup $\langle a\rangle=\{a^n$ | $n\in\mathbb{Z}\}$ divides the order of the group.

I know that for a finite, cyclic group $H=\langle b\rangle$ of order $k$, I have the following:

(i) The element $b^m$ generates $H$ if and only if $\gcd(m,k)=1$.

(ii) If $S$ is a subgroup of $H$, then $S=\langle b^m\rangle$ for some divisor $m$ of $k$.

(iii) If $d$ and $m$ are divisors of $k$, then $\langle b^d\rangle\subseteq\langle b^m\rangle$ if and only if $m|d$.

Can I use this, even if my group $G$ is only said to be finite? Can I draw something from this?

Any help/hints would be most welcome. ^_^

• Are you sure you are meant to do this generality? In that case, consider the cosets $g\left<a\right> = \{ga^n\mid n\in \mathbb{Z}\}$ and show that these are all either equal or disjoint and have the same sizes. – Tobias Kildetoft Sep 13 '13 at 7:35
• If you happen to know that $a^n = 1$ when $n = |G|$, then it is possible to do a proof from that using fairly elementary ideas. But I am not sure if that weaker version of Lagrange has a nice proof without proving the full theorem anyway. – Tobias Kildetoft Sep 13 '13 at 7:42

Hint: For two elements $g,h$, define $g \sim h$ if and only if $g^{-1}h \in \langle a \rangle$. You can check that this is an equivalence relation, hence $G$ decomposes into a disjoint union of equivalence classes $G = \coprod [g]$, where $[g]$ is the equivalence class of $g$.

Now determine how many elements there are in each of the equivalence classes and conclude.

(I assumed you are unfamiliar with Lagrange theorem)

• and you just proved Lagrange's Theorem. – Ittay Weiss Sep 13 '13 at 7:36
• @IttayWeiss: well, it was hard not to do it. I don't really get the point of doing this exercise only in the cyclic case :/ – zarathustra Sep 13 '13 at 7:38
• It's certainly a weird think to ask to show before establishing Lagrange. – Ittay Weiss Sep 13 '13 at 7:42

By Lagrange's theorem the order of any subgroup of a finite group divides the order of the big group. So certainly the order of a cyclic subgroup generated by any element divides the order of the big group.

• It seems unlikely that the OP is allowed to use lagrange. – Tobias Kildetoft Sep 13 '13 at 7:34
• I have to agree with the previous commenter. ^_^ – Desperate Fluffy Sep 13 '13 at 7:34
• I see. I don't know of any other way to prove it except to essentially prove Lagrange's Theorem first. – Ittay Weiss Sep 13 '13 at 7:37