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. ^_^

  • 1
    $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Sep 13 '13 at 7:35
  • 1
    $\begingroup$ 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. $\endgroup$ – 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)

  • $\begingroup$ and you just proved Lagrange's Theorem. $\endgroup$ – Ittay Weiss Sep 13 '13 at 7:36
  • $\begingroup$ @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 :/ $\endgroup$ – zarathustra Sep 13 '13 at 7:38
  • $\begingroup$ It's certainly a weird think to ask to show before establishing Lagrange. $\endgroup$ – 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.

  • 1
    $\begingroup$ It seems unlikely that the OP is allowed to use lagrange. $\endgroup$ – Tobias Kildetoft Sep 13 '13 at 7:34
  • $\begingroup$ I have to agree with the previous commenter. ^_^ $\endgroup$ – Desperate Fluffy Sep 13 '13 at 7:34
  • 1
    $\begingroup$ I see. I don't know of any other way to prove it except to essentially prove Lagrange's Theorem first. $\endgroup$ – Ittay Weiss Sep 13 '13 at 7:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.