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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. ^_^

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    $\begingroup$ 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
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    $\begingroup$ 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
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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)

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  • $\begingroup$ 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
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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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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