Order of an element in a cyclic group

Question

Without using Lagrange's Theorem or permutations, why must the order of an element in a cyclic group divide the order of the group?

I think it has something to do with the fact that every element in a cyclic group can be written as a form of the generator.

Use the good ol divison algorithm for integers. – Oiler Oct 04 '16 at 20:37 — Oiler, Oct 04 '16 at 20:37

M.U. · Accepted Answer · 2016-10-04T20:34:20.977

0

Let $z \in G$ be a generator of the cyclic group $G$ of order $n$. Let $x \in G$ be arbitrary, i.e. $x = z^k$ for some $0 < k < n$ and let $l = ord(x)$. You want $l | n$. So by euclidean division we can write $n = pl + q$ for $p,q$ integers and $0 \leq q < l$. Clearly we have $e = x^n = x^{pl+q} = x^{pl} x^q$. You have to show that $q = 0$, can you do it?

edited Oct 04 '16 at 20:34

answered Oct 04 '16 at 19:00

M.U.

2,013

Not really, I don't know what to do with that equation. Also what does p stand for? – Maria Oct 04 '16 at 19:46
@Maria ... see the edit. Is it now clearer? – M.U. Oct 04 '16 at 20:34
Yes! Thank you so much – Maria Oct 05 '16 at 03:49

Order of an element in a cyclic group

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