In a finite group, the equation $x^m=e$ has $m$ solutions for each positive $m$ that divides the order of the group

Question

Show that in a finite cyclic group $G$ of order $n$ written multiplicatively the equation $x^m=e$ has $m$ solutions $x\in G$ for each positive $m$ that divides $n$

I am having trouble understanding how to do this without using Lagrange theorem. I could really use some help here

Something is wrong with the statement. Check it. For example in the Klein four group the equation $x^2=e$ has four solutions, not two as you are claiming. — Jyrki Lahtonen, Sep 22 '14 at 07:16
As a general rule, you should put the question in the body of the post. The title should just indicate roughly what it is about. — Derek Holt, Sep 22 '14 at 16:08
@DerekHolt Actually, it's not a problem if the title indicates exactly what the question is. Saves people the trouble of opening it. But I agree the question should contain the question. — , Sep 22 '14 at 16:50
I wish I knew how to prove the case $m=n$ without using Lagrange's theorem. — lhf, Sep 22 '14 at 17:17
@JyrkiLahtonen You may have missed that the group is cyclic. — Yuval Filmus, Sep 23 '14 at 00:24
@YuvalFilmus, in the original version of the question that key bit was left out :-) — Jyrki Lahtonen, Sep 23 '14 at 05:04

score 1 · Answer 1 · answered Sep 23 '14 at 00:26

We can assume that $G = \mathbb{Z}_n$, the group of integers modulo $n$ under addition. So that we don't get confused, we represent the group operation of $\mathbb{Z}_n$ as addition. Your question asks for the number of solutions of the equation $mx \equiv 0 \pmod{n}$, where $0 \leq x < n$. Since $m\mid n$, we can write each such $x$ in the form $y(n/m)+z$, where $0 \leq z < n/m$ and $0 \leq y < m$. We have $mx = ny+mz \equiv mz \pmod{n}$, so the solutions are $z = 0$ and $0 \leq y < m$, of which there are exactly $m$.

In a finite group, the equation $x^m=e$ has $m$ solutions for each positive $m$ that divides the order of the group

1 Answers1