# A cyclic group of order “rs” where (r, s) = 1

I was given this question and I'm not really sure how to approach this...

Assume $(r,s) = 1$. Prove that If $G = \langle x\rangle$ has order $rs$, then $x = yz$, where $y$ has order $r$, $z$ has order $s$, and $y$ and $z$ commute; also prove that the factors $y$ and $z$ are unique.

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What does unique mean ? –  Amr Nov 17 '12 at 10:10
@Amr: Do you think he is trying to prove $y$ and $z$ are unique? Because the sentence has not looked like a a question! –  Babak S. Nov 17 '12 at 10:12
I don't know. But if he was trying to do so then what about $x=ya^{-1}az$ –  Amr Nov 17 '12 at 10:15
Yes, sorry, it wasn't very clear what the question is... I added "prove" in the correct place. –  amirbd89 Nov 17 '12 at 10:24
The question statement lacks clarity. First, there is the word "cyclic" in the header, but it is doesn't appear in the question statement itself. So it's confusing: is $G$ cyclic after all? Second, $x$ is undefined. Do we have to prove that for all $x$ in $G$ there exists a unique pair $(y,z)$ with the desired properties? –  Dan Shved Nov 17 '12 at 10:30

From $(r,s)=1$ we find integers $n,m$ with $nr+ms=1$. Let $y=x^{ms}$, $z=x^{nr}$. Then $yz=x^{ms+nr}=x$. The fact that $x$ and $y$ commute is trivial because the cyclic group $G$ is abelian. Also, we have $y^r=(x^m)^{rs}=1$, $z^s=(x^n)^{rs}=1$, hence the orders are at least divisors of $r$ and $s$, respectively. If the actual orders are $r'|r$ and $s'|s$, then $x^{r's'}=y^{r's'}z^{r's'}=1$, hence $r's'$ is a positive multiple of $rs$, hence at least $rs$. We conclude that $r'=r$, $s'=s$. Finally, assume we have another solution $x=y'z'$ with the required properties. Then $z^r=y^rz^r=x^r = y'^rz'^r=z'^r$ implies $z=z^{nr+ms}={z^r}^n{z^s}^m={z^r}^n={z'^r}^n={z'^r}^n{z'^s}^m=z'^{nr+ms}=z'$ and similarly $y=y'$.

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The cyclic group $G$ is isomorphic to the additive group of $\Bbb Z/rs\Bbb Z$, so this is just the Chinese remainder theorem for the coprime moduli $r$ and $s$ (in the statement for rings $\Bbb Z/n\Bbb Z$, but only considering their additive structure).

Concretely, the elements of $G$ of order dividing $r$ are generated by $x^s$ and vice versa, and among the $r$ elements of order dividing $r$ there is one, say $y$, such that $z=y^{-1}x$ has order dividing $s$. One has $x=yz$ and $y,z$ commute (they are both in the group $G$ generated by $x$); if either the order of $y$ were a strict divisor of $r$ or the order of $z$ were a strict divisor of $s$ then it would follows that the order of $x$ is a strict divisor of $rs$, which is false, so the orders of $y,z$ are respectively exactly $r,s$. Concretely you can find $y,z$ by writing $1=\gcd(r,s)=ar+bs$ using the extended Euclidean algorithm; then $y=x^{bs}$ and $z=x^{ar}$.

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Here is a hint: you can set $y=x^{sn}$ and $z=x^{rm}$. To find the appropriate $n$ and $m$ you can use Bezout's identity.

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