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.
 A: 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.
A: 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 $y=y'$.
A: 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 $z=x^{ar}$.
