Direct product of finite cyclic groups of coprime orders The Question is this:
How many generators are there of the group $G\times H$,  if $G$ and $H$ are cyclic groups of order $m$ and $n$, which are coprime?
Let's say that $G$ is generated by $g$, and $H$ by $h$.
Here I already proved that $(g,h)$ is generator of $G\times H$ but I can't come up with another one. So does $G\times H$ really have any other generator? 
Thanks!
 A: A group $G\times H$ is cyclic if and only if, given $|G| = m$ and $|H| = n$, $\gcd(m,n) = 1$. And if $\gcd(m,n) = 1$, then the order of $G\times H$ is $\text{lcm}(|G||H|) = mn$
A cyclic group is by definition, generated by a single element, so if you've found that $ (g, h) $ generates $G\times H$, then you've shown that $\langle (g, h)\rangle = G\times H$. 
Indeed, any integer relatively prime to the modulus of a cyclic group will additively generate the cyclic group. That is, the number of elements generating the cyclic group $G = \phi(m)$, the number of elements generating the cyclic group $H = \phi(n)$. So the number of elements generating $$G\times H = \phi(m)\cdot\phi(n)$$
A: Hint: If you've proven that $G\times H$ is cyclic, then it must be of order $mn$. What do the generators of a cyclic group of order $mn$ look like? Think of a generator with a relatively prime power to the order...
As a bonus, we can even say it has $\varphi(mn)=\varphi(m)\cdot\varphi(n)$ generators, where $\varphi$ is the Euler phi function.
A: Hint: If a group of order $n$ has a generator then it has precisely $\phi(n)$ generators.
