G x H: Proving Cyclicity with GCD = 1

[hsong9]

Homework Statement

Let G and H be cyclic groups, with |G| = m and |H| = n. If gcd(m,n) =1, show that G x H is cyclic.

The Attempt at a Solution

Let a = (g,h) in G x H. Then |a| = lcm (|g|,|h|).
Since gcd(m,n)=1, then lcm (m,n) = mn.
Thus lcm (|g|,|h|) = lcm (m,n) = mn.
so <a> = G x H has mn elements and a cyclic group.
Right?

 

[Dick]

Basically ok. But you want to be a little more careful with your choice of g and h. It's not true that for every element g in G that |g|=|G|. E.g. |e|=1. And G could have nontrivial subgroups. You'd better make sure that g and h are generators.

 

[hsong9]

so.. you mean I have to make sense that for every element g in G that |g|=|G|?
If so..
Actually, the problem has Hint that [ G=<g> and H = <h>, show |(g,h)| = mn].
If consider this hint, my answer is ok?
or any other thing needs to prove?

 

[Dick]

The point is that at least one g in G satisfies |g|=|G|. Otherwise, would it be a cyclic group? That's what the proof is missing.