$\mathbb{Z}_m \times \mathbb Z_n$ is cyclic if and only if $\gcd(m,n)=1$.
I know the question is a duplicate, I understand the proof for the case of two groups, but for some reason, I can't prove the general case of n groups by induction, and I couldn't find a proof either.
The proof in the case of two groups only uses the fact that gcd(a,b) divides a and b, but when we get to 3 numbers say a, b and c, I needed the fact that gcd(a,b)*gcd(a,c)*gcd(b,c) divides the product of any two of a,b,c, but that is not correct.
So, how to do the induction proof (I'm not interested in proving it by constructing an isomorphism, just using the same argument in the case of two groups)?
I'm stuck at the point where I need the fact that gcd(a,b)*gcd(a,c)*gcd(b,c) divides the product of any two of a,b,c. Is there a way to do the proof the same way as in the 2 groups case?
In the 2 groups case, we rais the generator to the power ab/gcd(a,b) and we get the identity, so we conclude that $ab\leq ab/gcd(a,b)$, so gcd(a,b)=1, where ab is the order of direct product group.
If you already have the result for 2 groups, then for 3 groups, you need two things:
Then, if $G_1, G_2, G_3$ are cyclic with pairwise relatively prime orders, so are $G_1\times G_2$ and $G_3$, and therefore $(G_1\times G_2)\times G_3$ is cyclic. And if the orders are not pairwise relatively prime, then either $G_1\times G_2$ is not cyclic (and so the product $(G_1 \times G_2)\times G_3$ is also not cyclic), or else $G_1\times G_2$ is cyclic but not of order relatively prime to $G_3$, so the product is agaian not cyclic.
The same logic here would then give a full induction proof for $n$ cyclic groups.
If you need help establishing the required fact about relatively prime numbers, the easiest way is to simply use the fundamental theorem of arithmetic to say that any prime dividing $c$ doesn't divide $a$ or $b$, and therefore doesn't divide $ab$.