Prove that $G \times H$ is cyclic $\iff$ $\gcd(m,n) =1$

Question

Let $G$ and $H$ be groups and let $a\in G$ and $b \in H$ such that $|a|=m$ and $|b|=n$ for some positive integers m and n. Now consider $(a,b) \in G\times H$.

(i) Prove that $$|(a,b)|=\text{lcm}(m,n).$$

(ii) If, in addition, $G$ and $H$ are cyclic with $G=<a>$ and $H=<b>$, so that $|G|=|<a>|=m$ and $|H|=|<b>|=n$. Prove that $$G\times H \text{ is cyclic }\iff \gcd(m,n)=1.$$

I need direction on how to get these proofs started and if there are any other, definitions or facts I could use to help prove both statements.

Answer 1

$(a,b)^p=(a^p,b^p)=(e_G,e_H)$ implies that $p$ is a multiple of m and n thus lcd(m,n) divides p.

If G =$<a>, H=<b>$ and gcd(m,n)=1, lcd(m,n)=mn is the cardinal of $G\times H$, the first part shows that the order of $(a,b)$ is lcd(m,n)=mn.

Suppose $G\times H$ is cyclic, its generator is $(a^u,b^v)$ there exists p such that $(a^u,b^u)^p=(a,e_H)$ this implies $pu=hm$ is a multiple of m and $pu=1+cn$ thus $hm=1+cn, hm-cn=1$ gcd(m,n)=1

Answer 2

For i), we have: $|G\cap H|$ is subgroup of $G$ and $H$, thus its cardinal is $\gcd(|G|,|H|)=\gcd(m,n)$

Now, using orbit-stabilizer theorem: $$|G\times H|=\frac{|G|\times|H|}{|G\cap H|}=\frac{mn}{\gcd(m,n)}=\text{lcm}(m,n)$$

For ii), the set $|G\times H|$ generated by $m$ elements of $G$ and $n$ elements of $H$ has at most $mn$ elements.

Suppose $(m,n)=1$, thus $|G\times H|=mn \Rightarrow$ all these elements are distinct. Thus, $G\times H=(<a>,<b>)$, thus cyclic.

If $G\times H$ is cyclic. Then proceed as @Tsemo Aristide's answer.