Let $\displaystyle G $ and $\displaystyle H $ be cyclic groups. If $\displaystyle |G| = m $ and $\displaystyle |H| = n $ and $\displaystyle (m,n) = 1 $, show $\displaystyle G \times H $ is cyclic. (Hint: show $\displaystyle |(g,h)| = mn $.)
Let $\displaystyle G $ and $\displaystyle H $ be cyclic groups. If $\displaystyle |G| = m $ and $\displaystyle |H| = n $ and $\displaystyle (m,n) = 1 $, show $\displaystyle G \times H $ is cyclic. (Hint: show $\displaystyle |(g,h)| = mn $.)
Suppose G=<g>, H=<h>, and |(g,h)|=k. Then $\displaystyle (e_G,e_H)=(g,h)^k=(g^k,h^k)$ so $\displaystyle g^k=e_G, h^k=e_H$. But then from the properties of the order, $\displaystyle m,n$ both divide $\displaystyle k$,
so k is a common multiple. Then what we want is the least common multiple of the orders, and that is $\displaystyle \dfrac{mn}{(m,n)}=mn=|G\times H|$ thus we have found the generator, and the result follows