Group Direct Product of Cyclic Groups

Theorem
Let $G$ and $H$ both be finite cyclic groups with orders $n = \left\vert{G}\right\vert$ and $m = \left\vert{H}\right\vert$ respectively.

Then their group direct product $G \times H$ is cyclic if and only if $g$ and $h$ are coprime, i.e. $g \perp h$..

Proof
Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively.

Necessary condition
Suppose:


 * $(1): \quad \left\vert{G}\right\vert = n, G = \left \langle {x} \right \rangle$
 * $(2): \quad \left\vert{H}\right\vert = m, H = \left \langle {y} \right \rangle$
 * $(3): \quad m \perp n$

Then:

But then $\left({x, y}\right)^{n m} = e_{G \times H} = \left({x^{n m}, y^{n m}}\right)$ and thus $k \mathop \backslash n m$.

So $\left\vert{\left({x, y}\right)}\right\vert = n m \implies \left \langle{\left({x, y}\right)}\right \rangle = G \times H$.

Sufficient condition
Suppose that $G \times H$ is cyclic.

Let $\left({ x,y }\right)$ be a generator of $G \times H$.

By Cardinality of Cartesian Product the order of $G \times H$ is $\left|{ G }\right|\cdot \left|{ H }\right| = gh$.

Therefore by Order of Generator is Order of Group $\left|{ \left({ x,y }\right) }\right| = gh$.

On the other hand, by Order of Group Element in Group Direct Product we have:
 * $\left|{ \left({ x,y }\right) }\right| = \operatorname{lcm}\left\{{(|x|,|y|}\right\}$

Next we claim that $x$ generates $G$.

Indeed, if $x' \in G$, then $\left({ x',e_H }\right) \in G \times H$, so there exists $k \in \N$ such that:
 * $\left({ x,y }\right)^k = \left({ x^k,y^k }\right) = \left({ x',e_H }\right)$

and therefore $x^k = x'$.

Thus the powers of $x$ generate the whole group $G$,

In the same way, it is seen that $y$ generates $H$.

Therefore by Order of Generator is Order of Group it follows that $\left\vert{ x }\right\vert = g$ and $\left\vert{ y }\right\vert = h$.

Thus we have that:
 * $gh = \left|{ \left({ x,y }\right) }\right| = \operatorname{lcm}\left\{{ g,h }\right\}$

Moreover by Product of GCD and LCM we have that:
 * $\operatorname{lcm}\left\{{ g,h }\right\} = \dfrac{gh}{\gcd\left\{{ g,h }\right\}}$

These two equalities imply that $\gcd\left\{{ g,h }\right\} = 1$.

That is, $g$ and $h$ are coprime.