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 $g$ and $h$ are coprime, that is, $g \perp h$.

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:
 * $\left({x, y}\right)^{n m} = e_{G \times H} = \left({x^{n m}, y^{n m}}\right)$

Thus:
 * $k \mathrel \backslash n m$

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:
 * $\left\vert{G}\right\vert \cdot \left\vert{H}\right\vert = g h$

Therefore by Order of Generator is Order of Group:
 * $\left\vert{ \left({x, y}\right) }\right\vert = g h$

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

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

Let $x' \in G$.

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:
 * $g h = \left\vert{\left({x, y}\right)}\right\vert = \operatorname{lcm}\left\{ {g, h}\right\}$

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

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

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