Group Direct Product of Cyclic Groups

Theorem
Let $G$ and $H$ both be finite cyclic groups with orders $n = \order G$ and $m = \order H$ respectively.

Then:
 * The group direct product $G \times H$ is cyclic


 * $n$ and $m$ are coprime, that is:
 * $n \perp m$
 * $n \perp m$

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

Necessary condition
Suppose:


 * $(1): \quad \order G = n, G = \gen x$
 * $(2): \quad \order H = m, H = \gen y$
 * $(3): \quad m \perp n$

Then:

But then:
 * $\tuple {x, y}^{n m} = e_{G \times H} = \tuple {x^{n m}, y^{n m} }$

Thus:
 * $k \divides n m$

So:
 * $\order {\tuple {x, y} } = n m \implies \gen {\tuple {x, y} } = G \times H$

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

Let $\tuple {x, y}$ be a generator of $G \times H$.

By Cardinality of Cartesian Product of Finite Sets the order of $G \times H$ is:
 * $\order G \cdot \order H = n m$

Therefore by Order of Cyclic Group equals Order of Generator:
 * $\order {\tuple {x, y} } = n m$

On the other hand, by Order of Group Element in Group Direct Product we have:
 * $\order {\tuple {x, y} } = \lcm \set {\tuple {\order x, \order y} }$

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

Let $x' \in G$.

Then:
 * $\tuple {x', e_H} \in G \times H$

so there exists $k \in \N$ such that:
 * $\tuple {x, y}^k = \tuple {x^k, y^k} = \tuple {x', e_H}$

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 Cyclic Group equals Order of Generator:
 * $\order x = n$
 * $\order y = m$

Thus we have that:
 * $n m = \order {\tuple {x, y} } = \lcm \set {n, m}$

Moreover by Product of GCD and LCM we have that:
 * $\lcm \set {n, m} = \dfrac {n m} {\gcd \set {n, m} }$

These two equalities imply that:
 * $\gcd \set {n, m} = 1$

That is, $n$ and $m$ are coprime.