Group Direct Product of Cyclic Groups

Theorem
Let $G$ and $H$ both be finite cyclic groups whose orders are coprime, i.e. $\left\vert{G}\right\vert \perp \left\vert{H}\right\vert$.

Then their group direct product $G \times H$ is cyclic.

Corollary
Let $n_1, n_2, \ldots, n_s$ be a sequence of integers, all greater than $1$, such that for any pair of them $n_i$ and $n_j$, $n_1 \perp n_j$.

Let $G_i$ be a cyclic group of order $n_i$ for each $i: 1 \le i \le s$.

Then $G_1 \times G_2 \times \cdots \times G_s$ is cyclic of order $n_1 n_2 \ldots n_s$.

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

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 $k \backslash n m$.

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

Proof of Corollary

 * When $s = 1$ the result is trivial.


 * Assume the result holds for $s = k$.

Then $H = G_1 \times G_2 \times \ldots \times G_k$ is cyclic of order $n_1 n_2 \ldots n_k$.

Applying the main result to $H \times G_{k+1}$ gives us the result for $s = k+1$.

The result follows by induction.