Group Direct Product of Cyclic Groups

Theorem
If $$G$$ and $$H$$ are both finite cyclic groups whose orders are coprime, i.e. $$\left|{G}\right| \perp \left|{H}\right|$$, then $$G \times H$$ 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) $$\left|{G}\right| = n, G = \left \langle {x} \right \rangle$$;
 * 2) $$\left|{H}\right| = m, H = \left \langle {y} \right \rangle$$;
 * 3) $$m \perp n$$.

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

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.