Group Direct Product of Infinite Cyclic Groups

Theorem
The group direct product of two infinite cyclic groups is not cyclic.

Proof
Let $G_1 = \left({G_1, \circ_1}\right)$ and $G_2 = \left({G_2, \circ_2}\right)$ be infinite cyclic groups.

Let $G = \left({G, \circ}\right) = G_1 \times G_2$.

Let $G_1 = \left \langle {g_1} \right \rangle, G_2 = \left \langle {g_2} \right \rangle$.

From Generators of Infinite Cyclic Group:
 * $g_1$ and $g_1^{-1}$ are the only generators of $G_1$;
 * $g_2$ and $g_2^{-1}$ are the only generators of $G_2$.

So a generator of $G$ must be of the form $\left({ g_1^{\pm 1}, g_2^{\pm 1} }\right)$ to have any hope to generate all of $G$.

Suppose that $G = \left \langle { \left({g_1, g_2}\right) } \right \rangle$; the other three cases are similar.

Let $e_1$ be the identity element of $G_1$, and let $x_2 \in G_2$.

From the definition of an infinite cyclic group, both $g_1$ and $g_2$ are of infinite order.

Suppose now $\left({e_1, x_2}\right) \in \left \langle { \left({g_1, g_2}\right) } \right \rangle$.

Then there is an $i \in \Z$ such that we have:
 * $\left({g_1, g_2}\right)^i = \left({g_1^i, g_2^i}\right) = \left({e_1, x_2}\right)$

However, as $g_1$ is of infinite order, it must be that $i = 0$.

Hence $x_2 = g_2^0 = e_2$, where $e_2$ is the identity element of $G_2$.

It follows that $\left({e_1, g_2}\right) \not \in \left \langle { \left({g_1, g_2}\right) } \right \rangle$.

Therefore $G \neq \left \langle { \left({g_1, g_2}\right) } \right \rangle$.

It follows that $G$ cannot be generated by one element.