Group Direct Product of Infinite Cyclic Groups

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

Intuitive Proof
Let $\left({G_1, +_1}\right), \left({G_2, +_2}\right)$ be infinite cyclic groups.

Let $\left({G, +}\right) = \left({G_1, +_1}\right) \times \left({G_2, +_2}\right)$.

Let $\left({G_1, +_1}\right) = \left \langle {g_1} \right \rangle, \left({G_2, +_2}\right) = \left \langle {g_2} \right \rangle$.

From Generators of Infinite Cyclic Group:
 * $\left \langle {g_1} \right \rangle$ and $\left \langle {-g_1} \right \rangle$ are the only generators of $\left({G_1, +_1}\right)$;
 * $\left \langle {g_2} \right \rangle$ and $\left \langle {-g_2} \right \rangle$ are the only generators of $\left({G_2, +_2}\right)$.

So a generator of $\left({G, +}\right)$ must be of the form $\left \langle {\pm g_1, \pm g_2} \right \rangle$.

However, consider the element $\left({g_1, 2g_2}\right) \in G$.

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

So you can't make $\left({g_1, 2g_2}\right)$ from products of $\left \langle {\pm g_1, \pm g_2} \right \rangle$.