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 $$\left \langle {\pm g_1, \pm g_2} \right \rangle$$.