Group Direct Product of Infinite Cyclic Groups

Theorem

Let $G_1 = \struct {G_1, \circ_1}$ and $G_2 = \struct {G_2, \circ_2}$ be infinite cyclic groups.

Let $G = \struct {G, \circ} = G_1 \times G_2$.

Let $G_1 = \gen {g_1}, G_2 = \gen {g_2}$.

$g_1$ 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 $\tuple { g_1^{\pm 1}, g_2^{\pm 1} }$ to have any hope to generate all of $G$.

Without loss of generality, suppose that $G = \gen {\tuple {g_1, g_2} }$.

Let $e_1$ be the identity element of $G_1$.

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 $\tuple {e_1, x_2} \in \gen {\tuple {g_1, g_2} }$.

Then there is an $i \in \Z$ such that we have:

$\tuple {g_1, g_2}^i = \tuple {g_1^i, g_2^i} = \tuple {e_1, x_2}$

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 $\tuple {e_1, g_2} \notin \gen {\tuple {g_1, g_2} }$.

Therefore:

$G \ne \gen {\tuple {g_1, g_2} }$

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

$\blacksquare$

1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 43 \delta$