Let $G_1, G_2$ be non-trivial groups. If $G_1 \times G_2$ is cyclic, then $G_1 \times G_2$ is finite.
I'm asked about the veracity of this statement and I did the following: suppose BWOC that $G_1 \times G_2$ is cyclic and $G_1 \times G_2$ is infinite. WLOG, one assumes that $G_1$ is infinite. Now, $G_1 \times G_2 = \langle(a,b)\rangle$ and $b\neq 1_{G_2}$, because $G_2$ is non-trivial. But now we should have that $(1_{G_1},b) \in \langle(a,b)\rangle$, but as $G_1$ is infinite, $a^k$=$1_{G_1}$ $\iff$ $k=0$, so we could only obtain $(1_{G_1},1_{G_2})$ based on the hypothesis that $G_1 \times G_2 = \langle(a,b)\rangle$ but not that $(1_{G_1},b) \in \langle(a,b)\rangle$. Then, $G_1 \times G_2$ cannot be a cyclic group but this contradiction derived from supposing that $G_1 \times G_2$ is infinte, so it must be that $G_1 \times G_2$ is finite. Is there something that I'm missing and/or ways to improve the argument?
This question is equivalent to: Prove that $\mathbb{Z}$ is not the product of two non-trivial groups. This follows from the fact that a nontrivial subgroup of $\mathbb{Z}$ is of the form $n\mathbb{Z}$ for some $n > 1$, and $\mathbb{Z}/n\mathbb{Z}$ is torsion. This is equivalent to Arturo's direct computation in the comments.
For a high-brow proof, we can further compact the argument using flatness of $\mathbb{Q}$: An injection $\mathbb{Z}^2 \to \mathbb{Z}$ of abelian groups would give an injection $\mathbb{Q}^2\to \mathbb{Q}$ of vector spaces, which is impossible for dimension reasons.