# 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.

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?

• It's correct, but casting it as a proof by contradiction is unnecessary. Let $(a,b)$ be a generator; then $a\neq 1$ and $b\neq 1$. If $a\neq 1$, then there exists $k\neq 0,1$ such that $(a,b)^k = (a,1)$; so $b^k=1$, so $b$ has finite order; and $a^{k-1}=1$; so $a$ has finite order. And therefore $(a,b)$ has finite order. Nov 3, 2023 at 16:55
• @Arturo, do you mean "$a\neq 1$ or $b\neq 1$" instead of and ?
– lhf
Nov 3, 2023 at 17:30
• @lhf I went back and forth. In fact, because the projections have to be nontrivial, both are nontrivial. But if you assume "or" then you also need to argue what to do if $k=1$. Nov 3, 2023 at 17:34

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.