Free group generated by two generators is isomorphic to product of two infinite cyclic groups

Question

Is the following statement true or not?

The free group generated by two generators is isomorphic to the direct product of two infinite cyclic groups.

I know that if the generated group is abelian then the statement is True but I don't know if it's not abelian. I think it's wrong but can't come up with a counter example.

Thanks in advance.

Answer 1

It is not true that the free group on two generators $F_2$ is isomorphic to $\mathbb{Z}\times\mathbb{Z}$.

To see this, suppose otherwise. Then $F_2$ is abelian. By the universal property of free groups, every group which can be generated by two elements is a homomorphic image of $F_2$. In particular, the symmetric group $S_5$ is a homomorphic image of $F_2$. As "being abelian" is preserved under homomorphic images, we have that $S_5$ is abelian. This is a contradiction.

(The above argument still works if you replace $S_5$ with your favourite two-generated, non-abelian group. For example, $S_3$, or any non-abelian simple group.)

EDIT: As Derek Holt pointed out in the comments to the question, this question is equivalent to asking whether all countable groups are abelian! This is because every countable group embeds into a two-generated group (which is a classical result: Higman, G. , Neumann, B. H. and Neuman, H. (1949), Embedding Theorems for Groups. Journal of the London Mathematical Society, s1-24: 247-254. doi:10.1112/jlms/s1-24.4.247).