$(\mathbb{Q} , + )$ and $( \mathbb{Q} \times \mathbb{Q} , +)$ are not isomorphic as groups

Question

Prove that $(\mathbb{Q}, +)$ and $(\mathbb{Q} \times \mathbb{Q} , +)$ are not isomorphic as groups.

This question was asked in my abstract algebra quiz( now over) and I couldn't prove it.

I have been following Emil Artin.

I don't have any ideas on how this question should be approached. I thought I should assume that a isomorphism $f$ exists in hope of getting some contradiction but I am unable to think of what exact proposition should I use and need some clues.

Answer 1

Suppose they are, then there is a bijective homomorphism $f:\mathbb{Q}\times\mathbb{Q}\to\mathbb{Q}$. Let's say $$f\big((1,0)\big)=\frac{a_1}{b_1}\text{ and }f\big((0,1)\big)=\frac{a_2}{b_2}.$$ Then since $(a_2b_1)f((1,0))=(a_1b_2)f((0,1))=a_1a_2$, we have $f((a_2b_1,0))=f((0,a_1b_2))$ by applying the properties of homomorphisms.

This contradicts $f$ being a bijection.

Answer 2

Well, I have one approach for "$(\mathbb{Q},+)$ is not isomorphic to $(\mathbb{Q} \times \mathbb{Q}, +)$ as a group".

That is, "every finitely generated subgroup of $(\mathbb{Q},+)$ is cyclic " but for $(\mathbb{Q} \times \mathbb{Q}, +)$ there exist some finitely generated subgroups which are not cyclic. For example, Subgroup generated by $S={(1,0),(0,1)}$ is finitely generated but the subgroup generated by the set $S$ is $\mathbb{Z}\times\mathbb{Z}$ which is not cyclic.