Can $(\Bbb Z[x], +)$ be isomorphic to $(\Bbb Q, +)$ as groups?
My attempt $:$ We know that $(\Bbb Z[x], +)$ is isomorphic to $(\Bbb Q_{\gt 0}, \cdot)$ as groups. Now if there exists a group isomorphism between $(\Bbb Z[x], +)$ and $(\Bbb Q, +)$ then we have a group isomorphism $\varphi : (\Bbb Q, +) \longrightarrow (\Bbb Q_{\gt 0}, \cdot).$ But then there exists $a \in \Bbb Q$ such that $\varphi (a) = 2 \implies \varphi \left (\frac {a} {2} + \frac {a} {2} \right ) = 2 \implies \varphi \left (\frac {a} {2} \right )^2 = 2 \implies \varphi \left (\frac {a} {2} \right ) = \pm \sqrt {2} \notin \Bbb Q_{\gt 0},\ $ a contradiction. Hence we conclude that there does not exist any group isomorphism (in fact onto group homomorphism) between $(\Bbb Z[x], +)$ and $(\Bbb Q_{\gt 0}, \cdot).$
Am I right? Can anybody please verify my argument above? Thanks for your time.
That's correct, but uses much more than it's needed.
For every $a\in\mathbb{Q}$, there exists $b\in\mathbb{Q}$ such that $a=2b$.
On the other hand, there is no $p(x)\in\mathbb{Z}[x]$ such that $2p(x)=1$.
This is essentially the same argument, but doesn't use the knowledge about $\mathbb{Z}[x]$ being isomorphic to $\mathbb{Q}>0$ which is not relevant.
Any abelian group that's not divisible cannot be isomorphic to $\mathbb{Q}$.