TITLE: Prove that the additive group $ℚ$ is not isomorphic with the multiplicative group $ℚ^*$.
QUESTION [24 upvotes]: Prove that the additive group $ℚ$ is not isomorphic with the multiplicative group $ℚ^*$.
Prove that $ℚ^*_{>0}$ is not isomorphic with $ℚ$.

REPLY [3 votes]: I give you two different proofs to each statement.

Reason a) Only $0$ has finite order in $\mathbb{Q}$, but ${-1,1}$ has finite order in $\mathbb{Q}^*$.
Reason b) $\mathbb{Q}$ is indecomposable, but $\mathbb{Q}^* = \mathbb{Q}^*_{>0}\oplus \mathbb{Z}_2$ is decomposable.
Reason a) $\mathbb{Q}^*_{>0}$ is a free abelian group on $\mathbb{N}$ and $\mathbb{Q}$ is not free abelian.
Reason b) $\mathbb{Q}$ is indecomposable, but $\mathbb{Q}^*_{>0}$ is decomposable (since it's free abelian on $\mathbb{N}$).