Prove that the additive group $ℚ$ is not isomorphic with the multiplicative group $ℚ^*$. 
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*Prove that the additive group $ℚ$ is not isomorphic with the multiplicative group $ℚ^*$.

*Prove that $ℚ^*_{>0}$ is not isomorphic with $ℚ$.
 A: For example, the equation $x+x=a$ has a solution in $ℚ$, unlike $x \cdot x=a$ in $ℚ^*_{>0}$.
A: I give you two different proofs to each statement.


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*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 $\mathbb{N}$).

A: If $t$ means the torsion subgroup of a group so $$t(\mathbb Q,+)=\{0\}\neq \{\pm1\}=t(\mathbb Q^*,\cdot)$$
A: In addition to other good answers, it might be worthwhile to observe that not only are these groups not exactly isomorphic, but, in fact, they are wildly different: the positive multiplicative group is free on the primes, while the additive group is (uniquely-) divisible, meaning that for every integer $n$ and $a$ in the additive group, $n\cdot x=a$ has a (unique) solution $x$. That is, the positive multiplicative group is projective (implied by free-ness), while the additive group is injective (implied by divisibility... this is Baer's criterion). So one really should feel that they are completely different.
A: If $\phi$ were such mapping (i.e. $\phi :\mathbb{Q} \to \mathbb{Q^*}$ is an isomorphism it means that it must satisfy all the properties of isomorphism), there would be a rational number $a$ such that $\phi(a)=-1$.
$$-1=\phi(a)=\phi \bigg(\frac{1}{2}a+\frac{1}{2}a\bigg)=\phi \bigg(\frac{1}{2}a\bigg)\phi \bigg(\frac{1}{2}a\bigg)=\bigg[\phi\bigg(\frac{1}{2}a\bigg)\bigg]^{2}$$
Is there any rational number whose square is $-1$?
If you think about it, it is the simplest solution by contradiction.
A: Hint: Count elements of finite order.
