TITLE: Why is Rationals w.r.t addition not an Isomorphism to Rationals w.r.t. multiplication?
QUESTION [7 upvotes]: Question states: Recall the additive groups Z,Q and R, and the multiplicative groups Q* and R* of non-zero numbers. show that: 
(b) Q is not isomorphic to Q*
(c) R is not isomorphic to R*
I can see why there would not be a Bijection If it where to restricted to only the positive numbers but am failing to see an example why these are not isomorphisms

REPLY [5 votes]: Using the morphism $$\mathrm{sign} : x \mapsto \left\{ \begin{array}{cl} +1 & \text{if} \ x >0 \\ -1 & \text{if} \ x<0 \end{array} \right.,$$ you can notice that $\mathbb{Q}^*$ (or $\mathbb{R}^*$) has a finite quotient, namely $\mathbb{Z}_2$.
On the other hand, $\mathbb{Q}$ (or $\mathbb{R}$) has no non-trivial finite quotient. Indeed, for any epimorphism $\varphi : \mathbb{Q}\twoheadrightarrow F$ onto a finite group $F$, say of order $n$, we have 
$$\varphi(q)= \varphi \left( n \frac{q}{n} \right)=n \cdot \varphi \left( \frac{q}{n} \right)=1.$$
Therefore, $\varphi$ is trivial, and because $\varphi$ is onto, $F$ has to be trivial. 
This argument comes from a more general result: a divisible group has no non-trivial finite quotient. In fact, the quotient of a divisible group has to be divisible itself.