Why is Rationals w.r.t addition not an Isomorphism to Rationals w.r.t. multiplication?

Question

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

Answer 1

You can do this by contradition. Let's assume that $(\Bbb Q, +)$ and $(\Bbb Q^*, \cdot)$ are isomorphic. Then there exists a isomorphism $\varphi: (\Bbb Q, +) \to (\Bbb Q^*, \cdot)$. Since $\varphi$ has to be surjective, there exists a $q \in \Bbb Q$, such thtat $\varphi(q) = -1$. Now we calculate using the homomorphism properties of $\varphi$ $$ -1 = \varphi(q) = \varphi\left( \frac q 2 + \frac q 2 \right) = \varphi\left( \frac q 2 \right) \varphi\left( \frac q 2 \right) = \left[ \varphi\left( \frac q 2 \right) \right]^2 \; ,$$ which is a contradiction, since there is no element $p \in \Bbb Q^*$, such that $p^2 = -1$.

You can use the same argumentation to show that $(\Bbb R, +)$ and $(\Bbb R^*, \cdot)$ are not isomorphic.

Answer 2

In the group of nonzero rationals with respect to multiplication there is an element which is its own inverse, namely $-1$. But no such element exists in the group of rationals with respect to addition. Such elements must be preserved by any isomorphism.

Answer 3

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.