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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

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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.

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  • $\begingroup$ Just to atract your attention, according to Machi's textbook Groups, $\mathbb{R}^+(*)$ is isomorphic to $\mathbb{R}$(+), p5. $\endgroup$ – user2820579 Mar 16 '18 at 19:16
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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.

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    $\begingroup$ I think another way to rephrase what @Lee Mosher said is that the number of elements which are their own inverses must be the same in both groups if you had an isomorphism and that is not the case. Very good argument. $\endgroup$ – D1811994 May 22 '15 at 16:05
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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.

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  • $\begingroup$ The book I'm reading says, Suppose that θ : G → G is an isomorphism. Then (d) the order of g in G is equal to the order of θ(g) in G. Is this just a more specific version of your last line? $\endgroup$ – Darren May 23 '15 at 11:36
  • $\begingroup$ In fact, I don't see the link between what you said and I wrote... Can you be more precise in your question? $\endgroup$ – Seirios May 23 '15 at 15:03

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