8
$\begingroup$

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

$\endgroup$
2
  • 8
    $\begingroup$ Note that $\mathbb R$ (under addition) is isomorphic to $\mathbb R^+$ (under multiplication). The isomorphism is $\exp$. $\endgroup$
    – wlad
    May 22, 2015 at 16:15
  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/1474283 $\endgroup$
    – Watson
    Dec 14, 2016 at 12:31

3 Answers 3

39
$\begingroup$

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.

$\endgroup$
2
  • 5
    $\begingroup$ 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, 2015 at 16:05
  • $\begingroup$ Non-trivial elt, yes? There should be $\pm 1 \in k^\times$ and $0 \in k$ $\endgroup$ Jun 19, 2022 at 14:18
22
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Just to atract your attention, according to Machi's textbook Groups, $\mathbb{R}^+(*)$ is isomorphic to $\mathbb{R}$(+), p5. $\endgroup$ Mar 16, 2018 at 19:16
5
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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, 2015 at 11:36
  • 1
    $\begingroup$ 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, 2015 at 15:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .