Prove that the additive group $ℚ$ is not isomorphic with the multiplicative group $ℚ^*$.
Prove that $ℚ^*_{>0}$ is not isomorphic with $ℚ$.
Prove that the additive group $ℚ$ is not isomorphic with the multiplicative group $ℚ^*$. [duplicate]
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asked Mar 19, 2013 at 11:52
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2$\begingroup$ Something to ponder: This question has four up-votes. It also has four answers. Now, I upvoted this question, but didn't answer it...so... $\endgroup$– user1729Mar 19, 2013 at 15:34
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6 Answers
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For example, the equation $x+x=a$ has a solution in $ℚ$, unlike $x \cdot x=a$ in $ℚ^*_{>0}$.
answered Mar 19, 2013 at 11:57
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2$\begingroup$ I've never thought about it this way before - it is really very, very elegant! $\endgroup$– user1729Mar 19, 2013 at 11:59
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5$\begingroup$ @user1729: Oh! This is an usual trick. $\endgroup$ Mar 19, 2013 at 14:01
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1$\begingroup$ @einpoklum: Thank you. This is more understandable. $\endgroup$ Mar 19, 2013 at 16:42
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If $t$ means the torsion subgroup of a group so $$t(\mathbb Q,+)=\{0\}\neq \{\pm1\}=t(\mathbb Q^*,\cdot)$$
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$\begingroup$ Thanks Don. Honestly, I read the Chapter which contain free abelian and pure and divisible groups in Rotman's book. And I know these points from there. All is for J. Rotman :-) $\endgroup$– MikasaMar 19, 2013 at 18:12
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$\begingroup$ Like Ittay Weiss' answer, this does not solve 2. $\endgroup$ Mar 20, 2013 at 8:41
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$\begingroup$ Could you provide a link to a proof that shows that two groups are not isomorphic if their torsion subgroups are not equal? $\endgroup$ Feb 14, 2015 at 5:44
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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.
answered Mar 19, 2013 at 17:13
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2$\begingroup$ +1 For a very insightful, nice answer...alas, it seems to be a little advanced for someone asking what the OP did. $\endgroup$ Mar 19, 2013 at 18:05
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4$\begingroup$ @DonAntonio, Yes, you are certainly correct, but I think it's not a bad thing to mention "fancy" things "prematurely", so that the next time a beginner hears/sees those words, they're just a touch less alien. And there's the point to be made that these groups are not different "on a mere technicality". And on some days I have the opinion that whenever I have a strong impulse to comment/rant, I should not entirely repress it, since it might be informative to someone else, if only as a bad example. :) $\endgroup$ Mar 19, 2013 at 18:11
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1$\begingroup$ For what it's worth, as someone who does topology mostly, my group theory has gotten rusty and I feel that answer like this is exactly what I needed to read. So, while the OP may not have benefited from this answer, I did. Thanks, and +1. $\endgroup$ Mar 19, 2013 at 19:22
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$\begingroup$ @JasonDeVito, Thx for the feedback! :) $\endgroup$ Mar 19, 2013 at 19:28
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$\begingroup$ Excellent answer @paulgarrett . I particularly like the observation that "the positive multiplicative group is free on the primes" $\endgroup$– magmaOct 7, 2013 at 0:00
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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.
answered Mar 19, 2013 at 13:30
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1$\begingroup$ This counterexample is ok but $\phi(a)=-1$ is not allowed as image is to be from positive rationals. By the way, $\phi(a)=2$ can work well. $\endgroup$ Nov 2, 2020 at 7:36
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Hint: Count elements of finite order.
answered Mar 19, 2013 at 11:53
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I give you two different proofs to each statement.
- 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 $\mathbb{N}$).
