# Prove that the additive group $ℚ$ is not isomorphic with the multiplicative group $ℚ^*$.

1. Prove that the additive group $ℚ$ is not isomorphic with the multiplicative group $ℚ^*$.

2. Prove that $ℚ^*_{>0}$ is not isomorphic with $ℚ$.

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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... –  user1729 Mar 19 '13 at 15:34

For example, the equation $x+x=a$ has a solution in $ℚ$, unlike $x \cdot x=a$ in $ℚ^*_{>0}$.

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I've never thought about it this way before - it is really very, very elegant! –  user1729 Mar 19 '13 at 11:59
@user1729: Oh! This is an usual trick. –  Boris Novikov Mar 19 '13 at 14:01
@einpoklum: Thank you. This is more understandable. –  Boris Novikov Mar 19 '13 at 16:42

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.

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+1 For a very insightful, nice answer...alas, it seems to be a little advanced for someone asking what the OP did. –  DonAntonio Mar 19 '13 at 18:05
@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. :) –  paul garrett Mar 19 '13 at 18:11
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. –  Jason DeVito Mar 19 '13 at 19:22
@JasonDeVito, Thx for the feedback! :) –  paul garrett Mar 19 '13 at 19:28
Excellent answer @paulgarrett . I particularly like the observation that "the positive multiplicative group is free on the primes" –  magma Oct 7 '13 at 0:00

If $t$ means the torsion subgroup of a group so $$t(\mathbb Q,+)=\{0\}\neq \{\pm1\}=t(\mathbb Q^*,\cdot)$$

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+1 This is the first approach I thought of. –  DonAntonio Mar 19 '13 at 18:04
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 :-) –  Babak S. Mar 19 '13 at 18:12
Nice, Babak! +1 :-) –  amWhy Mar 20 '13 at 0:13
Like Ittay Weiss' answer, this does not solve 2. –  Marc van Leeuwen Mar 20 '13 at 8:41

Hint: Count elements of finite order.

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This answers 1. but not 2. –  Marc van Leeuwen Mar 20 '13 at 8:39

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.

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I give you two different proofs to each statement.

1. 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.
2. 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 $\mathbb{N}$).
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