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It is well known that the exponential function induces an isomorphism between the additive group of real numbers and the multiplicative group $\mathbb{R}_{>0}$. I was wondering if there exists an isomorphism between the additive group of rationals and the multiplicative group $\mathbb{Q}_{>0}$.

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marked as duplicate by 23rd, Cameron Buie, Carl Mummert, user43208, Daniel Fischer Oct 20 '13 at 18:56

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There can't be. In the additive group, you can divide by $n$ for all $n \in \mathbb{Z}^+$, i.e. for all $x$ there is a $y$ with $x = n\cdot y \;(= \underbrace{y + y + \dotsb + y}_{n\text{ times}})$.

In the multiplicative group of positive rational numbers, that would correspond to the existence of $n$-th roots for all $n > 0$.

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I hope you know the definition of torsion subgroup $tG$. If $x\in(\mathbb Q^*,\cdot)$ such that $|x|\le\infty$, then $\exists n\in\mathbb N$, $x^n=1$ and so $x=\pm 1$. This means that $$t Q^*=\{1\}$$. By the similar way, you can see that $tQ=\{0\}$. So two groups cannot be isomorphic.

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  • $\begingroup$ Good observation/good point! +1 $\endgroup$ – Namaste Oct 21 '13 at 0:35
  • $\begingroup$ @amWhy: Thanks Amy. You always make me glad. ;-) $\endgroup$ – mrs Oct 21 '13 at 8:00

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