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Prove that the group of rationals under addition is not isomorphic to the multiplicative group of positive rational numbers.

What I tried: Suppose $\phi: (\mathbb{Q}, +) \rightarrow (\mathbb{Q}^+, *) $ is such an isomorphism. Then I tried to use $\phi(\frac{a}{b})\phi(\frac{c}{d}) = \phi(\frac{a}{b} + \frac{c}{d})$ and bijectivity to get a contradiction.

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Let $f:(\Bbb Q, +)\to (\Bbb Q^+,\cdot)$ be an isomorphism. Let $a=f(1)$. Then $f(1/n)=a^{1/n}$ for every natural $n$. Then the $n$th root of $a$ is rational for every $n$. This implies $f(1)=1$. But then $f(2)=f(1+1)=1\cdot 1=1$ and $f$ is not injective.

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    Just quick clarification question: why does $a = f(1)$ imply $f(1/n) = a^{1/n}$? – 2017-01-07
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    @setholopolus because $f(1/n)^n =f(n(1/n))=f(1)$. – 2017-01-07
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The equation $x+x=a$ has a solution for all $a$ for the additive group, but the equation (written multiplicatively) $x\cdot x=a'$ in $\mathbb{Q}^{+}$ does not. Hence the groups cannot be isomorphic.

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Essentially, one group is divisible when the other is not. For example, if $\phi(a)=2$ , then $\phi\left(\frac{a}{2}\right)$ should be a square root of two.

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$\newcommand{\Z}{\mathbb{Z}}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$In the first group, any two non-trivial cyclic subgroups $H, K$ have non-trivial intersection. Let $H = \Span{a}$ and $K = \Span{b}$, with $a, b \ne 0$, $a = x/y$ and $b = z/t$, with $x, y, z, t \in \Z$. Then $$ 0 \ne x z = (y z) \cdot a = (x t) \cdot b \in H \cap K. $$

In the second group, $H = \Span{2}$ and $K = \Span{3}$ have trivial intersection.

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A further comment how to see the difference (could be made into a precise proof, though). The positive multiplicative group is free on the primes, while the additive group is divisible, meaning that for every integer $n$ and $a$ in the additive group, $n\cdot x=a$ has a (unique) solution $x$. So the multiplicative group is projective (because of freeness), while the additive group is injective (because of divisibility). These are two very different properties for infinite groups.

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    I think it is valuable to recall this, but I do not feel that it is an answer in the current form, though it's close. To wit, alright the properties are very different but different does not mean/imply mutually exclusive and you do not assert in this answer that the one group does not *also* have the property of the other. – 2017-01-06
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    @quid Yes, you are right, of course. This should be perhaps only a remark. – 2017-01-06