Prove $(\mathbb{Q}, +) \ncong (\mathbb{Q}^+, *)$ 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. 
 A: 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.
A: 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.
A: 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.
A: $\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.
A: 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.
