Are the additive group of rationals and the multiplicative group of positive rationals isomorphic? This question is a little bit different from Group of positive rationals under multiplication not isomorphic to group of rationals since I was wondering if logarithmic function could solve this or not, thank you.
Consider two groups $(\mathbb Q^+,\cdot)$ and $(\mathbb Q,+)$, does an isomorphism exist between them?
My attempt: Let $\varphi:\mathbb Q^+\rightarrow\mathbb Q$ be the isomorphic function, then the below statement must hold true for all $a,b\in\mathbb Q^+$:
$$\varphi(a\cdot b)=\varphi(a)+\varphi(b)$$
So I guess maybe a logarithmic function would be fine here, since it's bijective, too.
But the problem is I can not show for a specific base like $10$ for example, $\log(\mathbb Q^+)=\mathbb Q$.
 A: So now you know they're not isomorphic, but you can go a bit further with what you already know. For example, $\mathbb{Q}^+$ has a subgroup whose elements are the powers $2^n$ of $2$, including $1$ and $1/2$, etc. Because the map $n \mapsto 2^n$ from $(\mathbb Z,+)$ is injective, this is a cyclic subgroup. You can do the same thing with the powers $3^n$ and see these two subgroups intersect only in $\{1\}$.
Now, you know every natural number has a unique factorization in terms of primes. Can you use that to say more about $(\mathbb Q^+,\cdot)$?
By comparison, you can check for any two elements $g$ and $h$ of $(\mathbb Q,+)$,there are integers $m$ and $n$ such that $mg = nh$. This means that, in contrast to the case of $(\mathbb Q^+,\cdot)$, any two cyclic subgroups have a nontrivial intersection. One says that $(\mathbb Q^+,\cdot)$ has rank equal to 1, meaning there's a $\mathbb Z$ subgroup, but only one independent such $\mathbb Z$ subgroup, in that all other isomorphic copies of $\mathbb Z$ meet it nontrivially.
A: HINT: For each $q\in\Bbb Q$ the equation $x+x=q$ has a solution; is it true that $x\cdot x=q$ has a solution for each $q\in\Bbb Q^+$?
