Is there an operational isomorphism from $(\mathbb{Z},+)$ to $(\mathbb{Q}^{+},\cdot)$? Let $\left(\mathbb{Z},+\right)$ and $\left(\mathbb{Q}^{+},\cdot\right)$ be groups (let the integers be a group with respect to addition and the positive rationals be a group with respect to multiplication). Is there a function $\phi\colon\mathbb{Z}\mapsto\mathbb{Q}^{+}$ such that:


*

*$\phi(a)=\phi(b) \implies a=b$ (injection)

*$\forall p\in\mathbb{Q}^{+} : \exists a\in\mathbb{Z} : \phi(a)=p$ (surjection)

*$\phi(a+b) = \phi(a)\cdot\phi(b)$ (homomorphism)


? If so, provide an example. If not, disprove.

No, because $\mathbb Z$ is generated by $1$ while $\mathbb Q^+$ as a group under multiplication is not finitely generated. If such a homomorphism $\phi$ existed, then we could write any element of $\mathbb Q$ as $\phi (n)=\phi(1)^n$, so $\mathbb Q$ would be generated by $\phi(1)$. But $\phi(1)=a/b$ for some $a,b\in\mathbb Z$, and clearly raising $a/b$ to an integral exponent wont give us $1/p$ for primes $p$ not dividing $b$.
