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.
 A: 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 $b$.
A: If such an isomorphism existed it would of course be onto so there would exist some $n \in \mathbb{Z}$ such that $\phi(n) = \frac{1}{2}$ for example. But then, $$\phi(n) = \phi(1+\cdots + 1) = \phi(1)\cdots \phi(1) = \phi(1)^n = \frac{1}{2}.$$ This implies $n=1$ since otherwise $\left(\frac{1}{2}\right)^{1/n} \notin \mathbb{Q}$. So, $\phi(1) = \frac{1}{2}$.  Thus, for any $n \in \mathbb{Z}$, $\phi(n) = \phi(1)^n = \frac{1}{2^n}$, so clearly $\phi$ is not onto since we only achieve powers of two in the image.  So such an isomorphism cannot exist. 
A: Every nontrivial subgroup of $\mathbb{Z}$ is infinite cyclic. But, $\mathbb{Q}^+$ contains a rank 2 abelian subgroup, namely all rational numbers of the form $2^m 3^n$ for integer values of $m,n$.
