Is $<\mathbb Q^+, \times>$ the free abelian group on countably infinitely many generators? It seems to make sense to me that it should be, with the generators being the set of primes. However, I'm not sure that my intuition is right.
Additionally, would this not be contradicted by the fact that $2\times3\times5\times7\times11\times13...$ is not a positive rational? 
 A: Yes, the prime numbers generate the positive rationals under multiplication: every element can be written as a product of powers of primes (corollary to existence of prime factorizations), and they satisfy no multiplicative relations (corollary to uniqueness of prime factorizations) so it is freely generated by them. The free abelian group generated by a set does not include infinite products of elements of the set, only finite products, so there is no contradiction in $2\times3\times5\cdots\not\in\Bbb Q$.
A: Unless I'm missing something obvious, this should be true. I'm not sure what you mean by $2 \times 3 \times 5 \times 7 \times 11 \times 13...$ not being a positive rational. In the group theory setting we only consider finite words in the generators. To put it another way, the free abelian group over some set $\Lambda$ is isomorphic to
$$\bigoplus_{\lambda \in \Lambda} \mathbb{Z}$$
Not 
$$\prod_{\lambda \in \Lambda} \mathbb{Z}$$
That is, the elements are tuples with only finitely many entries non-trivial.
If you are referring to the infinite product being evaluated by some sort of zeta-regularization argument (which yields $4\pi^2$), then this is not an issue because those types of arguments occur in a very different setting where we consider everything sitting in the complex analytic setting and then use various techniques to create somewhat exotic ways of evaluation series, but this is not an evaluation that makes sense even in the topologically enriched sense, and certainly not in the group sense. 
