There exist free powered groups for any set of primes and any size of generating set

Statement
Suppose $$\pi$$ is a set of primes and $$S$$ is a generating set. Then, there exists a free $$\pi$$-powered group on $$S$$, denoted $$F(S,\pi)$$ with $$S$$ identified as a subset of this group, satisfying the following conditions:


 * 1) This group is $$\pi$$-powered.
 * 2) It has no proper subgroup that contains $$S$$ and is also $$\pi$$-powered.
 * 3) Any set map from $$S$$ to a group $$G$$ extends to a unique group homomorphism from $$F(S,\pi)$$ to $$G$$.

Related facts

 * Every group is a subgroup of a divisible group
 * Powering-injective group need not be embeddable in a rationally powered group

In the variety formalism
We use the description of powered groups in terms of being a group powered over a unital ring, where the ring is $$\mathbb{Z}[\pi^{-1}]$$ (see also equivalence of definitions of powered group for a set of primes). The groups powered over a unital ring form a variety of algebras. We then look at the free algebra on a generating set in that variety. This guarantees existence. The only thing we need to check is that the elements of the generating set do not collapse. This is easily verified by noting that the rational vector space with basis the generating set is a quotient of such a group, and this quotient separates all the generators.

Explicit construction in terms of group presentations
Start with the abstract free group on $$S$$. In each iteration, do the following:


 * Adjoin $$p^{th}$$ roots (for all $$p \in \pi$$) of all the elements so far.
 * Take the free group generated by all these.
 * For every pair of elements that have the same $$p^{th}$$ power (for one or more $$p \in \pi$$), set them to be equal (i.e., quotient out by the relation of their being equal).

The group constructed at each stage has a natural homomorphism to it from the previous group. The direct limit of this sequence is the desired group.