Every pi-torsion-free nilpotent group can be embedded in a unique minimal pi-powered nilpotent group

Statement
Suppose $$\pi$$ is a set of prime numbers and $$G$$ is a $$\pi$$-torsion-free nilpotent group, i.e., $$G$$ is a nilpotent group and it is $$\pi$$-torsion-free -- it has no elements of order $$p$$ for any $$p \in \pi$$. Then, the following are true:


 * There exists a group $$K$$ containing $$G$$ such that $$K$$ is a $$\pi$$-powered nilpotent group and such that for any element $$x \in K$$, there exists a $$\pi$$-number $$n$$ (i.e., a number $$n$$ all whose prime divisors are in $$\pi$$) with $$x^n \in G$$.
 * Further, if $$L$$ is any $$\pi$$-powered nilpotent group containing $$G$$, there is a subgroup $$K_1$$ of $$L$$ containing $$G$$ such that the embeddings of $$G$$ in $$K$$ and $$K_1$$ are equivalent. Moreover, this $$K_1$$ is precisely the set of $$x \in L$$ such that there exists a $$\pi$$-number $$n$$ (i.e., a number $$n$$ all whose prime divisors are in $$\pi$$) with $$x^n \in G$$.

The localization functor

 * Every group admits an initial homomorphism to a pi-powered group

2 X 2 matrix of related ideas
The result can be understood in a 2 X 2 matrix of results: