Powering-injective group need not be embeddable in a rationally powered group

Statement
It is possible to have a group $$G$$ that is a group in which every power map is injective but such that there does not exist any rationally powered group $$K$$ for which $$G$$ is a subgroup of $$K$$.

Related facts
The result can be understood in a 2 X 2 matrix of results:

Proof
The proof idea is to take a free product of Baumslag-Solitar group:BS(1,2) with the group of integers identify the group of squares in the latter with a suitable subgroup of the former.