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

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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:

Group assumption on both the starting group and the big group Divisibility/powering/torsion assumption on the starting group Divisibility/powering/torsion assumption on the big group Is this always possible? Proof
nilpotent group none divisible group No nilpotent group need not be embeddable in a divisible nilpotent group
nilpotent group \pi-torsion-free group (equivalently, \pi-powering-injective group) \pi-powered group Yes every pi-torsion-free nilpotent group can be embedded in a unique minimal pi-powered nilpotent group
arbitrary group none divisible group Yes every group is a subgroup of a divisible group, every pi-group is a subgroup of a divisible pi-group
arbitrary group \pi-torsion-free group (equivalently, \pi-powering-injective group) \pi-powered group No Powering-injective group need not be embeddable in a rationally powered group

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.

References

  • Wreath products and p-groups by Gilbert Baumslag, Proceedings of the Cambridge Philosophical Society, Volume 55, Page 224 - 231(Year 1959): ungated PDFMore info, Page 230, example attributed to B. H. Neumann, but appears recorded for the first time in this paper.