Powering-injective group need not be embeddable in a rationally powered group
From Groupprops
Contents
Statement
It is possible to have a group that is a group in which every power map is injective but such that there does not exist any rationally powered group for which is a subgroup of .
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 | -torsion-free group (equivalently, -powering-injective group) | -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 | -torsion-free group (equivalently, -powering-injective group) | -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 PDF}^{More info}, Page 230, example attributed to B. H. Neumann, but appears recorded for the first time in this paper.