In this question, I asked whether there existed groups G with finitely presentable subgroups H such that gHg−1 is a proper subgroup of H for some ginG. Robin Chapman pointed out that the group of affine automorphisms of mathbbQ contains examples where HcongmathbbZ.
This leads me to the following more general question. A group Gamma is "coHopfian" if any injection GammahookrightarrowGamma is an isomorphism. To put it another way, Gamma does not contain any proper subgroup isomorphic to itself. The canonical example of a non-coHopfian group is a free group Fn on n letters. Chapman's example exploits the fact that F1congmathbbZ contains proper subgroups kmathbbZ isomorphic to mathbbZ.
Now let Gamma be a non-coHopfian group and let Gamma′subsetGamma be a proper subgroup with Gamma′congGamma. Question : does there exist a group Gamma″ such that GammasubsetGamma″ and an automorphism phi of Gamma″ such that phi(Gamma)=Gamma′? How about if we restrict ourselves to the cases where Gamma and Gamma″ are finitely presentable? I expect that the answer is "no", and I'd be interested in conditions that would assure that it is "yes".
If such a Gamma″ existed, then we could construct an example answering my linked-to question above by taking G to be the semidirect product of Gamma″ and mathbbZ with mathbbZ acting on Gamma″ via phi. This question thus can be viewed as asking whether Chapman's answer really used something special about mathbbZ.
No comments:
Post a Comment