In this question, I asked whether there existed groups GG with finitely presentable subgroups HH such that gHg−1gHg−1 is a proper subgroup of HH for some ginGginG. Robin Chapman pointed out that the group of affine automorphisms of mathbbQmathbbQ contains examples where HcongmathbbZHcongmathbbZ.
This leads me to the following more general question. A group GammaGamma is "coHopfian" if any injection GammahookrightarrowGammaGammahookrightarrowGamma is an isomorphism. To put it another way, GammaGamma does not contain any proper subgroup isomorphic to itself. The canonical example of a non-coHopfian group is a free group FnFn on nn letters. Chapman's example exploits the fact that F1congmathbbZF1congmathbbZ contains proper subgroups kmathbbZkmathbbZ isomorphic to mathbbZmathbbZ.
Now let GammaGamma 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.
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