We may say that two finitely generated groups $G$ and $H$ are generating-bijective when there exist homomorphisms $\phi:G\rightarrow H$ and $\psi:H\rightarrow G$ such that, for each ordered generating sets $\mathfrak g$ and $\mathfrak h$ of $G$ and $H$, resp. we can find ordered generating sets $\mathfrak h'$ and $\mathfrak g'$ of $H$ and $G$ resp. such that $$\phi(\mathfrak g')=\mathfrak h,\quad \psi(\mathfrak h')=\mathfrak g$$ Do there exist finitely generated groups $G$ and $H$ along with epimorphisms $\phi:G\rightarrow H$ and $\psi:H\rightarrow G$, such that $G$ and $H$ aren't generating-bijective?
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asked May 3, 2016 at 5:34
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$\begingroup$ "ordered" is not so important to emphasize: you just mean you can lift generating subsets (if it's ordered at the bottom, it becomes ordered at the top). $\endgroup$– YCorMay 3, 2016 at 9:27
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1$\begingroup$ Restatement: let's say that $f:G\to H$ is an LPE (Lifting Property Epimorphism) if every generating subset of $H$ lifts to a generating subset of $G$ (i.e. is the 1-to-1 image by $f$ of some generating subset of $G$). Your question is whether there exist f.g. groups $G,H$ with epimorphisms $G\to H\to G$ such that there does not exist any LPE $G\to H$. $\endgroup$– YCorMay 3, 2016 at 9:30
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1$\begingroup$ As a first basic remark, an example of epimorphism $G\to H$ such that there exists no LPE $G\to H$ is the projection $\mathbf{Z}\to \mathbf{Z}/5\mathbf{Z}$ (indeed, either $\{1\}$ or $\{2\}$ in $\mathbf{Z}/5\mathbf{Z}$ cannot be lifted to a generator of $\mathbf{Z}$). $\endgroup$– YCorMay 3, 2016 at 9:31
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$\begingroup$ @YCor: I emphasized "ordered" because the cardinal of a generating set is important in my problem, and this is considered in your new definition by "1-to-1". $\endgroup$– Meisam Soleimani MalekanMay 3, 2016 at 14:25
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