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We have a finitely generated group $G$ which is infinite. Does there exist such a finite generating set $S$, $G= \langle S\rangle$, that the corresponding Cayley graph $\mathrm{Cayley}(G,S)$ can be presented as a disjoint union of infinite Hamiltonian paths?

Alspash and others proved it in the case of Abelian groups for all generating sets, but I need the case of nonabelian groups. I'd be happy just for the existence of the corresponding generating set.

UPD. Sure, the case of groups without torsion is rother easy. The main interest is about groups with ''big torsion'' for example localy finite groups.

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  • $\begingroup$ The question isn’t quite clear: of course the whole Cayley graph can’t be a disjoint union of lines unless it is itself a line. Assuming you just want to cover the vertices, this is easy for any generating set containing an element $g$ of infinite order: the left cosets of $\langle g\rangle$ give the paths you want. In particular, such a generating set exists unless the group is torsion. $\endgroup$
    – HJRW
    Commented May 9, 2022 at 7:24
  • $\begingroup$ Sure, the main problem is for locally finite groups and groups which are simillar to them. For clear, one line -- is a disjoint union of itself :-) $\endgroup$
    – Andronick Arutyunov
    Commented May 9, 2022 at 9:36
  • $\begingroup$ Ah -- the question currently makes it sound like you might be interested in anything beyond the abelian case. Perhaps it would be a good idea to edit to explain that you want to understand the case of torsion groups? $\endgroup$
    – HJRW
    Commented May 9, 2022 at 9:39
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    $\begingroup$ Thank you, I edited the question. In fact for my (surely great) theorem I need information on other type on some kind of factor graph. So I need any information on this question :-) I was reading for several days different articles but most of them are defended to finite groups or to ideas about all generating sets, so they give an information just about specific groups. And I need an information about all groups, but I can take a specific ''good'' generating set. $\endgroup$
    – Andronick Arutyunov
    Commented May 9, 2022 at 15:18
  • $\begingroup$ Could you clarify what you mean by "infinite Hamiltonian path"? Are they allowed to be 2-way infinite, i.e. double rays? When you say that Cayley(G,S) can be presented as a disjoint union of such paths, do you mean edge-disjoint union? (If not the question makes no sense to me.) $\endgroup$
    – Agelos
    Commented Nov 2, 2022 at 9:06

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