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Let $G=\langle g_1,g_2\rangle$ denote the free group of rank 2. For a subgroup $H$ of $G$ consider the quotient graph with vertex set $H\setminus G$ of right cosets, where $Hg$ and $Hg'$ are connected by an edge if $Hg g_i^{\pm 1}=Hg'$ for $i=1$ or $i=2$.

Does there exist a transitive action on $H\setminus G$, which commutes with each of the mappings given by right-multiplication with $g_1^{\pm 1 }$ or $g_2^{\pm 1}$ on $H\setminus G$?

Clearly, the answer is yes if $H\setminus G$ is a group (and the action is given by left-multiplication). Can you think of other examples or related concepts?

EDIT: Unfortunately, I don't understand R W's answer. It seems that he says that - in general - the answer to my question is "no". (Please let me know if I'm wrong.)

On the other hand, the answer is "yes" for all normal subgroups (left and right multiplication commute). So, for which subgroups is the answer "yes"?

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  • $\begingroup$ This reads like an assignment question, and if it is, it is off-topic for MO. $\endgroup$ Aug 22, 2013 at 2:29
  • $\begingroup$ I came up with this question by myself. Since it looks like an assignment question, there should be some kind of "nice" answer? $\endgroup$
    – Jorinde
    Aug 22, 2013 at 2:42
  • $\begingroup$ I would add a group theory tag... $\endgroup$ Aug 22, 2013 at 5:19
  • $\begingroup$ @Jorinde - in that case, welcome to MO :-) $\endgroup$ Aug 22, 2013 at 6:38

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Indeed, there is a nice answer - which is "no". The point is that a necessary and sufficient condition for a graph to be equipped with a structure of a Schreier graph of the free group $F_2$ is that all its vertices have degree 4. You are asking about Schreier graphs whose group of automorphisms (as Schreier graphs, i.e., the automorphisms of the base graph which preserve Schreier labeling of edges) is transitive. Now, already your base graph need not have any automorphisms.

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  • $\begingroup$ I don't understand your answer. Do you agree that, if H is a normal subgroup, then a transitive action on the quotient group exists (left multiplication), and this action commutes with right-multiplication? (Unfortunately, my background on graphs is very limited.) $\endgroup$
    – Jorinde
    Aug 22, 2013 at 7:02
  • $\begingroup$ @R W: Could you give some explanation, how your answer applies if $H$ is normal? $\endgroup$
    – Jorinde
    Aug 23, 2013 at 4:24

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