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When is a Schreier Coset graph on a group $G$ with subgroup $H$ and symmetric generating set $S$(without identity) vertex transitive?

It is well known that when $H$ is normal, the Schreier coset graph corresponding is isomorphic to a Cayley graph and hence vertex-transitive. But, what is the characterization of $H$ and $S$ so that the graph be vertex-transitive. Typically, when is the Schreier coset graph is not vertex-transitive? Note that we neglect the self loops that may occur because of having elements in $S$ that also belong to $H$. Any hints? Thanks beforehand.

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  • $\begingroup$ as the action on (right, say) cosets is transitive, I would expect any such graph to be vertex-transitive. $\endgroup$
    – Dima Pasechnik
    Mar 14, 2022 at 9:49
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    $\begingroup$ @DimaPasechnik [almost] any regular graph is a Schreier graph, but not every regular graph is vertex-transitive; see e.g. this question for more references. $\endgroup$
    – ARG
    Mar 14, 2022 at 9:53
  • $\begingroup$ You should specify whether you consider the graph as labeled, whether you include self-loops, and whether (assuming the graph unlabeled), you identify multiples edges. $\endgroup$
    – YCor
    Mar 14, 2022 at 10:01
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    $\begingroup$ OK, I see. If I am not mistaken, one example of a vertex-transitive Schreier graph comes from S being a conjugacy class (or a union of such classes) in G, and H not containing any element of S. $\endgroup$
    – Dima Pasechnik
    Mar 14, 2022 at 10:14
  • $\begingroup$ @DimaPasechnik: Transitivity of the right action on right cosets tells you exactly that the Schreier graph is connected. It tells you nothing about the symmetries of the graph, because the right action on cosets (i.e. vertices) does not a priori extend to the edges. $\endgroup$
    – HJRW
    Mar 14, 2022 at 11:44

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