This question is pertaining to finite connected vertex-transitive graphs.

I recently read "Transitive permutation groups without semiregular subgroup" by Cameron, Giudici, Jones, Kantor, Klin, Marušič, Nowitz (publisher link; MSN review), where I found the concept of "elusive groups".

A permutation group $G$ acting on a set $X$ is called elusive if $G$ is transitive and contains no nontrivial semiregular subgroup (equivalently, no fixed-point-free element of prime order). I understand that transitive subgroups of elusive groups are elusive.

My question is: In light of the Polycirculant Conjecture, are the following statements is true?

Elusive groups cannot be the full automorphism group of any vertex-transitive graph.


Elusive groups cannot be any transitive subgroup of the full automorphism group of any vertex-transitive graph.

P.S. I am trying to understand how an elusive Group is a hindrance towards the Polycirculant Conjecture.

(Previously posted on MathSE.)

  • $\begingroup$ I have edited it, now. $\endgroup$
    – user52949
    Jul 14, 2019 at 14:10
  • 1
    $\begingroup$ If $G$ is transitive and elusive, it can just be viewed as transitive subgroup of the automorphism group of the complete graph (on the given set). So if I understand correctly, the second question has a negative answer. $\endgroup$
    – YCor
    Jul 14, 2019 at 14:37
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    $\begingroup$ Please wait more than a few hours before crossposting. $\endgroup$
    – verret
    Jul 14, 2019 at 21:01

1 Answer 1


Any elusive group of degree $n$ is a subgroup of the full automorphism group of the complete graph $K_n$, so your second statement is not true.

The Polycirculant Conjecture asserts that the full automorphism group of a digraph contains a derangement of prime order, i.e, is not elusive.

But while there are various families of elusive groups known, none of these groups are the full automorphism group of a digraph.

  • $\begingroup$ Thank you for your explanation. But does that mean we can not have a GRR on an Elusive group, say $M_{11}$? $\endgroup$
    – user52949
    Jul 14, 2019 at 17:02
  • $\begingroup$ No, it does not, I've already explained this to you (in a private communication). You are conflating abstract groups and permutation groups. Saying "$M_{11}$ is elusive" without specifying an action is meaningless. $M_{11}$ is only elusive in one particular action of degree $12$. All its other actions are non-elusive, including its regular one. $\endgroup$
    – verret
    Jul 14, 2019 at 21:04

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