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Let $G$ be a group and $\varphi$ be a transitive group action of $G$ on $\{1, 2, \dots, n\}$. Does a subset $A\subseteq G$ such that $|A|=n$ and $A$ "acts transitively" on $\{1, 2, \dots, n\}$ always exist? Since transitive group actions are usually defined only for groups what I mean by $A$ "acting transitively" is that $\forall i, j \in \{1, 2, \dots, n\}: \exists f \in A: \varphi(f,i) = j$.

I suspect that the answer is negative but I have not been able to find a counterexample so far.

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I think a not-terribly-easy counterexample can be found in this paper

Theo Grundhöfer and Peter Müller, Sharply 2-transitive sets of permutations and groups of affine projectivities. Beiträge Algebra Geom. 50 (2009), no. 1, 143–154.

In Theorem 1.7 one find the statement

Let $G$ be the Conway group Co3 in its doubly transitive action of degree 276. Then the stabilizer $G_\omega$ of degree 275 has no sharply transitive subset.

A sharply transitive subset should be exactly a set like the $A$ in the question.

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  • $\begingroup$ I don't quite see why $G_\omega$ acts transitively on the 275 element set $\Omega'$. Could you maybe point to the part of the article that clarifies this? $\endgroup$ – Pavel Madaj Jul 9 at 18:06
  • $\begingroup$ Since $G$ acts 2-transitively on $\Omega$, if $\omega \in \Omega$, the one-point stabiliser $G_\omega$ acts transitively on $\Omega \setminus \{\omega \} $. $\endgroup$ – Andreas Caranti Jul 9 at 18:21
  • $\begingroup$ Ah, of course. Thanks. $\endgroup$ – Pavel Madaj Jul 9 at 18:37
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$S_4$ has a transitive action of degree $6$, with point-stabiliser given by $\langle (1,2),(3,4)\rangle$. This has no regular subgroup (because the stabiliser does not have a complement, which is easy to see, since the only subgroups of order $6$ in $S_4$ are $S_3$'s, and these all intersect the point-stabiliser non-trivially).

The index $2$ subgroup $A_4$ is also transitive, and so also does not have a regular subgroup. These are the two smallest examples of transitive groups with no regular subgroups.

I just did a computer check, and they also don't have regular subsets.

Here's a short human-checkable proof of this fact in the case of $A_4$:

This action of $A_4$ on $6$ points is imprimitive, with three blocks of size $2$. The double transpositions preserve the blocks, fixing one point-wise and swapping the elements in the other two, the elements of order three permute the blocks.

We are looking for a regular subset $X$, which must have size $6$.

The eight $3$-cycles have only four distinct images for a particular point (since they always map a point outside its block), so there can't be more than four $3$-cycles in $X$.

This means that $X$ contains at least two non-$3$-cycles, but all these have images in common (the double transposition fix some points, so agree with the identity on those, and each pair of double transposition has a block in common which they swap), contradiction.

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    $\begingroup$ Looks good, and it's definitely better than my reference to an extremely more complicated example. $\endgroup$ – Andreas Caranti Jul 10 at 7:39

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