# Small “transitive” subsets of transitive groups

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.

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.

• 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? – Pavel Madaj Jul 9 at 18:06
• Since $G$ acts 2-transitively on $\Omega$, if $\omega \in \Omega$, the one-point stabiliser $G_\omega$ acts transitively on $\Omega \setminus \{\omega \}$. – Andreas Caranti Jul 9 at 18:21
• Ah, of course. Thanks. – Pavel Madaj Jul 9 at 18:37

$$S_4$$ 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.

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