Groups of derangements: what is known about subgroups of a symmetric group $S_{n}$ that contain only derangements (plus the identity)?

A derangement is a permutation that has no fixed points.

My question is . . .

What is known about subgroups of a symmetric group $$S_{n}$$ that contain only derangements (plus the identity)?

It is clear that elements of such groups would need to be a product of $$\frac{n}{k}$$ disjoint $$k$$-cycles.

It would be simple to see cyclic groups, but this is certainly not all possible.

For example, if $$n=6$$, I can generate in GAP a subgroup generated by $$(163)(245)$$ and $$(15)(23)(46)$$ which contains only derangements plus the identity.

Is any more known?

Such a group is sometimes called semiregular (or free, or fixed point free, although the later two are more often reserved for group actions than permutation groups), see https://en.wikipedia.org/wiki/Group_action#Types_of_actions

All its orbits have size equal to the size of the group. In the extremal case, the group has just one orbit, of size the order of the group, in which case it is called regular.

All (abstract) groups can occur as regular groups, this is Cayley's Theorem. In particular, nothing can be said about the (abstract) structure of such a group.

• Great, thank you! Commented Feb 26, 2019 at 2:21

Frobenius' theorem gives an example of a situation in which a set of derangements forms a group.

It says that if $$G$$ is a finite group that acts on a finite set $$\Omega$$, and if this action satisfies the following two assumptions:

1. the action is transitive
2. every non-identity element of $$G$$ fixes at most one point of $$\Omega$$

then the set containing the identity and every derangement in $$G$$ is actually a subgroup! An action which meets these two criteria is called a Frobenius action.

This is a rare case in which we are rewarded for being naïve. The set of derangements in a group of permutations can't possibly form a group, because the identity isn't a derangement. A naïve person might try to repair this problem by simply adjoining the identity to the set of derangements. In the case of Frobenius actions, this actually works!

Interestingly, every known proof of Frobenius' theorem uses the character theory of finite groups, which, at first blush, looks like it has nothing to do with something as concrete as groups of permutations.