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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?

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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.

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  • $\begingroup$ Great, thank you! $\endgroup$
    – B. Peet
    Feb 26 '19 at 2:21

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