Mathematics Asked by ycras on November 3, 2020

$S$ is a finite poset or lattice ; $A$ and $B$ two distinct elements. If there is at least one automorphism that maps $A$ to $B$, can I find one such automorphism that is an involution?

The set of automorphisms of $S$ is a subgroup of its permutation group, so any automorphism can be decomposed into products of cycles with disjoint supports. It seems to me that if I look for all the automorphisms of $S$ that map $A$ to $B$, if this set is not empty then I should be able to find one, $F$, with cycles of max lenght 2, in which case $F = F^{-1}$. But is this true?

In other terms, I’m looking for an automorphism that swaps A and B, and also swaps any pairs of elements as required by compatibility with the partial order (e.g., swap a cover of A with a cover of B), leaving all other elements unchanged. I believe that if any automorphism mapping A to B exists, then one such automorphism exists, but I’m stuck about how to prove it.

NB this is not homework (I’m close to 60) but amateur interest in lattices and posets, and I haven’t done any serious math since my PhD…. so thanks for being indulgent!

The answer is *no*; consider the following finite poset, where the two dashed segments are supposed to be two ends of the same segment:

Its automorphism group is cyclic of order $3$, so there are nontrivial automorphisms but no involutions.

In fact more is true; for every finite group $G$ there exists a finite poset $P$ with $operatorname{Aut}(P)cong G$. Moreover, given a set of generators of $G$ one can construct such a finite poset $P$ explicitly. For more details see this paper.

Answered by Servaes on November 3, 2020

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