# Is there an involutive automorphism mapping two given elements of a poset or lattice?

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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