Abstract
Let G,H be closed permutation groups on an infinite set X, with H a subgroup of G. It is shown that if G and H are orbitequivalent, that is, have the same orbits on the collection of finite subsets of X, and G is primitive but not 2transitive, then G=H.
Introduction
We consider closed permutation groups acting on an infinite set X; that is, subgroups G of Sym(X) which have the form Aut(M) for some relational structure M with domain X. Two permutation groups G,H on the set X are said to be orbitequivalent if, for every positive integer k, the groups G and H have the same orbits on the collection of unordered kelement subsets of X. This generalises a definition for finite permutation groups. Observe that if G,H are orbitequivalent, then they are each orbitequivalent to 〈G,H〉. Thus, to investigate such pairs, it suffices (in part) to consider G,H with H a subgroup of G. Easily, if G,H are orbitequivalent, then G is transitive (on X) if and only if H is transitive, and also G and H preserve the same systems of imprimitivity on X; so G is primitive on X (that is, preserves no proper nontrivial equivalence relation on X) if and only if H is primitive.
Our main theorem is the following. Our particular interest is in the case when X is countably infinite, but the proofs below do not use countability.
Theorem 1.1
Let G,H be orbitequivalent closed permutation groups on the infinite set X, with H≤G, and suppose that G is primitive but not 2transitive. Then H=G.
See Sect. 4 for a discussion of a possible strengthening, suggested by a referee, where the assumption H≤G is weakened, replaced by the assumption that H is closed and not 2transitive.
It is easily checked (see e.g. [3, Sect. 2.4]) that a subgroup of Sym(X) is closed in the sense defined above if and only if it is closed in Sym(X) in the topology of pointwise convergence on Sym(X) with respect to the discrete topology on X, where the basic open sets are cosets of pointwise stabilisers of finite sets; that is, basic open sets have the form \(U_{f_{0}}:=\{f\in\mathrm{Sym}(X): f\mbox{~extends~} f_{0}\}\) for some bijection f _{0} between finite subsets of X. Thus, if G≤Sym(X) is closed then H≤G is dense in G if and only if H and G have the same orbits on X ^{k} for all k>0. In particular, if H is a closed proper subgroup of G≤Sym(X), then for some k>0, some Gorbit on X ^{k} (the set of ktuples from X) breaks into more than one Horbit. The assumption in the theorem that G and H are closed seems essential; indeed, any subgroup H of Sym(X) is orbitequivalent to its closure, and, for example, the dense (and so orbitequivalent) subgroups of Sym(X) are exactly the subgroups of Sym(X) which are ktransitive for all positive integers k, and these seem hopelessly unclassifiable.
This paper takes its motivation from two sources. First, there is an extended literature on primitive orbitequivalent pairs of permutation groups on a finite set X; see for example [12, 21, 22]. Clearly, the symmetric and alternating groups Sym_{ n } and Alt_{ n }, in their natural actions on {1,…,n}, are orbitequivalent for n≥3. Also, if G is a permutation group on a finite set X and has a regular orbit U on the power set \(\mathcal{P}(X)\), and H is a proper subgroup of G, then H is intransitive on U, and so H is not orbitequivalent to G. It is shown in [4] that if X is finite then there are just finitely many primitive subgroups of Sym(X) which do not contain Alt(X) and have no regular orbit on \(\mathcal{P}(X)\) (and so could have an orbitequivalent proper subgroup). Such primitive groups G (with no regular orbit on X) are classified by Seress in [19], who then classifies all pairs of finite primitive orbitequivalent permutation groups (H,G) with H<G. There is further work on the finite imprimitive case in [20].
Let us say that two permutation groups G,H on a countably infinite set X are strongly orbitequivalent if they have the same orbits on the power set \(\mathcal{P}(X)\) (not just on the collection of finite subsets of X). As pointed out by the referee, it is known that if G,H are strongly orbitequivalent permutation groups on the countably infinite set X, with H≤G, and G is primitive and each orbit of a point stabiliser G _{ x } is finite, then H=G; there is no topological closure assumption here. Indeed, by [24, Corollary 6], G has a regular orbit on \(\mathcal{P}(X)\), from which the assertion follows.
The second source of motivation is more modeltheoretic, namely the study of homogeneous structures. Recall that a countable (possibly finite) structure M in a first order relational language is said to be homogeneous if every isomorphism between finite substructures of M extends to an automorphism of M. A natural generalisation, originally considered by Fraïssé in [9], is to say that the countable structure M is sethomogeneous if, whenever U,V are isomorphic finite substructures of M, there is g∈Aut(M) with U ^{g}=V. Finite sethomogeneous graphs are classified by Ronse in [18], and a very short proof was given by Enomoto in [8] that every finite sethomogeneous graph is homogeneous. There is a classification of finite sethomogeneous digraphs (allowing two vertices to be linked by an arc in each direction) in [10], building on a corresponding classification of finite homogeneous digraphs by Lachlan [13]. Also, there are initial results on countably infinite sethomogeneous structures, in particular graphs and digraphs, in [7] and [10]. The latter paper poses the following related question: given a homogeneous structure M, when does Aut(M) have a proper closed subgroup H which acts sethomogeneously on M, that is, has the same orbits as Aut(M) on the collection of unordered finite subsets of M? Equivalently, for which M does Aut(M) have a proper closed orbitequivalent subgroup? (Here, and throughout the paper, we use the same symbol M for a structure and for its domain.)
A countably infinite set X in the empty language is homogeneous, and has automorphism group Sym(X). By a theorem of Cameron [2], Sym(X) has just four orbitequivalent closed proper subgroups, namely Aut(X,<), Aut(X,B), Aut(X,C), and Aut(X,S). Here < is a dense linear order without end points on X, B is the (ternary) linear betweenness relation on X induced from <, C is the (also ternary) circular order on X induced from <, and S is the corresponding arity 4 separation relation. Observe that Aut(X,S)=〈Aut(X,B),Aut(X,C)〉 and is 3transitive but not 4transitive. Our conjecture below would strengthen Theorem 1.1 in the countable case by removing the ‘not 2transitive’ assumption.
Conjecture 1.2
Let G and H be distinct orbitequivalent primitive closed permutation groups on a countably infinite set X. Then G and H belong to the list Aut(X,<), Aut(X,B), Aut(X,C), Aut(X,S), Sym(X) described above.
Recall the following standard terminology, for a permutation group G on a set X, and an integer k>0: G is ktransitive if it is transitive on the ordered ksubsets of X; and G is khomogeneous if it is transitive on the unordered ksubsets of X. Also, if U is a subset of X then G _{{U}} and G _{(U)} denote, respectively, the setwise and pointwise stabilisers of U in G, and if x∈X then G _{ x }:={g∈G:x ^{g}=x}.
The proof of Theorem 1.1 splits into two cases:

(1)
G is primitive but not 2homogeneous;

(2)
G is 2homogeneous (and so primitive) but is not 2transitive.
Our main tool for both cases is the notion of local rigidity, introduced, so far as we know, in [7]. We shall say that a permutation group G acting on an infinite set X acts locally rigidly if for all finite U⊂X, there is some finite V⊂X such that U⊆V and the setwise stabiliser G _{{V}} of V fixes U pointwise. Likewise, a first order relational structure M is locally rigid if, for every finite substructure U of M, there is a finite substructure V of M containing U such that every automorphism of V fixes U pointwise. Clearly, if a relational structure M is locally rigid, then any subgroup of its automorphism group acts locally rigidly on M. Strengthening the notion of local rigidity, we shall later say that an infinite first order structure M is cofinally rigid if, for every finite substructure U of M, there is a finite substructure V of M with U⊆V such that the automorphism group of V is trivial. Here, ‘substructure’ is used in the standard modeltheoretic sense, corresponding to the graphtheoretic notion of ‘induced subgraph’.
Lemma 1.3
Let G,H be closed permutation groups on X, with H≤G. If G and H are orbitequivalent and G acts locally rigidly, then H=G.
Proof
It suffices to show that H has the same orbits as G on X ^{k} for all k. So let \(\overline{u}_{1}, \overline{u}_{2} \in X^{k}\) be in the same orbit of G; that is, there is g∈G such that \(\overline{u}_{1}^{g} = \overline{u}_{2}\). Let U _{1},U _{2}⊂X be enumerated by \(\overline{u}_{1}, \overline {u}_{2}\), respectively. Since G acts locally rigidly on X, there is finite V _{1}⊂X such that U _{1}⊆V _{1} and \(G_{\{V_{1}\}} \le G_{(U_{1})}\). Let \(V_{2} := V_{1}^{g}\). Then V _{1},V _{2} are in the same orbit of G on V _{1}sets, so by orbitequivalence there is some h∈H such that \(V_{1}^{h} = V_{2}\). Now \(gh^{1} \in G_{\{V_{1}\}}\), so in fact \(gh^{1} \in G_{(U_{1})}\). Thus \(\overline{u}_{1}^{h} = \overline{u}_{2}\) as required. □
In both cases (1) and (2) (G primitive, and either not 2homogeneous, or 2homogeneous but not 2transitive) we shall show that G acts locally rigidly on X. In fact, in the second case we show that G is a group of automorphisms of a cofinally rigid tournament. Our method to show the local rigidity of such actions stems from a similar result in [7], which we adapt. Formally, we view a graph Γ as a relational structure Γ=(X,R), where R is a symmetric irreflexive binary relation on X. Given a graph Γ, if x,y are vertices we write x∼y if x and y are adjacent, and let Γ(x):={v∈X:v∼x}, the neighbour set of x. We shall prove in Lemma 2.3 a strengthening of the following result.
Lemma 1.4
[7]
Let Γ be an infinite graph such that, for all distinct vertices x,y of Γ, the sets Γ(x)∖Γ(y) and Γ(y)∖Γ(x) are both infinite. Then Γ is locally rigid.
We draw attention to a basic Ramseytheoretic principle which is wellknown, for example in model theory, and used below in both the primitive not 2homogeneous case, and the 2homogeneous not 2transitive case.
Definition 1.5
Let L be a finite relational language, let M be a first order Lstructure, A a finite subset of the domain of M, and P _{1},…,P _{ r } disjoint subsets of M∖A, with P _{ i }:={p _{ i,0},…,p _{ i,n−1}} for each i=1,…,r. We say that P _{1}…,P _{ r } are mutually indiscernible over A if the following holds for any nonnegative integers e _{1},…,e _{ r }<n: for each j=1,…,r, let \(\bar{p}_{j},\bar{p}_{j}'\) be e _{ j }tuples from P _{ j }, each listed in increasing order (so if \(\bar {p}_{j}=(p_{j,i(1)},\ldots,p_{j,i(e_{j})})\), then i(1)<⋯<i(e _{ j })); then the map taking \(\bar{p}_{j}\) to \(\bar {p}_{j}'\) for each j, extended by the identity on A, is an isomorphism of Lstructures.
As a small example, let L consist of binary relations R and <, let A={a _{1},a _{2}}, P _{1}={p _{1i }:i∈ℕ}, and P _{2}={p _{2i }:i∈ℕ}, with a _{1},a _{2} and the p _{ ji } all distinct, and let M⊃A∪P _{1}∪P _{2}. Suppose that R is symmetric and irreflexive, so determines a graph structure on M, with P _{1} a complete subgraph with all its elements adjacent to a _{1}, and with no other adjacencies on A∪P _{1}∪P _{2}. Suppose also that p _{1i }<p _{1j } whenever i,j∈ℕ with i<j, and that < does not hold between other pairs from A∪P _{1}∪P _{2}. Then P _{1},P _{2} are mutually indiscernible over A.
Lemma 1.6
Let M,L,A be as in Definition 1.5 with M infinite, and let Q _{1},…,Q _{ r } be countably infinite disjoint subsets of M∖A. Let n be a positive integer. Then the following hold.

(i)
There are subsets P _{1}⊂Q _{1},…,P _{ r }⊂Q _{ r }, each of size n, such that P _{1},…,P _{ r } are mutually indiscernible over A (with respect to some indexing of each P _{ i }).

(ii)
If every relation of L is of arity at most 2, and P _{1},…,P _{ r } are as in (i), then for each i=1,…,r, either some relation of L induces a total order on P _{ i }, or every permutation of P _{ i }, extended by the identity on R _{ i }:=A∪⋃_{ j≠i } P _{ j }, is an automorphism of the induced Lstructure on R:=A∪P _{1}∪⋯∪P _{ r }.
Proof
(i) Let Q _{ i }:={q _{ i,j }:j∈ℕ} for each i=1,…,r. Let d be the maximum arity of a relation in L. Colour each subset {i _{1},…,i _{ rd }} of ℕ in such a way that given natural numbers i _{1}<⋯<i _{ rd } and k _{1}<⋯<k _{ rd }, the map
is an isomorphism over A if and only if {i _{1},…,i _{ rd }} and {k _{1},…,k _{ rd }} have the same colour. Finitely many colours are required; for since L is finite, there are finitely many possible isomorphism types of tuple of length r ^{2} d+A. By Ramsey’s Theorem, replacing ℕ by an infinite monochromatic subset if necessary, we may suppose that ℕ is monochromatic. Now let p _{ i,j }:=q _{ i,(i−1)(n+d)+j } for each i=1,…,r and j=0,…,n+d−1. Put \(P'_{i}:=\{p_{i,0},\ldots ,p_{i,n+d1}\}\) and P _{ i }:={p _{ i,0},…,p _{ i,n−1}} for each i=1,…,r.
We claim that P _{1},…,P _{ r } are mutually indiscernible over A. Indeed, let e _{1},…,e _{ r } be integers with 0≤e _{ j }<n for each n, and for each j let \(\bar{p}_{j}\) and \(\bar{p}_{j}'\) be e _{ j }tuples from P _{ j }, with \(\bar{p}_{j}=(p_{j,i_{j}(1)},\ldots,p_{j,i_{j}(e_{j})})\) for 0≤i _{ j }(1)<⋯<i _{ j }(e _{ j })<n, and \(\bar{p}_{j}'=(p'_{j,i_{j}(1)'},\ldots,p'_{j,i_{j}(e_{j})'})\) for 0≤i _{ j }(1)′<⋯<i _{ j }(e _{ j })′<n. We must show that the map fixing A pointwise and taking each \(\bar{p}_{j}\) to \(\bar{p}_{j}'\) is an isomorphism. Since the language has arity d, we may suppose that each \(\bar{p}_{j}\) has length at most d. Hence, by extending each \(\bar{p}_{j}\) and \(\bar{p}_{j}'\) by entries from \(P_{j}'\setminus P_{j}\) if necessary, we may suppose that each \(\bar{p}_{j}\) has length exactly d, that is, \(e_{j}=e_{j}'=d\) for each j=1,…,r. If we now write \(\bar {p}_{1}\ldots\bar{p}_{r}\) as a tuple of q _{ ij }, then the second indices are increasing: formally,
with i _{1}(1)<⋯<i _{1}(d)<⋯<(r−1)(n+d)+i _{ r }(1)<⋯<(r−1)(n+d)+i _{ r }(d). The same holds for \(\bar{p}_{1}'\ldots\bar{p}_{r}'\). It follows from the definition of the colouring that the map \(\bar{p}_{1}\ldots\bar {p}_{r}\to\bar{p}_{1}'\ldots\bar{p}_{r}'\) is indeed an isomorphism over A.
(ii) This is immediate from (i). Indeed, suppose that no binary relation of L induces a total order on P _{ i }. Then by the indiscernibility (applied to subsets of P _{ i } of size at most 3) each binary relation is symmetric on P _{ i }. It follows that any transposition (p _{ i,j },p _{ i,j+1}), extended by the identity on the rest of R, is an automorphism of R. Since such permutations generate a group fixing R _{ i } pointwise and inducing the symmetric group on P _{ i }, the result follows. □
The case of Theorem 1.1 when G is primitive but not 2homogeneous is handled in Sect. 2, and the 2homogeneous but not 2transitive case is treated in Sect. 3. Section 4 consists of some further observations, about bounds in local rigidity, approaches to Conjecture 1.2, and regular orbits on the power set. We also observe that our proofs give a slight strengthening of Theorem 1.1, namely Theorem 4.1.
G primitive but not 2homogeneous
In this section we prove the following.
Proposition 2.1
Let G be a primitive but not 2homogeneous permutation group on an infinite set X. Then the action of G on X is locally rigid.
The proposition follows rapidly from the following two lemmas. The first uses an argument in [16, Proposition 4.4].
Lemma 2.2
Let G be a primitive but not 2homogeneous permutation group on an infinite set X. Then there is a Ginvariant graph Γ with vertex set X such that for all distinct x,y∈X, the symmetric difference Γ(x)△Γ(y) is infinite.
Proof
Let U be any Gorbit on the collection of 2subsets of X. Then U is the edge set of a Ginvariant graph Γ _{0} with vertex set X, and as G is not 2homogeneous, Γ _{0} is not complete. For x∈X, write Γ _{0}(x) for the neighbour set of x in Γ _{0}. Define the equivalence relation ≡_{0} on X, putting x≡_{0} y if and only if Γ _{0}(x)△Γ _{0}(y) is finite. Then ≡_{0} is Ginvariant, so by primitivity ≡_{0} is trivial or universal. The lemma holds if ≡_{0} is trivial, so we shall suppose that ≡_{0} is universal.
Recall that a graph is locally finite if all of its vertices have finite degree.
Claim
Either Γ _{0} or its complement is locally finite.
Proof of Claim
Suppose not, and fix x∈X. Then both Γ _{0}(x) and X∖Γ _{0}(x) are infinite. If y∈Γ _{0}(x) then (as ≡_{0} is universal) Γ _{0}(y)∖Γ _{0}(x) is finite. Hence as G _{ x } has at most two orbits on Γ _{0}(x) there is k∈ℕ such that for all y∈Γ _{0}(x), we have Γ _{0}(y)∖Γ _{0}(x)≤k. Pick distinct z _{1},…,z _{ k+1}∈X∖({x}∪Γ _{0}(x)). Then as x≡_{0} z _{ i } for each i, each set Γ _{0}(z _{ i })∩Γ _{0}(x) is cofinite in Γ _{0}(x). Hence there is \(y\in\varGamma_{0}(x)\cap \bigcap_{i=1}^{k+1}\varGamma_{0}(z_{i})\). Then z _{1},…,z _{ k+1}∈Γ _{0}(y)∖Γ _{0}(x), so Γ _{0}(y)∖Γ _{0}(x)≥k+1, which is a contradiction. □
By the claim, replacing Γ _{0} by its complement if necessary, we may suppose that Γ _{0} is locally finite. By our original assumption that Γ _{0} is not complete or null (that is, an independent set), Γ _{0} has an edge. By primitivity, Γ _{0} is connected. Now let Γ be the graph on X whose edge set consists of the set of unordered pairs an even distance apart in Γ _{0}. Then Γ is also Ginvariant. Pick v _{0}∈X, and choose a Γ _{0}path v _{0}∼v _{1}∼v _{2}∼⋯ so that the distance d _{0}(v _{0},v _{ i }) between v _{0} and v _{ i } in Γ _{0} equals i for each i (this is certainly possible, for example by König’s Lemma). Then v _{2i }∈Γ(v _{0})∖Γ(v _{1}) for each i>0. Thus Γ(v _{0}) and Γ(v _{1}) have infinite symmetric difference. It follows that ≡ is not universal, where ≡ is the Ginvariant equivalence relation on X defined by putting x≡y whenever Γ(x)△Γ(y) is finite. Since G is primitive, ≡ is trivial, so Γ(x)△Γ(y) is infinite for all distinct x,y∈X. □
In the next lemma, and later in the paper, if A,B are sets we write A⊂_{ f } B if B∖A is infinite and A∖B is finite. The lemma below extends Lemma 1.4, since under the assumptions of that lemma, x<y (as defined below) never holds. If u,v,w are distinct vertices of the graph Γ, we say w separates u and v if w∈(Γ(u)△Γ(v))∖{u,v}, and call a collection of such vertices w a separating set for u and v.
Lemma 2.3
Let Γ=(X,R) be an infinite graph, and suppose that Γ(x)△Γ(y) is infinite for any distinct x,y∈X. Write x<y whenever Γ(y)⊂_{ f } Γ(x). Then the structure Γ _{<}=(X,R,<) is locally rigid.
Proof
We adapt the proof of Proposition 6.1 from [7]. So let U={u _{1},…,u _{ n }} be a finite subset of X. We aim to find finite V with U⊂V⊂X, such that Aut(V,R,<) fixes U pointwise.
For each u _{ i },u _{ j }∈U, with i<j, we find an infinite separating set Q _{ ij }⊂X∖U as follows: if u _{ i }<u _{ j }, then let Q _{ ij }⊂Γ(u _{ i })∖(Γ(u _{ j })∪{u _{ j }}); if u _{ j }<u _{ i }, then let Q _{ ij }⊂Γ(u _{ j })∖(Γ(u _{ i })∪{u _{ i }}); and if u _{ i }∥u _{ j } (that is, u _{ i },u _{ j } are incomparable under <), then let Q _{ ij }⊂Γ(u _{ i })∖(Γ(u _{ j })∪{u _{ j }}). By refining the Q _{ ij } further, we may suppose that if i<j and k<l with (i,j)≠(k,l), then Q _{ ij }∩Q _{ kl }=∅.
Let K be a positive integer, chosen sufficiently large for the argument below. By Lemma 1.6 with respect to the language L={R,<}, we can choose for each i<j a subset P _{ ij } of Q _{ ij } with P _{ ij }=K, such that the collection of all sets P _{ ij } is mutually indiscernible over U. Let W=U∪⋃(P _{ ij }:1≤i<j≤n). Then each P _{ ij } carries a complete or null induced graph structure, and for each x,y∈P _{ ij } and z∈W∖P _{ ij }, we have x∼z if and only if y∼z.
For any subset Y of X, define the equivalence relation ≈_{ Y } on Y, where, for x,y∈Y, x≈_{ Y } y if and only if (Γ(x)△Γ(y))∩Y⊆{x,y}. Then ≈_{ Y }classes always carry a complete or null induced subgraph structure. If Z is an ≈_{ Y }class, then for z _{1},z _{2}∈Z and y∈Y∖Z, we have y∼z _{1}⇔y∼z _{2}; in particular, all elements of Sym(Y)_{(Y∖Z)} induce automorphisms of (Y,R). Observe that if Y _{1}⊂Y _{2}⊆X and x,y∈Y _{1}, then \(x\approx_{Y_{2}} y \) implies \(x\approx_{Y_{1}} y\). We often identify such Y with the induced subgraph (Y,R∩Y ^{2}) of Γ which it carries. Thus, ≈_{ Y } is Aut(Y,R)invariant.
Now, each P _{ ij } lies in a ≈_{ W }class of W. Deleting some sets P _{ ij } if necessary (only where elements of distinct sets P _{ ij } are ≈_{ W }equivalent, and retaining the assumption that any two distinct elements of U are separated by some set of form P _{ ij }), we may suppose: no two elements x,y in distinct sets P _{ ij },P _{ kl } are ≈_{ W }equivalent. Also, ≈_{ W }classes contain at most one point of U; for if u _{ i },u _{ j }∈U with i<j then there is a nonempty set P _{ kl } whose elements separate u _{ i } and u _{ j }, so witness that \(u_{i}\not\approx_{W} u_{j}\). Let \(m={n\choose2}\), an upper bound on the number of distinct sets P _{ ij }. Adjusting the P _{ ij } and hence W further, we arrange the sizes of the P _{ ij } so that P _{ ij }≥2 for each i,j and distinct ≈_{ W }classes of W of size at least two all have different sizes, with size at most m+1∈ℕ; the number K was chosen sufficiently large for this to be possible, so \(K\geq \frac{(m+1)(m+2)}{2}1\), but due to a later application of the Pigeonhole Principle K will be chosen larger than this. Now every ≈_{ W }class of W of size greater than 1 consists of a set P _{ ij }, possibly together with an element of U. We will say that a set Y⊆X is huge if Y>m+1.
Any automorphism of (W,R) will preserve ≈_{ W }, and will fix setwise each ≈_{ W }class of size at least two (as these classes all have different sizes). Hence, if no element of U is ≈_{ W }equivalent to any element of any P _{ ij } (that is, elements of U lie in ≈_{ W }classes of size 1), then as the P _{ ij } separate the elements of U, any automorphism of W will fix U pointwise, as required. So the concern is that some ≈_{ W }class C in W of size at least two might consist of a set P _{ ij } together with some u∈U. In such a case there would be an automorphism of (W,R) mapping u to some vertex in C∖{u}, indeed, as observed above, any permutation of C, extended by the identity on W∖C, gives an automorphism of (W,R).
So suppose u∈C∩U as in the last paragraph. By the Pigeonhole Principle (retaining all the above reductions, so initially working with larger sets P _{ ij }) we may suppose for all such C,u that either uc (that is, u and c are incomparable with respect to <) for all c∈C∖{u}, or u<c for all c∈C∖{u}, or c<u for all c∈C∖{u}. For such u,c and C, we add a finite set S _{ cu } of additional vertices of Γ to Waccording to the following recipe.
If C is null, then for each c∈C∖{u} for which u≮c, the set Γ(c)∖Γ(u) is infinite, and we choose infinite \(S_{cu}'' \subset\varGamma(c) \setminus(\varGamma (u)\cup W)\). If C is complete, then for each c∈C∖{u} for which c≮u, the set Γ(u)∖Γ(c) is infinite, and we choose infinite \(S_{cu}'' \subset\varGamma(u) \setminus(\varGamma (c)\cup W)\). In other cases (C null and u<c for all c∈C∖{u}, or C complete and c<u for all c∈C∖{u}) we do not add any corresponding set \(S_{cu}''\). The sets \(S_{cu}''\) are chosen so that if (c,u)≠(c′,u′) then \(S_{cu}''\cap S_{c'u'}''=\emptyset\). We may suppose, again by the Pigeonhole Principle, that for each such c,u, either cx for all \(x\in S_{cu}''\), or c<x for all \(x\in S_{cu}''\), or x<c for all \(x\in S_{cu}''\). By Lemma 1.6 with respect to L={R,<}, we may replace each set \(S_{cu}''\) by a very large finite subset \(S_{cu}'\) in such a way that the collection of all such sets \(S_{cu}'\) is mutually indiscernible over W. Let V′ be the union of all such sets \(S_{cu}'\) and of W. Observe that each \(S_{cu}'\) carries a complete or null induced subgraph of Γ, and for each \(x,y \in S_{cu}'\) and \(z \in V' \setminus S_{cu}'\), we have x∼z if and only if y∼z. In particular, any two elements of a set \(S_{cu}'\) are ≈_{ V′}equivalent. Finally, since the \(S_{cu}'\) could be chosen as large (finite) as needed, we may choose huge subsets \(S_{cu}\subset S_{cu}'\) for all such c,u, so that the following holds, where V is the union of W and the S _{ cu }: all elements of V∖W lie in huge ≈_{ V }classes (which may meet W), and distinct huge ≈_{ V }classes have different sizes. It follows that each huge ≈_{ V }class is fixed setwise by any automorphism of (V,R).
We aim to show that every automorphism of (V,R,<) must fix U pointwise, which will complete the proof of the lemma. As a first step, observe that every huge ≈_{ V }class S contains some set S _{ cu }. We claim that no huge ≈_{ V }class meets U. For suppose S is a huge ≈_{ V }class, with a∈U∩S. There is u∈U and c∈W∖U, and a ≈_{ W }class C with c,u∈C, such that S⊃S _{ cu }. Clearly a=u, since otherwise, as S _{ cu } separates c from u, also a would separate c and u and lie in U⊂W, contradicting that c≈_{ W } u. Now if C is null then, by our rule for the process adding S _{ cu }, all vertices of S∖{u} are adjacent to c; hence c separates u from other elements of S, so \(u\not\in S\), a contradiction. Likewise, if C is complete, then all vertices of S∖{u} are nonadjacent to c, so again c separates u from the rest of S, so u∉S. This proves the claim.
Claim
Let g∈Aut(V,R,<). Then there is h∈Aut(V,R,<)_{(U)} such that gh fixes W setwise.
Proof of Claim
There are distinct (so differentsized) huge ≈_{ V }classes S _{ j } (for j∈J), each fixed setwise by g, such that V∖W⊆⋃_{ j∈J } S _{ j }. We may assume that W is not fixed setwise by g, as otherwise the claim is trivial. Hence, for some j∈J, we have (S _{ j }∩(V∖W))^{g}≠S _{ j }∩(V∖W).
First, we show that S _{ j }∩W=1. There are u∈U, and some ≈_{ W }class C containing distinct elements u,c of W, such that S _{ j }⊇S _{ cu }. We may suppose that C is null, and S _{ cu }⊂Γ(c)∖Γ(u), as the other case where C is complete and S _{ cu }⊂Γ(u)∖Γ(c) is similar. Now no element of W∖{u,c} could lie in S _{ j }, for otherwise it would separate u from c in W contradicting that u≈_{ W } c. Hence, since S _{ j } is a huge ≈_{ V }class so is disjoint from U, we have S _{ j }∩W⊆{c}. Thus, due to the existence of the element g, we have S _{ j }∩W={c}. In fact, S _{ j }=S _{ cu }∪{c}: for if c′∈C∖{u,c} then S _{ c′u }≠∅ but c′ separates elements of S _{ c′u } from c so elements of S _{ c′u } do not lie in S _{ j }; and if u′∈U∖{u}, then no set of form S _{ du′} could be a subset of S _{ j }, for otherwise c (in W) would separate d from u′ so the set S _{ du′} would not have been added.
By our assumption, there is v∈S _{ cu } such that v ^{g}=c. It is not possible that S _{ cu } is totally ordered by <; this follows easily from the facts that g induces an automorphism of (S _{ cu }∪{c},<), and the earlier assumption that either c<x for all x∈S _{ cu }, or x<c for all x∈S _{ cu }, or cx for all x∈S _{ cu }. It follows by Lemma 1.6(ii) that any permutation of S _{ cu }, extended by the identity on the rest of V, is an automorphism of (V,R,<). In particular distinct elements of S _{ cu } are <incomparable, so as v ^{g}=c, the set S _{ j } is an antichain with respect to <. Now it could not happen that there is some t∈V∖S _{ j } whose <relation to c is different from its <relation to all other elements of S _{ j }. For otherwise \(t^{g^{1}}\) would have a different <relation to v and to all other elements of S _{ j }, contradicting the mutual indiscernibility in the construction of S _{ cu }. It follows that if g′ is the inverse of g on S _{ j } and the identity on the rest of V, then g′∈Aut(V,R,<). The element h of the claim will be a product of elements of the form g′, each acting on a different huge ≈_{ V }class. □
To finish the proof of the lemma, let g∈Aut(V,R,<), and let h be as in the claim. We must show u ^{g}=u for all u∈U. Now by construction gh fixes W setwise, and we claim that gh fixes U setwise. Indeed, suppose for a contradiction that u∈U and \(u^{gh}\not\in U\). As the ≈_{ W }classes of W of size greater than one are all of different sizes, they are all fixed setwise by gh. Hence, as all elements of W∖U lie in ≈_{ W }classes of size greater than one, u ^{gh} and hence also u lie in some ≈_{ W }class C of size greater than one. Now, by the construction of V from U, either u is the greatest or least element of C with respect to <, or u and u ^{gh} are separated by some huge set of form \(S_{u,u^{gh}}\). The first case is impossible as gh preserves <. The second case is also impossible, since as the huge ≈_{ V }classes all have different sizes, they are fixed setwise by gh. Thus, as claimed, gh induces an automorphism of (W,R,<) which fixes U setwise. Hence gh fixes U pointwise; for any two distinct elements of U are separated by an ≈_{ W }class of size greater than one, and all such classes have different sizes, so are fixed setwise by gh. Thus, g fixes U pointwise. □
Remark 2.4
Careful inspection of the above proof shows that if U=n, then V may be chosen to have size at most O(n ^{8}). For in constructing W from U, if \(m={n\choose2}\) we add at most m sets P _{ ij }, each of size at least 2 and all of different sizes, so W=n+k where \(k:=W\setminus U\leq\frac{(m+1)(m+2)}{2}1\). Then in adding the sets S _{ cu } to obtain V, we add at most k such sets, each of size at least m+2, and all of different sizes. Thus, \(V\setminus W\leq(m+2)+(m+3)+\cdots+(m+k+1)=\frac{k}{2}(2m+k+3)\). Thus, \(V\leq\frac{1}{2}(2n+k(2m+k+5))\). This is used in Theorem 4.1 below.
Proof of Proposition 2.1
By Lemma 2.2, there is a Ginvariant graph Γ on X such that for all distinct x,y∈X, the set Γ(x)△Γ(y) is infinite. The partial order < defined in Lemma 2.3 is clearly also Ginvariant. The proposition thus follows immediately from that lemma. □
G 2homogeneous but not 2transitive
By Proposition 2.1, to complete the proof of Theorem 1.1 it suffices to prove the following.
Proposition 3.1
Let G be a 2homogeneous but not 2transitive permutation group on an infinite set X. Then the action of G on X is locally rigid.
Recall that a tournament is a directed loopless digraph (T,→) such that for any distinct vertices x,y, exactly one of x→y or y→x holds. A group which is 2homogeneous but not 2transitive has just one orbit on unordered 2sets, but two orbits on ordered pairs of distinct elements. Each of these orbits is the arc set of a Ginvariant tournament with vertex set X. Thus, to prove Proposition 3.1, we develop analogues of the methods of Sect. 2, but for tournaments.
Let → denote the arc relation in a tournament T=(X,→), and let G=Aut(T). For x∈X, we let Γ ^{+}(x):={y∈X:x→y}, the set of outneighbours of x. For x,y,z∈X, we say that z separates x,y if x→z→y or y→z→x. Furthermore Z⊂X separates x,y if each z∈Z separates x,y. We write x→Z if x→z for each z∈Z. We say that a tournament is rigid if it admits no nonidentity automorphisms.
Proposition 3.2
Let T=(X,→) be an infinite tournament such that for any distinct x,y∈X, the sets Γ ^{+}(x)∖Γ ^{+}(y) and Γ ^{+}(y)∖Γ ^{+}(x) are both infinite. Then T is cofinally rigid.
We first isolate an easy lemma, used to prove Proposition 3.2, in case it has other uses. It may be known.
Let T=(X,→) be a tournament. We will say that A⊂X is a nice set if A≠∅ and for all a _{1},a _{2}∈A and v∈X∖A, we have a _{1}→v if and only if a _{2}→v. (That is, all vertices in a nice set are related in the same way to vertices outside the nice set; equivalently, no vertex outside a nice set separates a pair of vertices inside the nice set.) Note that vacuously any singleton is a nice set, and X is nice. Furthermore, we will say that A⊂X is a good set, if A is totally ordered by → and is nice. We consider the maximal good subsets of X, that is, good sets A such that there is no good set A′⊂X with A′⊃A.
Lemma 3.3
If T=(X,→) is a tournament, then the maximal good subsets of X form a partition of X.
Proof
We claim that if A is good and B≠A is maximal good (where A,B⊆X), then either A⊂B or A∩B=∅. To see this, let d∈A∩B, and let C=A∪B. We show that C is good, which ensures B=C.
Let c _{1},c _{2}∈C, v∈X∖C. Now c _{1}→v if and only if d→v if and only if c _{2}→v. This holds because A and B are both nice and d∈A∩B. Hence C is nice. If C is not totally ordered, then there is some 3cycle c _{1}→c _{2}→c _{3}→c _{1} in C. Since A and B are both totally ordered, we must have at least one of these points in A∖B and one in B∖A. Suppose c _{1}∈A∖B and c _{2}∈B∖A (the other case is similar). Then if c _{3}∈A, then c _{2} separates c _{1},c _{3}, contradicting the fact that A is nice. Otherwise c _{3}∈B, then similarly c _{1} separates c _{2},c _{3}, contradicting the fact that B is nice. Hence C is totally ordered. Now B⊆C, and C is good, so A⊆B=C by maximality of B.
Observe also that if (A _{ i })_{ i∈I } is an increasing sequence of good sets (that is, I is totally ordered and A _{ i }⊆A _{ j } whenever i<j) then A:=⋃_{ i∈I } A _{ i } is good: indeed, if either clause in the definition of good is violated then it would be violated in some A _{ i }. Also, each singleton in X is a good set. The lemma now follows immediately from the claim, using Zorn’s Lemma if X is infinite. □
Proof of Proposition 3.2
Let U={u _{1},…,u _{ n }}⊂X. For any distinct i,j∈{1,…,n}, the set Γ ^{+}(u _{ i })∖Γ ^{+}(u _{ j })={v∈X:u _{ i }→v→u _{ j }} is infinite. Hence by Ramsey’s Theorem (indeed, a simple case of Lemma 1.6), there is U _{ ij }⊆Γ ^{+}(u _{ i })∖(Γ ^{+}(u _{ j })∪{u _{ j }}) with U _{ ij }=ℵ_{0}, such that U _{ ij } is totally ordered by →. Note that the sets U _{ ij },U _{ ji } both separate u _{ i },u _{ j } (since u _{ i }→U _{ ij }→u _{ j }, and u _{ j }→U _{ ji }→u _{ i }). We may choose the U _{ ij } so that if (i,j)≠(k,l) then U _{ ij }∩U _{ kl }=∅.
Claim 1
Let N be any positive integer. Then there are finite subsets V _{ ij } of U _{ ij } (for all distinct integers i,j with 1≤i,j≤n) of size N such that the following holds, where T′ is the induced subtournament of T with vertex set U∪⋃_{ i≠j } V _{ ij }: for any distinct i,j∈{1,…,n}, and for each x,y∈V _{ ij } and v∈T′∖V _{ ij }, we have x→v if and only if y→v.
Proof of Claim 1
This is an immediate application of Lemma 1.6. □
Provided we initially choose N large enough, we may cut the V _{ ij } down further, and so suppose that each set V _{ ij } has size exactly 2^{r} for some r≥2, and that distinct sets V _{ ij } and V _{ kl } have distinct sizes. Observe (for use in Theorem 4.1) that T′ has \(n+\sum_{i=2}^{m+1} 2^{i}\) vertices where \(m=2{n\choose2}\), that is, it has \(n+2^{n^{2}n+2}2\) vertices. We claim that T′ is rigid, which suffices to prove the lemma. Let V denote the vertex set of T′ (a union of U and the sets V _{ ij }).
The sets V _{ ij } are clearly all good, though possibly not maximal good. Hence, by Lemma 3.3, if B∩V _{ ij }≠∅ and B is maximal good, then V _{ ij }⊆B.
The idea of the proof is as follows. First observe that automorphisms of the subtournament (V,→) of T preserve the family of maximal good sets. We aim to show that by our construction of V, all nonsingleton maximal good sets in V have different sizes, so in fact each is fixed setwise, and hence pointwise, by any automorphism. We then show that if some automorphism α of (V,→) fixes pointwise all nonsingleton maximal good subsets of V, then α fixes V pointwise.
Claim 2
If A is a good subset of V, then A∩U≤1.
Proof of Claim 2
Suppose u _{1},u _{2}∈A∩U, with u _{1}≠u _{2}. We have u _{1}→V _{12}→u _{2}. Since A is good, we must have V _{12}⊂A: otherwise any y∈V _{12}∖A separates u _{1},u _{2}, contradicting the fact that A is nice. Similarly, we have u _{2}→V _{21}→u _{1}, and we must have V _{21}⊂A. But then we have {u _{1},u _{2}}∪V _{12}∪V _{21}⊆A, and u _{1}→V _{12}→u _{2}→V _{21}→u _{1}. But then A is not totally ordered by →, which contradicts the fact that A is good. □
Thus, maximal good sets are unions of sets V _{ ij } with at most one element of U added (this includes the case of a singleton point of U). Then by our choice of the sizes of the V _{ ij } in the construction, any two nonsingleton maximal good sets have different sizes. (For let the V _{ ij } have sizes n _{1},…,n _{ t }, say. These were chosen as distinct powers of 2, and so all numbers of the form \(n_{i_{1}} + \cdots+ n_{i_{s}}\) or \(n_{i_{1}} + \cdots+ n_{i_{s}} +1\) are distinct.) Hence any automorphism of V fixes each nonsingleton maximal good set setwise, and hence also pointwise since each is totally ordered and so rigid. Thus any automorphism fixes all elements of V∖U pointwise, and so also fixes U pointwise; indeed, for each pair of elements of U there is some Z⊂V∖U separating the pair, and so no automorphism can move points of U. □
Corollary 3.4
Let T be an infinite tournament with 2homogeneous automorphism group. Then T is cofinally rigid.
Proof
By Ramsey’s Theorem, there is a subtournament of T of the form {x _{ i }:i∈ℕ} with x _{ i }→x _{ j } if and only if i<j (or possibly with all arcs reversed). Clearly, if i<j, then Γ ^{+}(x _{ j })△Γ ^{+}(x _{ i })≥j−i−1. By 2homogeneity of Aut(T), there is d∈ℕ∪{ℵ_{0}} such that if x≠y then Γ ^{+}(x)△Γ ^{+}(y)=d. Hence, d≥n for each n∈ℕ, so d=ℵ_{0}.
We may suppose that T is not totally ordered by →; for otherwise, if U is a finite set of vertices then the induced subtournament on U is totally ordered and so rigid, and hence T is cofinally rigid (as in the definition of cofinal rigidity in the Introduction we may put V=U). By Proposition 3.2, the proof of the corollary now reduces to the following claim.
Claim
For all distinct x,y∈X, the sets Γ ^{+}(x)∖Γ ^{+}(y) and Γ ^{+}(y)∖Γ ^{+}(x) are both infinite.
Proof of Claim
Suppose that for some u,v∈X with u≠v, the set Γ ^{+}(u)∖Γ ^{+}(v) is infinite, but Γ ^{+}(v)∖Γ ^{+}(u) is finite. Now, using 2homogeneity, define an order relation < on X, such that x<y if and only if Γ ^{+}(x)∖Γ ^{+}(y) is infinite. This is a Aut(T)invariant partial order on X, containing comparable pairs. By 2homogeneity, it follows that < is a total order, and it or its reverse agrees with →. This contradicts the above assumption. □
Proof of Proposition 3.1
As noted above, there is a Ginvariant tournament T with vertex set X, whose arc set is a Gorbit on {(x _{1},x _{ x })∈X ^{2}:x _{1}≠x _{2}}. The proposition now follows immediately from 2homogeneity and Corollary 3.4. □
Proof of Theorem 1.1
This is immediate from Lemma 1.3 and Propositions 2.1 and 3.1. □
Further remarks
The proof of Theorem 1.1 yields the following, where, for any positive integer n, X ^{[n]} denotes the set of nelement subsets of X: there is a function f:ℕ→ℕ such that for any l∈ℕ, if H≤G are closed permutation groups on an infinite set X with G primitive but not 2transitive, and G and H have the same orbits on X ^{[n]} for all n≤f(l), then G and H have the same orbits on X ^{m} for all m≤l. An upper bound for f is given by the cardinality of V in terms of U in the definition in the Introduction of a group G acting locally rigidly. By the proofs of Propositions 2.1 and 3.1, we obtain the following slight strengthening of Theorem 1.1, probably far from best possible. Observe that with m and k as in Remark 2.4, \(\frac{1}{2}(2n+k(2m+k+5))\leq n+2^{n^{2}n+2}2\) for all n>1, so the bound in the proof of Proposition 3.1 dominates.
Theorem 4.1
Let G,H be closed permutation groups on the infinite set X, with G primitive but not 2transitive on X, and with H≤G. Let n∈ℕ, and suppose that G and H have the same orbits on the set X ^{[l]} for each \(l\leq n+2^{n^{2}n+2}2\). Then G and H have the same orbits on X ^{m} for each m≤n.
Theorem 1.1 requires the assumption of primitivity. For example, \(\mathrm{Aut}({\mathbb{Q}},<) {\rm Wr} C_{2}\) is orbitequivalent to \(\mathrm{Sym}({\mathbb{Q}}) {\rm Wr} C_{2}\) (in the natural imprimitive action). However, a proof of Conjecture 1.2 should yield a lot of information about the imprimitive case.
A proof of Conjecture 1.2, at least if via local rigidity, would appear to require arguments considerably more involved than those of this paper. As an example, suppose that G is 2primitive (that is, 2transitive and with primitive point stabilisers) but not 3homogeneous on the infinite set X. We conjecture that G acts locally rigidly. There is a Ginvariant 3hypergraph Γ on X, and we would like to show that Γ (possibly expanded by some other Ginvariant relations) is locally rigid. Given x∈X, there is an induced graph Γ _{ x } on X∖{x} on which G _{ x } acts primitively. However, it is not clear that local rigidity of Γ _{ x } transfers to local rigidity of Γ, or more generally that a straightforward induction on the degree of transitivity of G can be made to work. There may also be an approach to local rigidity of hypergraphs using [15, Lemma 2.5] and related results.
We cannot even prove the conjecture under the assumptions that X is countable and G is locally compact (that is, there is some finite F⊂X such that all orbits of G _{(F)} on X are finite). Even the case when G is countable is open. A first class to consider would be that of primitive groups with finite point stabiliser, for which Smith [23] gives a usefullooking version of the O’NanScott Theorem.
However, as evidence for the conjecture, we observe that a potential source of counterexamples, suggested by the family of closed supergroups of Aut(ℚ,<) listed in Conjecture 1.2, fails. Indeed, let (T,<) be any of the countable 2homogeneous trees (that is, semilinear orders) classified by Droste in [6]. There is a family of interesting primitive closed permutation groups associated with Aut(T,<), namely the primitive Jordan permutation groups with primitive Jordan sets classified in [1]: we have in mind Aut(T,<), the automorphism group of the ternary general betweenness relation on T induced from <, the automorphism group of the corresponding countable Cstructure, a structure whose elements are a dense set of maximal chains in (T,<), and the automorphism group of the corresponding Drelation (a quaternary relation on the set of ‘directions’ of the betweenness relation). It can be checked that each of these groups acts locally rigidly. We omit the details.
One referee asked whether, in Theorem 1.1, the assumption that H≤G can be replaced by the assumption that H is closed, orbitequivalent to G, and not 2transitive. The problem here is that we cannot apply the theorem to the group G as a subgroup of the (topological) closure of 〈G,H〉, as the latter may be 2transitive; and our proof of Theorem 1.1 depends essentially on Lemma 1.3, which requires the assumption H≤G. In fact, as the following example shows, the referee’s question has a negative answer.
Example 4.2
Let <_{1} be the natural order on ℚ, and let <_{2} be defined on ℚ by putting x<_{2} y if y<_{1} π<_{1} x, or if x<_{1} y and x,y both lie in ℚ∩(π,∞) or both in ℚ∩(−∞,π). Put G=Aut(ℚ,<_{1}) and H:=Aut(ℚ,<_{2}). Then G,H are distinct 2homogeneous but not 2transitive closed groups which are each orbitequivalent to Sym(ℚ), and so are orbitequivalent to each other. In fact the topological closure of 〈G,H〉 is the (2transitive) automorphism group of the natural circular order on ℚ.
Possibly this is essentially the only such example: in any such examples G,H must be 2homogeneous, not 2transitive permutation groups on X such that the two nondiagonal Gorbits on X ^{2} are not the same as those of H. So we modify the referee’s question, and ask the following.
Question 4.3
Is there an example of a pair of distinct orbitequivalent primitive but not 2transitive closed permutation groups G and H on an infinite set X, such that at least one of G,H does not preserve a total order on X?
In [5] a permutation group G on X is defined to be orbitclosed if there is no H≤Sym(X) which properly contains G and is orbitequivalent to G. Such G will be a closed permutation group, and Conjecture 1.2 asserts that if X is countably infinite then the only primitive closed permutation groups which are not orbitclosed are the proper subgroups of Sym(X) listed in that conjecture. In [5] the authors define G≤Sym(X) to be a relation group if there is a collection R of finite subsets of X such that
Clearly any relation group is orbitclosed. Also, by [5, Corollary 4.3], any finite primitive orbitclosed group is a relation group. We do not know whether this holds without finiteness, and in particular cannot answer the following question, to which Siemons drew our attention.
Question 4.4
Is Aut(ℚ,<) the only primitive but not 2transitive closed permutation group of countable degree which is not a relation group?
As a small example, let Γ _{3} be the universal homogeneous 2edgecoloured graph with edges coloured randomly red or green; that is, the unique countably infinite homogeneous 2edgecoloured graph such that for any three finite disjoint sets U,V,W of vertices, there is a vertex x not adjacent to any vertex in U, adjacent by a red edge to each element of V and by a green edge to each element of W. At first sight, G=Aut(Γ _{3}) is not a relation group, but in fact it is a relation group; for we may take R to consist of the 2sets joined by a red edge and the 3sets which carry a green triangle.
Our remarks in the Introduction suggest a further question. Again, for convenience, we shall consider actions on a countably infinite set X. A subset Y of X is a moiety of X if Y=X∖Y.
Question 4.5
Which primitive closed permutation groups G on a countably infinite set X have a regular orbit on moieties?
To say that G has a regular orbit on moieties of X is close to the condition, in the language of [14, 24] and elsewhere, that any first order structure M on X with G=Aut(M) has distinguishing number 2, that is, that G has a regular orbit on the power set \(\mathcal{P}(X)\). Some results on these conditions are obtained in [14]. For example, if M is a homogeneous structure such that the collection of finite structures which embed in it is a ‘free amalgamation class’, and Aut(M) is primitive but for some k is not ktransitive, then Aut(M) has a regular orbit on moieties, and so distinguishing number 2. In particular, this holds for the random graph, as follows already from [11, Theorem 3.1]. On the other hand, as noted in [14] it is easily seen that Aut(ℚ,<) has no regular orbit on \(\mathcal {P}({\mathbb{Q}})\). Indeed, clearly there is no regular orbit on finite sets; and if A is a moiety of ℚ whose setwise stabiliser is trivial, then A is dense and codense in ℚ, but the structure (ℚ,<,P), where P is a unary predicate naming a dense codense set, is homogeneous so admits \(2^{\aleph_{0}}\) automorphisms.
This suggests the following strengthening of orbitequivalence. Recall from the Introduction the notion of strong orbit equivalence. The following conjecture is implied by Conjecture 1.2, for it is easily seen that the five closed groups containing Aut(ℚ,<) all have different orbits on \(\mathcal {P}({\mathbb{Q}})\). For example, Aut(ℚ,<) has an orbit consisting of increasing subsets of order type ω with rational supremum, but this family of sets is not invariant under the automorphism groups of the induced circular order or linear betweenness relation.
Conjecture 4.6
Let G,H be strongly orbitequivalent closed permutation groups on the countably infinite set X. Then H=G.
Again, the assumption that the groups are closed is necessary. Stoller ([25], see also [17]) gives an example of a proper subgroup H of G=Sym(ℕ) which is strongly orbitequivalent to G; namely, let H consist of those permutations g of ℕ such that there are two partitions, dependent on g, of ℕ into finitely many sets A _{1},…,A _{ k } and B _{1},…,B _{ k } (so ℕ=A _{1}∪⋯∪A _{ k }=B _{1}∪⋯∪B _{ k }, each partitions) such that for each i=1,…,k, the element g induces an order isomorphism (A _{ i },<)→(B _{ i },<).
Finally, we mention a conjectural strengthening of Lemmas 1.4 and 2.3. It is a special case of a much stronger conjecture in [7].
Conjecture 4.7
Let Γ be an infinite graph such that for any distinct vertices x,y the set Γ(x)△Γ(y) is infinite. Then Γ is locally rigid.
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Lockett, D.C., Macpherson, H.D. Orbitequivalent infinite permutation groups. J Algebr Comb 38, 973–988 (2013). https://doi.org/10.1007/s1080101304340
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Keywords
 Primitive permutation group
 Orbitequivalent
 Sethomogeneous