Invariance groups of finite functions and orbit equivalence of permutation groups
Eszter K. Horváth, Géza Makay, Reinhard Pöschel, Tamás Waldhauser
aa r X i v : . [ m a t h . G R ] D ec Invariance groups of finite functions andorbit equivalence of permutation groups
Eszter K. Horv´ath [email protected]
G´eza Makay [email protected] Reinhard P¨oschel [email protected]
Tam´as Waldhauser [email protected]
Abstract
Which subgroups of the symmetric group S n arise as invariancegroups of n -variable functions defined on a k -element domain? It ap-pears that the higher the difference n − k , the more difficult it is toanswer this question. For k ≥ n , the answer is easy: all subgroupsof S n are invariance groups. We give a complete answer in the cases k = n − k = n −
2, and we also give a partial answer in the gen-eral case: we describe invariance groups when n is much larger than n − k . The proof utilizes Galois connections and the correspondingclosure operators on S n , which turn out to provide a generalization oforbit equivalence of permutation groups. We also present some com-putational results, which show that all primitive groups except for thealternating groups arise as invariance groups of functions defined ona three-element domain. This paper presents a Galois connection that facilitates the study of per-mutation groups representable as invariance groups of functions of several Bolyai Institute, University of Szeged, Aradi v´ertan´uk tere 1, H-6720 Szeged, Hungary Institut f¨ur Algebra, Technische Universit¨at Dresden, D-01062 Dresden, Germany Partially supported by the T ´AMOP-4.2.1/B-09/1/KONV-2010-0005 program of theNational DevelopmentAgency of Hungary. Partially supported by the Hungarian National Foundation for Scientific Researchunder grant no. K83219. Partially supported by the Hungarian National Foundation for Scientific Researchunder grant no. K77409. k := { , . . . , k } for someinteger k ≥
2. We say that an n -ary function f : k n → m is invariant undera permutation σ ∈ S n , if f ( x , . . . , x n ) = f ( x σ , . . . , x nσ )holds for all ( x , . . . , x n ) ∈ k n , and we denote this fact by σ ⊢ f . The in-variance group (or symmetry group) of f is the subgroup { σ ∈ S n | σ ⊢ f } of the full symmetric group S n . We will say that a group G ≤ S n is ( k, m ) -representable if there exists a function f : k n → m whose invariance groupis G . Furthermore, we call a group ( k, ∞ ) -representable if it is ( k, m )-representable for some natural number m . Note that ( k, ∞ )-representabilityis equivalent to being the invariance group of a function f : k n → N .A group G ≤ S n is (2 , f : { , } n → { , } ), anda group is ( k, ∞ )-representable if and only if it is the invariance group of apseudo-Boolean function (i.e., a function f : { , } n → R , cf. [CrHa11, Chap-ter 13]). Invariance groups of (pseudo-)Boolean functions are important ob-jects of study in computer science (see [ClKr91] and the references therein);however, our main motivation comes from the algebraic investigations ofA. Kisielewicz [Ki98]. Kisielewicz defines a group G to be m -representable if there is a function f : { , } n → m whose invariance group is G (equiv-alently, G is (2 , m )-representable), and G is defined to be representable ifit is m -representable for some positive integer m (equivalently, G is (2 , ∞ )-representable). It is easy to see that a group is representable if and onlyif it is the intersection of 2-representable groups (i.e., invariance groups ofBoolean functions). It was stated in [ClKr91] that every representable groupis 2-representable; however, this is not true: as shown by Kisielewicz [Ki98],the Klein four-group is 3-representable but not 2-representable. Moreover,it is also discussed in [Ki98] that it is probably very difficult to find anothersuch example by known constructions for permutation groups.In this paper we focus on ( k, ∞ )-representability of groups for arbitrary k ≥
2. It is straightforward to verify that a group is ( k, ∞ )-representable ifand only if it is the intersection of invariance groups of operations f : k n → k (cf. Fact 2.2). We introduce a Galois connection between operations on k and permutations on n , such that the Galois closed subsets of S n are exactlythe groups that are representable in this way. Our main goal is to charac-terize the Galois closed groups; as it turns out, the difficulty of the problemdepends on the gap d := n − k between the number of variables and the sizeof the domain. The easiest case is d ≤
0, where all groups are closed (see2roposition 3.3); for d = 1 the only non-closed groups are the alternatinggroups (see Proposition 3.4). The case d = 2 is considerably more difficult(see Proposition 5.1), and the general case, which includes representabilityby invariance groups of Boolean functions, seems to be beyond reach. How-ever, we provide a characterization of Galois closed groups for arbitrary d provided that n is much larger than d (more precisely, n > max (cid:0) d , d + d (cid:1) ;see Theorem 3.1.)Let us mention that our approach is also related to orbit equivalence ofgroups (see Section 2(A)). In the case k = 2, two groups have the sameGalois closure if and only if they are orbit equivalent, whereas the cases k > S n . Thus our Galois connection provides a parameterized version of orbitequivalence that could be interesting from the viewpoint of the theory ofpermutation groups.In Section 2 we formalize the Galois connection, we discuss its relationshipto orbit equivalence, and we recall some basic facts about subdirect productsof groups. We state our main result (Theorem 3.1) in Section 3, where weprove it in the special cases d ≤ d = 1, and we also make somegeneral observations about closures of direct and subdirect products. Weprove Theorem 3.1 in Section 4, and in Section 5 we present results of somecomputer experiments, which, together with Theorem 3.1, settle the case d = 2. Finally, in Section 6 we relate our approach to relational definabilityof permutation groups (cf. [Wi69]) and we formulate some open problems. Throughout the paper, n and k denote positive integers; we always assumethat n, k ≥
2, and we denote the difference n − k by d . As usual, S B and A B denote the symmetric and alternating groups, respectively, on an arbitraryset B , and S n stands for the symmetric group on the set n = { , . . . , n } . (A) A Galois connection for invariance groups In order to precisely state the problem that we study, first we introducesome terminology and notation. The correspondence ⊢ defined in Section 1induces a Galois connection between permutations of n and n -ary operationson k . More precisely, let O ( n ) k = { f | f : k n → k } denote the set of all n -ary3perations on k , and for F ⊆ O ( n ) k and G ⊆ S n let F ⊢ := { σ ∈ S n | ∀ f ∈ F : σ ⊢ f } , F ( k ) := ( F ⊢ ) ⊢ ,G ⊢ := { f ∈ O ( n ) k | ∀ σ ∈ G : σ ⊢ f } , G ( k ) := ( G ⊢ ) ⊢ . As for every Galois connection, the assignment G G ( k ) is a closureoperator on S n , and it is easy to see that G ( k ) is a subgroup of S n for everysubset G ⊆ S n (even if G is not a group). For G ≤ S n , we call G ( k ) the Galoisclosure of G over k , and we say that G is Galois closed over k if G ( k ) = G .Sometimes, when there is no risk of ambiguity, we will omit the referenceto k , and speak simply about (Galois) closed groups and (Galois) closures.Similarly, we have a closure operator on O ( n ) k ; the study of this closure oper-ator constitutes a topic of current research of the authors. However, in thispaper we focus on the “group side” of the Galois connection; more precisely,we address the following problem. Problem 2.1.
For arbitrary k, n ≥ , characterize subgroups of S n that areGalois closed over k . As we shall see, this problem is easy if k ≥ n , and it is very hard if n ismuch larger than k . Our main result is a solution in the intermediate case,when d = n − k > n . Complementing thisresult with a computer search for small values of n , we obtain an explicitdescription of Galois closed groups for n = k − n = k − n .Observe that if k ≥ k , then G ( k ) ≤ G ( k ) , hence if G is Galois closed over k , then it is also Galois closed over k . Thus we have the most non-closedgroups in the Boolean case (i.e., in the case k = 2), whereas for k ≥ n everysubgroup of S n is Galois closed (see Proposition 3.3).The following fact appears in [ClKr91] for k = 2, and it remains validfor arbitrary k . We omit the proof, as it is a straightforward generalizationof the proof of the equivalence of conditions (1) and (2) in Theorem 12 of[ClKr91]. Fact 2.2.
A group G ≤ S n is Galois closed over k if and only if G is ( k, ∞ ) -representable. (B) Orbits and closures The symmetric group S n acts naturally on k n : for a = ( a , . . . , a n ) ∈ k n and σ ∈ S n , let a σ = ( a σ , . . . , a nσ ) be the action of σ on a . We denote the4rbit of a ∈ k n under the action of the group G ≤ S n by a G , and we use thenotation Orb ( k ) ( G ) for the set of orbits of G ≤ S n acting on k n : a G := { a σ | σ ∈ G } , Orb ( k ) ( G ) := (cid:8) a G | a ∈ k n (cid:9) . Clearly, σ ⊢ f holds for a given σ ∈ S n and f ∈ O ( n ) k if and only if f is constant on the orbits of (the group generated by) σ . Therefore, for any G, H ≤ S n , we have G ⊢ = H ⊢ if and only if Orb ( k ) ( G ) = Orb ( k ) ( H ). Onthe other hand, from the identity G ⊢⊢⊢ = G ⊢ (which is valid in any Galoisconnection), it follows that G ⊢ = H ⊢ is equivalent to G ( k ) = H ( k ) . Thus wehave G ( k ) = H ( k ) ⇐⇒ Orb ( k ) ( G ) = Orb ( k ) ( H ) (1)for all subgroups G, H of S n .Two groups G, H ≤ S n are orbit equivalent , if G and H have the sameorbits on the power set of n (which can be identified naturally with n ), i.e.,if Orb (2) ( G ) = Orb (2) ( H ) holds [In84, SiWa85]. One can define a similarequivalence relation on the set of subgroups of S n for any k ≥ k if and only if it is the greatest group amongthose having the same orbits on k n (cf. Theorem 2.2 of [Ki98] in the Booleancase). Therefore, the Galois closure of G over k can be described as follows: G ( k ) = (cid:8) σ ∈ S n | ∀ a ∈ k n : a σ ∈ a G (cid:9) . (2)Orbit equivalence of groups has been studied by several authors; let usjust mention here a result of Seress [Se97] that explicitly describes orbitequivalence of primitive groups (see [SeYa08] for a more general result). Forthe definitions of the linear groups appearing in the theorem, we refer thereader to [DiMo96]. Theorem 2.3 ([Se97]) . If n ≥ , then two different primitive subgroupsof S n are orbit equivalent if and only if one of them is A n and the otherone is S n . For n ≤ , the nontrivial orbit equivalence classes of primitivesubgroups of S n are the following: (i) for n = 3 : { A , S } ;(ii) for n = 4 : { A , S } ;(iii) for n = 5 : { C , D } and { AGL (1 , , A , S } ;5iv) for n = 6 : { PGL (2 , , A , S } ;(v) for n = 7 : { A , S } ;(vi) for n = 8 : { AGL (1 , , AΓL (1 , , ASL (3 , } and { A , S } ;(vii) for n = 9 : { AGL (1 , , AΓL (1 , } , { ASL (2 , , AGL (2 , } and { PSL (2 , , PΓL (2 , , A , S } ;(viii) for n = 10 : { PGL (2 , , PΓL (2 , } and { A , S } . In our terminology, Theorem 2.3 states that for n ≥
11 every primitivesubgroup of S n except A n is Galois closed over , whereas for n ≤
10 theonly primitive subgroups of S n that are not Galois closed over are the oneslisted above (omitting the last group from each block, which is the closure ofthe other groups in the same block). (C) Direct and subdirect products In the sequel, B and D always denote disjoint subsets of n such that n = B ∪ D , and G × H stands for the direct product of G ≤ S B and H ≤ S D .In this paper we only consider direct products with the intransitive action,i.e., the two groups act independently on disjoint sets. Given permutations β ∈ S B and δ ∈ S D , we write β × δ for the corresponding element of S B × S D .Let π and π denote the first and second projections on the direct product S B × S D . Then we have π ( β × δ ) = β and π ( β × δ ) = δ for every β ∈ S B , δ ∈ S D , and σ = π ( σ ) × π ( σ ) for every σ ∈ S B × S D .Recall that a subdirect product is a subgroup of a direct product suchthat the projection to each coordinate is surjective. Hence, if G ≤ S B × S D and G = π ( G ), G = π ( G ), then G is a subdirect product of G and G .We denote this fact by G ≤ sd G × G , and by G < sd G × G we meana proper subdirect subgroup of G × G . According to Remak [Re30], thefollowing description of subdirect products of groups is due to Klein [Kl1890].(Of course, the theorem is valid for abstract groups, not just for permutationgroups. For an English reference, see Theorem 5.5.1 of [Ha76].) Theorem 2.4 ([Kl1890, Re30]) . If G ≤ sd G × G , then there exists a group K and surjective homomorphisms ϕ i : G i → K ( i = 1 , such that G = { g × g | ϕ ( g ) = ϕ ( g ) } . Note that in the above theorem we have G = G × G if and only if K isthe trivial (one-element) group. 6 The main result and some general observa-tions
Our main result is the following partial solution of Problem 2.1 for the case n ≫ d = n − k . Theorem 3.1.
Let n > max (cid:0) d , d + d (cid:1) and G ≤ S n . Then G is not Galoisclosed over k if and only if G = A B × L or G < sd S B × L , where B ⊆ n is such that D := n \ B has less than d elements, and L is an arbitrarypermutation group on D . Note that the set D in the theorem above is much smaller than B , thus B is a “big” subset of n , and L ≤ S D is a “little group”, hence the notation.The subdirect product G < sd S B × L is not determined by B and L , butin Proposition 3.10 we give a fairly concrete description of these groups.Proposition 3.11 shows that the groups given in Theorem 3.1 are indeed notGalois closed over k (and that their Galois closure is S B × L ). In Section 4we will prove that these are the only non-closed groups; however, already inthis section we present the proof for the case d = 1 (i.e., k = n − (A) The case k = n − S n ) a denotes the stabilizer of a ∈ k n under the action of S n . Notethat this stabilizer is the direct product of symmetric groups on the sets { i ∈ n | a i = j } , j ∈ k . Proposition 3.2.
For every G ≤ S n , we have G ( k ) = \ a ∈ k n ( S n ) a · G. Proof.
We reformulate the condition a σ ∈ a G of (2) for a ∈ k n , σ ∈ S n asfollows: a σ ∈ a G ⇐⇒ ∃ π ∈ G : a σ = a π ⇐⇒ ∃ π ∈ G : a σπ − = a ⇐⇒ ∃ π ∈ G : σπ − ∈ ( S n ) a ⇐⇒ σ ∈ ( S n ) a · G. Now from (2) it follows that σ ∈ G ( k ) if and only if σ ∈ ( S n ) a · G holds forall a ∈ k n . 7ith the help of Proposition 3.2, we can prove that all subgroups of S n are Galois closed over k if and only if k ≥ n . Proposition 3.3. If k ≥ n ≥ , then each subgroup G ≤ S n is Galois closedover k ; if ≤ k < n , then A n is not Galois closed over k .Proof. Clearly, if k ≥ n then there exists a tuple a ∈ k n whose componentsare pairwise different. Consequently, ( S n ) a is trivial and therefore G ( k ) ⊆ ( S n ) a · G = G for all G ≤ S n by Proposition 3.2. On the other hand, if k < n then there is a repetition in every tuple a ∈ k n , hence ( S n ) a containsa transposition. Therefore ( S n ) a · A n = S n for all a ∈ k n , thus A n ( k ) = S n byProposition 3.2.Now we can solve Problem 2.1 in the case k = n −
1, which is the simplestnontrivial case. The proof of the following proposition already contains thekey steps of the proof of Theorem 3.1.
Proposition 3.4.
For k = n − ≥ , each subgroup of S n except A n isGalois closed over k .Proof. If G ≤ S n is not Galois closed over k , then Proposition 3.2 shows thatfor all π ∈ G ( k ) \ G and for all a ∈ k n , we have π ∈ ( S n ) a · G , hence π = γσ forsome γ ∈ ( S n ) a and σ ∈ G . Therefore, γ = πσ − ∈ G ( k ) ; moreover, γ = idfollows from π / ∈ G . Thus we see that G ( k ) contains at least one non-identitypermutation from every stabilizer: G ( k ) = G = ⇒ ∀ a ∈ k n ∃ γ ∈ ( S n ) a \ { id } : γ ∈ G ( k ) . (3)Now fix i, j ∈ n , i = j , and let a = ( a , . . . , a n ) ∈ k n be a tuple suchthat a r = a s ⇐⇒ { r, s } = { i, j } or r = s . Then ( S n ) a = { id , ( ij ) } , where( ij ) ∈ S n denotes the transposition of i and j . Applying (3), we see that( ij ) ∈ G ( k ) for all i, j ∈ n , hence G ( k ) = S n . From Proposition 3.2 it followsthat G ( k ) ⊆ ( S n ) a · G ⊆ S n = G ( k ) , i.e., S n = ( S n ) a · G for every a ∈ k n .Choosing a as above, we have S n = { id , ( ij ) } · G , hence G is of index at most2 in S n . Therefore, we have either G = A n or G = S n ; the latter is obviouslyGalois closed, whereas A n is not Galois closed over k by Proposition 3.3.Clote and Kranakis [ClKr91] define a group G ≤ S n to be weakly rep-resentable , if there exist positive integers k, m with 2 ≤ k < n and 2 ≤ m such that G is the invariance group of some function f : k n → m (equiva-lently, G is ( k, ∞ )-representable for some k < n ). Proposition 3.3 shows thatthe restriction k < n is important; allowing k = n would make all groupsweakly representable. Proposition 3.4 yields a complete description of weaklyrepresentable groups. 8 orollary 3.5. All subgroups of G ≤ S n except for A n are weakly repre-sentable.Proof. According to Fact 2.2, a subgroup of S n is weakly representable ifand only if it is Galois closed over k for some k < n . This is equivalent tobeing Galois closed over n − , as the closures for k = 2 , , . . . , n − S n are Galois closed over n − except for A n . (B) Closures of direct and subdirect products The following proposition describes closures of direct products, and, as acorollary, we obtain a generalization of [Ki98, Theorem 3.1].
Proposition 3.6.
For all G ≤ S B and H ≤ S D , we have G × H ( k ) = G ( k ) × H ( k ) .Proof. For notational convenience, let us assume that B = { , . . . , t } and D = { t + 1 , . . . , n } . If a = (1 , . . . , , , . . . , ∈ k n with t ones followed by n − t twos, then the stabilizer of a in S n is S B × S D . Hence from Proposition 3.2it follows that G × H ( k ) ≤ ( S B × S D ) · ( G × H ) = S B × S D , i.e., every elementof G × H ( k ) is of the form β × δ for some β ∈ S B , δ ∈ S D . For arbitrary a =( a , . . . , a n ) ∈ k n , let a B = ( a , . . . , a t ) ∈ k t and a D = ( a t +1 , . . . , a n ) ∈ k n − t .It is straightforward to verify that a β × δ ∈ a G × H if and only if a βB ∈ a GB and a δD ∈ a HD . Thus applying (2), we have β × δ ∈ G × H ( k ) ⇐⇒ ∀ a ∈ k n : a β × δ ∈ a G × H ⇐⇒ ∀ a ∈ k n : (cid:16) a βB ∈ a GB and a δD ∈ a HD (cid:17) ⇐⇒ (cid:16) ∀ a B ∈ k t : a βB ∈ a GB (cid:17) and (cid:0) ∀ a D ∈ k n − t : a δD ∈ a HD (cid:1) ⇐⇒ β ∈ G ( k ) and δ ∈ H ( k ) ⇐⇒ β × δ ∈ G ( k ) × H ( k ) . Corollary 3.7.
For all G ≤ S B and H ≤ S D , the direct product G × H isGalois closed over k if and only if both G and H are Galois closed over k .Proof. The “if” part follows immediately from Proposition 3.6. For the “onlyif” part, assume that G × H is Galois closed over k . From Proposition 3.6we get G × H = G ( k ) × H ( k ) , and this implies G = G ( k ) and H = H ( k ) .9 emark 3.8. If n < m , then any subgroup G of S n can be naturally embed-ded into S m as the subgroup G × (cid:8) id m \ n (cid:9) . From Proposition 3.6 it followsthat G × (cid:8) id m \ n (cid:9) ( k ) = G ( k ) × (cid:8) id m \ n (cid:9) , i.e., there is no danger of ambiguityin not specifying whether we regard G as a subgroup of S n or as a subgroupof S m . Remark 3.9.
Proposition 3.6 and Corollary 3.7 do not generalize to subdi-rect products. It is possible that a subdirect product of two Galois closedgroups is not Galois closed. For example, let G = { id , (123) , (132) , (12) (45) , (13) (45) , (23) (45) } < sd S { , , } × S { , } ;then G (2) = S { , , } × S { , } , hence G is not Galois closed over . It is alsopossible that a subdirect product is closed, although the factors are not bothclosed: let G = { id , (13) (24) , (1234) (56) , (1432) (56) } < sd h (1234) i × h (56) i ;then G is Galois closed over , but the 4-element cyclic group is not Galoisclosed over (its Galois closure is the dihedral group of degree 4).Next we determine (the closures of) the special subdirect products in-volving symmetric and alternating groups that appear in Theorem 3.1. Proposition 3.10.
Let | B | > max ( | D | , and L ≤ S D . If G ≤ sd A B × L ,then G = A B × L . If G ≤ sd S B × L , then either G = S B × L , or there existsa subgroup L ≤ L of index , such that G = ( A B × L ) ∪ (cid:0) ( S B \ A B ) × ( L \ L ) (cid:1) . (4) Proof.
Suppose that G ≤ sd A B × L , and let K and ϕ , ϕ be as in Theorem 2.4(for G = A B and G = L ). Since A B is simple, the kernel of ϕ is either { id B } or A B . In the first case, K is isomorphic to A B ; however, this cannotbe a homomorphic image of L , as | L | ≤ | S D | < | A B | . In the second case, K is trivial and G = A B × L . If G ≤ sd S B × L , then there are three possibilitiesfor the kernel of ϕ , namely { id B } , A B and S B . Just as above, the first caseis impossible, while in the third case we have G = S B × L . In the secondcase, K is a two-element group, hence by letting L be the kernel of ϕ , weobtain (4). Proposition 3.11.
Let | D | < d ≤ n − d and let G be any one of the subdirectproducts considered in Proposition 3.10. Then G ( k ) = S B × L . roof. Since k = n − d > | D | , all subgroups of S D are closed by Proposi-tion 3.3, hence L ( k ) = L . On the other hand, k < | B | implies that A B is notclosed; in fact, we have A B ( k ) = S B . Therefore A B × L ( k ) = A B ( k ) × L ( k ) = S B × L , and also S B × L ( k ) = S B × L . It remains to consider the case when G is of the form (4). Then we have A B × L ≤ G ≤ S B × L , thus S B × L = A B × L k ) ≤ ∗ G ( k ) ≤ S B × L ( k ) = S B × L. (5)Moreover, G ( k ) contains ( S B \ A B ) × ( L \ L ), and this shows that the firstcontainment in (5) (marked with asterisk) is strict. However, S B × L is ofindex 2 in S B × L , therefore we can conclude that G ( k ) = S B × L . The proof of Theorem 3.1 is based on the same idea as that of Proposition 3.4:1) first we use (3) with specific tuples a to show that G ( k ) must be a“large” group (see Subsection 4(A) below), and then2) we prove that G is of “small” index in G ( k ) (see Subsection 4(B) below).For the first step, we will need to apply (3) for several groups acting ondifferent sets, hence, for easier reference, we give a name to this property. Definition 4.1.
Let Ω ⊆ n be a nonempty set, and let us consider thenatural action of S Ω on k Ω for a positive integer k ≥
2. We say that H ≤ S Ω is k -thick , if ∀ a ∈ k Ω : ∃ γ ∈ ( S Ω ) a \ { id Ω } : γ ∈ H. We will use thickness with two types of tuples a ∈ k Ω . First, let a containonly one repeated value, which is repeated exactly d + 1 times, say at thecoordinates i , . . . , i d +1 ∈ Ω (note that such a tuple exists only if | Ω | ≥ d + 1).Then the stabilizer of a is the full symmetric group on { i , . . . , i d +1 } , therefore k -thickness of H implies that ∃ γ ∈ S { i ,...,i d +1 } \ { id } : γ ∈ H. (6)Next, let d values be repeated in a , each of them repeated exactly twotimes, say at the coordinates i , j ; i , j ; . . . ; i d , j d (here we need | Ω | ≥ d ). Then the stabilizer of a is the group generated by the transpositions( i j ) , ( i j ) , . . . , ( i d j d ). Thus k -thickness of H implies that ∃ γ ∈ h ( i j ) , ( i j ) , . . . , ( i d j d ) i \ { id } : γ ∈ H. (7)11he first paragraph of the proof of Proposition 3.4 can be reformulatedas follows: Fact 4.2. If G ≤ S n is not Galois closed over k , then G ( k ) is k -thick. (A) The closures of non-closed groups The goal of this subsection is to prove the following description of theclosures of non-closed groups.
Proposition 4.3.
Let n > d + d . If G ≤ S n is not Galois closed over k ,then G ( k ) is of the form S B × L , where B ⊆ n is such that D := n \ B hasless than d elements, and L is a permutation group on D . Throughout this subsection we will always assume that
G < G ( k ) ≤ S n with n > d + d , where d = n − k ≥
1. We consider the action of G ( k ) on n (not on k n ), and we separate two cases upon the transitivity of thisaction. First we deal with the transitive case, for which we will make use ofthe following theorem of Bochert [Bo1889] (see also [DiMo96, Wi64]). Theorem 4.4 ([Bo1889]) . If G is a primitive subgroup of S Ω not containing A Ω , then there exists a subset I ⊆ Ω with | I | ≤ | Ω | such that the pointwisestabilizer of I in G is trivial. Lemma 4.5.
Let Ω ⊆ n such that | Ω | > max (2 d, d ) . If H is a transitive k -thick subgroup of S Ω , then H = A Ω or H = S Ω .Proof. Assume for contradiction that H satisfies the assumptions of thelemma, but H does not contain A Ω . If H is primitive, then let us con-sider the set I given in Theorem 4.4. Since | Ω \ I | ≥ | Ω | > d , we can find d + 1 elements i , . . . , i d +1 in Ω \ I . Since H is k -thick and | Ω | ≥ d + 1, wecan apply (6) for i , . . . , i d +1 , and we obtain a permutation γ = id in thepointwise stabilizer of I in H , which is a contradiction.Thus H cannot be primitive. Since it is transitive, there exists a nontrivialpartition Ω = B ˙ ∪ · · · ˙ ∪ B r (8)with | B | = · · · = | B r | = s and r, s ≥ H preserves this partition. We will prove by contradiction that r ≤ d and s ≤ d . First let us assume that r > d ; let B = { i , j , . . . } , . . . , B d +1 = { i d +1 , j d +1 , . . . } , and let γ be the permutation provided by (6). Since γ = id,there exist p, q ∈ { , . . . , d + 1 } , p = q such that γ ( i p ) = i q . On the other12and, we have γ ( j p ) = j p , and this means that γ does not preserve the par-tition (8). Next let us assume that s > d ; let B = { i , . . . , i d +1 , . . . } , B = { j , . . . , j d +1 , . . . } , and let γ be the permutation provided by (7). Since γ = id,there exists p ∈ { , . . . , d } such that γ ( i p ) = j p . On the other hand, we have γ ( i d +1 ) = i d +1 , and this means that γ does not preserve the partition (8).We can conclude that r, s ≤ d , hence we have | Ω | = rs ≤ d < | Ω | , acontradiction. Lemma 4.6. If G ( k ) is transitive, then G ( k ) = S n .Proof. Since n > d + d , we have n > max (2 d, d ). Thus from Fact 4.2 andLemma 4.5 it follows that either G ( k ) = A n or G ( k ) = S n . However, A n is notGalois closed over k by Proposition 3.3, because n > k .Now let us consider the intransitive case. The first step is to prove thatin this case there is a unique “big” orbit. Lemma 4.7. If G ( k ) is not transitive, then it has an orbit B such that D = n \ B has less than d elements.Proof. We claim that G ( k ) has at most d orbits. Suppose to the contrary,that there exists d + 1 elements i , . . . , i d +1 ∈ n , each belonging to a differentorbit. If γ ∈ G ( k ) is the permutation given by (6), then there exist p, q ∈{ , . . . , d + 1 } , p = q such that γ ( i p ) = i q , and this contradicts the fact that i p and i q belong to different orbits of G ( k ) . Now, the average orbit size isat least nd > d , therefore there exists an orbit B = { i , . . . , i d , . . . } of size atleast d . We will show that the complement of B has at most d − d elements j , . . . , j d outside B . With the help of (7) we obtain a permutation γ ∈ G ( k ) for which thereexists p ∈ { , . . . , d } such that γ ( i p ) = j p . This is clearly a contradiction,since i p belongs to the orbit B , whereas j p belongs to some other orbit.At this point we know that G ( k ) ≤ S B × S D . Using the the notation G = π (cid:0) G ( k ) (cid:1) and L = π (cid:0) G ( k ) (cid:1) for the projections of G ( k ) , we have G ( k ) ≤ sd G × L . Lemma 4.8. If G ( k ) is not transitive and B is the big orbit given in Lemma 4.7,then G ( k ) = S B × L for some L ≤ S D .Proof. First we show that G inherits k -thickness from G ( k ) . Let b ∈ k B , andextend b to a tuple a ∈ k n such that the components a i ( i ∈ D ) are pairwisedifferent (this is possible, since | D | < k ). The k -thickness of G ( k ) implies that13here exists a permutation γ ∈ ( S n ) a ∩ G ( k ) \ { id } , and from G ( k ) ≤ sd G × L it follows that γ = β × δ for some β ∈ G , δ ∈ L . The construction of thetuple a ensures that δ = id D , hence we have id B = β ∈ ( S B ) b ∩ G , and thisproves that G is a k -thick subgroup of S B .Since B is an orbit of G ( k ) , the action of G on B is transitive. From n > d + d it follows that | B | = n − | D | > n − d ≥ max (2 d, d ), henceLemma 4.5 shows that G ≥ A B . This means that either G ( k ) ≤ sd A B × L or G ( k ) ≤ sd S B × L . Now with the help of Proposition 3.10 and Proposition 3.11we can conclude that G ( k ) = S B × L . (Note that the assumption | B | > d = 1 and n ≤
4. However, d = 1 implies D = ∅ , what contradicts the intransitivity of G ( k ) .)Combining Lemmas 4.6 and 4.8, we obtain Proposition 4.3, q.e.d. (B) The non-closed groups In this subsection we prove the following Proposition 4.9. It describesthe groups G with G ( k ) = S B × L and therefore completes also the proof ofTheorem 3.1. Proposition 4.9.
Let n > max (cid:0) d , d + d (cid:1) , let B ⊆ n and D = n \ B suchthat | D | < d , and let L ≤ S D . If G ≤ S n is a group whose Galois closureover k is S B × L , then G ≤ sd A B × L or G ≤ sd S B × L . Throughout this subsection we will assume that n > max (cid:0) d , d + d (cid:1) ,where d = n − k ≥
1, and G ( k ) = S B × L , where B and L are as in theproposition above. Let G = π ( G ) ≤ S B and G = π ( G ) ≤ S D ; then wehave G ≤ sd G × G . As in Subsection 4(A), we begin with the transitivecase (i.e., D = ∅ ), and we will use the following well-known result (see, e.g.,[Wi64, Exercise 14.3]). Proposition 4.10. If n > and H is a proper subgroup of S n different from A n , then the index of H is at least n . Lemma 4.11. If G ( k ) = S n , then G = A n or G = S n .Proof. Let a ∈ k n be the tuple which was used to obtain (7); then we have( S n ) a = h ( i j ) , ( i j ) , . . . , ( i d j d ) i . From Proposition 3.2 we obtain S n = G ( k ) ⊆ ( S n ) a · G, hence we have ( S n ) a · G = S n . Since | ( S n ) a | = 2 d , the index of G in S n is atmost 2 d < n , and therefore Proposition 4.10 implies that G ≥ A n if n > n ≤
4, then d = 1, thus we can apply Proposition 3.4.14 emma 4.12. If G ( k ) = S n × L , then G ≥ A n and G = L .Proof. Clearly, G ≤ G × G implies S B × L = G ( k ) ≤ G × G k ) = G k ) × G k ) by Proposition 3.6. It follows that G ( k )1 = S B , and thus Lemma 4.11yields G ≥ A n . (One can verify that the inequality n > max (cid:0) d , d + d (cid:1) holds if we write | B | in place of n and | B | − k in place of d .) On the otherhand, k > | D | implies that G k ) = G by Proposition 3.3, hence G ≤ S B × L = G ( k ) ≤ G k ) × G k ) = G k ) × G . Applying π to these inequalities, we obtain G ≤ L ≤ G , and this proves G = L .Since G ≤ sd G × G , Lemma 4.12 immediately implies Proposition 4.9,q.e.d. The Galois closures of a group G ≤ S n over k for k = 2 , , . . . form a nonin-creasing sequence, eventually stabilizing at G itself: G (2) ≥ G (3) ≥ · · · ≥ G ( n − ≥ G ( n ) = G ( n +1) = · · · = G. (9)We computed the Galois closures of all subgroups of S n for 2 ≤ k ≤ n ≤ G (i.e., G is Galois closed over ), and for all other groups (9)consists only of two different groups (namely G (2) and G ). Table 1 showsthe list of groups corresponding to the latter case, up to conjugacy. For eachgroup, the first column gives the smallest n for which G can be embedded into S n (here we mean an embedding as a permutation group, not as an abstractgroup; cf. Remark 3.8). We also give the largest k such that G ( k ) = G , i.e.,(9) takes the form G (2) = . . . = G ( k ) > G ( k +1) = . . . = G .Some of the entries in Table 1 may need some explanation. Using thenotation of Theorem 2.4, each subdirect product in the table correspondsto a two-element quotient group K : for symmetric groups S n we take thehomomorphism ϕ : S n → K with kernel A n (cf. Proposition 3.10), whereasfor the dihedral group D we take the homomorphism ϕ : D → K whosekernel is the group of rotations in D . The group S ≀ S is the wreathproduct of S and S (with the imprimitive action); equivalently, it is thesemidirect product ( S × S ) ⋊ S (with S acting on the direct product by15 ≤ S n G ( k ) n = 3 , k = 2 A S n = 4 , k = 3 A S n = 4 , k = 2 C D n = 5 , k = 4 A S n = 5 , k = 2 AGL (1 , S n = 5 , k = 2 S × sd S S × S n = 5 , k = 2 A × S S × S n = 5 , k = 2 C D n = 6 , k = 5 A S n = 6 , k = 2 PGL (2 , S n = 6 , k = 3 S × sd S S × S n = 6 , k = 3 A × S S × S n = 6 , k = 2 S × sd S S × S n = 6 , k = 2 A × S S × S n = 6 , k = 2 D × sd S D × S n = 6 , k = 2 C × S D × S n = 6 , k = 3 ( S ≀ S ) ∩ A S ≀ S n = 6 , k = 2 S ≀ sd S S ≀ S n = 6 , k = 2 R ( (cid:28) ) S ( (cid:28) )Table 1: Nontrivial closures for n ≤ S ≀ sd S we mean the “subdirect wreathproduct” ( S × sd S ) ⋊ S . Finally, the groups S ( (cid:28) ) and R ( (cid:28) ) denote thegroup of all symmetries and the group of all rotations (orientation-preservingsymmetries) of the cube, acting on the six faces of the cube.Combining these computational results with Theorem 3.1, we get thesolution of Problem 2.1 for the case d = 2. Proposition 5.1.
For k = n − ≥ , each subgroup of S n except A n and A n − (for n ≥ ) and C (for n = 4 ) is Galois closed over k .Proof. If n >
6, then we can apply Theorem 3.1, and we obtain the excep-tional groups A n and A n − from the direct product A B × L with | D | = 0 and | D | = 1, respectively. If n ≤
6, then the non-closed groups can be read fromTable 1.We have also examined the linear groups appearing in Theorem 2.3 bycomputer, and we have found that all of them are Galois closed over . Thuswe have the following result for primitive groups. Proposition 5.2.
Every primitive permutation group except for A n ( n ≥ is Galois closed over . We have introduced a Galois connection to study invariance groups of n -variable functions defined on a k -element domain, and we have studied thecorresponding closure operator. Our main result is that if the difference d = n − k is relatively small compared to n , then “most groups” are Galoisclosed, and we have explicitly described the non-closed groups. The boundmax (cid:0) d , d + d (cid:1) of Theorem 3.1 is probably not the best possible; it remainsan open problem to improve it. Problem 6.1.
Determine the smallest number f ( d ) such that Theorem 3.1is valid for all n ≥ f ( d ) . For fixed d , the inequality n > max (cid:0) d , d + d (cid:1) fails only for “small”values of n , so one might hope that these cases can be dealt with easily.However, our investigations indicate that there is a simple pattern in theclosures if n is much larger than d , and exactly those exceptional groups cor-responding to small values of n are the ones that make the problem difficult.(We can say that the Boolean case is the hardest, as in this case n is just d + 2.) We have fully settled only the cases d ≤
2; perhaps it is feasible toattack the problem for the next few values of d .17 roblem 6.2. Describe the (non-)closed groups for d = 3 , , . . . . The chain of closures (9) for the groups that we investigated in our com-puter experiments has length at most two: for all k ≥
2, we have either G ( k ) = G (2) or G ( k ) = G . This is certainly not true in general; for example,we have A × · · · × A t ( k ) = A × · · · × A k × S k +1 × · · · × S t , hence G (2) > G (3) > · · · > G ( t − > G ( t ) = G holds for G = A × · · · × A t . Itis natural to ask if there exist groups with long chains of closures that are notdirect products of groups acting on smaller sets. As Proposition 5.2 shows,we cannot find such groups among primitive groups. Problem 6.3.
Find transitive groups with arbitrarily long chains of closures.
The closure operator defined in Section 2(A) concerns the Galois closurewith respect to the Galois connection induced by the relation ⊢ ⊆ S n × O ( n ) k ,based on a natural action of S n on k n (see Section 1). In permutation grouptheory also another closure operator, called k -closure is used, which wasintroduced by H. Wielandt ([Wi69, Definition 5.3]). This notion describesthe Galois closures with respect to the Galois connection induced by therelation ⊲ ⊆ S n × P ( n k ). Here σ ∈ S n acts on r = ( r , . . . , r k ) ∈ n k accordingto r σ := ( r σ, . . . , r k σ ), and, for a k -ary relation ̺ ⊆ n k , we have σ ⊲ ̺ if andonly if σ preserves ̺ , i.e., r σ ∈ ̺ for all r ∈ ̺ . For G ⊆ S n , the k -closure ( G ⊲ ) ⊲ is denoted by Aut Inv ( k ) G ([P¨oKa79]), or by G ( k ) = gp( k -rel G ) ([Wi69]). Agroup G ≤ S n is k -closed if and only if it can be defined by k -ary relations,i.e., if there exists a set R of k -ary relations on n such that G consists of thepermutations that preserve every member of R . The following propositionestablishes a connection between the two notions of closure. Proposition 6.4.
For every G ≤ S n and k ≥ , the Galois closure G ( k +1) iscontained in the k -closure of G . In particular, every k -closed group is Galoisclosed over k + .Proof. The proof is based on a suitable correspondence between n k and( k + ) n . Let r = ( r , . . . , r k ) ∈ n k be a k -tuple whose components arepairwise different. We define κ ( r ) = ( a , . . . , a n ) ∈ ( k + ) n as follows: a i = ℓ, if i = r ℓ ; k + 1 , if i / ∈ { r , . . . , r k } . κ is a partial map from n k to ( k + ) n , and it is straightforward toverify that κ is injective, and κ ( r ) σ − = κ ( r σ ) holds for all σ ∈ S n and r ∈ n k with mutually different components. (Here κ ( r ) σ − refers to theaction of S n on ( k + ) n by permuting the components of n -tuples, while r σ refers to the action of S n on n k by mapping k -tuples componentwise.)Now let G ≤ S n and π ∈ G ( k +1) ; we need to show that r π ∈ r G forevery r ∈ n k . We may assume that the components of r are pairwise distinct(otherwise we can remove the repetitions and work with a smaller k ). From π ∈ G ( k +1) it follows that κ ( r ) π − ∈ κ ( r ) G . Therefore, we have κ ( r π ) = κ ( r ) π − ∈ κ ( r ) G = κ (cid:0) r G (cid:1) , and then the injectivity of κ gives that r π ∈ r G . Note that the proposition above implies that each group that is not Galoisclosed over k (such as the ones in Theorem 3.1) is also an example of apermutation group that cannot be characterized by ( k − Mathieu group M is Galois closed over (since it is the automorphism group of a hypergraph), but it is not 5-closed(since it is 5-transitive, and this implies that the 5-closure of M is the fullsymmetric group S ). In some sense, this is a worst possible case, as it isnot difficult to prove that if a subgroup of S n is Galois closed over , then itis (cid:4) n (cid:5) -closed (in particular, M is 6-closed). Problem 6.5.
Determine the smallest number w ( n, k ) such that every sub-group of S n that is Galois closed over k is also w ( n, k ) -closed. Acknowledgement
The authors are grateful to Keith Kearnes, Erkko Lehtonen and S´andor Rade-leczki for stimulating discussions, and also to P´eter P´al P´alfy who suggestedthe example mentioned before Problem 6.3.
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