On 2-local automorphisms of symmetric groups

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Introduction and results
Transformations of algebraic structures which 2-locally belong to a given class of maps have been the subject of considerable interest since their introduction in Šemrl's 1997 paper [12]. The concept is a play on the earlier notion of linear local maps, defined by Kadison [5], and Larson and Sourour [6], independently. Given any algebraic structure A, a 2-local automorphism φ of A is a self-map with the property that for any a, b ∈ A, there exists an automorphism φ a,b of A such that φ(a) = φ a,b (a) and φ(b) = φ a,b (b) (1) hold. The notions of 2-local derivations, 2-local isometries, etc. can be defined on appropriate structures analogously. The context in which said types of transformations are generally studied is that of operator algebras, function algebras, and various S292

B. Gyenti
Considering the results cited above, it is natural to study the problem of whether the collection of all automorphisms of a given group is reflexive with respect to the group operation. Due to the novelty of this concept, it is not surprising that such endeavours are yet to be pursued, with the exception of [8,Proposition 3 and Theorem 4], where the case of GL(n, C) and the unitary group of B(H ) are examined. More interestingly, neither has the older (and simpler) problem of determining all 2-local automorphisms of a given group received much attention. As far as the author knows, the only paper concerning this question is [9], where the unitary group and the general linear group of B(H ) are studied.
The purpose of this paper is to present results about the multiplicative reflexive closure of the automorphism groups of symmetric groups, and possibly to motivate further research of this problem formerly in the realms of functional analysis in a more algebraic setting. The choice of symmetric groups for such an end may be justified by the fact that the proposed problem can be regarded as a discrete analogue of Molnár's [7,Theorem 2]. Moreover, one might say that it is natural to examine whether symmetric groups possess a given nontrivial group property.
Throughout the rest of this paper, will be a countable (that is, finite or countably infinite) set and S will denote the symmetric group acting on from the right. In order to stay consistent with this convention, conjugation of group elements will be understood as g h = h −1 gh. It is well known that all automorphisms of S are inner, provided that | | = 6 [1, Section 8.2]. The first result of the paper reads as follows. Theorem 1.1 Let be a countable set and let φ : S → S be an arbitrary function with the property that for any σ, τ ∈ S there exists σ,τ ∈ S such that φ(σ ) φ(τ ) = (σ τ ) σ,τ (3) holds. Then there exists ∈ S such that either The converse statement also holds.
In other words, the reflexive closure of the group of inner automorphisms of S (which is the full automorphism group if | | = 6) with respect to the group operation consists of all inner automorphisms and inner antiautomorphisms of S . (An inner antiautomorphism is an inner automorphism composed with the operation of taking inverse.) As a corollary to this result, 2-local automorphisms of symmetric groups can be easily determined. The symmetric group acting on six elements, S 6 , differs from all other symmetric groups in that it has outer automorphisms. It is well known that Out(S 6 ) is isomorphic to the cyclic group of order 2. The next theorem states that the full automorphism group of S 6 turns out to be almost multiplicatively reflexive, a result analogous to Theorem 1.1.

Theorem 1.3
Let φ : S 6 → S 6 be an arbitrary function with the property that for any σ, τ ∈ S 6 there exists an automorphism α σ,τ of S 6 such that Then φ is necessarily an automorphism or an antiautomorphism. The converse statement also holds.
Again, 2-local automorphisms of S 6 can be determined using this result.

Corollary 1.4
The automorphism group of S 6 is 2-reflexive.
We remark that the feature of the symmetric groups established in the above theorems, namely, that the reflexive closures of their respective automorphism groups with respect to the group operation are small, is not a property shared by all groups. In fact, counterexamples are quite easily found. If G is a group whose automorphism group is transitive on non-unit elements, then every self-map φ of G with the property that φ(g −1 ) = φ(g) −1 holds for all g ∈ G falls into the multiplicative reflexive closure of Aut(G). Such groups are, amongst others, the additive group of any vector space, specifically every cyclic group of prime order, and every group with exactly two conjugacy classes.

Proofs
We fix some, more or less standard (see [1]), notational and naming conventions which will be used subsequently.
As stated before, is a countable set and S is the symmetric group acting on from the right. Elements of will be referred to as points, elements of S as permutations, and e will stand for the unit element of S . The point in to which the point x ∈ is mapped by the permutation σ ∈ S will be denoted by x σ . We will say that σ moves x if x σ = x, otherwise σ fixes x and x is a fixed point of σ . The set of fixed points of σ will be denoted by Fix(σ ). We will say that two permutations S294 B. Gyenti σ, τ ∈ S are disjoint if \ Fix(σ ) and \ Fix(τ ) are disjoint, that is, there is no point in which is moved by both σ and τ . The order of σ ∈ S will be denoted by o(σ ). A permutation with exactly one orbit of length at least two will be called a cycle. As it is well known, if is finite, any permutation in S can be uniquely written as a product of pairwise disjoint cycles. We will use the familiar one-line notation to represent permutations. In the case of infinite , an infinite cycle σ will be denoted by, for example, (. . . x 0 x 1 x 2 . . . ), that is, a sequence of points in infinite in both directions, where x σ 0 = x 1 , etc. A permutation of an infinite set can of course still be decomposed into pairwise disjoint cycles, although perhaps into infinitely many of them. Hence we will consider a permutation on infinitely many elements simply a collection (instead of product) of disjoint cycles, and make sense of the notion that a permutation σ contains a cycle τ , which we will denote by τ ∈ σ . Admittedly, this convention may be inconsistent with the earlier definition of disjointness of permutations, but, hopefully, this will cause no confusion.
The following simple observation will be used during the course of the subsequent proofs frequently and without explicit reference.  An analogous observation can be made about the product of an infinite cycle and a transposition.
We will denote the fact that two permutations σ and τ are conjugate by σ ∼ τ . With this notation, (3) reads as We remark that although the following proof of Theorem 1.1 is completely elementary, the overall strategy is somewhat reminiscent of the proof of [7, Theorem 2].
Proof of Theorem 1.1 An inner automorphism of S clearly satisfies (5). If φ is an inner antiautomorphism, then thus the converse statement is clear.
The main part of the theorem will be proved in multiple steps. Let φ be a self-map of S satisfying (5).
A permutation σ ∈ S is called an involution if σ 2 = e. It is clear from (5) that φ preserves involutions in both directions, that is, σ is an involution if and only if φ(σ ) is. Furthermore, φ is injective on the set of involutions, otherwise e = φ(σ ) φ(ϑ) ∼ σ ϑ = e would hold for some involutions σ, ϑ ∈ S , σ = ϑ. Suppose that φ(σ ) = e for a permutation σ of order 2 and let x ∈ \ Fix(σ ), y ∈ \ {x, x σ }. Then would hold, which is an obvious contradiction, since the permutation on the left is an involution, while the permutation on the right either contains the cycle (x x σ y), if y ∈ Fix(σ ), or the cycle (x x σ yy σ ), if y / ∈ Fix(σ ). Hence φ maps permutations of order 2 to permutations of order 2, consequently, if is finite, φ(e) = e necessarily holds.
For | | = ℵ 0 , assume that φ(e) = e, that is, φ(e) is a permutation of order 2 and e is not in the range of φ. Let = {a i : i ∈ Z}, and consider the permutations Then σ ϑ is the product of two disjoint infinite cycles and has no fixed points, hence the same is true for Let then x 2k+1 be the element of Let then x 2k+2 be the element of thus φ(σ )φ(ϑ) contains an infinite cycle for every fixed point of φ(ϑ), and a similar argument can be made for fixed points of φ(σ ). Consequently, |Fix(φ(σ ))| + |Fix(φ(ϑ))| 2.
The zero, one, or two fixed points can be divided amongst φ(σ ) and φ(ϑ) in six ways, and we proceed to show that every one of these possibilities entails a contradiction, meaning that the original assumption that φ(e) = e must be false.
It is clear that for every permutation σ of order 2, the order of φ(e) φ(σ ) is 2 due to (5). Consequently, if (ab) is a transposition in φ(e) and σ is an arbitrary permutation of order 2, then exactly one of the following options must hold true: The same can be said with the roles of φ(e) and φ(σ ) reversed. Assume now that φ(σ ) has two fixed points, then φ(ϑ) has none. Necessarily, has no fixed points, φ(e) must move both x 0 and y 0 , thus (x 0 y 0 ) ∈ φ(e). Since (x 0 x 1 ), (y 0 y 1 ) ∈ φ(ϑ) and (x 0 y 0 ) ∈ φ(e), we must have (x 1 y 1 ) ∈ φ(e) as well. Proceeding similarly, it is clear that The permutation φ(σ ) φ( ) is the product of two disjoint infinite cycles and has no fixed points, and the permutation φ(ϑ) φ( ) is the product of two disjoint 3-cycles, due to (5). Putting in place of ϑ in the preceding arguments, it is clear that since φ(σ ) is assumed to have two fixed points, φ( ) does not have any.
Finally, suppose that both φ(σ ) and φ(ϑ) are fixed-point-free. Then Indeed, (7) holds for t = 0 by assumption. If it holds for some even t, then, due that is, (7) holds for t + 1. If (7) holds for an odd t, then, due to the fact that , that is, (7) holds for t + 1. Putting t = l − k in (7) yields (x k+l x k+l+1 ) ∈ φ(e), implying that either φ(e) φ(σ ) or φ(e) φ(ϑ) (depending on the parity of k + l) fixes x k+l and x k+l+1 , contradictory to the fact that both of these permutations are fixed-point-free due to (5). A similar argument can be made for l + 1 < k also. Suppose that there is a transposition contained in φ(e) of the form (x 2k x 2l ), where k < l. Then it follows by induction that for all t ∈ N, 0 t 2l − 2k, Hence the assumption that φ(σ ) and φ(ϑ) are both fixed-point-free led to a contradiction, thus we conclude that φ(e) = e must be false.
Step 2 For all x ∈ there exists a unique s(x) ∈ such that for every transposition τ in S , if τ moves x, then φ(τ ) moves s(x).
Let a ∈ be an arbitrary element, then for any pair b, c of points in \ {a}, holds, thus there must be an s(a) ∈ that both φ((ab)) and φ((ac)) move. We show that the map a → s(a) is well defined. This is obvious for | | = 3, as in this case every element of is moved by exactly two transpositions. To prove that the above map is indeed well defined it only has to be shown that by choosing one of b or c differently, we get the same s(a). Let then d ∈ \ {a, b, c} be arbitrary, and x, y ∈ such that φ((ab)) = (s(a)x) and φ((ac)) = (s(a) y). It is clear that x, y, and s(a) are three distinct points in . Since the transposition φ((ad)) must move either s(a) or x. Assume the latter, that is, φ((ad)) = (xz) with some z ∈ . Then leads to a contradiction, since the permutation on the left is a transposition, namely, either (s(a)x) or (s(a) y). In conclusion, the assumption that out of s(a) and x, the permutation φ((ad)) moves the latter cannot hold, thus we get that φ((ad)) moves s(a) as well. This means that the map a → s(a) is well defined.
Step 3 s is bijective.
Suppose that s is not injective and let a, b ∈ be such that a = b and s(a) = s(b). It follows by the injectivity of φ on transpositions that s(x) = s(a) holds for all x ∈ . For finite this implies that there are | | − 1 transpositions in the image of φ, contradicting the fact that φ is injective on transpositions. If is infinite, then for any four distinct points r , t, u, v ∈ ,
For infinite let x 0 ∈ be arbitrary, we proceed to show that x 0 is in the range of s. Let σ ∈ S be a permutation whose cycle decomposition consists of exactly one n-cycle for every n 2 (and does not contain an infinite cycle and has no fixed point). Then the same can be said about φ(σ ) due to φ(σ ) ∼ σ , thus precisely one of the cycles of φ(σ ), the (n + 1)-cycle, say, moves x 0 , that is, there exist distinct points x 1 , . . . , x n ∈ \ {x 0 } such that (x 0 . . . x n ) ∈ φ(σ ). Let (y 0 . . . y n ) be the (n + 1)-cycle in σ with y 0 , . . . , y n ∈ . If n 2, then there are precisely n + 1 transpositions τ in S such that the permutation φ(σ )τ has one fixed point and two n-cycles, these are (x 0 x 1 ), . . . , (x n−1 x n ), (x n x 0 ). On the other hand, φ((y 0 y 1 )), . . . , φ((y n−1 y n )), φ((y n y 0 )) are n + 1 distinct transpositions with this property, since Consequently (x 0 x 1 ) = φ((y i y i+1 )) holds for some 0 i n (or possibly (x 0 x 1 ) = φ((y n y 0 ))), hence x 0 is in the range of s. It follows by the arbitraryness of x 0 that s is bijective. We can argue similarly if n = 1, then (x 0 x φ(σ ) 0 ) is the only transposition τ in S with the property that φ(σ ) τ has two fixed points. Let (y 0 y 1 ) be the transposition in σ , then φ((y 0 y 1 )) has said property, hence (x 0 x φ(σ ) 0 ) = φ((y 0 y 1 )), thus x 0 is in the range of s.
It follows that s ∈ S . Since the function sφ( · )s −1 : S → S satisfies (5), we may and we will assume henceforth that φ acts as the identity on transpositions.
Let σ be an infinite cycle with infinitely many fixed points. If x ∈ \ Fix(σ ), then σ (x x σ 3 ) is the product of an infinite cycle and a 3-cycle which are disjoint. By (5), the same is true for φ(σ )(x x σ 3 ), which entails that φ(σ ) moves both x and x σ 3 . It follows that \ Fix(σ ) ⊆ \ Fix(φ(σ )), and the converse of this inclusion is shown similarly, implying that σ and φ(σ ) move the same points of . Consequently, (10) now means that either x φ(σ ) = x σ or x σ φ(σ ) = x holds whenever x ∈ \ Fix(σ ). The argument in the preceding paragraph can be repeated verbatim to show that either φ(σ ) = σ or φ(σ ) = σ −1 holds in this case as well.
The function φ( · ) −1 : S → S satisfies (5), thus we may and we will assume henceforth that φ acts as the identity on 3-cycles as well. The theorem is now proved for | | = 3, hence we will also suppose that | | 4.
Step 6 φ fixes every cycle in S .
Let ϑ ∈ S be a finite or cofinite cycle and suppose that φ(ϑ) = ϑ −1 . Then for arbitrary would hold, which is a clear contradiction. Indeed, if ϑ is cofinite, then the cycle on the left has two more fixed points than the cycle on the right, and if ϑ is finite, then the cycle on the left is shorter than the cycle on the right. Let ϑ be a cycle that is not finite and not cofinite and assume again that φ(ϑ) = ϑ −1 . Let σ be a fixed-point-free cycle such that for every x ∈ \ Fix(ϑ) there exists n ∈ N such that x σ n = x ϑ . Then we can write σ as where x ∈ \ Fix(ϑ) and x ( j) i ∈ Fix(ϑ), and it was already shown that φ(σ ) = σ holds. Then that is, σ ϑ is the product of two disjoint infinite cycles and has no fixed points. On the other hand, is the product of infinitely many pairwise disjoint cycles, contradicting (5). Thus necessarily φ(ϑ) = ϑ holds in this case also.
Let ϑ be a permutation of order 2 which is not a transposition. Let x, y ∈ \ Fix(ϑ) be two points such that x ϑ = y. Then The permutation on the right contains a 4-cycle, hence so does the permutation on the left. It follows that φ(ϑ) moves both x and y, and x φ(ϑ) = y. This implies that every transposition contained in ϑ is contained in φ(ϑ) as well. Reversing the role of ϑ and φ(ϑ) in this argument yields the convere statement, consequently, ϑ and φ(ϑ) contain the same transpositions, hence they are equal.
Step 8 Let ϑ be a permutation of order 2 and σ a cycle disjoint from ϑ which is not a transposition. Then φ(ϑσ ) = ϑσ .
For any permutation ∈ S and n ∈ N ∪ {∞} let n denote the permutation consisting of the n-cycles of . Applying the argument in the previous step it is easy to see that φ(ϑσ ) 2 = ϑ holds whenever σ is a finite cycle. When σ is an infinite cycle, then for any x ∈ \ Fix(φ(ϑσ ) ∞ ), the permutation φ(ϑσ )(x x φ(ϑσ ) 4 ) contains a 4-cycle in spite of φ(ϑσ ) 2 not moving x and x φ(ϑσ ) 4 , thus the same reasoning does not suffice. But the points of moved by φ(ϑσ ) ∞ can be easily recognised, indeed, φ(ϑσ ) ∞ moves x if and only if there exists y ∈ (namely, y = x φ(ϑσ ) 5 ) such that the permutation φ(ϑσ )(x y) contains a 5-cycle. Thus φ(ϑσ ) 2 = ϑ holds after all.
Assume that σ is an infinite cycle and ϑσ is cofinite. Let x ∈ \ Fix(σ ), then x, x σ ∈ Fix(φ(ϑσ ) 2 ) and φ(ϑσ )(x x σ ) ∼ ϑσ (x x σ ). The permutation on the right has one more fixed points than φ(ϑσ ) has, hence either x φ(ϑσ ) = x σ or x σ φ(ϑσ )) = x. One can argue the same way as in the previous paragraph to show that φ(ϑσ ) = ϑσ holds in this case as well.
Assume that σ is an infinite cycle and ϑσ has infinitely many fixed points. Let η be a permutation of order 2 moving precisely the points in Fix(ϑσ ). By Step 7, φ(η) = η. Since φ(ϑσ )η ∼ ϑσ η holds and the permutation on the right is fixed-point-free, η must move all fixed points of φ(ϑσ ), that is, Fix(φ(ϑσ )) ⊆ \ Fix(η) = Fix(ϑσ ). Choosing an η which moves precisely the fixed points of φ(ϑσ ), the reverse inclusion follows similarly, thus Fix(φ(ϑσ )) = Fix(ϑσ ). Let η be a permutation of order 2 moving exactly the points of this set. It is clear by now that φ(ϑσ ) ∞ and σ move the same elements of , let x be such an element. Then the permutation ϑσ η(x x σ ) has one fixed point, hence, due to the same is true for φ(ϑσ )η(x x σ ). Keeping in mind that η fixes both x and x σ , either x φ(ϑσ ) = x σ or x σ φ(ϑσ ) = x follows. The usual induction argument yields that either φ(ϑσ ) = ϑσ or φ(ϑσ ) = ϑσ −1 . As the latter would imply the contradiction seen in (12), the statement of this step is proved.
Let ϑ ∈ S be a cofinite permutation and x ∈ \ Fix(ϑ) arbitrary. Then and since the permutation on the right has more fixed points than φ(ϑ) has, necessarily either x φ(ϑ) = x ϑ or x ϑφ(ϑ) = x follows. The same can be said for x ϑ n in place of x, where n ∈ Z is arbitrary. The usual induction argument yields that if σ denotes the cycle in ϑ which moves x, then either σ ∈ φ(ϑ) or σ −1 ∈ φ(ϑ). If σ is not a transposition, then the latter would imply that which is a contradiction, as |Fix(φ(ϑ)σ )| = |Fix(ϑσ )| + o(σ ). Thus every cycle which is contained in ϑ is contained in φ(ϑ) as well, and since they both have the same (finite) number of fixed points, φ(ϑ) = ϑ follows.
Let ϑ ∈ S be a permutation with infinitely many fixed points. It can be shown the same way as in the previous Step that Fix(φ(ϑ)) = Fix(ϑ). Let η be a permutation of order 2 moving precisely the points of this set. Then for any x ∈ \ Fix(ϑ), holds, which implies that either x φ(ϑ) = x ϑ or x ϑφ(ϑ) = x. This holds with x ϑ n in place of x for every n ∈ Z, thus by denoting the cycle in ϑ which moves the point x by σ , either σ ∈ φ(ϑ) or σ −1 ∈ φ(ϑ) holds. In case of the latter, if o(σ ) = 2, then φ(ϑ)ησ = φ(ϑ) φ(ησ ) ∼ ϑησ would be a contradiction as the permutation on the left has o(σ ) fixed points, while the permutation on the right has none. It follows that every cycle contained in ϑ is contained in φ(ϑ) as well, and reversing the role of φ(ϑ) and ϑ in this argument yields the converse statement. In conclusion, φ(ϑ) = ϑ.
The proof made use of the fact that is countable on several occasions. It might be interesting to consider the same problem for uncountable .
Proof of Corollary 1.2 Every 2-local (inner) automorphism of S satisfies (3), hence, by Theorem 1.1, it is necessarily an (inner) automorphism or an (inner) antiautomorphism. Clearly any automorphism is a 2-local automorphism, thus one needs only to see whether antiautomorphisms are 2-local automorphisms as well.
(i) Trivial (and lengthy) computation shows that if | | < 6, then every antiautomorphism of S is in fact a 2-local automorphism. It suffices to verify that for any σ, ϑ ∈ S , there exists ∈ S such that σ −1 = σ and ϑ −1 = ϑ . (ii) In the | | = 6 case it is easy to see that there is no η ∈ S such that both Turning to the proof of Theorem 1.3, the following terminology is introduced. For a permutation σ moving finitely many points we say that σ is of type r where t ∈ N and r i , β i ∈ N for all 1 i t, if the cycle decomposition of σ consists of exactly β 1 cycles of length r 1 , . . . , exactly β t cycles of length r t . Furthermore,6 will stand for an arbitrary set of cardinality 6.
For permutations σ, τ ∈ S 6 , σ ≈ τ will denote the fact that for some automorphism α of S 6 , α(σ ) = τ holds. With this notation, (4) reads as The proof of Theorem 1.3 is even more involved than that of Theorem 1.1, but it goes along a similar route and contains no novelty. We have, for this reason, attempted to keep it brief.
Proof of Theorem 1. 3 It is clear that every automorphism of S 6 satisfies (13). If φ is an antiautomorphism, then φ( · ) −1 is an automorphism, hence holds for every σ, ϑ ∈ S 6 . Thus the converse statement is clear.
The other direction will again be proved in multiple steps. Let φ be a function which satisfies (13).
It follows from (13) that φ is a bijective transformation of the set of involutions in S 6 . Suppose that φ(σ ) = e for some involution σ = e. If σ = (x y) with x, y ∈6, then holds, where z ∈6 \ {x, y}. This is a contradiction, since the permutation on the left is of order 2, while the permutation on the right is of order 3. If σ is the product of two or three pairwise disjoint transpositions, then with x, y ∈6 \ Fix(σ ), x σ = y, holds, which is a contradiction, since the permutation on the right is of order 4. Necessarily, φ(e) = e follows.
Step 2 φ maps all transpositions to permutations of the same type.
If φ maps all transpositions to permutations of type 2 3 , then for an arbitrary outer automorphism α of S 6 the function α • φ : S 6 → S 6 satisfies (13) and maps transpositions to transpositions. Thus we may and we will assume henceforth that φ maps transpositions to transpositions.
Step 3 For all x ∈6 there exists a unique s(x) ∈6 such that for every transposition τ in S 6 , if τ moves x, then φ(τ ) moves s(x).
Possibly by working with sφ( · )s −1 , we may and we will assume that φ acts as the identity on the set of transpositions.
The same argument previously seen in Step 5 of the proof of Theorem 1.1 suffices.
We may and we will assume henceforth that φ fixes all 3-cycles.
The argument in Step 6 of the proof of Theorem 1.1 proves the statement.
Step 7 φ fixes all other types of permutations as well.