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On Sillke's bijection
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Theoretical Computer Science ••• (••••) •••–•••
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On Sillke’s bijection R. Cori a , C. Reutenauer b,∗ a b
Labri, Université Bordeaux 1, France Département de Mathématiques, Université du Québec à Montréal, Canada
a r t i c l e
i n f o
Article history: Received 29 September 2015 Received in revised form 1 April 2016 Accepted 10 April 2016 Available online xxxx
a b s t r a c t Indecomposable permutations in S n+1 , subgroups of index n of the free group on two generators and doubly pointed hypermaps of cardinality n are equinumerous. We give here a proof of a bijection, due to Sillke, between these three sets. © 2016 Elsevier B.V. All rights reserved.
Keywords: Bijections Indecomposable permutations Subgroups of the free group Hypermaps
1. Introduction
Antonio Restivo is a prominent figure in the world of Combinatorics of words. Some of his contributions contain also beautiful bijections. We dedicate to him this article which sheds light on a wrongly neglected bijection on permutations. There has been recently a lot of interest in finding bijections between the set of indecomposable permutations and subgroups of finite index of the free group on two generators, see the article by Ossona de Mendez and Rosenstiehl [8], the article of the first-named author [4] and that of Bacher with the second-named author [1]. The fact that these objects are counted by the same numbers is known since long time: formulas for the number of subgroups were given by Hall [7] and for the number of indecomposable permutations by Comtet [2]; the formula is the same, so that the existence of a bijection follows. The formula in [2] for this number cn is recursive: c 1 = 1, cn = n! − 1≤ p ≤n−1 c p (n − p )!. A bijection between these two sets was first given by Dress and Franz [5,6]. Another bijection, which has interesting symmetry properties with respect to the inverse permutation, was given by Sillke, first in an article without proof [10], but with an illuminating example, which we reproduce below, and then, with proof, in his PhD thesis [9], written in German. The aim of the present article is to give a proof of Sillke’s bijection, using some new ideas. We hope to give more audience to this beautiful bijection, which we consider to be less well-known as it should be. We also point out the relation to maps and hypermaps, which was already noted in [5,6]. The authors thank the two referees for many suggestions and simplifications of the proofs, and for sharing pictures. The second-named author was supported by an NSERC grant (Canada).
*
Corresponding author. E-mail addresses: [email protected] (R. Cori), [email protected] (C. Reutenauer).
http://dx.doi.org/10.1016/j.tcs.2016.04.015 0304-3975/© 2016 Elsevier B.V. All rights reserved.
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Doctopic: Algorithms, automata, complexity and games
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R. Cori, C. Reutenauer / Theoretical Computer Science ••• (••••) •••–•••
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Fig. 1. The permutation θ = 527163984.
Fig. 2. A left-to-right maximum and a right-to-left minimum.
2. Left-to-right maxima and right-to-left minima of permutations Let θ ∈ S n (the symmetric group on [n] = {1, . . . , n}). A left-to-right maximum of θ is a value y ∈ [n] such that for θ(x) = y, one has
∀i ≤ x, θ(i ) ≤ y .
(1)
We call x the position of the left-to-right maximum and y its value. We may identify the left-to-right maximum to the point (x, y ) of the graph of θ ; this graph is represented in the Cartesian plane, see for example Fig. 11 : the left-to-right maxima of θ = 527163984 are y = 5, 7, 9. Similarly, a right-to-left minimum of θ is a value y ∈ [n] such that for θ(x) = y, one has
∀i ≥ x, θ(i ) ≥ y . The right-to-left minima of θ above are y = 4, 3, 1. On the graph of θ , left-to-right maxima and right-to-left minima are seen as in Fig. 2: A is a left-to-right maximum, B is a right-to-left minimum and the shaded areas contain no point of the graph, except A and B. Denote by A k , . . . , A 2 , A 1 the k left-to-right maxima of θ , in increasing order of x-coordinate (equivalently y-coordinate); similarly, the l right-to-left minima are B l , . . . , B 2 , B 1 , in increasing order of y-coordinate (equivalently x-coordinate). Note the inverse order of indices, which is motivated by the inverse Silke bijection. One has A k = (1, θ(1)), A 1 = (θ −1 (n), n), B l = (θ −1 (1), 1), B 1 = (n, θ(n)). Note also that one may have k = 1 and l = 1: k = 1 (resp. l = 1) if and only if θ(1) = n (resp. θ(n) = 1). For two successive left-to-right maxima A i and A i −1 , the shaded area in Fig. 3 does not contain any point of the graph, except these two points. We say that θ ∈ S n is indecomposable if there is no p ∈ [n − 1] such that θ([ p ]) ⊂ [ p ]. Clearly, a permutation is decomposable if and only if for some p ∈ [n − 1], its graph is contained in [ p ] × [ p ] ∪ { p + 1, . . . , n} × { p + 1, . . . , n}. For later use, we slightly generalize this result.
1
The running example of the present article is taken from Sillke [10].
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Fig. 3. Two successive left-to-right maxima.
Fig. 4. The left-to-right maxima and the right-to-left minima.
Lemma 2.1. A permutation θ ∈ S n is decomposable if and only if there exist p , q ∈ [n − 1] such that its graph is contained in [ p ] × [q] ∪ { p + 1, . . . , n} × {q + 1, . . . , n}. Proof. The condition is necessary, by taking p = q. Conversely, suppose that such p , q exist. Recall that each column (resp. row) of integer x-coordinate (resp. y-coordinate) between 1 and n contains exactly one point of the graph. Considering the columns of x-coordinate between 1 and p, we see that we must have q ≥ p. Arguing with rows, we see that also p ≥ q. Thus p = q and θ is decomposable. 2 Thus, when θ is indecomposable, its graph looks like Fig. 4, where the points of the graph are all inside the polygon delimited by the A’s and the B’s, or on it. By Lemma 2.1, the two polygonal lines A k . . . A 1 and B l . . . B 1 do not intersect. Note that the A’s (resp. B’s) are the maximal north–west (resp. south–east) points of the graph. 3. Doubly ordered sets and permutations We begin by motivating the introduction of doubly ordered sets. Let G be the graph of a permutation θ ∈ S n . On G, there are naturally two total orders. The first one is denoted ≤x ; it is defined for any points M , N in G by: M ≤x N if and only if x( M ) ≤ x( N ), where x( P ) is the x-coordinate of point P . Similarly the order ≤ y of G is defined by: M ≤ y N if and only if y ( M ) ≤ y ( N ). Thus we have two total orders on G. Call doubly ordered set a triple ( E , ≤1 , ≤2 ), where E is a finite set and ≤1 , ≤2 are total orders on E. An isomorphism of two doubly ordered sets ( E , ≤1 , ≤2 ) and ( F , ≤1 , ≤2 ) is a bijection φ : E → F which is increasing for the orders ≤i , that is: for any e , e in E and for i = 1, 2, e ≤i e implies φ(e ) ≤i φ(e ). Permutations are a set of representatives of the isomorphism classes of doubly ordered sets. Indeed, a permutation θ ∈ S n defines the doubly ordered set (G , ≤x , ≤ y ) as above. Conversely, if ( E , ≤1 , ≤2 ) is any doubly ordered set of cardinality n, we obtain a permutation as follows: consider first the example of E = {a, b, c , d}, ≤1 = {b <1 a <1 d <1 c } and ≤2 = {d <2 a <2 c <2 b}; the associated permutation is then θ = 3241, obtained by identifying 1234 = badc. Formally, let ωi , i = 1, 2, be the unique increasing order isomorphism ( E , ≤i ) → [n]; then the set G = {(ω1 (e ), ω2 (e )) | e ∈ E } is the graph of the permutation θ = ω2 ◦ ω1−1 and (G , ≤x , ≤ y ) is isomorphic with ( E , ≤1 , ≤2 ): the isomorphism φ : E → G is φ(e ) = (ω1 (e ), ω2 (e )).
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Doctopic: Algorithms, automata, complexity and games
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Fig. 5. Position of points.
Fig. 6. Right-to-left minima.
Moreover, if the graphs of two permutations, viewed as doubly ordered sets, are isomorphic, then the permutations are equal, as the reader is invited to verify. Lemma 3.1. Let ( E , ≤1 , ≤2 ) be a doubly ordered set and θ the associated permutation. (i) θ is indecomposable if and only if E has no nontrivial proper upper order ideal for both orders. (ii) A left-to-right-maximum of θ corresponds to an element f of E such that for any e in E, one has e ≤1 f ⇒ e ≤2 f . (iii) A right-to-left-minimum of θ corresponds to an element f of E such that for any e in E, one has e ≤2 f ⇒ e ≤1 f . Proof. We may assume that ( E , ≤1 , ≤2 ) = (G , ≤x , ≤ y ), where G is the graph of θ . Then (i) follows directly from the definition of indecomposability. For (ii), we note that if y is a left-to-right maximum of θ , then the point f = (x, y ) of G satisfies: for any point e = (i , j ) of G with e ≤x f , one has i ≤ x and therefore j = θ(i ) ≤ y by Eq. (1) and thus e ≤ y f . The proof of the converse, and of (iii), is quite similar. 2 4. A property of left-to-right maxima and right-to-left minima In an ordered set ( E , ≤) we denote by [e ≤] the principal upper ideal {x ∈ E | e ≤ x}. Lemma 4.1. Let θ be a permutation in S n and take the notation of Section 2. Let i = 1, . . . , k, j = 1, . . . , l. Let
E i j = [ A i ≤x ] \ [ B j ≤ y ], F i j = [ B j ≤ y ] \ [ A i ≤x ].
(2)
(i) If E i j is nonempty, then max≤x ( E i j ) = B j +1 . (ii) If F i j is nonempty, then max≤ y ( F i j ) = A i +1 . Proof. It is enough to prove (i). We represent the set [ A i ≤x ] by the non-shaded area, at the right of the vertical line x = x( A i ), including this line, as in Fig. 5. Similarly for the complement of the set [ B j ≤ y ]. We assume that E i j is nonempty; hence j < l since [ B l ≤ y ] is the whole set G. Since there are no points in the shaded area indicated in Fig. 6, and since E i j is nonempty, we see that B j +1 must be at the right of the vertical line x = x( A i ) or on it. Hence we have Fig. 7, where the unshaded area is E i j ; thus its maximum for the order ≤x is B j +1 . 2 5. Cyclic form of a permutation We want to associate each indecomposable permutation θ with two permutations σ and α on the set of points of the graph G of θ , as shown for example in Fig. 8 for the permutation θ = 527163984. The permutation σ associates each point M of G with the point having x-coordinate equal to x( M ) + 1, if such a point exists and if it is not a left-to-right maximum. Otherwise σ ( M ) is the left-to right-maximum A j where A j is the minimal index such that x( A i ) ≤ x( M ). Symmetrically, the permutation α associates each point M with the point having y-coordinate equal to y ( M ) + 1, if such point exists and if it is not a right-to left minimum. Otherwise, α ( M ) is the right-to-left minimum B j where j is the minimal index such that y ( B j ) ≤ y ( M ).
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Doctopic: Algorithms, automata, complexity and games
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Fig. 7. Region.
Fig. 8. The cycles of
σ and α .
The permutations σ and α may be completely defined by the following condition: given a permutation τ of any set E, call cyclic form of τ each well-parenthesis word w, which represents τ in cyclic form, including cycles of length 1:
w = (h1 , τ h1 , τ 2 h2 , . . . , τ i 1 −1 h1 )(h2 , τ h2 , . . . , τ i 2 −1 h2 ) . . . The lengths of the cycles are therefore i 1 , i 2 , . . . . For given τ , such a form is of course not unique. Given such a word w, we associate with it the total order on E, whose list from the smallest to the largest is obtained by removing the parenthesis. Define the two permutations σ and α of the graph G of θ by the condition that they have respectively a cyclic form u, v as follows
u = ( A k , . . .) . . . ( A 2 , . . .)( A 1 , . . . , B 1 ), v = ( B l , . . .) . . . ( B 2 , . . .)( B 1 , . . . , A 1 ), and such that the associated orders satisfy ≤u =≤x and ≤ v =≤ y . This definition of σ , α is somewhat undirect but it defines them completely. Equivalently, one can define them as follows. Let A 0 = B 0 = (n + 1, n + 1). Then one has for any M ∈ G:
• If x( A i +1 ) ≤ x( M ) < x( A i ) then – either x( M ) + 1 < x( A i ) and then σ ( M ) = (x( M ) + 1, θ(x( M ) + 1)), – or x( M ) + 1 = x( A i ) and then σ ( M ) = A i +1 . • If y ( B j +1 ) ≤ y ( M ) < y ( B j ) then – either y ( M ) + 1 < y ( B j ) and then α ( M ) = (θ −1 ( y ( M ) + 1), y ( M ) + 1), – or y ( M ) + 1 = y ( B j ) and then α ( M ) = B j +1 . Lemma 5.1. If i = k, . . . , 2, then x(α ( A i )) ≥ x( A i −1 ). Proof. The statement is trivially true if α ( A i ) = A i −1 . If y (α ( A i )) = y ( A i ) + 1, then Fig. 9, left part, shows that α ( A i ) can only be on the right of A i −1 . Otherwise, one has α ( A i ) = B j (and hence y ( A i ) + 1 = y ( B j −1 )) and Fig. 9, right part, shows that B j must be on the right of A i −1 , or θ would be indecomposable by Lemma 2.1. 2 Recall that the action of a group on a set is called transitive if there is a unique orbit. Lemma 5.2. The group generated by σ and α acts transitively on G.
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Doctopic: Algorithms, automata, complexity and games
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Fig. 9. The mapping
α.
Proof. Note that B 1 is the maximum of G for the order ≤x . Thus is enough to show that for any M in G, with M = B 1 , either σ ( M ) >x M or ασ ( M ) >x M. If x(σ ( M )) = x( M ) + 1, then σ ( M ) >x M. Otherwise σ (m) = A i for some i ≥ 2 because M = B 1 . Using Lemma 5.1, x(ασ ( M )) = x(α ( A i )) ≥ x( A i −1 ) > x( M ), hence ασ ( M ) >x M. 2 6. Doubly pointed hypermaps and Sillke’s theorem (first form) Consider the set of quintuples ( E , σ , α , A , B ), where E is a finite set, σ , α are permutations of E which act transitively on E (that is, the subgroup they generate acts transitively), and A , B are distinct elements of E such that σ ( B ) = A and α ( A ) = B. An isomorphism between two such quintuples ( E , σ , α , A , B ) and ( E , σ , α , A , B ) is a bijection τ : E → E
such that τ ( A ) = A , τ ( B ) = B , τ σ = σ τ and that τ α = α τ . A doubly pointed hypermap is an isomorphism class of such quintuples. Sillke’s mapping is the mapping that associates to each indecomposable permutation θ a doubly pointed hypermap (G , σ , α , A 1 , B 1 ), following the construction of the previous sections. Note that if a permutation is indecomposable, so is its inverse. Theorem 6.1. Sillke’s mapping is a bijection between indecomposable permutations in S n and doubly pointed hypermaps of cardinality n. If θ is mapped onto the class of ( E , σ , α , A , B ), then θ −1 is mapped onto the class of ( E , α , σ , B , A ). The proof rests on two technical lemmas. The first one is essentially Sillke’s inverse algorithm, which is implicitly described in the proof below. Lemma 6.1. Given a quintuple ( E , σ , α , A , B ) as above, there exist unique cyclic forms for σ and α , respectively u = ( A k , . . .)( A k−1 , . . .) . . . ( A 1 , . . . , B 1 ) and v = ( B l , . . .)( B l−1 , . . .) . . . ( B 1 , . . . , A 1 ), with A 1 = A, B 1 = B, such that, if one puts E i j = [ A i ≤u ] \ [ B j ≤ v ], F i j = [ B j ≤ v ] \ [ A i ≤u ], then: (i) If E i j is nonempty, then max≤u ( E i j ) = B j +1 . (ii) If F i j is nonempty, then max≤ v ( F i j ) = A i +1 . Proof. Suppose that there are other cyclic forms u , v for
σ , α respectively:
u = ( A k , . . .)( A k −1 , . . .) . . . ( A 1 , . . . , B 1 ) and
v = ( B l
, . . .)( B l
−1 , . . .) . . . ( B 1 , . . . , A 1 ). We have A 1 = A = A 1 and B 1 = B = B 1 by hypothesis. Moreover, conditions (i) and (ii) hold for the sets E i j = [ A i ≤u ] \
[ B j ≤ v ] and F i j = [ B j ≤ v ] \ [ A i ≤u ]. Suppose that for some s, t ≥ 1 we have A 1 = A 1 , A 2 = A 2 , . . . , A s = A s , B 1 = B 1 , B 2 = B 2 , . . . , B t = B t . We show that these equalities may be extended to the case s + 1 or t + 1, unless s = k and t = l. For this, consider the sets E st and F st . If both are empty, then [ A s ≤u ] = [ B t ≤ v ]; this set is therefore invariant under σ and α (because this set is a union of cycles of both σ and α ), thus equal to E, since they act transitively; thus s = k and t = l. Assume that E st = ∅. Then B t +1 = max≤u ( E st ), by hypothesis. Moreover, E st = E st by the inductive hypothesis, and B t +1 = max≤u ( E st ). Thus B t +1 = B t +1 . This proves uniqueness in the Lemma, by symmetry.
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In order to prove existence, observe that what precedes proves also the following fact (which mimics Sillke’s algorithm): there exist cyclic forms for σ and α , respectively
u = ( A k , . . .)( A k−1 , . . .) . . . ( A 1 , . . . , B 1 ) and
v = ( B l , . . .)( B l−1 , . . .) . . . ( B 1 , . . . , A 1 ), with B 1 = B , A 1 = A and a sequence of pairs (1, 1) = (s0 , t 0 ) < (s1 , t 1 ) < . . . < (sk+l−2 , tk+l−2 ) = (k, l), where < is the natural order of N2 , and such that for any h = 0, . . . , k + l − 3
• either (sh+1 , th+1 ) = (sh + 1, th ), F sh th = ∅ and max≤v ( F sh th ) = A sh +1 ; • or (sh+1 , th+1 ) = (sh , th + 1), E sh th = ∅ and max≤u ( E sh th ) = B th +1 . In particular, (i) and (ii) hold for each pair (i , j ) = (sh , th ), h = 0, . . . , k + l − 2. We prove conditions (i) and (ii) for all pairs (i , j ). For this we use the easy claim: let H be a totally ordered set and E an nonempty subset of H . If the subset E is obtained from E by adding elements which are smaller than those of E, then max( E ) = max( E ). If the nonempty subset E
is obtained from E by removing a lower order ideal, then max( E ) = max( E ). We deduce the following facts:
• If E i j is nonempty, then max≤u ( E i +1, j ) = max≤u ( E i j ), because E i +1, j = E i j ∪ ([ A i +1 , A i [≤u \ [ B j ≤ v ]) and the elements of E i j are all ≥u A i ; • If both E i j and E i −1, j are nonempty, then max≤u ( E i j ) = max≤u ( E i −1, j ), because E i −1, j = E i j \ [ A i , A i −1 [≤u . Since E i j ⊂ E i +1, j , we deduce that, j being fixed, if for some (s, j ), E sj is nonempty and if max≤u ( E sj )) = B j +1 , then (i) holds for each (i , j ). Now, observe that the sequence of pairs
(s0 , t 0 ) < (s1 , t 1 ) < . . . < (sk+l−2 , tk+l−2 ) = (k, l) is a discrete path from (1, 1) to (k, l); moreover, for any j, there exists s such that (s, j ) is on this path and that E sj is nonempty: indeed, if (s, j ) = (k, l) is on the path and E sj = ∅, then by 2), F sj = ∅ and (s + 1, j ) is on the path; moreover, E kj is nonempty. It follows that (i) holds for all pairs (i , j ). The proof of (ii) is similar. 2 Lemma 6.2. Under the hypothesis of the previous lemma, one has: (i) There is no nontrivial proper upper ideal for both orders ≤u and ≤ v ; (ii) ∀i = 1, . . . , k, M ≤u A i ⇒ M ≤ v A i ; (iii) ∀ j = 1, . . . , l, M ≤ v B j ⇒ M ≤u B j . Proof. (i) Suppose that H is a nontrivial proper upper ideal for both orders. One has A 1 = max≤ v ( E ), thus A 1 ∈ H since H is a nonempty upper ideal for ≤ v . Thus [ A 1 ≤u ] is contained in H , since H is an upper ideal for ≤u . We deduce that there is a maximum i such that [ A i ≤u ] ⊂ H . Similarly there is a maximum j such that [ B j ≤ v ] ⊂ H . If we had E i j = ∅, then, since E i j ⊂ H , we would obtain by Lemma 6.1 that B j +1 ∈ H ; thus [ B j +1 ≤ v ] ⊂ H , a contradiction. Thus E i j = ∅. Similarly F i j = ∅. It follows that [ A i ≤u ] = [ B j ≤ v ]. This implies that σ and α do not act transitively, since the latter set is invariant under them, except if we have i = k; but this cannot happen since H = E = [ A k ≤u ]. (ii) For i = 1 this is clear since A 1 = max≤ v ( E ). Suppose that 2 ≤ i < k. Then F il = ∅: indeed, otherwise [ B l ≤ v ] \ [ A i ≤u ] = ∅, which implies i = k, because [ B l ≤ v ] = E; but we have assumed i < k. Thus by Lemma 6.1, A i +1 = max≤v ( E \ [ A i ≤u ]). Let M
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Doctopic: Algorithms, automata, complexity and games
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7. Pointed hypermaps, subgroups of finite index of the free group on two generators and Sillke’s theorem (second form) Consider quadruples ( E , σ , α , A ), where E is a finite set, σ , α are two permutations of E acting transitively on it and A is an element of E. An isomorphism between two such quadruples ( E , σ , α , A ) and ( E , σ , α , A ) is a bijection τ : E → E
such that τ ( A ) = A , τ σ = σ τ and τ α = α τ . A pointed hypermap is an isomorphism class of such quadruples. As observed by Sillke, there is a bijection between doubly pointed hypermaps of cardinality n + 1 and pointed hypermaps of cardinality n: let the doubly pointed hypermap be represented by the quintuple ( E , σ , α , A , B ); then in σ and α represented in cycle notation, remove the element B. This construction induces the desired bijection. Let F 2 be the free group on two generators a, b. Let H be a subgroup of index n of F 2 . By letting F 2 act on the set E of left cosets of H , we obtain a pair (σ , α ) of permutations of this set E (namely the permutations of E induced by a, b respectively), which act transitively on E, together a distinguished element of E (namely the coset H ). In other words, we obtain a quadruple as above. It is easy to see that this construction induces a bijection between subgroups of finite index n and pointed hypermaps of cardinality n. Theorem 7.1. There is a bijection between the set of indecomposable permutations in S n+1 and subgroups of index n in the free group F 2 , or equivalently, pointed hypermaps of cardinality n. In this bijection, if θ ∈ S n+1 corresponds to the subgroup H and the pointed hypermap represented by ( E , σ , α , A ), then θ −1 corresponds to the image of H under the automorphism of F 2 exchanging the generators a and b, and to the pointed hypermap represented by ( E , α , σ , A ). Motivated by the last assertion, we may ask what happens for other natural involutions defined either on indecomposable permutations or on subgroups of F 2 . For example, to which involution of subgroups of F 2 does the conjugation of indecomposable permutations by the longest permutation n + 1, n, . . . , 2, 1 (which preserves indecomposability) correspond? Or to which involution of indecomposable permutations does the involution of subgroups induced by the anti-automorphism a → a−1 , b → b−1 of F 2 correspond? One may also wonder if it is possible to identify directly on indecomposable permutations conjugate subgroups of F 2 , and normal subgroups; partial answers have been obtained by Chekkal [3]. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
R. Bacher, C. Reutenauer, Number of right ideals and a q-analogue of indecomposable permutations, Canad. J. Math. (2016), in press. n L. Comtet, Sur les coefficients de l’inverse de la série formelle n!t , C. R. Acad. Sci. Paris Ser. A 2 (75) (1972) 569–572. A. Chekkal, Yamanouchisation, PhD thesis, Université du Québec à Montréal, 2010. R. Cori, Indecomposable permutations, hypermaps and labeled Dyck paths, J. Combin. Theory Ser. A 116 (2009) 1326–1343. A.W.M. Dress, R. Franz, Parametrizing the subgroups of finite index in a free group and related topics, Bayreuth. Math. Schr. 20 (1985) 1–8. A.W.M. Dress, R. Franz, Zur Parametrisierung von Untergruppen freier Gruppen, Beitr. Algebra Geom. 20 (1987) 125–134. M. Hall, Subgroups of finite index in free groups, Canad. J. Math. 1 (1949) 187–190. P. Ossona De Mendez, P. Rosenstiehl, Transitivity and connectivity of permutations, Combinatorica 24 (2004) 487–501. T. Sillke, Zur Kombinatorik von Permutationen, in: Séminaire Lotharingien de Combinatoire, vol. B21c, 1989, 8 pp. T. Sillke, Eine Bijektion zwischen Untergruppen freier Gruppen und Systemen konnexer Permutationen (A bijection between subgroups of free groups and systems of connex permutations), Dissertation, Universität Bielefeld Fakultät für Mathematik, 1992 (in German).
Doctopic: Algorithms, automata, complexity and games
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Theoretical Computer Science ••• (••••) •••–•••
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Theoretical Computer Science www.elsevier.com/locate/tcs
On Sillke’s bijection R. Cori a , C. Reutenauer b,∗ a b
Labri, Université Bordeaux 1, France Département de Mathématiques, Université du Québec à Montréal, Canada
a r t i c l e
i n f o
Article history: Received 29 September 2015 Received in revised form 1 April 2016 Accepted 10 April 2016 Available online xxxx
a b s t r a c t Indecomposable permutations in S n+1 , subgroups of index n of the free group on two generators and doubly pointed hypermaps of cardinality n are equinumerous. We give here a proof of a bijection, due to Sillke, between these three sets. © 2016 Elsevier B.V. All rights reserved.
Keywords: Bijections Indecomposable permutations Subgroups of the free group Hypermaps
1. Introduction
Antonio Restivo is a prominent figure in the world of Combinatorics of words. Some of his contributions contain also beautiful bijections. We dedicate to him this article which sheds light on a wrongly neglected bijection on permutations. There has been recently a lot of interest in finding bijections between the set of indecomposable permutations and subgroups of finite index of the free group on two generators, see the article by Ossona de Mendez and Rosenstiehl [8], the article of the first-named author [4] and that of Bacher with the second-named author [1]. The fact that these objects are counted by the same numbers is known since long time: formulas for the number of subgroups were given by Hall [7] and for the number of indecomposable permutations by Comtet [2]; the formula is the same, so that the existence of a bijection follows. The formula in [2] for this number cn is recursive: c 1 = 1, cn = n! − 1≤ p ≤n−1 c p (n − p )!. A bijection between these two sets was first given by Dress and Franz [5,6]. Another bijection, which has interesting symmetry properties with respect to the inverse permutation, was given by Sillke, first in an article without proof [10], but with an illuminating example, which we reproduce below, and then, with proof, in his PhD thesis [9], written in German. The aim of the present article is to give a proof of Sillke’s bijection, using some new ideas. We hope to give more audience to this beautiful bijection, which we consider to be less well-known as it should be. We also point out the relation to maps and hypermaps, which was already noted in [5,6]. The authors thank the two referees for many suggestions and simplifications of the proofs, and for sharing pictures. The second-named author was supported by an NSERC grant (Canada).
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Corresponding author. E-mail addresses: [email protected] (R. Cori), [email protected] (C. Reutenauer).
http://dx.doi.org/10.1016/j.tcs.2016.04.015 0304-3975/© 2016 Elsevier B.V. All rights reserved.
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Fig. 1. The permutation θ = 527163984.
Fig. 2. A left-to-right maximum and a right-to-left minimum.
2. Left-to-right maxima and right-to-left minima of permutations Let θ ∈ S n (the symmetric group on [n] = {1, . . . , n}). A left-to-right maximum of θ is a value y ∈ [n] such that for θ(x) = y, one has
∀i ≤ x, θ(i ) ≤ y .
(1)
We call x the position of the left-to-right maximum and y its value. We may identify the left-to-right maximum to the point (x, y ) of the graph of θ ; this graph is represented in the Cartesian plane, see for example Fig. 11 : the left-to-right maxima of θ = 527163984 are y = 5, 7, 9. Similarly, a right-to-left minimum of θ is a value y ∈ [n] such that for θ(x) = y, one has
∀i ≥ x, θ(i ) ≥ y . The right-to-left minima of θ above are y = 4, 3, 1. On the graph of θ , left-to-right maxima and right-to-left minima are seen as in Fig. 2: A is a left-to-right maximum, B is a right-to-left minimum and the shaded areas contain no point of the graph, except A and B. Denote by A k , . . . , A 2 , A 1 the k left-to-right maxima of θ , in increasing order of x-coordinate (equivalently y-coordinate); similarly, the l right-to-left minima are B l , . . . , B 2 , B 1 , in increasing order of y-coordinate (equivalently x-coordinate). Note the inverse order of indices, which is motivated by the inverse Silke bijection. One has A k = (1, θ(1)), A 1 = (θ −1 (n), n), B l = (θ −1 (1), 1), B 1 = (n, θ(n)). Note also that one may have k = 1 and l = 1: k = 1 (resp. l = 1) if and only if θ(1) = n (resp. θ(n) = 1). For two successive left-to-right maxima A i and A i −1 , the shaded area in Fig. 3 does not contain any point of the graph, except these two points. We say that θ ∈ S n is indecomposable if there is no p ∈ [n − 1] such that θ([ p ]) ⊂ [ p ]. Clearly, a permutation is decomposable if and only if for some p ∈ [n − 1], its graph is contained in [ p ] × [ p ] ∪ { p + 1, . . . , n} × { p + 1, . . . , n}. For later use, we slightly generalize this result.
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The running example of the present article is taken from Sillke [10].
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Fig. 3. Two successive left-to-right maxima.
Fig. 4. The left-to-right maxima and the right-to-left minima.
Lemma 2.1. A permutation θ ∈ S n is decomposable if and only if there exist p , q ∈ [n − 1] such that its graph is contained in [ p ] × [q] ∪ { p + 1, . . . , n} × {q + 1, . . . , n}. Proof. The condition is necessary, by taking p = q. Conversely, suppose that such p , q exist. Recall that each column (resp. row) of integer x-coordinate (resp. y-coordinate) between 1 and n contains exactly one point of the graph. Considering the columns of x-coordinate between 1 and p, we see that we must have q ≥ p. Arguing with rows, we see that also p ≥ q. Thus p = q and θ is decomposable. 2 Thus, when θ is indecomposable, its graph looks like Fig. 4, where the points of the graph are all inside the polygon delimited by the A’s and the B’s, or on it. By Lemma 2.1, the two polygonal lines A k . . . A 1 and B l . . . B 1 do not intersect. Note that the A’s (resp. B’s) are the maximal north–west (resp. south–east) points of the graph. 3. Doubly ordered sets and permutations We begin by motivating the introduction of doubly ordered sets. Let G be the graph of a permutation θ ∈ S n . On G, there are naturally two total orders. The first one is denoted ≤x ; it is defined for any points M , N in G by: M ≤x N if and only if x( M ) ≤ x( N ), where x( P ) is the x-coordinate of point P . Similarly the order ≤ y of G is defined by: M ≤ y N if and only if y ( M ) ≤ y ( N ). Thus we have two total orders on G. Call doubly ordered set a triple ( E , ≤1 , ≤2 ), where E is a finite set and ≤1 , ≤2 are total orders on E. An isomorphism of two doubly ordered sets ( E , ≤1 , ≤2 ) and ( F , ≤1 , ≤2 ) is a bijection φ : E → F which is increasing for the orders ≤i , that is: for any e , e in E and for i = 1, 2, e ≤i e implies φ(e ) ≤i φ(e ). Permutations are a set of representatives of the isomorphism classes of doubly ordered sets. Indeed, a permutation θ ∈ S n defines the doubly ordered set (G , ≤x , ≤ y ) as above. Conversely, if ( E , ≤1 , ≤2 ) is any doubly ordered set of cardinality n, we obtain a permutation as follows: consider first the example of E = {a, b, c , d}, ≤1 = {b <1 a <1 d <1 c } and ≤2 = {d <2 a <2 c <2 b}; the associated permutation is then θ = 3241, obtained by identifying 1234 = badc. Formally, let ωi , i = 1, 2, be the unique increasing order isomorphism ( E , ≤i ) → [n]; then the set G = {(ω1 (e ), ω2 (e )) | e ∈ E } is the graph of the permutation θ = ω2 ◦ ω1−1 and (G , ≤x , ≤ y ) is isomorphic with ( E , ≤1 , ≤2 ): the isomorphism φ : E → G is φ(e ) = (ω1 (e ), ω2 (e )).
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Fig. 5. Position of points.
Fig. 6. Right-to-left minima.
Moreover, if the graphs of two permutations, viewed as doubly ordered sets, are isomorphic, then the permutations are equal, as the reader is invited to verify. Lemma 3.1. Let ( E , ≤1 , ≤2 ) be a doubly ordered set and θ the associated permutation. (i) θ is indecomposable if and only if E has no nontrivial proper upper order ideal for both orders. (ii) A left-to-right-maximum of θ corresponds to an element f of E such that for any e in E, one has e ≤1 f ⇒ e ≤2 f . (iii) A right-to-left-minimum of θ corresponds to an element f of E such that for any e in E, one has e ≤2 f ⇒ e ≤1 f . Proof. We may assume that ( E , ≤1 , ≤2 ) = (G , ≤x , ≤ y ), where G is the graph of θ . Then (i) follows directly from the definition of indecomposability. For (ii), we note that if y is a left-to-right maximum of θ , then the point f = (x, y ) of G satisfies: for any point e = (i , j ) of G with e ≤x f , one has i ≤ x and therefore j = θ(i ) ≤ y by Eq. (1) and thus e ≤ y f . The proof of the converse, and of (iii), is quite similar. 2 4. A property of left-to-right maxima and right-to-left minima In an ordered set ( E , ≤) we denote by [e ≤] the principal upper ideal {x ∈ E | e ≤ x}. Lemma 4.1. Let θ be a permutation in S n and take the notation of Section 2. Let i = 1, . . . , k, j = 1, . . . , l. Let
E i j = [ A i ≤x ] \ [ B j ≤ y ], F i j = [ B j ≤ y ] \ [ A i ≤x ].
(2)
(i) If E i j is nonempty, then max≤x ( E i j ) = B j +1 . (ii) If F i j is nonempty, then max≤ y ( F i j ) = A i +1 . Proof. It is enough to prove (i). We represent the set [ A i ≤x ] by the non-shaded area, at the right of the vertical line x = x( A i ), including this line, as in Fig. 5. Similarly for the complement of the set [ B j ≤ y ]. We assume that E i j is nonempty; hence j < l since [ B l ≤ y ] is the whole set G. Since there are no points in the shaded area indicated in Fig. 6, and since E i j is nonempty, we see that B j +1 must be at the right of the vertical line x = x( A i ) or on it. Hence we have Fig. 7, where the unshaded area is E i j ; thus its maximum for the order ≤x is B j +1 . 2 5. Cyclic form of a permutation We want to associate each indecomposable permutation θ with two permutations σ and α on the set of points of the graph G of θ , as shown for example in Fig. 8 for the permutation θ = 527163984. The permutation σ associates each point M of G with the point having x-coordinate equal to x( M ) + 1, if such a point exists and if it is not a left-to-right maximum. Otherwise σ ( M ) is the left-to right-maximum A j where A j is the minimal index such that x( A i ) ≤ x( M ). Symmetrically, the permutation α associates each point M with the point having y-coordinate equal to y ( M ) + 1, if such point exists and if it is not a right-to left minimum. Otherwise, α ( M ) is the right-to-left minimum B j where j is the minimal index such that y ( B j ) ≤ y ( M ).
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Fig. 7. Region.
Fig. 8. The cycles of
σ and α .
The permutations σ and α may be completely defined by the following condition: given a permutation τ of any set E, call cyclic form of τ each well-parenthesis word w, which represents τ in cyclic form, including cycles of length 1:
w = (h1 , τ h1 , τ 2 h2 , . . . , τ i 1 −1 h1 )(h2 , τ h2 , . . . , τ i 2 −1 h2 ) . . . The lengths of the cycles are therefore i 1 , i 2 , . . . . For given τ , such a form is of course not unique. Given such a word w, we associate with it the total order on E, whose list from the smallest to the largest is obtained by removing the parenthesis. Define the two permutations σ and α of the graph G of θ by the condition that they have respectively a cyclic form u, v as follows
u = ( A k , . . .) . . . ( A 2 , . . .)( A 1 , . . . , B 1 ), v = ( B l , . . .) . . . ( B 2 , . . .)( B 1 , . . . , A 1 ), and such that the associated orders satisfy ≤u =≤x and ≤ v =≤ y . This definition of σ , α is somewhat undirect but it defines them completely. Equivalently, one can define them as follows. Let A 0 = B 0 = (n + 1, n + 1). Then one has for any M ∈ G:
• If x( A i +1 ) ≤ x( M ) < x( A i ) then – either x( M ) + 1 < x( A i ) and then σ ( M ) = (x( M ) + 1, θ(x( M ) + 1)), – or x( M ) + 1 = x( A i ) and then σ ( M ) = A i +1 . • If y ( B j +1 ) ≤ y ( M ) < y ( B j ) then – either y ( M ) + 1 < y ( B j ) and then α ( M ) = (θ −1 ( y ( M ) + 1), y ( M ) + 1), – or y ( M ) + 1 = y ( B j ) and then α ( M ) = B j +1 . Lemma 5.1. If i = k, . . . , 2, then x(α ( A i )) ≥ x( A i −1 ). Proof. The statement is trivially true if α ( A i ) = A i −1 . If y (α ( A i )) = y ( A i ) + 1, then Fig. 9, left part, shows that α ( A i ) can only be on the right of A i −1 . Otherwise, one has α ( A i ) = B j (and hence y ( A i ) + 1 = y ( B j −1 )) and Fig. 9, right part, shows that B j must be on the right of A i −1 , or θ would be indecomposable by Lemma 2.1. 2 Recall that the action of a group on a set is called transitive if there is a unique orbit. Lemma 5.2. The group generated by σ and α acts transitively on G.
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Fig. 9. The mapping
α.
Proof. Note that B 1 is the maximum of G for the order ≤x . Thus is enough to show that for any M in G, with M = B 1 , either σ ( M ) >x M or ασ ( M ) >x M. If x(σ ( M )) = x( M ) + 1, then σ ( M ) >x M. Otherwise σ (m) = A i for some i ≥ 2 because M = B 1 . Using Lemma 5.1, x(ασ ( M )) = x(α ( A i )) ≥ x( A i −1 ) > x( M ), hence ασ ( M ) >x M. 2 6. Doubly pointed hypermaps and Sillke’s theorem (first form) Consider the set of quintuples ( E , σ , α , A , B ), where E is a finite set, σ , α are permutations of E which act transitively on E (that is, the subgroup they generate acts transitively), and A , B are distinct elements of E such that σ ( B ) = A and α ( A ) = B. An isomorphism between two such quintuples ( E , σ , α , A , B ) and ( E , σ , α , A , B ) is a bijection τ : E → E
such that τ ( A ) = A , τ ( B ) = B , τ σ = σ τ and that τ α = α τ . A doubly pointed hypermap is an isomorphism class of such quintuples. Sillke’s mapping is the mapping that associates to each indecomposable permutation θ a doubly pointed hypermap (G , σ , α , A 1 , B 1 ), following the construction of the previous sections. Note that if a permutation is indecomposable, so is its inverse. Theorem 6.1. Sillke’s mapping is a bijection between indecomposable permutations in S n and doubly pointed hypermaps of cardinality n. If θ is mapped onto the class of ( E , σ , α , A , B ), then θ −1 is mapped onto the class of ( E , α , σ , B , A ). The proof rests on two technical lemmas. The first one is essentially Sillke’s inverse algorithm, which is implicitly described in the proof below. Lemma 6.1. Given a quintuple ( E , σ , α , A , B ) as above, there exist unique cyclic forms for σ and α , respectively u = ( A k , . . .)( A k−1 , . . .) . . . ( A 1 , . . . , B 1 ) and v = ( B l , . . .)( B l−1 , . . .) . . . ( B 1 , . . . , A 1 ), with A 1 = A, B 1 = B, such that, if one puts E i j = [ A i ≤u ] \ [ B j ≤ v ], F i j = [ B j ≤ v ] \ [ A i ≤u ], then: (i) If E i j is nonempty, then max≤u ( E i j ) = B j +1 . (ii) If F i j is nonempty, then max≤ v ( F i j ) = A i +1 . Proof. Suppose that there are other cyclic forms u , v for
σ , α respectively:
u = ( A k , . . .)( A k −1 , . . .) . . . ( A 1 , . . . , B 1 ) and
v = ( B l
, . . .)( B l
−1 , . . .) . . . ( B 1 , . . . , A 1 ). We have A 1 = A = A 1 and B 1 = B = B 1 by hypothesis. Moreover, conditions (i) and (ii) hold for the sets E i j = [ A i ≤u ] \
[ B j ≤ v ] and F i j = [ B j ≤ v ] \ [ A i ≤u ]. Suppose that for some s, t ≥ 1 we have A 1 = A 1 , A 2 = A 2 , . . . , A s = A s , B 1 = B 1 , B 2 = B 2 , . . . , B t = B t . We show that these equalities may be extended to the case s + 1 or t + 1, unless s = k and t = l. For this, consider the sets E st and F st . If both are empty, then [ A s ≤u ] = [ B t ≤ v ]; this set is therefore invariant under σ and α (because this set is a union of cycles of both σ and α ), thus equal to E, since they act transitively; thus s = k and t = l. Assume that E st = ∅. Then B t +1 = max≤u ( E st ), by hypothesis. Moreover, E st = E st by the inductive hypothesis, and B t +1 = max≤u ( E st ). Thus B t +1 = B t +1 . This proves uniqueness in the Lemma, by symmetry.
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In order to prove existence, observe that what precedes proves also the following fact (which mimics Sillke’s algorithm): there exist cyclic forms for σ and α , respectively
u = ( A k , . . .)( A k−1 , . . .) . . . ( A 1 , . . . , B 1 ) and
v = ( B l , . . .)( B l−1 , . . .) . . . ( B 1 , . . . , A 1 ), with B 1 = B , A 1 = A and a sequence of pairs (1, 1) = (s0 , t 0 ) < (s1 , t 1 ) < . . . < (sk+l−2 , tk+l−2 ) = (k, l), where < is the natural order of N2 , and such that for any h = 0, . . . , k + l − 3
• either (sh+1 , th+1 ) = (sh + 1, th ), F sh th = ∅ and max≤v ( F sh th ) = A sh +1 ; • or (sh+1 , th+1 ) = (sh , th + 1), E sh th = ∅ and max≤u ( E sh th ) = B th +1 . In particular, (i) and (ii) hold for each pair (i , j ) = (sh , th ), h = 0, . . . , k + l − 2. We prove conditions (i) and (ii) for all pairs (i , j ). For this we use the easy claim: let H be a totally ordered set and E an nonempty subset of H . If the subset E is obtained from E by adding elements which are smaller than those of E, then max( E ) = max( E ). If the nonempty subset E
is obtained from E by removing a lower order ideal, then max( E ) = max( E ). We deduce the following facts:
• If E i j is nonempty, then max≤u ( E i +1, j ) = max≤u ( E i j ), because E i +1, j = E i j ∪ ([ A i +1 , A i [≤u \ [ B j ≤ v ]) and the elements of E i j are all ≥u A i ; • If both E i j and E i −1, j are nonempty, then max≤u ( E i j ) = max≤u ( E i −1, j ), because E i −1, j = E i j \ [ A i , A i −1 [≤u . Since E i j ⊂ E i +1, j , we deduce that, j being fixed, if for some (s, j ), E sj is nonempty and if max≤u ( E sj )) = B j +1 , then (i) holds for each (i , j ). Now, observe that the sequence of pairs
(s0 , t 0 ) < (s1 , t 1 ) < . . . < (sk+l−2 , tk+l−2 ) = (k, l) is a discrete path from (1, 1) to (k, l); moreover, for any j, there exists s such that (s, j ) is on this path and that E sj is nonempty: indeed, if (s, j ) = (k, l) is on the path and E sj = ∅, then by 2), F sj = ∅ and (s + 1, j ) is on the path; moreover, E kj is nonempty. It follows that (i) holds for all pairs (i , j ). The proof of (ii) is similar. 2 Lemma 6.2. Under the hypothesis of the previous lemma, one has: (i) There is no nontrivial proper upper ideal for both orders ≤u and ≤ v ; (ii) ∀i = 1, . . . , k, M ≤u A i ⇒ M ≤ v A i ; (iii) ∀ j = 1, . . . , l, M ≤ v B j ⇒ M ≤u B j . Proof. (i) Suppose that H is a nontrivial proper upper ideal for both orders. One has A 1 = max≤ v ( E ), thus A 1 ∈ H since H is a nonempty upper ideal for ≤ v . Thus [ A 1 ≤u ] is contained in H , since H is an upper ideal for ≤u . We deduce that there is a maximum i such that [ A i ≤u ] ⊂ H . Similarly there is a maximum j such that [ B j ≤ v ] ⊂ H . If we had E i j = ∅, then, since E i j ⊂ H , we would obtain by Lemma 6.1 that B j +1 ∈ H ; thus [ B j +1 ≤ v ] ⊂ H , a contradiction. Thus E i j = ∅. Similarly F i j = ∅. It follows that [ A i ≤u ] = [ B j ≤ v ]. This implies that σ and α do not act transitively, since the latter set is invariant under them, except if we have i = k; but this cannot happen since H = E = [ A k ≤u ]. (ii) For i = 1 this is clear since A 1 = max≤ v ( E ). Suppose that 2 ≤ i < k. Then F il = ∅: indeed, otherwise [ B l ≤ v ] \ [ A i ≤u ] = ∅, which implies i = k, because [ B l ≤ v ] = E; but we have assumed i < k. Thus by Lemma 6.1, A i +1 = max≤v ( E \ [ A i ≤u ]). Let M
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7. Pointed hypermaps, subgroups of finite index of the free group on two generators and Sillke’s theorem (second form) Consider quadruples ( E , σ , α , A ), where E is a finite set, σ , α are two permutations of E acting transitively on it and A is an element of E. An isomorphism between two such quadruples ( E , σ , α , A ) and ( E , σ , α , A ) is a bijection τ : E → E
such that τ ( A ) = A , τ σ = σ τ and τ α = α τ . A pointed hypermap is an isomorphism class of such quadruples. As observed by Sillke, there is a bijection between doubly pointed hypermaps of cardinality n + 1 and pointed hypermaps of cardinality n: let the doubly pointed hypermap be represented by the quintuple ( E , σ , α , A , B ); then in σ and α represented in cycle notation, remove the element B. This construction induces the desired bijection. Let F 2 be the free group on two generators a, b. Let H be a subgroup of index n of F 2 . By letting F 2 act on the set E of left cosets of H , we obtain a pair (σ , α ) of permutations of this set E (namely the permutations of E induced by a, b respectively), which act transitively on E, together a distinguished element of E (namely the coset H ). In other words, we obtain a quadruple as above. It is easy to see that this construction induces a bijection between subgroups of finite index n and pointed hypermaps of cardinality n. Theorem 7.1. There is a bijection between the set of indecomposable permutations in S n+1 and subgroups of index n in the free group F 2 , or equivalently, pointed hypermaps of cardinality n. In this bijection, if θ ∈ S n+1 corresponds to the subgroup H and the pointed hypermap represented by ( E , σ , α , A ), then θ −1 corresponds to the image of H under the automorphism of F 2 exchanging the generators a and b, and to the pointed hypermap represented by ( E , α , σ , A ). Motivated by the last assertion, we may ask what happens for other natural involutions defined either on indecomposable permutations or on subgroups of F 2 . For example, to which involution of subgroups of F 2 does the conjugation of indecomposable permutations by the longest permutation n + 1, n, . . . , 2, 1 (which preserves indecomposability) correspond? Or to which involution of indecomposable permutations does the involution of subgroups induced by the anti-automorphism a → a−1 , b → b−1 of F 2 correspond? One may also wonder if it is possible to identify directly on indecomposable permutations conjugate subgroups of F 2 , and normal subgroups; partial answers have been obtained by Chekkal [3]. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
R. Bacher, C. Reutenauer, Number of right ideals and a q-analogue of indecomposable permutations, Canad. J. Math. (2016), in press. n L. Comtet, Sur les coefficients de l’inverse de la série formelle n!t , C. R. Acad. Sci. Paris Ser. A 2 (75) (1972) 569–572. A. Chekkal, Yamanouchisation, PhD thesis, Université du Québec à Montréal, 2010. R. Cori, Indecomposable permutations, hypermaps and labeled Dyck paths, J. Combin. Theory Ser. A 116 (2009) 1326–1343. A.W.M. Dress, R. Franz, Parametrizing the subgroups of finite index in a free group and related topics, Bayreuth. Math. Schr. 20 (1985) 1–8. A.W.M. Dress, R. Franz, Zur Parametrisierung von Untergruppen freier Gruppen, Beitr. Algebra Geom. 20 (1987) 125–134. M. Hall, Subgroups of finite index in free groups, Canad. J. Math. 1 (1949) 187–190. P. Ossona De Mendez, P. Rosenstiehl, Transitivity and connectivity of permutations, Combinatorica 24 (2004) 487–501. T. Sillke, Zur Kombinatorik von Permutationen, in: Séminaire Lotharingien de Combinatoire, vol. B21c, 1989, 8 pp. T. Sillke, Eine Bijektion zwischen Untergruppen freier Gruppen und Systemen konnexer Permutationen (A bijection between subgroups of free groups and systems of connex permutations), Dissertation, Universität Bielefeld Fakultät für Mathematik, 1992 (in German).