We find an explicit expression for the generating function of the number of
permutations in S_n avoiding a subgroup of S_k generated by all but one simple
transpositions. The generating function turns out to be rational, and its
denominator is a rook polynomial for a rectangular board

Mansour, Toufik

Vainshtein, Alek

English

arXiv

International audienceWe find an explicit expression for the generating function of the number of permutations in S_n avoiding a subgroup of S_k generated by all but one simple transpositions. The generating function turns out to be rational, and its denominator is a rook polynomial for a rectangular board

HAL: Hyper Article en Ligne

Avoiding maximal parabolic subgroups of S_k

Hal-Diderot

HAL Descartes

We find an explicit expression for the  generating function of the number of permutations in  S  n avoiding a subgroup of  S  k generated by all but one simple  transpositions. The generating function turns out to be  rational, and its denominator is a rook polynomial for a  rectangular board

Toufik Mansour

Alek Vainshtein

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Avoiding maximal parabolic subgroups of  S  k

We find an explicit expression for the generating function of the number of permutations in S_n avoiding a subgroup of S_k generated by all but one simple transpositions. The generating function turns out to be rational, and its denominator is a rook polynomial for a rectangular board

Episciences.org

Discrete Mathematics and Theoretical Computer Science4, 2000, 67–77Avoiding maximal parabolic subgroups of SkToufik Mansour1†and Alek Vainshtein2Department of Mathematics and Department of Computer Science, University of Haifa, Haifa, Israel 31905.1 Email: tmansur@study.haifa.ac.il 2 Email: alek@mathcs.haifa.ac.ilreceived 4 Jul 2000, accepted 16 Nov 2000.We find an explicit expression for the generating function of the number of permutations inSn avoiding a subgroup ofSk generated by all but one simple transpositions. The generating function turns out to be rational, and its denominatoris a rook polynomial for a rectangular board.Keywords: permutations, forbidden patterns, parabolic subgroups, Laguerre polynomials, rook polynomials1 Introduction and Main ResultLet [p] = {1, . . . , p} denote a totally ordered alphabet onp letters, and letα = (α1, . . . ,αm) ∈ [p1]m,β = (β1, . . . ,βm) ∈ [p2]m. We say thatα is order-isomorphicto β if for all 1 ≤ i < j ≤ m one hasα i < α jif and only if βi < β j . For two permutationsπ∈ Sn andτ ∈ Sk, anoccurrenceof τ in π is a subsequence1≤ i1 < i2 < .. . < ik ≤ n such that(πi1, . . . ,πik) is order-isomorphic toτ; in such a contextτ is usuallycalled thepattern. We say thatπ avoidsτ, or is τ-avoiding, if there is no occurrence ofτ in π. Patternavoidance proved to be a useful language in a variety of seemingly unrelated problems, from stack sorting[Kn, Ch. 2.2.1] to singularities of Schubert varieties [LS]. A natural generalization of single patternavoidance issubset avoidance; that is, we say thatπ∈ Sn avoids a subsetT ⊂ Sk if π avoids anyτ ∈ T.The set of all permutations inSn avoidingT ⊂ Sk is denotedSn(T). A complete study of subset avoidancefor the casek = 3 is carried out in [SS]. Fork > 3 the situation becomes more complicated, as the numberof possible cases grows rapidly. Recently, several authors have considered the case of generalk whenThas some nice algebraic properties. Paper [BDPP] treats the case whenT is the centralizer ofk−1 andkunder the natural action ofSk on [k] (see also Sec. 3 for more detail). In [AR],T is a Kazhdan–Lusztig cellof Sk, or, equivalently, the Knuth equivalence class (see [St, vol. 2, Ch. A1]). In this paper we considerthe case whenT is a maximal parabolic subgroup ofSk.Let si denote the simple transposition interchangingi andi +1. Recall that a subgroup ofSk is calledparabolic if it is generated bysi1, . . . ,sir . A parabolic subgroup ofSk is calledmaximalif the number ofits generators equalsk−2. We denote byPl ,m the (maximal) parabolic subgroup ofSl+m generated bys1, . . . ,sl−1,sl+1, . . . ,sl+m−1, and by fl ,m(n) the number of permutations inSn avoiding all the patterns inPl ,m. In this note we find an explicit expression for the generating function of the sequence{ fl ,m(n)}.†Partially supported by HIACS.1365–8050c© 2000 Maison de l’Informatique et des Mathématiques Discr̀etes (MIMD), Paris, France68 Toufik Mansour and Alek VainshteinTo be more precise, we prove the following more general result. Let us denoteσ = s1s2 . . .sk−1, that is,σ = (2,3, . . . ,k,1) (written in one-line notation), and leta be an integer, 0≤ a≤ k−1 (here and in whatfollows k = l +m). We denote byf al ,m(n) the number of permutations inSn avoiding the left cosetσaPl ,m;in particular, f 0l ,m(n) coincides withfl ,m(n). Let Fal ,m(x) denote the generating function of{al ,m(n)},Fal ,m(x) = ∑n≥0f al ,m(n)xn.Recall that theLaguerre polynomial Lαn(x) is given byLαn(x) =1n!exx−αdndxn(e−xxn+α),and therook polynomialof the rectangulars× t board is given byRs,t(x) = s!xsLt−ss (−x−1)for s≤ t and byRs,t(x) = Rt,s(x) otherwise (see [Ri, Ch. 7.4]).Main Theorem. Letλ = min{l ,m}, µ= max{l ,m}, thenFal ,m(x)Rl ,m(−x) =λ−1∑r=0xr r!r∑j=0(−1) j(lj)(mj)(rj) +(−1)λxλλ!µ−λ−1∑r=0xr r!(µ− r −1λ),or, equivalently,Fal ,m(x) =k−1∑r=0xr r!−(−1)λxµλ!Lµ−λλ (x−1)λ−1∑r=0(k+ r)!xrλ∑j=r+1(−1) j(lj)(mj)(k+rj) ,where k= l +m= λ +µ.The proof of the Main Theorem is presented in the next section.As a corollary we immediately get the following result (see [Ma, Theorem 1]).Corollary 1 Let0≤ a≤ k−1, thenf a1,k−1(n) ={(k−2)!(k−1)n+2−k for n≥ kn! for n < k.Proof. SinceR1,k−1 = 1+(k−1)x, the Main Theorem impliesFa1,k−1(x) =1−∑k−3r=0 xr+1(k− r −2)r!1− (k−1)x=xk−2(k−2)!1− (k−1)x+k−3∑r=0xr r!,and the result follows. ✷Another immediate corollary of the Main Theorem gives the asymptotics forf al ,m(n) asn→ ∞.Corollary 1.2. f al ,m(n) ∼ cγn, where c is a constant depending on l and m, andγ is the maximal root ofLµ−λλ ; in particular, γ≤ k−2+√1+4(l −1)(m−1).Proof. Follows from standard results in the theory of rational generating functions (see e.g. [St, vol. 1,Ch. 4]) and the fact that all the roots of Laguerre polynomials are simple (see [Sz, Ch. 3.3]). The upperbound onγ is obtained in [IL]. ✷Avoiding maximal parabolic subgroups of Sk 692 ProofsFirst of all, we make the following simple, though useful observation.Lemma 2 For any natural a, l, m, n such that1≤ a≤ l +m−1 one has fal ,m(n) = fm+l−am,l (n).Proof. Denote byρn andκn the involutionsSn → Sn that take(i1, i2, . . . , ik) to (ik, . . . , i2, i1) (reversal) to(n+1− i1,n+1− i2, . . . ,n+1− in) (complement), respectively. It is easy to see that for anyT ⊂ Sk, theinvolutionsρn andκn provide natural bijections between the setsSn(T) andSn(ρkT), and betweenSn(T)andSn(κkT), respectively. It remains to note thatρkκkσaPl ,m = σl+m−aPm,l . ✷From now on we assume that≥ 0, l ≥ 1, m≥ 1 are fixed, and denoteb = n−m+a. It follows fromLemma 2 that we may assume thata≤m, and henceb≤ n. This means, in other words, thatτ ∈Sk belongsto σaPl ,m if and only if (τ1, . . . ,τl ) is a permutation of the numbersa+ 1, . . . ,a+ l . In what follows weusually omit the indicesa, l , mwhenever appropriate; for example, instead off al ,m(n) we write just f (n).For anyn≥ k and anyd such that 1≤ d ≤ n, we denote bygn(i1, . . . , id) = gan;l ,m(i1, . . . , id) the numberof permutationsπ∈ Sn(σaPl ,m) such thatπj = i j for j = 1, . . . ,d. It is natural to extendgn to the cased = 0 by settinggn( /0) = f (n).The following properties of the numbersgn(i1, . . . , id) can be deduced easily from the definitions.Lemma 3(i) Let n≥ k and1≤ i ≤ n, thengn(. . . , i, . . . , i, . . .) = 0.(ii) Let n≥ k and a+1≤ i j ≤ b for j = 1, . . . , l, thengn(i1, . . . , i l ) = 0.(iii) Let n ≥ k, 1≤ r ≤ d ≤ l, and a+1≤ i j ≤ b for j = 1, . . . ,d, j 6= r, thengn(i1, . . . , id) ={gn−1(i1−1, . . . , ir−1−1, ir+1−1, . . . , id −1) if 1≤ ir ≤ agn−1(i1, . . . , ir−1, ir+1, . . . , id) if b+1≤ ir ≤ n.Proof. Property (i) is evident. Let us prove (ii). By (i), we may assume that the numbersi1, . . . , i l aredistinct. Take an arbitraryπ∈ Sn such thatπj = i j for j = 1, . . . , l . Evidently, for anyr ≤ a there exists aposition jr > l such thatπjr = r; the same is true for anyr ≥ b+1. Therefore, the restriction ofπ to thepositions 1,2, . . . , l , j1, j2, . . . , ja, jb+1, jb+2, . . . , jn (in the proper order) gives an occurrence ofτ ∈ σaPl ,min π. Hence,π /∈ Sn(σaPl ,m), which means thatgn(i1, . . . , i l ) = 0.To prove (iii), assume first that 1≤ ir ≤ a. Let π∈ Sn andπj = i j for j = 1, . . . ,d. We defineπ∗ ∈ Sn−1byπ∗j =πj −1 for 1≤ j ≤ r −1,πj+1−1 for j ≥ r andπj+1 > ir ,πj+1 for j ≥ r andπj+1 < ir .(1)We claim thatπ∈ Sn(σaPl ,m) if and only if π∗ ∈ Sn−1(σaPl ,m). Indeed, the only if part is trivial, sinceany occurrence ofτ ∈ σaPl ,m in π∗ immediately gives rise to an occurrence ofτ in π. Conversely, any70 Toufik Mansour and Alek Vainshteinoccurrence ofτ in π that does not includeir gives rise to an occurrence ofτ in π∗. Assume that thereexists an occurrence ofτ in π that includesir . Sincer ≤ d ≤ l , this occurrence ofτ containsa entriesthat are situated to the right ofir and are strictly less thanir . However, the wholeπ contains onlya−1such entries, a contradiction. It now follows from (1) that property (iii) holds for 1≤ ir ≤ a. The caseb+1≤ ir ≤ n is treated similarly with the help of the transformation(π∈ Sn) 7→ (π◦ ∈ Sn−1) given byπ◦j =πj for 1≤ j ≤ r −1,πj+1−1 for j ≥ r andπj+1 > ir ,πj+1 for j ≥ r andπj+1 < ir .✷Now we introduce the quantity that plays the crucial role in the proof of the Main Theorem. Forn≥ kand 1≤ d ≤ l we putA(n,d) = Aal ,m(n,d) =b∑i1,...,id=a+1gn(i1, . . . , id).As before, this definition is extended to the cased = 0 by settingA(n,0) = gn( /0) = f (n).Theorem 4 Let n≥ k+1 and1≤ d ≤ l −1, thenA(n,d+1) = A(n,d)− (m−d)A(n−1,d)−dA(n−1,d−1). (2)Proof. First of all, we introduce two auxiliary sums:B(n,d) = Bal ,m(n,d) =b+1∑i1,...,id=a+1gn(i1, . . . , id),C(n,d) = Cal ,m(n,d) =b∑i1,...,id=agn(i1, . . . , id),whereb = n−m+a; once again,B(n,0) = C(n,0) = f (n).Let us prove three simple identities relating together the sequences{A(n,d)}, {B(n,d)}, {C(n,d)}.Lemma 5 Let n≥ k and1≤ d ≤ l, then:A(n,d) = A(n,d−1)− (m−a)B(n−1,d−1)−aC(n−1,d−1),(m−a)A(n,d) = (m−a)B(n,d)− (m−a)dB(n−1,d−1),aA(n,d) = aC(n,d)−adC(n−1,d−1).Proof. To prove the first identity, observe that by definitions and Lemma 3(iii) for the caser = d, one hasA(n,d−1)−A(n,d) =b∑i1,...,id−1=a+1n∑id=1gn(i1, . . . , id)−A(n,d)Avoiding maximal parabolic subgroups of Sk 71=b∑i1,...,id−1=a+1(a∑id=1gn(i1, . . . , id)+n∑id=b+1gn(i1, . . . , id))=b∑i1,...,id−1=a+1(agn−1(i1−1, . . . , id−1−1)+(m−a)gn−1(i1, . . . , id−1))= aC(n−1,d−1)+(m−a)B(n−1,d−1),and the result follows.The second identity is trivial fora = m, so assume that 0≤ a≤ m−1 and observe that by definitionsand Lemma 3(ii) and (iii), one hasB(n,d) =b∑i1,...,id=a+1gn(i1, . . . , id)+d∑j=1(b∑i1,...,ı̂ j ,...,id=a+1gn(i1, . . . , i j−1,b+1, i j+1 . . . , id))= A(n,d)+d∑j=1(b∑i1,...,ı̂ j ,...,id=a+1gn−1(i1, . . . , i j−1, i j+1, . . . , id))= A(n,d)+dB(n−1,d−1),and the result follows.Finally, the third identity is trivial fora = 0, so assume that 1≤ a≤ m and observe that by definitionsand Lemma 3(ii) and (iii), one hasC(n,d) =b∑i1,...,id=a+1gn(i1, . . . , id)+d∑j=1(b∑i1,...,ı̂ j ,...,id=a+1gn(i1, . . . , i j−1,a, i j+1 . . . , id))= A(n,d)+d∑j=1(b∑i1,...,ı̂ j ,...,id=a+1gn−1(i1−1, . . . , i j−1−1, i j+1−1, . . . , id −1))= A(n,d)+dC(n−1,d−1),and the result follows. ✷Now we can complete the proof of Theorem 4. Indeed, using twice the first identity of Lemma 5, onegetsA(n,d+1) = A(n,d)− (m−a)B(n−1,d)−aC(n−1,d−1),dA(n−1,d) = dA(n−1,d−1)−d(m−a)B(n−2,d−1)−daC(n−2,d−1).Next, the other two identities of Lemma 5 implyA(n,d+1)−dA(n−1,d) = A(n,d)−dA(n−1,d−1)−((m−a)B(n−1,d)− (m−a)dB(n−2,d−1))−(aC(n−1,d)−adC(n−2,d−1))= A(n,d)−dA(n−1,d−1)− (m−a)A(n−1,d)−aA(n−1,d),and the result follows. ✷The next result relates the sequence{A(n,d)} to the sequence{ f (n)}.72 Toufik Mansour and Alek VainshteinTheorem 6 Let n≥ k and1≤ d ≤ l, thenA(n,d) =d∑j=0(−1) j j!(mj)(dj)f (n− j).Proof. LetD(n,d) = Dal ,m(n,d) denote the right hand side of the above identity. We claim that forn≥ k+1and 1≤ d ≤ l −1, D(n,d) satisfies the same relation (2) asA(n,d) does. Indeed,D(n−1,d) =d∑j=0(−1) j j!(mj)(dj)f (n−1− j)= −d∑j=1(−1) j( j −1)!(mj −1)(dj −1)f (n− j)+(−1)dd!(md)f (n−d−1),andD(n−1,d−1) =d−1∑j=0(−1) j j!(mj)(d−1j)f (n−1− j)= −d∑j=1(−1) j( j −1)!(mj −1)(d−1j −1)f (n− j),and henceD(n,d)− (m−d)D(n−1,d)−dD(n−1,d−1) = f (n)+(m−d)(−1)d+1d!(md)f (n−d−1)+d∑j=1(−1) j j!((mj)(dj)+m−dj(mj −1)(dj −1)+dj(mj −1)(d−1j −1))f (n− j)= f (n)+d∑j=1(−1) j j!(mj)(d+1j)f (n− j)+(−1)d+1(d+1)!(md+1)f (n−d−1) = D(n,d+1).It follows thatD(n,d) (as well asA(n,d)) are defined uniquely forn≥ k and 1≤ l ≤ d by initial valuesD(k,d), D(n,0), andD(n,1) (A(k,d), A(n,0), andA(n,1), respectively). It is easy to see that forn ≥ kone hasA(n,0) = D(n,0) = f (n). Next, the first identity of Lemma 5 ford = 1 givesA(n,1) = A(n,0)− (m−a)B(n−1,0)−aC(n−1,0) = f (n)−m f(n−1) for n≥ k.On the other hand, by definition,D(n,1) = f (n)−m f(n−1) for n≥ k,and henceA(n,1) = D(n,1). Finally, a simple combinatorial argument shows thatA(k,d) = d!(ld)(k−d)!− l !m! for 1 ≤ d ≤ l .Avoiding maximal parabolic subgroups of Sk 73On the other hand,D(k,d) =d∑j=0(−1) j j!(mj)(dj)(k− j)!− l !m!,since f (r) = r! for 1 ≤ r ≤ k−1 and f (k) = k!− l !m!. To proveA(k,d) = D(k,d) it remains to check thatd∑j=0(−1) j j!(mj)(dj)(k− j)! = d!(ld)(k−d)!,which follows from Lemma 7 below. ✷Finally, we are ready to prove the Main Theorem stated in Sec. 1. First of all, by Lemma 3(ii),A(n, l) =0 for n≥ k. Hence, by Theorem 6,l∑j=0(−1) j j!(mj)(lj)f (n− j) = 0 for n≥ k,or, equivalently,l∑j=0(−1) j j!(mj)(lj)x j f (n− j)xn− j = 0 for n≥ k.As was already mentioned,f (r) = r! for 1 ≤ r ≤ k−1, therefore, summing overn≥ k yieldsl∑j=0(−1) j j!(mj)(lj)x j(Fal ,m(x)−k− j−1∑i=0xi i!)= 0. (3)Recall that the rook polynomial of the rectangulars× t board,s≤ t, satisfies relationRs,t(x) =s∑j=0j!(tj)(sj)x j(see [Ri, Ch. 7.4]). Hence, (3) is equivalent toFal ,m(x)Rλ,µ(−x) =l∑j=0(−1) j j!(mj)(lj)x jk− j−1∑i=0xi i! =k−1∑r=0xr r!r∑j=0(−1) j(lj)(mj)(rj) .Let us divide the external sum in the above expression into three parts: the sum fromr = 0 to λ −1, thesum fromr = λ to µ−1, and the sum fromr = µ to k−1. By Lemma 7 below, the third sum vanishes,while the second sum is equal toµ−1∑r=λxr r!(−1)λ(r −λ)!(k− r −1)!(µ− r −1)!r!,and the first expression of the Main Theorem follows. The second expression is obtained easily from (3)and relation between rook polynomials and Laguerre polynomials given in Sec. 1. ✷It remains to prove the following technical result, which is apparently known; however, we failed tofind a reference to its proof, and decided to present a short proof inspired by the brilliant book [PWZ].74 Toufik Mansour and Alek VainshteinLemma 7 Let 1≤ s≤ t and letM(s, t) =s∑i=0(−1)i(si)(ti)(ni) .Then:M(s, t) =(n−ts )(ns)if n≥ s+ t,0 if t ≤ n≤ s+ t −1,(−1)s(s+t−n−1s )(ns)if s≤ n≤ t −1.Proof. Direct check reveals thatM(s, t) is a hypergeometric series; to be more precise,M(s, t) = 2F1[−t,−s−n ;1].Since−s is a nonpositive integer, the Gauss formula applies (see [PWZ, Ch. 3.5]), and we getM(s, t) = limz→nΓ(−z+ t +s)Γ(−z)Γ(t −z)Γ(s−z).Recall thatΓ(x)Γ(1−x) =πsinπx. (4)If n≥ s+ t, we apply (4) forx = −z+ t +s, x = t −z, x = s−z, x = −z, and getM(s, t) = −Γ(n− t +1)Γ(n−s+1)Γ(n− t −s+1)Γ(n+1)limz→nsinπ(t −z)sinπ(s−z)sinπzsinπ(t +s−z)=(n−ts)(ns) .If t ≤ n≤ s+ t −1, we apply (4) forx = t −z, x = s−z, x = −z, and getM(s, t) = −Γ(n− t +1)Γ(n−s+1)Γ(s+ t −n)Γ(n+1)limz→nsinπ(t −z)sinπ(s−z)sinπz= 0.Finally, if s≤ n≤ t −1, we apply (4) forx = s−z, x = −z, and getM(s, t) = −Γ(s+ t −n)Γ(n−s+1)Γ(t −n)Γ(n+1)limz→nsinπ(s−z)sinπz= (−1)s(s+t−n−1s)(ns) .✷3 Concluding remarksObserve first, that according to the Main Theorem,Fal ,m(x) does not depend ona; in other words,|Sn(Pl ,m)|=|Sn(σaPl ,m)| for anya. We obtained this fact as a consequence of lengthy computations. A natural questionwould be to find a bijection betweenSn(Pl ,m) andSn(σaPl ,m) that explains this phenomenon.Avoiding maximal parabolic subgroups of Sk 75Second, it is well known that rook polynomials (or the corresponding Laguerre polynomials) are relatedto permutations with restricted positions, see [Ri, Ch.7,8]. Laguerre polynomials also arise in a naturalway in the study of generalized derangements (see [FZ] and references therein). It is tempting to find acombinatorial relation between permutations with restricted positions and permutations avoiding maximalparabolic subgroups, which could explain the occurrence of Laguerre polynomials in the latter context.Finally, one can consider permutations avoiding nonmaximal parabolic subgroups ofSk. The first natu-ral step would be to treat the case of subgroups generated byk−3 simple transpositions. It is convenientto denote byPl1,l2,l3 (with l1 + l2 + l3 = k) the subgroup ofSk generated by all the simple transpositionsexcept forsl1 andsl1+l2; further on, we setfl1,l2,l3(n) = |Sn(Pl1,l2,l3)|, andFl1,l2,l3(x) = ∑n≥0 fl1,l2,l3(n)xn.It is easy to see thatFl1,l2,l3(x) = Fl3,l2,l1(x), so one can assume thatl1 ≤ l3. This said, the main result of[BDPP] can be formulated as follows: letk≥ 3, thenF1,1,k−2(x) =k−3∑r=1xr r! +(k−3)!xk−42(1− (k−1)x−√1−2(k−1)x+(k−3)2x2).To the best of our knowledge, this is the only known instance ofFl1,l2,l3(x). It is worth noting that even inthis, simplest case of nonmaximal parabolic subgroup, the generating function is no longer rational.References[AR] R. Adin and Yu. Roichman,Shape avoiding permutations, (1999), preprint math.CO/9912119.[BDPP] E. Barcucci, A. Del Lungo, E. Pergola, and R. Pinzani,Permutations avoiding an increasingnumber of length-increasing forbidden subsequences, Di crete Math. Theor. Comput. Sci.4(2000), 31–44.[FZ] D. Foata and D. Zeilberger,Laguerre polynomials, weighted derangements, and positivity, SIAMJ. Discr. Math.1 (1988), 425–433.[IL] M. Ismail and X. Li, Bounds on the extreme zeros of orthogonal polynomials, Proc. AMS115(1992), 131–140[Kn] D. Knuth, The Art of Computer Programming, vol. 1, Addison Wesley, Reading, MA, 1968.[LS] V. Lakshmibai and B. Sandhya,Criterion for smoothness of Schubert varieties inSl(n)/B, Proc.Indian Acad. Sci.100 (1990), 45–52.[Ma] T. Mansour,Permutations containing and avoiding certain patterns, Proc. 12th Conference onFormal Power Series and Algebraic Combinatorics (Moscow, 2000), 2000.[PWZ] M. Petkov̌sek, H. Wilf, and D. Zeilberger,A=B, A. K. Peters, Wellesley, MA, 1996.[Ri] J. Riordan,An Introduction to Combinatorial Analysis, Wiley, New York, 1967.[SS] R. Simion and F. Schmidt,Restricted permutations, Europ. J. Comb.6 (1985), 383-406.[St] R. Stanley,Enumerative Combinatorics, vols 1, 2, Cambridge University Press, Cambridge,1997, 1999.[Sz] G. Szego,Orthogonal Polynomials, AMS, Providence, RI, 1967.76 Toufik Mansour and Alek Vainshtein

Avoiding maximal parabolic subgroups of S_k

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