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DATE DUE DATE DUE DATE DUE 1/98 chlHCJDfiDuepes-p.“ THE KAZHDAN PROPERTY OF THE MAPPING CLASS GROUP OF CLOSED SURFACES AND THE FIRST COHOMOLOGY GROUP OF THEIR COFINITE SUBGROUPS By Feraydoun Taherkhani A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1999 ABSTRACT THE KAZHDAN PROPERTY OF THE MAPPING CLASS GROUP OF CLOSED SURFACES AND THE FIRST COHOMOLOGY GROUP OF THEIR COFINITE SUBGROUPS By Feraydoun Taherkhani This dissertation studies the mapping class group of closed surfaces Mg with respect to the property T or the Kazhdan property. The motivation for this work finds its root in [Iv4]. N. Ivanov asks in this preprint if every cofinite subgroup in ./\/i9 has a vanishing first cohomology group. In the same preprint, he addresses the more general question of whether .A/Ig is a Kazhdan group or not. If a group has cofinite subgroups having a non-trivial first cohomology group, then it cannot be a Kazhdan group. In the following we first consider M2, the mapping class group of a closed surface of genus 2, and Show that it does not satisfy the Kazhdan property by constructing subgroups of finite index having a non-vanishing first cohomology group. As a matter of fact there are a lot of subgroups with this property (see 3.3). We also tried to follow the same approach for a genus 3 surface, which turned out to be hopeless (see section 3.1). We adopted a new method that was originally introduced by R. Gilman to study the automorphism group of a free group, and after minor changes we were able to apply it to M9. In this way we were able to construct some subgroups of finite index in the mapping class group of a genus 3 surface. We also managed to calculate their first cohomology groups, which all turned out to be trivial. Unfortunately, most of the subgroups we Obtained this way contain the Torelli subgroup and are of no use to our problem (see section 3.2). In only one case did we find a cofinite subgroup that did not contain the Torelli subgroup, and it turned out to be of a trivial first cohomology group. We also used the same method to construct some more cofinite subgroups for M2, and all of them turned out to have \ 4w. T. i‘wowmrH‘TES. . "'3': . v ‘7. PEP-FL"..I.A. $7.... .3) .v . " a non-trivial first cohomology group (see 3.3). The subgroups constructed for A43 could neither respond to Ivanov’s question in the negative, nor could they deliver any substantiated evidence in favor of the mapping class group being a Kazhdan group. The current capacity of our computers have set the boundary for further subgroups to be calculated. Maybe if we manage to construct a variety of more subgroups in the future using better and more advanced machines, we would be able to shed some more light upon the questions studied in this dissertation. We have done some of the calculations in this paper by hand, but the bulk of the calculations have been carried out by the aid of a computer using the programming language GAP (see [Sc]). 1 1GAP Groups and Algorithms is a software package developed at RWTH-Aachen, that primarily deals with algorithms concerning finite groups, finitely presented groups and their presentations. There are several versions of CAP available. We have used version 3 release 4.4 on April 18“ 1997. GAP is free and can be obtained by an anonymous ftp to the following server: samson.math.rwth-aachen.de TO Sofia F ermina Taherkhani (my lovely daughter) iv ACKNOWLEDGMENTS First of all I wish to thank my academic advisor, Dr. N. Ivanou, for his guid- ance, help and encouragement throughout my studies at Michigan State University. I am especially grateful for his excellent lectures and the freedom he granted me in preparing my thesis. I also wish to express my gratitude to Dr. R. Phillips and Dr. J. D. McCarthy for their patience and valuable help through our conversations in answering my questions. Many thanks go to the other members of my committee, Dr. R. Fintushel for his excellent lectures in Algebraic Topology and Guage Theory, and Dr. C'. Rotthaus for teaching me Commutative Algebra. I My special gratitude goes to Dr. T. Breuer from RWTH—Aachen not only for providing some part of the GAP programs in advance, but also for his precious help and valuable support while I was writing my own GAP programs. I also wish to thank the German Academic Exchange Fund (DAAD) for their support in year 1998. Not at last I would like to thank my friend and colleague, Eric K uennen for his meticulous effort in reading my thesis and correcting it. Contents 1 Introduction 2 Constructing Examples For g = 2 3 Constructing Examples For g = 3 3.1 3.2 3.3 3.4 The Procedure to Construct 439"; .................... Computing a Presentation for H ..................... More Examples for g = 2 ......................... Final Remarks ............................... vi 10 13 21 26 29 List of Tables 2.1 Subgroups of low index in M2 ...................... 7 vii List of Figures 1.1 3.1 3.2 3.3 Dehn twists generating Mg ....................... 2 The generators of 7rg ........................... 14 The bounding pair generating 7; .................... 15 The action of Dehn twists ........................ 16 viii Chapter 1 Intro duct ion Let 89 be a closed surface of genus g, 7rg = 7r1(Sg) its fundamental group and M9 the corresponding mapping class group, i.e. the group of the isotopy classes of orientation preserving difieomorphisms of 89. The connection between Mg and combinatorial group theory was established by Nielsen towards the end of the twenties (see [Ni]). If we drop the assumption of orientation preserving and consider the group of isotopy classes of all diffeomorphisms, the so-called extended Mapping Class Group M3, then from a pure algebraic point of view M; is isomorphic to Out(7rg), the outer automor- phism group of 7rg. Hence Mg, as a subgroup of index two in M3, can be identified with a subgroup of index two in Out(7rg) that we denote by Out+(1rg). The group M9 is generated by special types of difieomorphisms called Dehn twists around simple closed curves (see [De]). Unaware of Dehn’s result, Lickorish finds a generating set of 39 — 1 elements (see [Lil] and [Li2]). The minimal number of Dehn twists generating M9 is 29+ 1 and was determined by Humphries (see [Hu1]). The Dehn twists around the simple closed curves a1, D1, 02, ,32, .., 019,39, 6 (see Figure 1) will be our choice for a generating set and we will denote them by D0,,D31, D0,, D5,, .., Dag, Dfig, D5. Figure 1.1: Dehn twists generating M9 The question of the existence of a finite presentation was settled by Birman and Hilden for the case of g = 2 (see [BH]) and by McCool (see [Mc]) for g 2 3. A simple presentation was determined by Wajnryb in 1983 that was slightly corrected in 1994 (see [le] and [BW]) based on results obtained by Hatcher and Thurston [HT]. Directly from the presentation we can establish the well-known result that M9 is a perfect group for g 2 3 (see [P]): H1(Mg,Z) :{ Z10 if g: 2 0 if g 2 3. Another property of M9 is its residual finiteness, which was prOved by Grossman [Gr]. Residual finiteness means that the intersection of all normal subgroups of fi- nite index in A49 is trivial (see [KMS] page 116). In the mid-eighties the analogy between Mg and the arithmetic groups was established (see [Ivl], [Iv2] and [Iv3]). For the definition of an arithmetic group we refer to [Huj], however for our purposes the only arithmetic group of interest would be the symplectic group over the integers, S p29(Z ), for g 2 2. In the following theorem we will list some of its properties: Theorem 1 Let Fg(m) be the kernel of the canonical epimorphism (1)9077.) : Sp29(Z) _) Sp2g(zm) a) Every cofinite subgroup of Sp29(Z) ( a subgroup of finite index} contains one of the congruence subgroups I‘g(m). b) Every non-trivial normal subgroup in Sp29(Z) different from the center contains a congruence subgroup and hence is of finite index. c) Every cofinite subgroup U of Sp29(Z) has a vanishing first cohomology group, i.e. H1(U) = 0. Since Sp29(Z) is finitely presented, this property is equivalent to saying that U / U ’ is finite. Proof: Parts a and b of the theorem are proved by Mennicke for the symplectic group (see [Me]) and generalized by Bass, Milnor and Serre in [BMS]. Part c of the theorem is a consequence of part b and the following proposition: Proposition 1 Let H g G be cofinite. Then if [H : H’] is finite, so is [G : G’]. Proof : Since H’ S G’, therefore H’ g G' D H and consequently G’ n H will be cofinite in G, but G’ H H S G’, which means that G’ is cofinite in G. q.e.d. Now we can prove part c of the theorem. We may assume that U is a normal subgroup of G = Spgg(Z). Otherwise we can pass to CoreG(U) = 096G g'lUg, which is a cofinite normal subgroup in G contained in U. Using the proposition, it suffices to prove the statement for CoreG(U). Since U is a normal subgroup of G, therefore U’ as a characteristic subgroup of U will be a normal subgroup of G. Using part b of the theorem, U’ is cofinite in G hence cofinite in U. q.e.d. Let N 31 7rg be a cofinite characteristic subgroup of 7rg. Then the canonical map Aut(7rg) —> Aut(7rg/N) factors through the outer automorphism group and after the restriction to My (recall that My 2 0ut+(7rg)) we obtain a homomorphism ‘I’g,N 2 Mg —> 0Ut(7Tg/N) whose kernel AM is a cofinite normal subgroup in M9. We call these subgroups the congruence subgroups, in analogy with arithmetic groups. N. Ivanov asks the following questions about My ( see [Iv4]). 1. Congruence Subgroup Problem for M9: Does any cofinite subgroup U 3 Mg contain one of the Ag”? 2. Does every cofinite subgroup U of Mg have a vanishing first cohomology group? 3. Does My satisfy the Kazhdan property? For an introduction to the Kazhdan property (or property T) we refer to [Kz], [HV] and [Lu]. The only important fact we need to know is that a T-group (a group satisfying the property T) has no cofinite subgroups having a non-vanishing first cohomology group. This means a negative response to the second question would answer the third question negatively as well. Theorem 2 M9 the mapping class group of a closed surface of genus g = 2 is not a K azhdan group. In the following we prove Theorem 2 by constructing examples of subgroups having a non-trivial first cohomology group. Chapter 2 Constructing Examples For g = 2 A49 acts on the first homology groups H1(Sg, Z) and H1(Sg, Zm) for every m E Z. This action preserves the symplectic form and gives rise to the homomorphisms 99 2 Mg —-) Spgg(Z) and Og(m) : Mg —> Sp29(Zm), which are known to be surjective. Let 7; = K ernel((~)g) be the Torelli-subgroup of Mg and 7;(m) = Kernel (Og(m)) the preimage of the congruence subgroup Fg(m) in M9. In this way we obtain a lot of cofinite normal subgroups of My all containing 7;. In particular, in the case of g = 2 and m = 2 we get 92(2) 1M2 —> SP4(Z2) '—‘—’ S6- Sp4(Z2) is isomorphic to 56, the symmetric group on six elements, which has order 720, therefore 73(2) 51 M; will be a normal subgroup of index 720 in M2. The normal subgroup 73(2) is generated by the squares of the Dehn twists around the simple closed curves and normally generated by the square of only one of the Dehn twists such as al, the first generator of M2 (see [Hu2]). Using the Schreier-Reidemeister method (see [J11] and [112]) we can calculate a presentation for 73(2) using GAP. The simplest presentation we can construct after all the possible reductions using Tietze transformations contains 14 generators and 388 relations of total length 8622. As a subgroup, 73(2) is generated by the following 14 elements: a1”, biz, a2”, 12;”, d”. —2 —1 —2 —1 —2 —1 —2 —1 —2 —1 —1 —2 —1 -1 —2 -1 —1 —1 blagbgzaglbfl, blagbgd'zbglag’lbfl. By writing the 388 relations in an additive form we obtain a matrix of 388 rows and 14 columns that we refer to as the relation matrix of the presentation. From the presentation we can compute H1(73(2)) (the commutator factor group of 73(2)) by applying the Gaufl-algorithm to this matrix to evaluate its invariant divisors. The divisors are 0,0, 0,0,0, 0, 0,0, 0, 2, 2, 2, 2, 4, meaning H1(73(2)) = 9% EB 4Z2 EB Z4. As a byproduct, we see that 14 is the lowest cardinality for a generating set Of this group ( since none of the invariant divisors is 1). This example takes care of the genus = 2. In addition, using an algorithm called ”Low Index Subgroups” in GAP we can calculate a complete list of the conjugacy classes of all subgroups of finite index bounded by a given number p. (The algorithm is efficient only for small indices. For example, p = 20 is already a huge index for M2). We have tabulated all the conjugacy classes of subgroups of M2 for p = 10 together with their commutator factor group in Table 2.1. Actually, H6 is the commutator subgroup of M2, which is a perfect group, and H7 corresponds to the only subgroup (up to conjugation) Of order 72 in Se. Hereby we have found the smallest index subgroup with non-trivial first cohomology group. Index H H/H’ 1 H1 Z10 2 H2 Z5 3 - - 4 _ - 5 H3 Zg 6 H4 210 6 H5 Z80 7 _ - 8 - - 9 - - 10 H6 0 10 H7 Z EB Z2 Table 2.1: Subgroups of low index in M2 Using the same method, we can find a generating set together with a presentation for H 72 H7 = (b1, 02—2, b2, agblalbgagd-lbglbflagl). Since H7 has a relatively small index and is generated by only four elements, the pro- cedure of finding a presentation can even be done by hand. The simplest presentation we can construct for H7 consists of 4 generators and 25 relations of total length 564. In the following, we write down this presentation explicitly and give the number of occurrences of each generator in the 25 relations. Generators: 1) g1 197 occurrences 2) 92 81 occurrences 3) 93 174 occurrences 4) 94 112 occurrences. Relations: 13) 14) 15) 16) 17) 18) 19) 95194-19394, 94’1929492", 92919291" ’92“ 191’ 1, -1 41—1 —1 93 92939193 92 9391 1 949194— 1919; 191‘ 19491“ 1, 95192939293192‘19392", 9491-193-194—193—19193919319391949f193191—193, 939193919491"1927191‘191’191939194'191493191"1 921‘ 19193919.? 191’ 193‘ ‘91” 1949193919491“ 193‘ 191’ 1, 9394939193919; 191’ 192:294‘191949391'19I‘93' 191‘ 1, 93949194—1914931g1"1932941195191—1939493919391, 9491' 193‘ 191‘ 1939; 193" 1.919391911914932 9119193949391, —-1 —1 —1 —1 —1 —1 939194 91 93 9291 949192 93919491 9319291493919171, 91'1939193919491’293‘191‘193‘lgigs 9191119393191“, 9295192—19391929119192939191 19:? l919:1 9191719119119491‘1971931, —1 —1 —1 —1 —1 —1 929193919491 93 91 92 94 93 929193 9194‘ 191' 195191— 193295 ‘93, 91719;]9491’195192949392‘193919Z’92" 919293191” 1949391“ 193191-194, 91939194-191’195’19291’191195291‘ 1949391 gglgaglgtgrlgglgrlgz, -1 —1 —2 —l —-l —l 91 9493 929392 91 92 93949194 91 9291—193— 1919491'193191‘19392‘ ‘94, 20) 21) 22) 23) 24) 25) 91‘ 1939493917191‘19293“29£{1929193919;191493”2 929391949? 19'3" 1929? 193‘ ’92" ’94" ’93.“ 1, 9391949149319?1971917193191939195191’1 931929192"193919Z19f19Z193’19192931. 949393491“93939?193191'1939492‘191‘194, 92949193919319;193—192—191—192—29491—1 94921927 19293919491“ 193191—192— 193191— 19492 91—1937ng 1919394919; 194—19193, 9519492939;‘91‘192‘191‘1917‘9491‘1997191'1 9392—1919491—19294939491—19319I193192939491 9.?193‘19719191‘191‘1929;1919392’1919491‘1, 929214919491"193’191'1939391‘195191‘193919491‘ 1 93192-19f19493919491-193-19291492493919? 91-19349? 19519291939194” 191‘ 1931917194- 191, 9391949? 19:? 191-2949; 193—292939194— 193939, 9.1491493292935119J191’193’191’193‘191729491‘1 949393—192939;?191-19491“19293919391949?195292—193. At the end we calculate the commutator factor group Of H7. The invariant divisors of the relation matrix are meaning or 1, 1, 2,0, H1(H7) = Z2 EB Z1 H1(H7) = z. This proves Theorem 2. Chapter 3 Constructing Examples For g =2 3 The result for genus g = 2 is not very surprising, because of the exceptional status Of g = 2. There are many properties that all surfaces of g 2 3 share, but a surface of g = 2 does not ( see [Jd1]). Therefore, the interesting examples would be for surfaces of higher genera. The case g = 3 is not only much more difficult to handle but also quite different in nature. The first reason is the following theorem proved by J. McCarthy (see [Ma]). Theorem 3 Let F 3 Mg be a subgroup of finite index containing the Torelli subgroup 7;, then H1(I‘) = 0. The proof uses a result Of D. Johnson in [M2] and the fact that the image of F in S p29(Z ) contains some congruence subgroup. According to this theorem, if a cofinite subgroup Of Mg with non-trivial first cohomology group exists, it has to be found among those that do not contain 7;. The residual finiteness of Mg assures us the existence of subgroups not containing the Torelli subgroup. The main problems we encounter for the construction of these subgroups are the following three: 1. How to find cofinite subgroups Of Mg? 2. How to check whether they contain 7; or not? 3. How to calculate the first cohomology groups of these subgroups? 10 11 Problem 1: In order to construct a whole series of cofinite subgroups of M; we have adopted a method that was originally introduced by R. Gilman (see [Gi]) to study the automorphism groups of free groups. We have modified this method and have applied it to M; as follows: Let Q be an arbitrary finite group, and G an arbitrary finitely presented group. Two epimorphisms gal and (152 from G onto Q have the same kernel if and only if they differ by an automorphism of Q, i.e 451 = wdg, where 1b 6 Aut(Q). If we denote the set of all epimorphisms of G onto Q by £pim(G, Q), then we have the following bijection: M: = {N s GIG/N ,_._, Q} <——> spe'mw, Q)/Aut(Q)- Now let G = n;. The automorphism group of 1r; acts on NQ as a permutation group. Let k be the cardinality of NQ and 31e the symmetric group on k elements. We obtain a homomorphism (ng I Aut(7r;) —> 5);. The inner automorphisms act trivially on Nq, hence ;,Q factors through the outer automorphism group of 7rg. After restriction to M9, we Obtain a homomorphism 4,9,6) 2 Mg —-> 51;. 29,62, the kernel of ;,Q, will be a cofinite normal subgroup Of Mg. In this way we obtain a lot of different normal subgroups of M; for different choices of Q. Remark 1 Let N E Nq, then H := 0361411911,) ¢(N) will be a cofinite characteristic subgroup of 11;, and A9,}; S Egg, i.e each of the subgroups 2;,Q contains at least one congruence subgroup. Problem 2: Based on Johnsons’ results on the Torelli subgroup [Jdl] we will give a method to determine whether 7; 3 299 or not. This will be discussed later, when we construct some subgroups (see section 3.1). Problem 3: This is the most difficult part Of the calculation to deal with. In order 12 to determine H1(U), a presentation of U has to be given. The algorithms we use to determine a presentation for a subgroup of a finitely presented group are based on the following classical theorem from combinatorial group theory: Theorem 4 (Schreier-Reidemeister) Let G be a finitely presented group with n generators and m relations. In addition, let H S G be a subgroup of finite index with [G : H] = p. Then H is also finitely presented, and there is an algorithm to construct a presentation for H on pn — p + 1 generators and at most mp relations. There are generally two different methods to construct such a presentation: Schreier— Reidemeister and Todd-Coxeter (see [J11] and [J12]). A modified version of both algorithms has been implemented on GAP. As we notice from the theorem, the com- plexity of the presentation increases with the index p. In addition, the length of the defining relations for H depends on p as well and can become eventually very large. (See the presentation that was constructed for 73(2) in the previous chapter. For instance, the first relation is a word of length 4, while the length of the last relation is 50). Therefore, constructing presentations for subgroups of huge indices can become an infeasible task (for instance p = 1000 is already huge in the case Of M2(2)). In the following sections we consider some examples for different choices of Q. At first we show that abelian Q’s are of no help for our purposes. Example 1 Q = Z2. Aut(Z2) is trivial, hence there are as many subgroups of index 2 in H3 2 «1(83) as different epimorphisms from n3 onto Z2. Therefore there are exactly 26 — 1 = 63 subgroups Of index 2 in 7r3. Consequently we obtain the following homomorphism: (P13, 3M3 —> 563- The action of M3 on these subgroups coincides with the symplectic action of M3 on the subspaces of codimension 1 in (Z2)6 2 H1(S3,Z2). Therefore 2,z, fac- tors through Sp6(%) and gives us a faithful permutation representation of S 174%), 1 lVldu:..In l1 , ‘UU: “.3... . 13 which is also transitive (see [Hup] page 221). This implies that 23%, = 73(2) is a normal subgroup of index |Sp6(%)|=1451520 that contains 7;, and consequently H1(23,%) = 0- Theorem 5 For every Abelian Q, H1(E;,Q) = 0. Proof: 7; acts trivially on H1(S;, Z). Since Q is Abelian, every N g 7r; with 7rg/N 2 Q contains «9. As a result 7; acts trivially on {N |1r;/N 2 Q}, i.e 7; 3 Egg. q.e.d. In the following section, Q will be always a non-abelian group, and we will describe the procedure to construct the homomorphism ;,Q and its kernel Egg. 3.1 The Procedure to Construct (1);,Q We proceed as follows: 1. We need to know Aut(Q). In general there is no algorithm to calculate the automorphism group of a finite group, unless the group belongs to a certain category such as P-groups or more generally nilpotent groups (see [Sc]). But for small Q this might be done by hand. 2. To find the set £pim(7r;, Q), we consider all tuples q1,q2, ..,ng E Q generating Q and satisfying the only defining relation of fig: 7'0 = [91192ll93194lu-[92g—1192g] = 1. Aut(Q) acts on the set of all these tuples. Let us choose an orbit representative for this action. This yields us the set £pim(7r;, Q) /Aut(Q). 3. For every d E £pim(7r;, Q), we determine N4, = Kernel (qt) as the normal closure of a finite set of elements of 7r; in the following manner: We construct a presentation for Q on generators q1,q2, .., Q29. Besides r0, the only relation of 7r;, this presentation of Q will satisfy some more relations such as r1,r2, ..,rk. If we rewrite r1,r2, ..,rk as words in generators of 7r;, we will obtain a normally generating set for N. 14 4. M; 2 0ut+(1r;) acts on the conjugacy classes of fig. Via a calculation carried out by hand we determine the action of D0,, , D5, , .., Dag , D39, D5 (see figure 3.2) on the conjugacy classes of the generators of 1r; (see figure 3.1). We choose a fixed presentation of 1r; as follows: 7r; : ( a1,b1, ..a;,b; | [a1,b1][a2,b2]..[a;,b;]=1). In figure 3.1 we have drawn 2g loops, with the appropriate orientation, whose isotopy classes represent the 2g generators of 1r; satisfying the only relation of 1r;. Hereby [a, b] = aba‘lb‘1 is the commutator of a and b, and we multiply curves fiom left to right, i.e. the curve ab is obtained by traversing first curve a then curve b. Figure 3.1: The generators Of 7r; In addition to D0”, D3,, .., Dag, D59, D5 we need also to know the action of D,, (see figure 3.2). The reason is the following fact established by D. Johnson [J d1]. The Torelli subgroup 7; is generated by all the elements of the form D51); 1, where 6 and 17 represent a bounding pair, i.e. a pair of disjoint simple closed 15 curves, representing the same non—trivial Z—homology class. Johnson further defines the genus of a bounding pair to be the smaller of the genera of the two pieces of the surface cut by the two curves, and proves that 7; is normally generated by any genus 1 bounding pair. Our choice in figure 3.2 is of genus 1. This result will be used to settle problem 2. That~ is, in order to determine if a certain normal subgroup of M; such as Egg contains 7; or not, all we have to do is to determine if ;,Q(D5D;1) is the trivial permutation or not, a fact that can easily be checked. Figure 3.2: The bounding pair generating 7; In the following we list the action of the Dehn twists on the generators of 7r;. Every Dehn twist changes only some of the generators, so we will not list those generators that remain invariant. We will need some new elements of 7r;, defined as follows: —1 —1 —1 - 01: b1 , 01,-: a,b, a,- b,_1, for z = 2, ..,g E,- = a,, for i = 1, ..g 62b2, nzaglbgl. These elements of Hg, namely (ai, 9,, 15,17), are denoted by the same letters as the simple closed curves used in the presentation of M; because they all happen to be simple closed curves representing the same isotopy classes. In figure 3.3 we have represented these new elements. For the right Dehn twists we obtain: Figure 3.3: The action of Dehn twists Dal I a1 Dot.- 01—1 bi—l at lDfir bi 05 I 0.2 I),7 Z (12 b2 as ba lll H 0101 04-10; 01151—101— 2 = 2, --19 (1101' (>191 026—1 (1277 77.15277 17.40377 v‘lbsn- We will also need the action of the left Dehn twists: .11.} u: . Dilute. .1... . . 17 1);}: 0.1 H alal—l D_.12 (1,21 H a,_1a,- 0;}: b,- H 1).-pf i=1,..,g D‘s-12 a2 i—i 0.25 Dgli 0,2 e—> 0277—1 52 H W277—1 aa H vaav“1 ()3 H flbgn-l. 5. Let w = w(a1, b1, ..,a;, b;) be some word in the generators of ng, and let Do, be some Dehn twist. Then we define Da(w) = w(Da(a1), Da(b1), .., Da(a;), Da(b;)). Now if N is the normal closure of n1,n2, ..,n,, where n,- are some words in a1,b1, .., a;, b;, then Da(N) will be a normal subgroup normally generated by Da(n1), .., Da(n1). Now let d be an element of £pim(7r;, Q) and N = N), be the corresponding kernel of 4’). If we apply Do, to N, we will Obtain another N¢e for some other d)’ E Epim(7r;, Q). Since we know all qb’s together with their kernels, we can find the appropriate normal subgroup that N gets mapped to under the action of D0,. In this way we construct the homomorphism (Pg’QI Mg —> Sk. In the following example we choose Q to be the smallest non-abelian finite group 5'3. First of all we need some facts from classical group theory. 18 Definition 1 Let G be a group and H a permutation group on a. finite set D with |f2| = n. The wreath product G 2 H of G with H is defined to be the set GIH :{(911921--19mh) ] giE Gah E H} with the group multiplication (91,921--,9mh)(9’119§1~19:uh’) = (919i(1)1929i.(2)1 --19n9i1(n)1 M1,)- Theorem 6 Let G and H be defined as in definition 1. Then G I H has a normal subgroup isomorphic to the direct product of n copies of G and is isomophic to a semidirect product of this normal subgroup by H. Hence the order of G 2 H will be [02H] = |G|"|H|- See [Hup] page 95 for a proof. Definition 2 A permutation group G on the set D is imprimitive if there exists a proper subset A C D such that for all g E G either gA = A or gA O A = (D. A is called an imprimitivity region of G. A transitive permutation group that is not imprimitive is called primitive. Theorem 7 Let G be an imprimitive permutation group on Q with imprimitivity region A, and let H = {g E G | gA = A} be the stabilizer of A. Let further R be a coset representative of G / H . Then a) Q = UrER TA- b) If IQ] = n is finite, then [(2] = [A] [G : H]. Let [A] = k and [G : H] = p, then G is isomorphic to a subgroup of the wreath product of S), 2 Sp. c) H acts transitively on A. The proof can be found again in [Hup] page 146. Theorem 8 Let G = (91,92, .., gm | r1,r2, ..rk) be a finitely presented group and ¢:G—>Sn 19 be a transitive permutation representation of G on n elements. Let us denote the image of d) by H. Then we can construct a presentation for H on h1 = 43(91), .., hm = ¢(g,,) (see section 3.4), H = (h1,h2,..hm | r’1,r;,..,r;c,), and N, the kernel of (t, is normally generated by r[, r’2, .., rib (Note that r1, .., rL. are the same words used in the presentation of H but written in the generators of G ) Let further {31, 32, ..st} be a set of representatives of the preimages of all elements of H under 4). Then a generating set for N as a subgroup will be N =(sf1rg-s, | i = 1,..,t, j = 1,..,k’). If U is a subgroup of H (such as the stabilizer of one point), generated by u1,..,ul, and w1,..,w, are their preimage representatives in G, then ¢‘1(U) is generated by ¢‘1(U)= 52520- 53 is solvable with Sf, 2 Z3. As mentioned earlier in example 1 there are exactly 63 subgroups M1, M2, .., M63 in 7rl(83) with Zz-quotient. Each of these normal subgroups M,- turns out to contain N,,1,N,-,2, --1Ni,40 normal subgroups with Z; as a quotient such that each Niu' is also normal in rr1(83) and M (83) /N,-,,- 2 S3. Let’s denote the set of these subgroups by N5, = {Ni.j}- In this way we obtain a partitioning of these 2520 normal subgroups into 63 blocks, A1, .., A53, each containing 40 normal 20 subgroups. The action of M3 on N 33 turns out to be transitive but not primitive. The imprimitivity regions are exactly the blocks A1, .., A63. The permutation of the A23 is determined by the action of M3 on the normal subgroups M1, M2, .., M53, which gives us a faithful transitive permutation representation of SP6(%) on 63 elements as we saw in example 1. The stabilizer of one of the Ags, sach as A1 for instance, under the action of M3 acts transitively on A1 (see theorem 7), therefore the restriction of this action on A1 will be a transitive permutation representation on 40 elements. I; Let us denote the image of this representation by U. We can also compute a set of generators for U. The following six permutations generate U: (5, 23, 32)(6, 24, 34)(7, 25, 33)(8, 26, 38)(9, 27, 40)( 10, 28, 39) (11, 29, 35)(12, 3o, 37)(13, 31, 36), (14, 32, 23)(15, 33, 24)(16, 34, 25)(17, 35, 26)(18, 36, 27)(19, 37, 28)(20, 38, 29) (21, 39, 30)(22, 40, 31), (2, 3, 4)(8, 9, 10)(11, 13, mm, 18, 19)(20, 22, 21)(26, 27, 28)(29, 31, 30) (35,36, 37)(38,40, 39), (1, 2, 4)(6, 11, 10)(7, 8, 12)(15, 20, 19)(16, 17, 21)(24, 29, 28)(25, 26, 30) (33,38, 37)(34,35, 39), (2, 3, 4)(8, 9, 10)(11, 13, 12)(17, 18, 19)(20, 22, 21)(26, 27, 28)(29, 31, 30) (35,36, 37)(38,40, 39), (2, 22, 19)(3, 21, 17)(4, 20, 18)(5, 25,34)(6, 23,33)(7, 24, 32)(11, 37,31) (12,36, 29)(13,35, 30). U turns out to be a simple group of order 25920. The only simple group of this order is PS p4(Z3), (see [AFG]). Thus we proved the following: Im(<1>3,s.) '2 PSp4(Za) 2 SP6(Z2), 21 and the order of this image is (25920)63 x 1451520 2 10284. It’s needless to mention that computing a presentation for E = 23,53 is a hopeless task, but we are not actually interested in computing this presentation, but rather knowing if H1(2) is finite or not. We are going to look at H = @3239}(Stab33.ss(M3)(N1,1)). As mentioned above the image of M3 under @353 turns out to be transitive, so the stabilizer of one of the {NM}: such as Nu, will be a subgroup of index 2520, and its preimage H will be a subgroup of the same index in M3. What seems to be realistic to handle is computing a presentation for H. On the other hand, 23,33 2 GoreM3(H) = 096,143 g‘ng, meaning that 73 _<_ 2353 if and only if 73 S H. 3.2 Computing a Presentation for H Even 2520 turns out to be a huge index, and we (our computer) won’t be able to compute a presentation for H. Therefore, we will break down the calculations in two steps. At first we look at K = 3ZI%(Stab¢3’zg(M3)(M1)), which will be Of index 63 in M3. Since K fixes M1, it acts on {N1,1, --,N1,4o}- Therefore (P353 can be restricted to K and we will get a homomorphism (that we denote by the same letter) (P353 2 K —> S40. In the previous example we denoted the image of K by U. Now we look at H : @313(Stabsa'53(x)(N1,1)), which will be a subgroup of index 40 in K, and a subgroup Of index 40 x 63 = 2520 in M3. Using Theorem 8 we can find a generating set for K. Then we construct a presentation for K on this generating set (see section 3.4). The advantage of this specific representation of K is that every element of K (written as a word in its generators) can be directly rewritten as a word in the generators of M3. This means that every subgroup of K can be directly realized as a subgroup of M3, and any homomorphism from M3 onto any permutation group can be restricted on K and easily evaluated. We can also find a generating set (using again Theorem 8) 22 for H as a subgroup of K together with a presentation. Using the presentation we can evaluate its first homology group, and taking advantage of the special presentation of K we can realize H as a subgroup of M3. The most time consuming part of the calculations involves the computation of a presentation for H. Here it would be reasonable to check at first if H contains the Torelli subgroup or not. The smallest generating set we can find for K consists of the following 8 elements: alrblaa172,b21013:b3rd—21a2b2db2—1a2-1- Then we calculate a presentation for K, and apply the Tietze transformations to simplify it. The simplest presentation we get has 8 generators and 242 relations of total length 66790. H is generated by the following nine elements: (11, b1) 09-23 b2: b3) d—Zr a2b2db2—la;la —1 —1 —1 a3b2 a3,a3b3b2 a3 , and turns out not to contain the Torelli subgroup. H has a presentation on 94 generators and 9401 relations of total length 2120026. At the end we determine the commutator factor group of H. The non-trivial invariant divisors. are 2, 6, 12. Hence H1(H) 2 ZzeBZMBZn. Although H is a subgroup that does not contain the Torelli subgroup, its first coho- mology group turns out to be zero. The next attempt will be undertaken using the quaternion group Q8 instead of 5'3. Example 3 Q = Q8 Q3 is a solvable group with Q’8 2 Z; and Qg/Q'S 2 Z2 63 Z2. We have exactly W = 651 subspaces of codimension two in (Z3)6, meaning there are exactly 651 normal subgroups M1, .., M651 in «1(83) with Z; EB Z2 as quotient. The action of 23 M3 on these subspaces won’t be transitive. There are exactly two types of subspaces of codimension two, and the action of M3 will be transitive on each of these families. One of them is called isotropic, on which the symplectic form vanishes, and the other one is called hyperbolic, on which the symplectic form is non-degenerate (see [Hup]). In our situation we will be only interested in the hyperbolic ones and there will be 315 of those. Each of these hyperbolic subgroups Mi’s contains exactly 16 normal subgroups {N,~,1, --,N.',16} such that their quotient in «1(53) is isomorphic to Q8. In this way we Obtain 315 x 16 = 5040 normal subgroups of quotient Q3. The homomorphism (1)393 1 M3 '9 55040 will have an image isomorphic to U 2 L, where L is a transitive permutation repre— sentation of Sp6(Z2) on 315 elements and U is a solvable permutation group of order 9216 on 16 elements. The order of this image will be (9216)315 >< 1451520 2 101255. Again we look at K = (P312262,(Stab¢s,z,ezq(Ms)(Mlll and at its subgroup H = {58(Stab33'qs(x)(N1,1)). K is a subgroup of index 315 containing the Torelli sub- group, and H is a subgroup of index 5040 that turns out not to contain the latter group. K is generated by the following 10 elements : 0'1) b1) 02—2) b2) 01—3-21 b3) d—2) “3939;710:311 agbzdbgla; 1, a2b1a1b2a2b1a3b2da2’lbglaglbl‘laglbg'laflbl’la;1, and has a presentation with 10 generators and 1140 relations of total length 285918. H is generated by the following 13 elements: 4 4 4 —1 —l alablra21b2ra3rb3ad)a3b3b2 a3 a —1 —-1 —2 2 —2 -1 2 —1 —1 —1 —1 -1 —1 —1 —1 —1 —2 —2 2 02b1albzazblasbzd02 92 a3 91 “2 b2 a1 91 “2 102 b1 ‘12, 24 and has a presentation with 86 generators and 18105 relations of total length 3935640. At the end we look at the commutator factor group of H. The non-trivial invariant divisors are: 2,4,8. Hence Therefore H1(H) = 0. E In the following we will list the result of our calculations for all the other choices for Q that we have been able to handle. Although in each case the calculations are slightly different and need to be treated separately, we will not get into the details and will just state the results. Example 4 Q 2 D3 There are 15120 normal subgroups of quotient D8 (the dihedral group of order 8) in M3. The action of M3 on these 15120 subgroups is not transitive and has 2 orbits of equal length of 7560 elements. The stabilizer of one of the elements in the first orbit is a subgroup H1 of index 7560 in M3 generated by 10 elements.- A presentation on 126 generators and 27169 relations of total length 6069283 can be constructed for H1. The non—trivial invariant divisors of H 1 / H i are: 2, 2, 2, 4,4, 4. Hence H1(H1) = 0. As a matter of fact H1 contains the Torelli subgroup. The stabilizer of one of the elements in the second orbit is a subgroup H2 of index 7560 in M3 generated by 10 elements. A presentation on 126 generators and 27168 relations of total length 6060990 can be constructed for H2. The non-trivial invariant divisors of Hg/Hé are: 2, 2, 4, 8. 25 Hence H1(H2) = 0. It turns out that H2 also contains the Torelli subgroup. Example 5 Q 2 D10 There are 9828 normal subgroups of quotient D10 (the dihedral group of order 10) in M3 falling into 9%§ = 156 blocks. The action of M3 on these 9828 subgroups is an imprimitive transitive group. The stabilizer Of one Of its element is a subgroup H of index 9828 in M3 generated by 11 elements. A presentation on 259 generators and 34795 relations of total length 5206667 can be constructed for H. The non-trivial invariant divisors of H / H’ are: 2, 4, 4. Hence H1(H) = 0. Here again H contains the Torelli subgroup. Example 6 Q 2 D12 There are 78120 normal subgroups of quotient D12 (the dihedral group of order 12) in M3. The action of M3 on these subgroups is not transitive and falls into 4 orbits. We were not able to calculate a generating set much less a presentation for the stabilizer of any of these subgroups. Example 7 Q 2 D14 There are 25200 subgroups of quotient D14 (the dihedral group of order 14) in M3 falling into 35% = 400 blocks. The action of M3 on theses subgroups is an imprim- itive transitive group. The stabilizer of one of its element is a subgroup H of index 25200 in M3 generated by 11 elements. A presentation on 701 generators and 89308 relations of total length 12995940 can be constructed for H. However, we were not able to calculate the invariant divisors of H / H ’ . But we could establish the fact that H contains the Torelli subgroup, therefore H1(H) = 0. 26 Example 8 Q = A4 For A4, the alternating group of order 12, we were not even able to calculate a full list of normal subgroups in M3 having an A4-quotient. Example 9 Q = T12 Besides D12 and A; there is another non-abelian group of order 12 that we name T12. In the following we will give a presentation: T12 = (11,12 | t§,tg,tlt;1t112) and a permutation representation for T12: T12 = ((1,2,4)(3,6,5)(7,8,9)(10, 12, 11), (1,3, 7, 10)(2,5,8, 11)(4,6,9, 12)) There are 80640 subgroups of quotient T12 in M3 falling into % = 1280 blocks. The action of M3 on these group is an imprimitive transitive group. The stabilizer of one of its element is a subgroup H of index 80640 in M3 generated by 24 elements. We were able to construct a presentation on 2336 generators and 287680 relations of total length 40604058 for H. Although we were not able to calCulate the invariant divisors, we could establish that H contains the Torelli subgroup, therefore H1(H) = 0. The choices for Q, considered in the examples above, were the only cases we were able to handle for g = 3. However, for g = 2 we were able to construct more subgroups. 3.3 More Examples for g = 2 In the following we will list some more subgroups we have been able to construct for M2 using the method we described in this chapter. Example 10 Q = $3 27 There are 60 normal subgroups of quotient S3 in M2. The action of M2 on these 60 subgroups is a transitive permutation group. The stabilizer of one of these subgroups is a subgroup H of index 60 in M2 generated by 7 elements. A presentation on 11 generators and 165 relations of total length 3496 can be constructed for H. The non-trivial invariant divisors of H / H ’ are: 2, 2, 0,0. Hence H1(H) = ZED Z. Example 11 Q 2 D3 There are 180 normal subgroups of quotient D8 in M2. The action of M2 on these 180 subgroups has 2 orbits of equal length 90. The stabilizer of one of these subgroups in the first orbit is a subgroup H1 of index 90 in M2 generated by 8 elements. A presentation on 35 generators and 391 relations of total length 6675 can be constructed for H1. The non-trivial invariant divisors of H 1 / H i are: 2, 2, 2, 0,0,0. Hence H1(H1) = Zea Z69 Z. The stabilizer Of one of the subgroups in the second orbit is a subgroup H2 of index 90 in M2 generated by 7 elements. A presentation on 36 generators and 393 relations of total length 6682 can be constructed for H2. The non-trivial invariant divisors of Hg/Hé are: 2, 2, 8, 0,0. Hence H1012): 233 Z- Example 12 Q = Q3 _J 28 There are 60 normal subgroups of quotient Q3 in M2. The action of M2 on these 60 subgroups is a transitive permutation group. The stabilizer of one of these subgroups is a subgroup H of index 60 in M2 generated by 11 elements. A presentation on 24 generators and 263 relations of total length 4562 can be constructed for H. The non-trivial invariant divisors Of H / H ’ are: 2, 2, 4, 0. Hence H1(H) = Z. Example 13 Q 2 D10 There are 90 normal subgroups Of quotient D10 in M2. The action of M2 on these 90 subgroups is a transitive permutation group. The stabilizer of one of these subgroups is a subgroup H of index 90 in M2 generated by 7 elements. A presentation on 15 generators and 245 relations of total length 5063 can be constructed for H. The non-trivial invariant divisors of H / H ’ are: 2, 2,4, 0,0. Hence H1(H) = ZED Z. Example 14 Q = T12 There are 480 normal subgroups of quotient T12 in M2. The action of M2 on these 480 subgroups is a transitive permutation group. The stabilizer of one of the these subgroups H is a subgroup of index 480 in M2 generated by 11 elements. A presen- tation on 68 generators and 1370 relations of total length 23395 can be constructed for H. The non-trivial invariant divisors of H / H’ are: 2, 2, 2, 0,0, 0. Hence H1(H) = ZED Zea Z. 29 3.4 Final Remarks As mentioned in the introduction, almost all of the calculations have been carried out using the programming language GAP. All of the programs have been written by the author, except for the following two cases, where some new programs, that at the time of writing this paper were not yet implemented in GAP’s library, were needed in: 1. calculating a presentation for a finite permutation group (see Theorem 8), 2. producing a presentation on a given set of generators (see section 3.2). The author’s special thanks goes to T. Brauer from RWTH-Aachen, not only for providing the programs in advance, but also for his valuable help and advice. The first machine we have used for the purpose of our calculation has been an Ultrasparc 1 Model 170 with 170 MGHz cpu and 128 MG RAM. The calculations for the groups 53, Q8 and D8 have been carried out using this machine. The three major parts of the calculations involve 1. calculating £pim(rr;, Q) (see section 3.1), 2. evaluating ;,Q (see section 3.1), 3. producing the presentations for H and K (see section 3.2). Part 3 is the most time consuming. The first two parts together take only about 20 percent of the total computing time. In the following we list the time used for the first three examples discussed in the paper: 1. Q 2 S3, 3142319 milliseconds (2 1 hour), 2. Q = Q3, 17031669 milliseconds (2 5 hours), 3. Q 2 D3, 24913769 milliseconds (2 7 hours). The other groups have been treated by a faster machine ( a Dell Power Edge, Pentium II processor with 400 MGHz cpu and 256 MG RAM). Although we have been able 30 to construct only a few number of subgroups inside M3, all of them have turned out to have a trivial first cohomology group. Unfortunately, the fact that almost all of them (except for Q = 5'3) contain the Torelli subgroup makes it very difficult for us to guess one way or the other, if M3 is a Kazhdan group. Considering that in our calculation Q has always been a solvable group, the choice of a simple group for Q would be an interesting case that deserves attention. 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