<span class="mathjax-formula">$(S_3,S_6)$</span>-Amalgams IV

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(S

3

; S

6

)-Amalgams IV.

WOLFGANGLEMPKEN(*) - CHRISTOPHERPARKER(**) PETERROWLEY(***)

Introduction.

This paper in the fourth in a series of seven papers devoted to the study of (S3;S6)-amalgams. We continue the section numbering of the first three parts [LPR1], and refer the reader to Sections 1 and 2 for notation and background material. However for the readers' convenience we summarize some of the main points from these sections.

Adopting the philosophy of Goldschmidt [Go] our study of (S₃; S₆)- amalgams proceeds by examining the action of a group G (which is a certain free amalgamated product) on a certain tree G. Now G has two orbits on V(G); the vertices of G, and one orbit on the edges of G. Let a₁;a₂;2V(G) be adjacent vertices. Then the properties of (S₃;S₆)-amalgams translate into this situation as follows:

1)G hGa1;Ga2i;

2)Ga1\Ga2Ga1a2 contains no non-trivial normal subgroup ofG;

3)G_a₁_a₂2Syl₂(G_a₁) \Syl₂(G_a₂);

4)C_G_a_i(O₂(G_a_i))O₂(G_a_i) fori1;2;and 5)G_a₁=O₂(G_a₁)S₃ andG_a₂=O₂(G_a₂)S₆.

The overall aim is to determine the group theoretic structure of the vertex stabilizersGa1andGa2.

Ford2V(G);we set

D(d) fl2V(G)jd(d;l)1g;

whered(;) in the standard graph theoretic distance onG. Also for i2N;

(*) Indirizzo dell'A.: Institute for Experimental Mathematics, University of Essen, Ellernstrasse 29, Essen, Germany.

(**) Indirizzo dell'A.: School of Mathematics, University of Birmingham, Edgbaston, Birmingham B15 2TT, United Kingdon.

(***) Indirizzo dell'A.: School of Mathematics, The University of Manchester P.O. Box 88, Manchester M60 1QD, United Kingdom.

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we let

D^[i](d) fl2V(G)jd(d;l)ig:

The two orbits ofGonV(G) are, in fact,a^G₁ anda^G₂;ford2V(G) we use d2O(S₃) (respectivelyd2O(S₆)) to mean thatd2a^G₁ (respectivelyd2a^G₂).

For d2V(G) put QdO2(Gd): Various normal subgroups of Gd contained inQdwill be analyzed extensively. We begin withZdwhich reappears over and over again in our arguments and is defined by

Z_d hV₁(Z(G_dl))jl2D(d)i:

Lettingk2Nwe further define

G^[k]_g hZljl2D^[k](d)i:

We shall concentrate on these subgroups mostly fork4, and we use the following abbreviations

VdG^[1]_d ; U_dG^[2]_d ; and WdG^[3]_d :

The first phase of our investigations, which amounts to a very con- siderable step in pinning down the possible structure for Ga1 and Ga2; consists of bouding the parameter b. This parameter, called the critical distance, is defined by

bminfbjm2V(G)g where

bmminfd(m;l)jl2V(G);Zm6Qlg:

We note thatb1. Fora;a⁰2V(G);ifd(a;a⁰)bandZa6Qa⁰;then we say that (a;a⁰) is a critical pair and denote the set of critical pairs byC.

Suppose that (d;d⁰)2 C:SinceGis a tree, there is a unique path inGbe- tweendandd⁰which we label in one of the following ways.

d d1 d2 db 2 db 1 d⁰

. . .

d⁰ b1 d⁰ 2 d⁰ 1

Often we use d 1 and d⁰ 1 to stand for, respectively, an arbitrary vertex ofD(d)n fd1gandD(d⁰)n fd⁰ 1g. When using (a;a⁰)2 Cwe shall

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always setba1:Let (a;a⁰)2 C:If [Za;Za⁰]61;then we say that (a;a⁰) is a non-commuting critical pair and if [Za;Za⁰]1; (a;a⁰) is called a commuting critical pair.

This paper marks the beginning of our work on the commuting case for (S₃;S₆)-amalgams ± the non-commuting case so far as boundingb, being covered in [LPR1]. Here we consider commuting critical pairs (a;a⁰) with a2O(S6)-those commuting critical pairs witha2O(S3) are the subject of parts V, VI and VII and the determinations of the structure of the (S3;S6)- amalgams once the critical distance in bounded can be found in [LPR2].

Our main conclusion here is contained in the following result.

THEOREM. Suppose that for (a;a⁰)2 C we have [Za;Za⁰]1 and a2O(S6):Then b2 f1;3g:

We conclude this introduction with some comments on the proof of this theorem as well as discussing some module facts.

Section 8 contains three preliminary results. One of these is the Core Argument given in Lemma 8.3. This is used frequently to lure subgroups into theG_a-cores ofZ_aandZ_a (see Section 8 for the definition ofZ_a). The structure ofZ_a, dealt with in Lemma 8.3, is also important in many of our later deliberations.

The majority of Section 9 is taken up with the proof of Theorem 9.2 which proves that [Z_a:Z_a \Z_a2]62 so long asb>1:Most of this proof is concerned with examining a particular Goldschmidt subamalgam (H_a;H_b) chosen so that it acts upon an FF-moduleV₀:This gives us access to results of Goldschmidt and Chermak which classify the possible amalgams and FF-modules. Our task then becomes the elimination of each of these possibilities. The most stubborn resistance is offered by the cases Hb S6andHb G2(2) (whereH hHa;H_biandHb H=C_H(V0)). To deal with these cases we need to make use of another result of Chermak's [Ch2].

The last section of this paper investigates the case [Z_a:Z_a \Z_a2]62:

Here we make greater use of the treeG. An important step in our analysis is Lemma 10.2 which says that our critical pairs are, in a certain sense, symmetric. Then in the next lemma we discover that, in fact, Z_aZa: Lemma 10.4 observes certain facts about commutators and Lemma 10.5 describes some properties ofSp₄(2) which are used in Lemma 10.6. Both Lemmas 10.7 and 10.8 focus uponU_aand, with these results to hand, the proof of the above theorem follows quickly.

If, ford2V(G);MdNdQdwithMdandNdboth normal inGd;then h(G_d;N_d=M_d) denotes the number of non-central G_d-chief factors in

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Nd=Md:When looking atGd-invariant sections such asNd=Mdwe find that we need to know many details about irreducibleGF(2)-modules forS3and S6: The former group has just one non-trivial irreducible GF(2)-module which has dimension 2. For HS₆ there are (up to isomorphism) four irreducibleGF(2)-modules the most important for us being the two of dimension 4. Either of these modules will be referred to as a natural S6- module (and sometimes denoted by 4). These two modules are related by the graph automorphism of Sp4(2)S6: The 6-dimensional permutation module V for S₆ is indecomposable and has an irreducible composition factor of dimension 4. Calculations inVare easy and enable us to find out all we need to now about natural modules. LetUbe aGF(2)H-module. We call U an orthogonal module ifU is indecomposable of dimension 5 and U=CU(H) is a natural module. Such a module is sometimes denoted by 4 whileW 1 1

4 indicates thatW is an indecomposable module of dimension 5 with [W;H] a natural module.

For an extensive (and up to date) account of amalgams the reader may consult [PR].

The authors thank Bernd Stellmacher for suggestions (too numerous to mention) which have substantially improved and shortened an earlier version of this paper.

8. Three Lemmata.

We begin by stating the following:

HYPOTHESIS8.1. If(a;a⁰)2 C;then[Za;Za⁰]1;a2O(S6)and b>1:

Before investigating the consequences of this hypothesis we state some well-known general observations on the commuting case.

LEMMA8.2. Let(a;a⁰)2 Cand assume[Za;Za⁰]1:

(i) bb_a is odd and b_b b1:

(ii) Z_bV₁(Z(G_ab))V₁(Z(G_b))<Z_a;Z(G_a)1and C_G_a(Z_a)Q_a: Alsoh(Ga;Za)1and, if b>1;h(Ga⁰;Va⁰=Za⁰)1

If, additionally,a2O(S6);then (iii) G_ab QaQ_b;and

(iv) if D(a⁰) fa⁰ 1;t₁;t₂g; then Z_a⁰ ₁; Z_t₁ and Z_t₂ are pairwise distinct, Z_a is transitive onft₁;t₂gand Z_t₁ and Z_t₂ are conjugate by an element of Z_a:

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PROOF. Since Ga⁰=Qa⁰ S3 orS6 andZa 6Qa⁰; Ga⁰Ga⁰a⁰ 1CG_a0(Za⁰) whence Za⁰ V1(Z(Ga⁰a⁰ 1))Z(Ga⁰): Then Za⁰ V1(Z(Ga⁰a⁰ 1))

V₁(Z(Ga⁰)):Ifb0(2);thenZaV₁(Z(G_ab))Z_bQa⁰, a contradiction.

So b1(2) and (i) holds. Supposeb>1 and h(G_a⁰;V_a⁰=Z_a⁰)0: Then, as Z_a⁰Z_a⁰ ₁V_a⁰;we obtainO²(G_a⁰)N_G_a₀(Z_a⁰ ₁);against Lemma 1.1 (ii).

Thus h(Ga⁰;Va0=Za⁰)1; the remainder of (ii) is a consequence of Lem- ma 1.1 (ii).

Clearly Qa⁰ 1 6Q_a0 and so as a⁰2O(S3) by (i) and a2O(S6) we have (iii). We note thatG_a⁰ is generated by two distinct edge stabilizers and so

(iv) follows easily. p

For the rest of this paper we assume Hypothesis 8.1 holds (with the exception of Lemma 10.2 where we do not requireb>1). Let (a;a⁰)2 C:

We chooseZ_ato be a minimal normal subgroup ofG_acontained inZ_a:By Lemma 8.2 (ii),Z_a is an irreducibleG_a=Q_a-module. Now fort2O(S₆);we define subgroupsZ_t :(Z_a)^g; whereg2Gis such that agt: This is easily seen to be well-defined. Forl2O(S3) and t2D(l) we also define V_l hZ^G_t ^li;Y_tcore_G_t(Z_l) andYt core_G_t(Zl):

LEMMA 8.3. (The Core Argument) Let t2O(S₃) and D(t) fl₁;l₂;l₃g:Suppose that HG_l₁_tand H6Q_t:Then

(i) Y_tZ_l_i\Z_l_j and YtZli\Zlj;for i6j2 f1;2;3g;

(ii) H is transitive onfl2;l3g;and

(iii) if H normalizes a subgroup Mof Z_l_i(respectively M of Z_l_i) for some i2 f2;3g, then MY_t(respectively MY_t).

PROOF. Note that Gtlk normalizes Z_l_i\Z_l_j and Zli\Zlj where fi;j;kg f1;2;3g: So, since [Z_l_i;Qli][Zli;Qli]1; Z_l_i\Z_l_j and Z_l_i\Z_l_j are both normalized byhQ_l_i;G_tl_ki Gt;and (i) holds. Part (ii) is

clear while (iii) follows from (i) and (ii). p

LEMMA8.4. Let(a;a⁰)2 C:

(i) If[Z_a:Z_a\Z_a2]2;thenh(Gb;V_b)1:

(ii) Qa62Syl₂(hQ^Ga^bi):

(iii) Z_a is natural G_a=Q_a-module.

PROOF. (i) Assume that [Z_a:Z_a\Z_a2]2: Then [Z_a:Y_b]2: By Lemma 1.1 (ii)Z_a<hZa^G^bi V_band so, asjV_b=Y_bj 2³;h(Gb;V_b)1:

(ii) Suppose that Qa2Syl₂(hQ^Ga^bi): From Lemma 8.2 (iii) Xb: hQ^Ga^bi coversG_b=Q_b:SetQQ_a\Q_b:ThenQO₂(X_b):Now Lemma 1.1 implies

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that no non-trivial characteristic subgroup ofQais normal inXb:Appealing to 3.2 yields that

(8.4.1) h(X_b;Q)1and

hV₁(Z(Q))^xjx2Aut(Q_a)and x has odd orderi is a normal subgroup of Xb contained in Q.

Noting that Qcore_G_b(Qa); we see that Q centralizes hZ^Ga^bi V_b: Hence, as b3; V_bV₁(Z(Q)): So (8.4.1) implies that U_a

hV_b^O²^(G^a⁾i Q: Since h(X_b;Q)1h(X_b;V_b) and V_b U_a Q_a; we have U_a is normal in X_b which is impossible by Lemma 1.1 (ii). Thus Qa62Syl₂(hQ^Ga^bi):

(iii) First we consider the case whenh(Gb;V_b)2:ThenV_a0\Qb6Qa: Since (a;a⁰)2 C and a⁰2O(S₃);Za acts transitively on D(a⁰)n fa⁰ 1g

ft;rgand henceZ_a\Q_rC_G_a0(V_a0):So we have [Z_a :CZ_a(V_a0\Qb)][Z_a :Z_a\Qb]2⁴: (8:4:2)

Now Proposition 2.9 (i) and (8.4.2) show thatZ_ais not isomorphic to the 16- dimensional irreducibleGF(2)S6-module, whenceZ_ais a natural module by Lemma 2.2 (i). So we may suppose that h(G_b;V_b)1: Thus [V_b;Q_b]

[Z_a;Q_b]:IfZ_a is the 16-dimensional Steinberg module forG_a G_a=Q_a; then, by Proposition 2.9, [Z_a;Q_b]E(2¹⁵) andC_Z_a(j)E(2⁸) for each in- volution j of Ga: Therefore, CGab([V_b;Qb])Qa: Set HbCGb([V_b;Qb]):

ThenHb/ Gb and soQa2Syl₂(Hb): But, asHb hQ^Ga^bi;this contradicts

part (ii), so completing the proof of (iii). p

9. The case[Z_a :Z_a\Z_a2]2.

Suppose that (a;a⁰)2 C:Our main result in this section is Theorem 9.2 in which we show that the case [Z_a:Z_a\Z_a2]2 cannot occur. If [Z_a:Z_a\Z_a2]2;then from Lemma 8.4h(G_b;V_b)1 andZ_aE(2⁴) is a naturalGa=Qa-module. Moreover we have thatY_b [V_b;Qb][Z_a;Qb] E(2³):We shall use these facts without further references in this section.

Set C_bC_G_b(V_b); K_b h[V_b;Qa]^G^bi and H_bC_G_b(Y_b): Observe that H_bhQ^Ga^bi:Our first lemma looks at the structure ofK_b andV_b:

LEMMA 9.1. Assume that [Z_a:Z_a\Z_a2]2: Then jK_bj 2³; Kb\Z_aCZ_a(Gab)and Kb[V_b;O²(Hb)]:In particular,jV_bj 2⁶:

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PROOF. Since [Z_a:Y_b]2[Z_a2:Y_b]; we have [Z_a2;Qa\Qb] Zb\Z_a:Therefore,Qaacts as a group of order 2 onV_b=(Zb\Z_a) which centralizes the subgroup [V_b;Qa]Z_a=(Z_b\Z_a):Since this latter group has index 2 inV_b=(Z_b\Z_a) and [V_b;Q_a]6Y_b;we havej[V_b;Q_a] (Z_b\Z_a)j 2²: IfK_bhas order 2²;thenK_bZ_aV_bwhich then gives [V_b;Q_a]1;contrary to h(Gb;V_b)1: Thus Kb has order 2³ with jKbY_b=Y_bj 2²: Suppose V_bKbY_b: Then [V_b;Qa]Z_a and, as [V_b;Qa] in normal in Gab; the uniseriality of Z_a as aG_ab-module, implies that [V_b;Qa]Y_b which is of course impossible. ThusjV_b=Y_bj 2³ andK_b\Z_aK_b\Y_bZ_a\Z_b

C_Z_a(G_ab):

THEOREM9.2. If(a;a⁰)2 C;then[Z_a:Z_a\Z_a2]62:

PROOF. Suppose that [Z_a :Z_a\Z_a2]2 for (a;a⁰)2 C and put QQ_a\Q_b: Then we have that H_b has a Sylow 2-subgroup T where Z₂T=Q_a Z(G_ab=Q_a) with T=Q_a acting as a transvection on Z_a: Let t2O2(Hb)nQ:ThenO2(Hb)QhtiandT htiQa:

We identifyGab=Qawithh(5 6);(1 3)(2 4);(1 2)iand assume thatZ_ais the natural module which admits (5 6) as a transvection. ThentQ_a(5 6):Let I f1;2;3;4gand letd_i2G_a;i2I;be such that

d₁Q_a(1 5 6); d₂Q_a(2 5 6); d₃Q_a (3 5 6); d₄Q_a(4 5 6):

ThentQa invertsd_iQafor eachi2I:SetI fhd_iQaiji2Ig:Fori2Iwe define the following subgroups:

Hi;a hT;dii;

H_i hHb;H_i;ai;

N_icore_H_i(T);

V_iV₁(Z(N_i)) and U_i[V_i;H_i]:

Note that for eachi2I;O₂(H_i;a)Q_aand thatH_i;a=Q_aS₃:Fori2I;we additionally set

He_iH_i=N_i; He_i;aH_i;a=N_i and He_i;b H_b=N_i: Obviously we have

(9.2.1) for i2I,hH_i;a;G_abi Ga:

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Since Hb=O2(Hb)S3 and Hi;a=QaS3;HfiHgi;a T~ Hgi;b is a Gold- schmidt amalgam. Because Gab=Qa permutes the set fhdiQai ji2Ig transitively and normalizes H_b;the type of the Goldschmidt amalgam is independent ofi2I:

Note that N_iO₂(H_i;a)\O₂(H_b)Q_a \O₂(H_b)Q and so, by Lemma 8.4 (ii),

(9.2.2) O₂(O²(H_b))6Q_a:

Let i2I: Recalling that tacts as a central transvection onZ_a; we have Z_i:C_Z_a(hd_i;ti)C_Z_a(d_i)E(2²) with Z_iC_Z_a(t)Y_b: Hence, Z_iZ(H_i) and soZ_iN_i:Therefore,jZf_aj 2²:We next show that (9.2.3) Z_a N_ifor each i2I:

We suppose that (9.2.3) is false, and seek a contradiction. IfjZ_aN_i=N_ij 2;

then Z_a h(Z_a \N_i)^H^i;ai N_i: So we must have jZ_aN_i=N_ij 2² with h(He_i;a;fZ_a)1:Also, from (9.2.2),h(Hg_i;b;O2(Hg_i;b))1:Let~bbe the critical distance of the amalgam (Hg_i;a;Hg_i;b):Thenb3 forces~b3 which then yields, using Theorem 3.5,~b3 andHfiis of typeG5orG¹₅. Both of these amalgams have the property thatjZ(O²(Hg_i;b))j 2:Now~b3 means that there existsh2Hf_isuch thatZf_aHe^h_i;bbutZf_a6O₂(He^h_i;b):

Since [N_i;Z_a]1;this leads to [N_i;O²(H_b)]1:Hence using (9.2.2), [N_i;hO²(H_b);O²(H_i;a)i]1:

Combining (9.2.1) and Lemma 1.1 (ii) gives

(9.2.3.1) N_icontains no non-trivial G_ab-invariant subgroups.

Letg2Gab:SinceGab normalizes O²(Hb);[N^g_i;O²(Hb)]1 and so, as jZ(O²(He_i;b))j 2;it follows thatN_iN_i^gN_iZ_a:Therefore N_iZ_a/ G_ab and consequentlyN_iZ_a/ G_aby (9.2.1). BecauseQ_anormalizesN_i;[N_iZ_a;Q_a]

[N_i;Q_a] is a normal subgroup ofG_acontained inN_i:So [N_iZ_a;Q_a]1 by (9.2.3.1). Clearly,F(NiZ_a)Niand thus, using (9.2.3.1) again, we obtain N_iZ_aV1(Z(Qa)): Since Qa=N_iZ_a is a Ga-invariant section of Qa with h(H_i;_a;Q_a=N_iZ_a)60;h(G_a;Q_a=N_iZ_a)60:Now, by Theorem 3.5,jTj e 2⁷ which then shows that Q_a=N_iZ_aE(2⁴) is a natural G_a=Q_a-module. As [V_b;Q_a]61;V_b 6N_iZ_a:HenceQ_a is generated by involutions and thus, appealing to Lemma 3.11, Qa is elementary abelian. But then Qa=Ni is

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elementary abelian which is not the case as Qa=Ni contains subgroups isomorphic toZ4Z4:With this contradiction we have verified (9.2.3).

From (9.2.3), Z_a Vi and therefore V_b hZa^H^bi Vi for all i2I.

Since

N_iC_H_b(V_i)C_b QT and N_iC_H_i;a(V_i)C_H_i;a(Z_a)QaT; we deduce that C_H_i;_a(V_i)C_H_b(V_i): Hence N_iC_H_i;a(V_i)C_H_b(V_i): Be- cause hH_i;a;G_abi G_a by (9.2.1), H_i;_a does not normalize V_b and so CHi(Vi)Hi;a6CHi(Vi)Hb: Consequently, putting HciHi=CHi(Vi);

Hc_iHb_i;_a T^ Hbb is a Goldschmidt amalgam with Hb_i;_aHe_i;_a and Hb_bHe_i;_b:

The first part of the next claim has been discussed a moment ago.

(9.2.4) For i2I;

(i) V_bViand NiCb;and

(ii) J(Cb)6N_i (J(Cb) being the elementary abelian version of the Thompson subgroup of C_b).

For (ii) note thatJ(C_b)N_iimplies thatJ(C_b)J(N_i) is normalized by hH_i;G_bi Gand, asJ(C_b)61;this contradicts Lemma 1.1.

It follows from (9.2.4) (ii) thatJ(Cd_b) contains an offender onV_i:Further, ash(Hb_b;O₂(H))b 1 andJ(C_b)/ H_b;Theorem 3.6 ([Table II and Corollary 3, Ch1]) yields that HciS6 or G2(2) and all the following additional in- formation.

(9.2.5) For i2I

(i) (He_1;a;He_i;b)is of type G¹₃or G¹₄ and is independent of i2I;

(iI) Hc_iS₆ or G₂(2);

(iii) if Hc_iS₆; then U_i is isomorphic to either the natural or orthogonal S6module and Cb=Niacts as a transvection on Ui; and

(iv) ifHc_iG2(2);then U_i=CUi(H_i)is isomorphic to the natural 6- dimensional G₂(2)-module andjU_ij 2⁷. Furthermore, C_b=N_ihas order 2³and is an offender on U_i.

We will need the following result about amalgams of typeG¹₃orG¹₄: (9.2.6) Let i2I and suppose We /Hei: If either We \Hei;a61 or We \Heb61;then O²(He_i)W:e

If, say, w2We where hwi 2Syl₃(He_b) then w;O2(He_b) We: Since

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(Hei;a;Heb) is of type G¹₃ or G¹₄,w;O2(Heb) 6O2(Hei;a) which forces v2We wherehvi 2Syl₃(Hei;a):This yieldsO²(Hei)W:e So we may supposeWe \

\He_bO₂(He_b) and likewise that We \He_i;aO₂(He_i;a). But then 16We \

\He_bWe \He_i;a/ He_i;a contradiction.

Note thatG_ab=Qa permutes the setI transitively by conjugation. As- sume thatg2G_ab andhd_iQ_ai^g hd_jQ_ai:Then, asQ_aT;we have

H^g_i;aHj;a

and, because G_ab normalizes H_b, we conclude that H^g_i H_j; N_i^gN_j; V_i^gV_j andU_i^gU_j:Set, fori;j2Iwithi6j;

H_ij hH_i;H_ji N_ijcore_H_ij(T) V_ijV₁(Z(N_ij)) and

H_ijH_ij=N_ij:

Note that G_ab has two orbits on the pairs fi;jg I: One orbit has length two and consists offf1;2g;f3;4ggand the second orbit has length four and consists of all the other pairs.

SetM T

i2IN_i:ThenMis normalized byG_ab:If [N₁;d_i]Mwere to hold then, M/ hd1;Gab;Hbi G which yields V_bM1: Thus [N₁;d₁]6M and so there exist j2I n fIg such that [N₁;d₁]6N_j:Since there is an element in G_ab which exchanges 1 and j, we then have

N_j;d_j 6N₁. The action ofG_ab=Q_aon pairs inIshows that (9.2.7) there exists j2 f2;3gsuch that[N1;d1]6N_j:

Note that ifi;j2I;i6j;and [N_i;d_i]N_j;thenN_ijN_i\N_j: (9.2.8) Suppose that j2I and[N1;d1]6Nj:Then

(i) C_T(Q_a)Q_a;and

(ii) for k2 f1;jg;C_H_b(N_k)N_k;C_H_k;a(N_k)N_k;h(H_k;a;O2(H_k;a))2 andh(H_b;O₂(H_b))2:

If (i) is false, thend1would centralizeQ_a:Hence, sinceN1jCbQa;d1

normalizesCbwhenceCb/hGab;d1i Ga;a contradiction. Thus (i) holds.

Suppose C_H_b(N_k)6N_k (where k2 f1;jg). Applying (9.2.6) with

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We C_Hg_k(Nk) yields that dk centralizes Nk: So [Nk;dk]N1jN1\Nj

which is not the case. ThereforeC_H_b(Nk)Hkand likewiseC_H_k;a(Nk)Nk

fork1;j:Now, appealing (9.2.5) (ii), the remainder of (ii) follows.

In the language of Chermak [Ch2], (9.2.7) and (9.2.8) say that, provided [N₁;d₁]6N_j; fTH_1;a;H_j;a;H_bg is a ``Large Triangular Amalgam''.

SinceO2(H1;a)QaO2(Hj;a);we may apply [Theorem A, Ch2] to obtain the following result.

(9.2.9) Assume that [N1;d1]6Nj: Then, for k2 f1;jg; Nk=N1j is elementary abelian, contains exacitly one non-central Hk-chief factor and N1\N_jis not normal in H_k:

(9.2.10) Suppose that[N1;d1]6Nj:Then H1and Hjhave no non-central chief factors within N_1j:

Suppose thatCT(N1j)N1j:ThenV1VjV1j:Define m min

A2A(Cb)fjAjA6N1orA6N2g and

K(C_b) fB2 A(C_b)j jBj<mg:

Then, by the definition of m, hK(C_b)i N₁\N_j and consequently is invariant under the conjugation action of H₁ and H_j: Therefore, hK(C_b)i N₁_j:By (9.2.4) (ii),J(C_b)6N_ifori2I;som>1:Furthermore, as there is an element ofGabwhich interchanges 1 andjwe know that there isA2 A(Cb) withA6N1andjAj m:From among all suchAselect one withjA\V₁jmaximal. Then by the Thompson Replacement TheoremAis an offender onV₁:So, since [C_b :N₁]2 in the caseHc₁S₆and by [ON]

whenHc₁G₂(2);we have

jAN1=N1j jV1=C_V₁(A)j 2 ifHc1S6

2³ ifHc1G2(2) 8<

:

In particular, this yields that B(A\N1)V12 A(Cb): Clearly, as V1N1j; B N1 and jBj <jAj: HenceB2 K(Cb) and thus BN1j:In particular,A\N1N₁_j and so

jV_1j=C_V_1j(A)j jV₁_j=V_1j\Aj jA=A\N₁_jj 2 if Hc1S6

2³ if Hc1G2(2): 8<

:

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Therefore V1jCV1j(A)V1: Since V1j=V1 admits H1 and A6N1; using (9.2.6) gives V1Vj/ H1 which contradicts (9.2.9). From this contradiction we deduce that C_T(N₁_j)6N₁_j: If C_T(N_1j)N₁\N_j; then C_T(N₁_j)

C_N₁(N₁_j)C_N_j(N₁_j)/H; whence C_T(N₁_j)N₁_j: Thus C_T(N₁_j)6N_i wherei2 f1;jg:Appealing to (9.2.6) we infer thatO²(H_b) centralizesN₁_j and then, by (9.2.6) again, we have proved (9.2.10).

One consequence of (9.2.5), (9.2.9) and (9.2.10) is thatU₁6N₁_jand that U1N1j=N1j contains the unique non-centralH1-chief factor inN1=N1j: (9.2.11) Z_aZ_aand Z_bC_Z_a(G_ab):

Since every H1;a-chief factor of N1 is contained in U1 and h(H1;a;C_U₁(Qa))1;h(H1;a;V1(Z(Qa)))1:Hence, asZ(Ga)1; Lemma 2.2 implies thatV₁(Z(Q_a)) is either a natural or a dual orthogonal module forS₆:In particular, (9.2.11) holds.

(9.2.12) Suppose that[N₁;d₁]6N_j andgb.d₁2D(a):Then

(i) [Z_a;d₁]K_bU₁;j[Z_a;d₁]K_bj 2⁴and Z_a\[Z_a;d₁]K_b [Z_a;d₁];

(ii) [K_g;C_b]6Z_a;and

(iii) U1Nj UjN1N1NjCb:

Sinceh(H₁;N₁)1;[Z_a;d₁][N₁;d₁]U₁:Now [Z_a;d₁]Z_bZ_g and so ZbU1:Now Lemma 9.1 implies that Kb[Za;d1]^H^b U1 and that j[Za;d1]Kbj 2⁴:So (i) holds.

Becauseh(G_g;K_g)1 andC_b6Q_g from the structure of the amalgam (H_1;a=N₁;H_b=N₁);part (ii) holds. Suppose that U₁N_j: Then [N₁;d₁] U₁N_j;a contradiction. Hence (iii) holds by (9.2.5) (iii) and (iv).

(9.2.13) Hc₁6S₆:

Suppose that Hc1S6: Then [U1;C_b] has order 2. Assume that [N₁;d₁]6N₃:Then, by (9.2.12) (iii) and by symmetry,

[U₃;C_b][U₁;U₃][U₁;C_b]:

Since also [U₁;d₁]6N₄;we have

[U1;Cb][U1;U4][U4;Cb]:

Therefore, [U1;Cb] is normalized by h(1 4)(2 3);(3 4);(5 6)i Gab=Qa: Hence, as j[U₁;C_b]j 2; [U₁;C_b]Z_b by (9.2.11) and this contradicts (9.2.12) (ii).

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Next suppose [N1;d1]6N2and that [N1;d1]N3:Applying (1 3)(2 4) to [N1;d1]6N2 gives [N3;d3]6N4: If U36N1; then, as jCb=N1j 2;

[U₁;U₂][U₁;C_b][U₁;U₃] is normalized by h(1 2);(1 3)(2 4);(5 6)i

G_ab=Q_a and we have the same contradiction as above. Therefore, U₃N₁ and so U₃U₁ is normalized by H₁ and, recalling that [N3;d3]6N4;

[U₃;N₁][U₃;U₄]:

If [U3;N1]1;then N3N1 and so from the action ofGab onIwe get N1N2N3N41; which is impossible. Therefore, [U3;N1]

[U3;U4] and conjugating by (1 2) we have [U3;U4] is normalized by h(3 4);d₁;d₂i S₄2:In particular, [U₁;U₄]Z_a and we have a contradiction to (9.2.12) (ii). We conclude thatHc₁6S₆:

By (9.2.5) (ii) and (9.2.13), to complete the proof of Theorem 9.2 it re- mains to eliminate the possibilityHc1G2(2):So assume thatHc1G2(2) and that [N1;d1]6N_j: Then from [ON], [U1 :CU1(Cb)]2³ and so jC_U₁(C_b)j 2⁴ by (9.2.5) (iv). Now (9.2.12) (ii) implies that C_U₁(C_b) K_b[Z_a;d₁]: Set gb:d₁: Then 16[K_g;C_b][K_g;U_j]: Since U_j does not admit transvections fromcH_j;we have thatj[K_g;C_b]j 2²by Lemma 9.1. It follows thatZ_g[U₁;U_j] and hence

CU1(Uj)[U1;Uj]CUj(U1):

But then [Z_a;d_j][Z_a;d₁]C_U₁(C_b) and this contradictsjZ_a\C_U₁(C_b)j 2²: p 10. Determination of b.

Again, let (a;a⁰)2 C:From Theorem 9.2 we have that [Z_a :Z_a\Z_a2]>2:

In this section we show that this situation leads to b3;so establishing our main theorem. So, by Lemma 8.3 (i), we have [Z_a :Y_b]>2:First we observe

LEMMA 10.1. Suppose t2O(S3) and l2D(t): Then no element in GtnG_lcentralizes a hyperplane of Z_l:

PROOF. This follows from [Z_a:Y_b]>2 and Lemma 8.3 (iii). p By Lemma 8.4 (iii),Z_a is a naturalGa=Qa-module-this fact will be used without reference. The next lemma is also needed when we later in- vestigate theb1 andb3 cases.

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LEMMA10.2. If b1;then

(i) for eacha⁰12D(a⁰)n fa⁰ 1g;Z_a016Qb; (ii) V_a0 6Q_b;and

(iii) Z_a6Q_a⁰:

PROOF. First we prove part (i), so we leta⁰12D(a⁰)n fa⁰ 1g:As- sume that Z_a01Qb; and argue for a contradiction. Since Za6Qa⁰; Lemma 10.1 gives

[Z_a01:Z_a01\Q_a]4:

So jZ_a01Qa=Qaj 4 and hence [Z_a=(Z_a\Qa⁰1)][Z_a :CZ_a(Z_a01)]4 because a fours subgroup ofS6does not centralize any hyperplane of the natural module (see Proposition 2.5).

SetR[Z_a\Q_a⁰;Z_a01]:Clearly,RZ_a\Z_a01:Since [Z_a:C_Z_a(Z_a01)]

4;R61. AlsoRZ_a\Z_a01C_Z_a01(Z_a) givesRZ_a0 1\Z_a01Y_a0; by Lemma 8.3. Hence jRj 2 or 4. First we examine the case jRj 2: If Z_aQa⁰;then we get [Z_a:Z_a\Qa⁰1]2 whereas [Z_a:Z_a\Qa⁰1]4:So Z_a6Qa⁰:Now chooset2Z_anQa⁰:Looking atZ_a01acting onZ_awe have that Z_a01 leavesRinvariant and, by Proposition 2.5 (ii),j[Z_a;Z_a01]j 4:Hence Z_a01acts (quadratically) on the 3-spaceZ_a=Rwithj[Z_a=R;Z_a01]j 2:As a consequence there exists XZ_a01 with [Z_a01:X]2 and such that [t;X]R:ThereforetnormalizesX R, whenceX RY_a0by Lemma 8.3 (iii) which contradictsjY_a0j 4:ThusjRj 62 and so we havejRj 4:Because jZ_a01Q_a=Q_aj 4;j[Z_a;Z_a01]j 4 by Proposition 2,5 (ii) which, asjRj 4;

forcesR[Z_a;Z_a01]:Therefore

[Z_a;Z_a01]RZ_a0 1\Z_a01Y_a0:

ThusZ_anormalizesZ_a01and soZ_a Q_a⁰ by Lemma 8.3 (iii). Since R[Z_a\Q_a⁰;Z_a01]Z_a\Z_a01C_Z_a01(Z_a) andZa 6Qa⁰;Lemma 10.1 andjRj 4 imply that

[Z_a\Q_a⁰;Z_a01]RZ_a:

So Z_a01 centralizes a hyperplane of Za=Z_a and therefore, because jZa⁰1Qa=Qaj 4;h(Ga;Za=Z_a)0:

BecauseZ_a Qa⁰ and (a;a⁰)2 C; Z_a<Za: However, combining Lem- mas 8.2 (ii) and 2.4, we then have Za 1

4 :But by Proposition 2.6 (ii), hCZa(Gab)^G^ai 6Zacontrary to the definition ofZa:With this contradiction we have completed the proof of part (i).