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

(S

, S

)-Amalgams V.

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

Introduction.

In part IV of this present series the commuting case for (S₃;S₆)- amalgams was examined when a2O(S₆) for (a;a⁰) a critical pair. This paper and the succeeding two parts are devoted to the commuting case when, for (a;a⁰) a critical pair,a2O(S3). In fact the bulk of our work is concerned with the situation h(G_b;V_b)ˆ1, where b2O(S₆). Unlike Part IV, for this situation, there appear to be no subamalgams which can be exploited early on. Although a very precise description ofV_bis obtained in Theorem 12.1 for our subsequent analysis we need to consider five sub- cases. This subdivision, given in Section 12, is done according to the size of core_GaV_b(a2D(b)) and also whetherV_b=Z_bis an orthogonal module or not.

Of the five possibilities three, as it were, are the ``mainline'' cases - Cases 1, 3 and 4. We briefly discuss and compare each of these cases.

Case 3 is concerned with the smallest possibility for coreGaVb, namely core_GaVbˆZa. This case has the most complicated possibilities forVb=Zb, withVb=Zb being a quotient of

4 1

1. For this case the core argument (Lemma 9.9) is especially valuable as it often enables us to restrict the size of certain commutators. Our scrutiny of Case 3 takes place in Part VI - the end result being that this case cannot arise! For all the other cases we obtain bounds on the parameterbwhich are pursued further in [LPR2]. At the other end of the scale we have Case 1 with [V_b :core_GaV_b]ˆ2 (and then Vb=Zb is a natural module). This, from the outset, is a very tight configuration (for example,Vbacts as a central transvection onVa⁰=Za⁰for

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

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

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

(a;a⁰)2C), which nevertheless requires a delicate analysis. By contrast with Case 3 the core argument is no use here whatsoever. Case 4 (when [V_b:core_GaV_b]ˆ2² and V_b=Z_b is a natural module) lies somewhere be- tween Cases 1 and 3. The core argument is of little use and, initally, the configuration has a greater degree of freedom.

We remark that central transvections make their presence felt in a big way in Parts V, VI and VII. Indeed, without such transvections Cases 1, 2 and 3 would be virtually non-existent.

For ease of reference we continue the section numbering started in [LPR1]. We make the most use of material in Sections 1 and 2, and it is there we refer the reader to for notation and background results. Briefly the contents of this paper are as follows. Section 11 is, mostly, a warm-up for the work in Section 12 where the above mentioned structure ofV_b is determined. Cases 1 and 2 are the subject of Section 13, the main con- clusions being given in Theorems 13.1 and 13.11.

11. Some preliminary results.

For this paper and Parts V and VI we assume the following hypothesis.

HYPOTHESIS11.0. If(a;a⁰)2C, then[Z_a;Z_a⁰]ˆ1anda2O(S₃).

Some elementary observations on this hypothesis are gathered in our first result of this section.

LEMMA11.1. Letm2O(S3)andl2D(m).

(i) [G_lm;Z_m]ˆZ_lˆV₁(Z(G_lm))ˆV₁(Z(G_l)).

(ii) CZm(Gm)ˆ1, GGm(Zm)ˆQm and ZmˆZl1Zl2 whenever l1;l22D(m)withl16ˆl2.

(iii) b1(2)and b>1.

(iv) If XG_lmand X6Q_m, then[X;Z_m]ˆZ_lˆC_Zm(X).

(v) If XQ_lis such that[X;V_l]6ˆ1, then[X;V_l]ˆZ_l. (vi) [Ql;Vl]Zl.

(vii) G is transitive on paths(l1;l2;l3)of length2wherel12O(S6).

PROOF. Let (a;a⁰)2C. Then [Z_a;Z_a⁰]ˆ1 implies Z_a⁰ Z(G_a⁰) and henceV₁(Z(G_a⁰ _1a⁰))ˆZ_a⁰ Z_j for allj2D(a⁰). Thereforeb1(2). Since Gab acts as an involution on Za, [Gab;Za]CZa(Gab)ˆV1(Z(Gab))ˆZb. Likewise [Gaa 1;Za]Za 1. Hence Za ˆZbZa 1ˆ[Gab;Za]Za 1. By

Lemma 11.1(ii) CZa(Ga)ˆ1 and thus Zb\Za 1ˆ1. Hence, by orders, Zb ˆ[Gab;Za]. SinceGacts edge-transitively uponG, we have verified (i) and (ii).

Assume bˆ1 holds. Since, by part (ii), [Z_a;Q_b]Z_b this then gives h(G_b;Q_b)ˆ0, contrary to the hypothesisC_Gb(Q_b)Q_b. Thereforeb>1.

For part (iv) we haveGlmˆXQmand so (iv) follows easily from (i).

Because [X;Vl]6ˆ1 there exists j2D(l) such that [X;Zj]6ˆ1. Thus X6Q_jby (ii) and now (iv) implies (v). Part (vi) is an easy consequence of (v). While (vii) follows from G being edge-transitive and, as l₂2O(S₃), G_l1l2being transitive onD(l₂)nfl₁g.

Our next lemma will be put to use after we have established Theo- rem 12.1.

LEMMA 11.2. If V_b=Z_b contains either a natural or orthogonal S6- module, then Q_aQ_b ˆG_ab and[V_b;Q_b]ˆZ_b.

PROOF. LetV_bXZ_bbe such thatX=Z_bis a natural or orthogonal S₆-module, and supposeQ_aQ_b6ˆG_ab. Then we haveQ_aQ_band soQ_jQ_b for allj2D(b) by Lemma 1.1(i), and henceVbZ(Qb). If [Vb;Qb]6ˆZb, then, by Lemma 11.1(v), we also haveVbZ(Qb). So, if the lemma is false, we may view V_b as aGF(2) (G_b=Q_b)-module. Appealing to Lemma 2.2(iv) givesXˆYC_X(G_b) withY4 or

4 1

. In the former caseC_X(G_b)ˆZ_b and in the latter [Zb:CX(Gb)]ˆ2. In either case (using Proposition 2.5(i) forY

4 1

) we obtainV1(Z(Gab))>Zb, against Lemma 11.1(i). There- foreQaQbˆGab and [Vb;Qb]ˆZb, as required.

LEMMA 11.3. Let(a;a⁰)2C and put CˆCVb(O²(Gb)). Then Za\CˆZb. PROOF. Since C_Za(Ga)ˆ1,Za is a direct sum of 2-dimensional irre- ducible G_a=Q_a-modules. IfZ_a\C>Z_b were to hold, then Z_a\C would contain a non-zeroG_a=Q_a-submodule ofZ_a, against Lemma 1.1(ii).

LEMMA 11.4. Let (d;d⁰)2C. If h(G_d‡1;V_d‡1)>1, then [V_d‡1\

\Q_d⁰;V_d⁰]6ˆ1.

PROOF. Suppose the result is false and, without loss of generality, we take aˆd and a⁰ˆd⁰. So h(G_b;V_b)>1 and [V_b\Q_a⁰;V_a⁰]ˆ1. Since [V_b:V_b\Q_a⁰]2³, [V_b :C_Vb(V_a⁰)]2³. If [V_a⁰ :V_a⁰\Q_b]2², then h(Gb;Vb)1. So [Va⁰ :Va⁰\Qb]2. Moreover,Va⁰\QbQaimplies that

[Va⁰ :CV_a0(Za)]2, contrary to h(Ga⁰;Va⁰)>1. Therefore Va⁰\Qb6Qa

and, similarly,Va⁰6Qb.

jZ_{a 1}j ˆ2 (11:4:1)

By Lemma 11.1(iv) CZa(Va⁰\Qb)ˆZb. Hence Lemma 11.1(ii) implies CZa1(Va⁰\Qb)ˆ1. So

1ˆZa 1\V_b\Qa⁰ ˆZa 1\Qa⁰:

Since [V_a⁰ :V_a⁰\Q_b]2, we have [V_a⁰:C_Va0(Z_{a 1})]2². Therefore, as h(G_a⁰;V_a⁰)>1, we getjZ_{a 1}Q_a⁰=Q_a⁰j ˆ2, which yields (11.4.1).

Forx2Vb; [Va⁰ :CV_a0(x)]2²: (11:4:2)

Let x2V_b. From V_a⁰\Q_b6Q_a and Lemma 11.1(v) [V_a⁰\Q_b;V_b]ˆZ_b. Thus j[Va⁰\Qb;x]j 2 by (11.4.1) from which, as [Va⁰ :Va⁰\Qb]2, (11.4.2) follows.

If jV_bQ_a⁰=Q_a⁰j 2², then V_b contains an element y, y62Q_a⁰, which is not a transvection on some non central G_a⁰-chief factor in V_a⁰ whence h(G_a⁰;V_a⁰)1 by (11.4.2). Thus [V_b:V_b\Q_a⁰]ˆ2 and so [Vb:CVb(Va⁰)]2. But then since Va⁰ 6Qb, this gives h(Gb;Vb)1, a contradiction. So the lemma holds.

COROLLARY11.5. Let(d;d⁰)2C and assumeh(G_d‡1; V_d‡1)>1. Then (i) [V_d‡1\Q_d⁰;V_d⁰]ˆZ_d⁰,[V_d⁰\Q_d‡1;V_d‡1]ˆZ_d‡1;and

(ii) V_d⁰ 6Q_d‡1.

PROOF. Lemmas 11.1(v) and 11.4 yield [Vd‡1\Q_d⁰;V_d⁰]ˆZ_d⁰. Suppose V_d⁰ Qd‡1. Using Lemma 11.1(vi) gives

Z_d⁰ˆ[V_d‡1\Q_d⁰;V_d⁰][V_d‡1;V_d⁰]Z_d‡1

and thus [Vd‡1;V_d⁰]ˆZ_d⁰ by orders. Then h(G_d⁰;V_d⁰)ˆ0 which is im- possible. Therefore V_d⁰ 6Qd‡1. Hence we may find r2D(a⁰) such that (r;d‡1)2C , and Lemmas 11.1(v) and 11.4 give the remaining part of (i).

LEMMA 11.6. Let (a;a⁰)2C and assume that h(G_b;V_b)>1. If jV_a⁰Q_b=Q_bj 2², thenjV_bQ_a⁰=Q_a⁰j ˆ2ˆ jV_a⁰Q_b=Q_bj.

PROOF. By Corollary 11.5 we may find l2D(b) such that [Z_l;V_a⁰\Q_b]6ˆ1. Lemma 11.1(iv) then gives C_Zl(V_a⁰\Q_b)ˆZ_b and so C_Zl\Qa0(V_a⁰\Q_b)ˆZ_b. Since [V_a⁰ :C_Va0(Z_l)]2[V_a⁰:V_a⁰\Q_b]2³ and

h(Ga⁰;Va⁰)>1, we infer that [Zl:Zl\Qa⁰]2. Now Va⁰\Qb acts as an involution onZl\Qa⁰and hence

j[Zl\Qa⁰;Va⁰\Qb]j ˆ[Zl\Qa⁰ :Zb] jZbj=2:

Because [Zl\Qa⁰;Va⁰\Qb]Za⁰\Zbwe see that [Zb:Za⁰\Zb]2. Let x2V_b and put Va⁰ ˆVa⁰=Za⁰. Then j[Va⁰\Q_b;x]j jZ_bj 2 (note that Z_b V_a⁰). Therefore [V_a⁰ :C_Va0(x)]2[V_a⁰ :V_a⁰\Q_b]2³ which, as h(G_a⁰;V_a⁰)>1, forces jV_bQ_a⁰=Q_a⁰j 2. From Corollary 11.5 there exists r2D(a⁰) such that (r;b)2C and so, asjV_bQ_a⁰=Q_a⁰j ˆ2, we may repeat the above argument to obtainjVa⁰Qb=Qbj ˆ2. This proves the lemma.

LEMMA 11.7. Suppose that b>3,(a;a⁰)2C andh(G_b;V_b)>1. Then U_aG_a⁰.

PROOF. If Ua6Qa⁰ 2, then there exists a 22D^[2](a) such that (a 2;a⁰ 2)2C. Applying Corollary 11.5 to (a 2;a⁰ 2) gives

Z_{a 1}ˆ[V_a⁰ ₂\Q_{a 1};V_{a 1}]V_a⁰ ₂Q_a⁰;

asb>3. ThenZ_aˆZ_{a 1}Z_bQ_a⁰, a contradiction. ThereforeU_aQ_a⁰ ₂. Now, by Corollary 11.5, Za⁰ ˆ[Vb\Qa⁰;Va⁰]Vb and so [Ua;Za⁰]ˆ1.

Therefore

UaC_Ga0 1(Za⁰)ˆG_a⁰ _1a⁰; as required.

LEMMA11.8. Let(a;a⁰)2C and suppose b>3andh(G_b;V_b)>1. Then [U_a;V_a⁰]6U_a.

PROOF. Supposing [Ua;Va⁰]Ua we seek to uncover a contradiction.

SettingP_bˆ hG_ab;Va⁰iandVb_bˆV_b=C_Vb(O²(G_b)) we first show that (11.8.1) (i) P_b=Q_bS₄Z₂

(ii) h(G_b;V_b)ˆ2 and Vb_b is isomorphic to a direct sum of two isomorphic naturalS₆-modules

(iii) jZaj ˆ2⁴ andjZbj ˆ jZa⁰j ˆ2². (iv) jZaQa⁰=Qa⁰j ˆ2 andj[Za;Va⁰=Za⁰]j ˆ2²

(v) C_Za(Va⁰\G_ab)ˆZ_b

From [U_a;V_a⁰]U_awe have thatP_bnormalizesU_aand henceP_b6ˆG_b by Lemma 1.1 (ii). By the parabolic argument (Lemma 3.10)

[Va⁰ :Va⁰\Gab]2, and so [Va⁰ :Va⁰\Qa]2². Thus [Va⁰ :CV_a0(Za⁰)]2². Thereforeh(Ga⁰;Va⁰)ˆ2 withjZaQa⁰=Qa⁰j ˆ2, [Va⁰:CV_a0(Za)]ˆ2²and both non central chief factors ofVa⁰ are isomorphic naturalS6-modules. Clearly V_a⁰ 6G_ab and so we conclude that P_b=Q_bS₄Z₂. Combining Lemmas 2.16 and 11.1(i), (vi) gives (ii) forVb_a⁰, so proving part (ii). Because Vbbˆ hZb^Ga^bi it follows from part (ii) that jZbaj ˆ2². Hence jZaj ˆ2⁴ and jZbj ˆ2² by Lemmas 11.1(ii) and 11.3, and we have (iii). Also, by Lemma 11.1(vi), Za acts as an involution upon Va⁰=Za⁰. By Lemma 11.1(i) h(G_a⁰;V_a⁰=Z_a⁰)ˆ2 and so [V_a⁰=Z_a⁰ :C_Va0=Z_a0(Z_a)]ˆ2². Hence j[Z_a;V_a⁰=Z_a⁰]j ˆ2², so establishing (iv). Finally we prove (v). If V_a⁰\G_ab Q_a, then [V_a⁰:C_Va0(Z_a)]2 which is impossible. So Va⁰\Gab 6Qa and thusZbˆCZa(Va⁰\Gab).

(11.8.2)P_b=Q_bis the parabolic with restriction 2

2

on the non-central chief factors inVb_b.

Suppose that (11.8.2) is false. Then, by (11.8.1)(i),(ii), P_b=Q_b is the parabolic with restriction 1

21

!

on all the (isomorphic) naturalS₆-modules inVb_b. From Lemma 11.1(i)bZ_aCbVb(G_ab). Letdbe an element of order 3 inP_b. SincedcentralizesCVbb(G_ab), we infer thatZ_aC_Vb(d). Thus

Z_a/hG_ab;di ˆP_b:

In particular,V_a⁰normalizesZ_a⁰and so [V_a⁰;Z_a]Z_a. From (11.8.1)(iv),(v) jZa\Qa⁰j ˆ2³ and CZa(Va⁰\Gab)ˆZb and consequently [Za\Qa⁰;Va⁰\

\Gab]6ˆ1. So [Za\Qa⁰;Va⁰]6ˆ1 and an appeal to Lemma 11.1(ii) gives [Za\Qa⁰;Va⁰]ˆZa⁰. Combining this with j[Za;Va⁰=Za⁰]j ˆ2² (since jZ_a⁰j ˆ jZ_bj) yieldsj[Z_a;V_a⁰]j ˆ2⁴. But then

Zaˆ[Za;Va⁰]Va⁰ Qa⁰;

contrary to (a;a⁰)2C. With this contradiction we have proved (11.8.2).

jV_bQ_a⁰=Q_a⁰j 2²: (11:8:3)

If jV_bQ_a⁰=Q_a⁰j 62², then V_bQ_a⁰ ˆZ_aQ_a⁰ whence [V_b;V_a⁰=Z_a⁰]ˆ

ˆ[Z_a;V_a⁰=Z_a⁰]. SinceZ_a⁰ [V_b;V_a⁰] by Corollary 11.5, using (11.8.1)(iii),(iv) we see thatj[V_b;V_a⁰]j ˆ2⁴. Hence, asjZ_bj ˆ2²,j[V_a⁰;V_b=Z_b]j ˆ2². Conse- quentlyVa⁰acts as a transvection upon each of the non-central chief factors in Vbb. Together (11.8.2) and Proposition 2.5(viii) imply thatO2(Pb)=Qb is the unique quadratically acting (upon each of the non-central chief factors inVb_b) E(2³)-subgroup of each Sylow 2-subgroup ofP_b=Q_b. Then Proposition 2.5(iii)

forcesVa⁰ O2(Pb)Gab, contradicting (11.8.1)(i). Thus we conclude that jVbQa⁰=Qa⁰j 2².

(11.8.4) A contradiction

By Corollary 11.5 there exists r2D(a⁰) such that (r;b)2C. If jV_a⁰Q_b=Q_bj 2², then applying Lemma 11.6 to (r;b) yieldsjV_bQ_a⁰=Q_a⁰j ˆ2, against (11.8.3). Therefore, asV_a⁰ is elementary abelian,jV_a⁰Q_b=Q_bj ˆ2³, and, of course, Va⁰ acts quadratically on Vb. Again, using (11.8.2) and Proposition 2.5(viii), we deduce the untenableVa⁰ O2(Pb)Gab. This is the desired contradiction which completes the proof of Lemma 11.8.

12. The structure ofVb/Zb.

THEOREM12.1. Let(a;a⁰)2C. Then

(i) h(G_b;V_b)ˆ1 with V_b=Z_b isomorphic to a quotient of 4

1

1;

and

(ii) jZ_aj ˆ2²,jZ_bj ˆ2.

PROOF. First we suppose thatb>3 andh(Gb;Vb)>1 and argue for a contradiction. So Lemmas 11.7 and 11.8 are available to giveUaGa⁰and [U_a;V_a⁰]6U_a. Also, from Corollary 11.5, [V_b;V_a⁰]Z_a⁰. So if U_aQ_a⁰ ˆ

ˆV_bQ_a⁰, we then haveU_aˆV_b(U_a\Q_a⁰) whence, using Lemma 11.1(vi), [Ua;Va⁰]ˆ[Vb;Va⁰]VbUa, a contradiction. ThereforeUaQa⁰ 6ˆVbQa⁰. SinceUa is elementary abelian we infer thatjVbQa⁰=Qa⁰j 2². Now Lem- ma 11.6 and Corollary 11.5(ii) give

jV_bQ_a⁰=Q_a⁰j ˆ jV_a⁰Q_b=Q_bj ˆ2:

(12:1:1)

PutV_b ˆV_b=Z_b.

(12.1.2) (i) (V_a⁰\Q_b)Q_a ˆG_ab.

(ii) h(G_b;V_b)ˆ2 and there exists X_b/G_b with Z_b X_bV_b such thatVb ˆXbC_Vb(Gb) andXbis isomorphic to a quotient of

4 1

4 1

. Moreover, the two 4's inXbare isomorphic naturalS₆-modules.

(iii) jZaj ˆ2⁴,jZbj ˆ jZa⁰j ˆ2²andh(Ga;Za)ˆ2.

By (12.1.1) [Va⁰ :Va⁰\Qb]ˆ2 and so, ash(Ga⁰;Va⁰)>1,Va⁰\Qb6Qa. Thus (i) holds, and [Va⁰ :CV_a0(Za)]ˆ2² with h(Ga⁰;Va⁰)ˆ2. Applying

Lemmas 2.16 and 2.2(iv) toVa⁰ˆVa⁰=Za⁰yields (ii) forVa⁰, so proving (ii).

Using Lemmas 11.1(ii), 11.3, part (ii) and the fact thatZ(Ga)ˆ1 gives (iii).

SetD(a)ˆ fb;a 1;lg.

[Va⁰ :Va⁰\Qa 1]2⁴: (12:1:3)

From (12.1.1) we have ZaQa⁰ˆVbQa⁰ and henceVa 1Qa⁰ VbQa⁰. By (12.1.2), UaQa⁰ˆVa 1Qa⁰. Since UaQa⁰ 6ˆVbQa⁰, jVa 1Qa⁰=Qa⁰j 2² and hence, by (12.1.2)(ii) and Proposition 2.5(iii), [Xa⁰ :C_Xa0(Va⁰ 1)]2⁴. If [X_a⁰\Q_{a 1};V_{a 1}]6ˆ1, then by Lemma 11.1(v)

Z_{a 1}ˆ[X_a⁰\Q_{a 1};V_{a 1}]V_a⁰ Q_a⁰

whence ZaˆZa 1Zb Qa⁰. Hence [Xa⁰\Qa 1;Va 1]ˆ1 from which (12.1.3) follows.

V_a⁰ ₂Q_{a 1} and [U_a;V_a⁰ ₂]ˆ1:

(12:1:4)

Suppose V_a⁰ ₂6Q_{a 1}. Then there exists r2D(a⁰ 2) such that (r;a 1)2C. Applying Corollary 11.5 to (r;a 1) gives Z_{a 1} [V_a⁰ ₂;V_{a 1}]V_a⁰ ₂Q_a⁰, since b>3. This is against (a;a⁰)2C , and therefore Va⁰ 2Qa 1. If [Va⁰ 2;Va 1]6ˆ1, then Lemma 11.1(v) gives Za 1ˆ[Va⁰ 2;Va 1] which again contradicts (a;a⁰)2C. Thus [Va⁰ 2;Va 1]ˆ1 and likewise [Va⁰ 2;V_l]ˆ1, so we have (12.1.4).

[V_a⁰ :V_a⁰\V_a⁰ ₂]2⁴: (12:1:5)

This follows from (12.1.3) and (12.1.4).

(12:1:6)j[U_a;V_a⁰\Q_b]j2⁴[V_a⁰:V_a⁰\V_a⁰ ₂] and (hence)j[U_a;V_a⁰\Q_b]j2⁸: In view of (12.1.3)Va⁰\Qb\Qa must act as at least a fours group on Va 1=Za 1, and thusj[Va⁰\Qb\Qa;Va 1]j 2⁴ by (12.1.2)(ii) and Propo- sition 2.5(iii). Further, we observe, as [Va⁰\Q_b\Qa;Va 1]Va⁰ and V_a⁰\Q_b interchangeslanda 1, that

[V_a⁰\Q_b\Q_a;V_{a 1}]ˆ[V_a⁰\Q_b\Q_a;V_l]V_{a 1}\V_l:

Let t2V_a⁰\Q_b be such that V_a⁰\Q_bˆ(V_a⁰\Q_b\Q_a)hti. Then, as t normalizes Va 1Vl and Va 1\Vl, we have j[Va 1Vl=Va 1\Vl;t]j ˆ

ˆ[Va 1:Va 1\Vl]. Now

[V_a⁰\Q_b;V_{a 1}V_l]ˆ[V_a⁰\Q_b\Q_a;V_{a 1}][t;V_{a 1}V_l] from which we deduce that

j[Va⁰\Qb;Va 1Vl]j 2⁴[Va 1:Va 1\Vl]:

So, by Lemma 11.1(vii),

j[V_a⁰\Q_b;U_a]j 2⁴[V_a⁰ :V_a⁰\V_a⁰ ₂]:

This, together with (12.1.5), yieldsj[V_a⁰\Q_b;U_a]j 2⁸. U_a doesn⁰t act quadratically uponX_a⁰=Z_a⁰: (12:1:7)

Suppose (12.1.7) is false. Then, combining (12.1.2)(ii), (iii) and Propo- sition 2.5(ii), we have j[Ua;Xa⁰]j 2⁸. Noting that [Ua;Xa⁰]ˆ [Ua;Va⁰], (12.1.6) forces

[U_a;V_a⁰]ˆ[U_a;V_a⁰\Q_b]U_a; contradicting Lemma 11.8. Thus (12.1.7) holds.

h(Ga;Ua)4:

(12:1:8)

Set U_a⁽ⁱ⁾ˆ[U_a;Q_a;i] and V_b⁽ⁱ⁾ˆ[V_b;Q_a;i]. From (12.1.2)(ii) and Lemma 11.2,G_abˆQ_aQ_b. HenceV_b⁽³⁾6ˆ1 by Proposition 2.5(i). Our aim is to show that U⁽¹⁾_a 6ˆV_b⁽¹⁾ andUa(2)6ˆV_b⁽²⁾. As a consequences (see Lemma 1.2(v))U_a=U_a⁽¹⁾,U_a⁽¹⁾=U_a⁽²⁾,U_a⁽²⁾=U_a⁽³⁾all contain at least one non-central chief factor for G_a. Because G_ab.V_b⁽³⁾6ˆ1, V_b⁽³⁾\Z_b6ˆ1 and so h(G_a;U_a⁽³⁾)6ˆ0 by Lemma 1.1(ii), we then obtain (12.1.8).

Suppose U_a⁽¹⁾ˆV_b⁽¹⁾. Then [V_a⁰;Q_a⁰ ₁]ˆ[V_a⁰ ₂;Q_a⁰ ₁] and so, by (12.1.4), Ua centralizes [Va⁰;Qa⁰ 1]ˆ[Xa⁰;Qa⁰ 1], contrary to (12.1.7). If Ua(2)ˆVb(2), then likewise Ua centralizes [Xa⁰;Qa⁰ 1;Qa⁰ 1] which again contradicts (12.1.7). ThereforeUa(1)6ˆV_b⁽¹⁾andUa(2)6ˆV_b⁽²⁾, as desired.

(12.1.9) UaQa⁰=Qa⁰ is the non-quadratic E(2³)-subgroup of Ga⁰ 1a⁰=Qa⁰

acting onX_a⁰=Z_a⁰.

Since [Z_a:Z_a\Q_a⁰]ˆ2, (12.1.2)(i), (iii) imply that [Z_a\Q_a⁰;V_a⁰\

\Q_b]6ˆ1. So there exists r2D(a⁰) for which [Z_a\Q_a⁰;Z_r]6ˆ1. Thus Za\Qa⁰ 6Qr. Note that [Zr:Zr\Qb]2 by (12.1.1). Hence Zr\Qb 6Qa for Zr\QbQa would imply [Zr:CZr(Za\Qa⁰)]2 and thenceh(Gr;Zr)1, contrary to (12.1.2)(iii).

Now

[U_a:C_Ua(Z_r\Q_b)][U_a:U_a\Q_r]2[U_a:U_a\Q_a⁰]

and therefore (12.1.8) forces [Ua:Ua\Qa⁰]ˆ2³. Thus, by (12.1.7), we have established (12.1.9).

We are now in a position to deduce the desired contradiction.

Combining (12.1.9), (12.1.2)(ii) with Proposition 2.5(viii) we get

[Xa⁰:CX_a0(Ua)]2⁶. Therefore [Va⁰ :Va⁰\Va⁰ 2]2⁶ by (12.1.4). Hence (12.1.6) forces j[Ua;Va⁰\Qb]j 2¹⁰. Since jZa⁰j ˆ2² and Za⁰[Ua;Va⁰], (12.1.2)(ii) impliesj[Ua;Va⁰]j 2¹⁰. Consequently

[U_a;V_a⁰]ˆ[U_a;V_a⁰\Q_b]U_a;

again contradicting Lemma 11.8. This completes the proof that h(Gb;Vb)ˆ1 when b>3. Now we examine the situation h(Gb;Vb)>1 whenbˆ3.

(12.1.10) (i) Z_a‡2[V_b;V_a⁰];

(ii) either jVbQa⁰=Qa⁰j ˆ jVa⁰Qb=Qbj ˆ2 or jVbQa⁰=Qa⁰j ˆ

ˆ jVa⁰Q_b=Q_bj ˆ2³;

(iii) h(G_b;V_b)ˆ2 and, for iˆ1;2, there exists X_b⁽ⁱ⁾/G_b with Z_bX_b⁽ⁱ⁾ such that V_bˆX_b⁽¹⁾X_b⁽²⁾ and X_b⁽ⁱ⁾=Z_b4 or

4 1

. Further the 4's inX⁽¹⁾_b andX⁽²⁾_b are isomorphic.

Part (i) follows from Corollary 11.5(i) while Corollary 11.5(ii) and Lemma 11.6 imply (ii). For (iii) we first note that (i) gives [V_b;G_b]ˆV_b. If jVbQa⁰=Qa⁰j ˆ jVa⁰Qb=Qbj ˆ2, then [Va⁰ :CV_a0(Za)]ˆ2² and so h(Ga⁰;Va⁰)ˆ2 with both non-central chief factors being isomorphic nat- ural modules. Using Lemma 2.16 and [Va⁰;Ga⁰]ˆVa⁰ yields the desired structure of V_b in this case. Turning to the latter possibility given in (ii) we deduce, just as in Lemma 11.6, that for all x2V_b [V_a⁰ :C_Va0(x)]2⁴. Hence h(G_a⁰;V_a⁰)ˆ2 and, as V_b acts as a quadraticE(2³)-group on V_a⁰, both non central chief factors are isomorphic natural modules. Appealing to Lemma 2.4 and Proposition 2.7(ii) completes the proof of (iii).

By (12.1.10)(iii) Vb=Zb is a quotient of 4

1

4

1

. Put Ybˆ

ˆC_Vb(O²(G_b)) and, foriˆ1;2, letY_b⁽ⁱ⁾be such thatY_b Y_b⁽ⁱ⁾,Y_b⁽ⁱ⁾/G_band Y_b⁽ⁱ⁾=Y_b4 with V_b=Y_bˆY_b⁽¹⁾=Y_bY_b⁽²⁾=Y_b. Let Y_b⁽³⁾ denote the inverse image in Vb of the diagonal submodule. Clearly VbˆY_b⁽ⁱ⁾Y_b^(j) for i6ˆj2 f1;2;3g so, without loss of generality, we may assume that Y_a⁽¹⁾0 6Q_b andY_a⁽²⁾0 6Q_b.

Assume for the moment thatjV_bQ_a⁰=Q_a⁰j ˆ jV_a⁰Q_b=Q_bj ˆ2. SoY_a⁽¹⁾0 acts as a transvection upon each of the non-central chief factors withinV_b. If [Y_a⁽¹⁾0 \Qb;Vb]6ˆ1, then E(2²)Zb[Y_a⁽¹⁾0 \Qb;Vb] whence [Y_a⁽¹⁾0 ;Vb]

E(2⁴). However consideringV_b acting uponY_a⁽¹⁾0 yields, asV_b acts as a transvection uponY_a⁽¹⁾0 =Z_a⁰, that [V_b;Y_a⁽¹⁾0 ]E(2³). Thus we conclude that [Y_a⁽¹⁾0 \Q_b;V_b]ˆ1 and, similarly, [Y_a⁽²⁾0 \Q_b;V_b]ˆ1. So Y_a⁽ⁱ⁾0 \Q_bˆ

ˆC_Y(i)

a0(V_b) for iˆ1;2 and, in particular, Y_a⁽ⁱ⁾0 \Q_b Y_b. Since

[Y_a⁽ⁱ⁾0 :Y_a⁽ⁱ⁾0 \Qb]ˆ2ˆ[Va⁰ :Va⁰\Qb] and, for iˆ1;2, Y_a⁽ⁱ⁾0 6Qb, we see thatVa⁰\Qbˆ(Y_a⁽¹⁾0 \Qb)(Y_a⁽²⁾0 \Qb)Y_a⁽³⁾0 . In particularY_a⁽³⁾0 Qband, since h(Ga⁰;Y_a⁽³⁾0 )ˆ1, Y_a⁽³⁾0 6Qa. Therefore Z_b ˆ[Za;Y_a⁽³⁾0 ]Y_a⁽³⁾0 . But then V_a⁰ˆY_a⁽³⁾0 which is impossible. So now we only need consider the possibility jV_bQ_a⁰=Q_a⁰j ˆ jV_a⁰Q_b=Q_bj ˆ2³. Recalling that Z_a⁰[V_b;V_a⁰], Propositio- n 2.5(ii) implies that

[Va⁰;Vb]Za⁰

X_a⁽¹⁾0 ;Ga⁰a⁰ 1;2

X_a⁽²⁾0 ;Ga⁰a⁰ 1;2 : Since [Va⁰;Vb]CV_a0(Vb) we see that

[Va⁰;V_b]ˆZa⁰

X_a⁽¹⁾0 ;Ga⁰a⁰ 1;2

X_a⁽²⁾0 ;Ga⁰a⁰ 1;2 :

Because [Va⁰ :Va⁰\Qb]ˆ2³ we have Va⁰\Qb ˆ hxi[Va⁰;Vb] where xˆx1x2withx_i2X_a⁽ⁱ⁾0. Since [Va⁰\Qb;Vb]ˆZbZa‡2, [x;Vb]Za‡2.

Set V_a⁰ˆV_a⁰=Y_a⁰. Then V_a⁰ˆX_a⁽¹⁾0 X_a⁽²⁾0 44 and by Lemma 11.3 Za‡2ˆ(Za‡2\X_a⁽¹⁾0 )(Za‡2\X⁽²⁾_a0 )E(2²). SinceX⁽ⁱ⁾_a0 /Ga⁰, we infer that [xi;Vb]Za‡2\X_a⁽ⁱ⁾0 (iˆ1;2). Without loss of generality we may suppose x162[X_a⁽¹⁾0 ;Ga⁰a⁰ 1;2] (becauseVa⁰\Qb6ˆ[Va⁰;Vb]). Sohx1i[X⁽¹⁾_{a 1};Ga⁰a⁰ 1;2]

E(2³) with

V_b;hx₁i

X_a⁽¹⁾0 ;G_a⁰_a⁰ ₁;2

Z_a‡2\X_a⁽¹⁾0 :

But Proposition 2.5(ix) shows this to be impossible and so we have ruled out the possibility h(G_b;V_b)>1 and bˆ3. Therefore we have shown that h(G_b;V_b)ˆ1. The remainder of the theorem follows from Proposition 2.9(i) and Lemmas 2.2(iv), 11.3.

LEMMA12.2. Let(a;a⁰)2C andD(a)ˆ fl1;l2;l3g.

(i) V1(Z(Gab))ˆZbZ2 andV1(Z(Qa))ˆZaˆZliZlj E(2²), 1i<j3.

(ii) G_ab ˆQ_aQ_b and[V_b;Q_b]ˆZ_b. (iii) core_GaV_bˆV_li\V_lj, i6ˆj.

(iv) Z_b V₁(Z(Q_b)) 1

1

.

PROOF. (i) is mostly a restatement of Theorem 12.1; V1(Z(Qa))ˆZa

follows sinceZa is a projectiveGa=Qa-module and V1(Z(Qab))ˆZb Za. Part (ii) follows from Theorem 12.1 and Lemma 11.2.

Iffi;j;kg ˆ f1;2;3g, then clearlyG_alknormalizesV_li\V_lj. Also, [V_li\V_lj;Q_li][V_li;Q_li]Z_liZ_aV_li\V_lj:

Using part (ii) we now see thatVli\Vlj/Ga, so proving (iii).

Turning to (iv), we setMˆV1(Z(Qb)). Observing thatM is aGb=Qb- module with soc(M)ˆZ_b we haveM,!P(1). SincejV₁(Z(G_ab))j ˆ2, Pro- position 2.5(i) implies that M doesn't contain submodules of type

4 1

. HenceM

11

by Lemma 2.4.

We will make frequent use of Lemma 12.2, often without specific re- ference.

LEMMA 12.3. Let (a;a⁰)2C and suppose that core_GaV_b>Z_a⁰. Then Vb=Zbis isomorphic to4or

4 1

.

PROOF. Since core_GaV_b>Za, [core_GaV_b;Qa]6ˆ1 by Lemma 12.2(i).

Clearly [core_GaV_b;Q_a]/G_a and so Lemma 12.2(i) implies that Z_a [core_GaV_b;Q_a]. Hence

Z_a[core_GaV_b;Q_a][V_b;G_b]

and therefore V_b ˆ[V_b;G_b]. Now the lemma is a consequence of Theo- rem 12.1.

LEMMA12.4. Let(a;a⁰)2C. Then

(i) [Qb;O²(Gb)]6Qa; in particular Qa\Qa6/Gb. (ii) G_bˆO²(G_b)Q_a.

PROOF. (i) PutRb:ˆ[Qb;O²(Gb)] and assume thatRbQa. Clearly, Rb Q:ˆQa\Qb and so Q/Gb with QˆC_Qb(Za)ˆC_Qb(Vb). Since [Q_b:Q]ˆ2 we get 2² jV1(Z(Q_b))j jC_Vb(Q_b)j 

jV_bj

p and hence

jV_bj 2⁴, a contradiction. This proves (i).

(ii) If L_b :ˆO²(G_b)Q_a6ˆG_b, then [G_b:L_b]ˆ2 with Q_a2SyI₂L_b; so Qa\QbˆO2(Lb)/Gb, but this contradicts (i).

Let (a;a⁰)2C and put YbˆCVb(O²(Gb)). Assume that jYbj ˆ2². We now define

Faˆ hY_ljl2D(a)i ˆDY_b^GaE and

Hbˆ hFmjm2D(b)i ˆ FaGb

:

ClearlyFa/GaandHb/Gb. Some less obvious properties of these groups are given in the next lemma.

LEMMA12.5.

(i) h(Ga;Fa)ˆ2; and (ii) h(G_b;H_b)2.

PROOF. (i) First we observe that ZaFa and, as jYbj ˆ2², that h(Ga;Fa)2. Ifh(Ga;Fa)ˆ1, thenFaˆYbZaand hence

[Q_a;F_a]ˆ[Q_a;Y_bZ_a]ˆ[Q_a;Y_b]Z_b <Z_a:

Since [Q_a;F_a]/G_a, we then getF_aV₁(Z(Q_a))ˆZ_a, which is impossible.

Thush(G_a;F_a)ˆ2.

(ii) SinceZa Fa, clearlyVb Hb. So if (ii) is false, thenHbˆFaVb. Hence [FaVb;Qb]/Gb and

[F_aV_b;Q_b]ˆ[F_a;Q_b][V_b;Q_b]ˆ[F_a;Q_b]Z_b:

By G_abˆQ_aQ_b and (i) j[F_a;Q_b]j 2² and so [F_aV_b;Q_b]ˆ[F_a;Q_b].

Because F_a=Z_a2 or 21, we see that j[F_a;Q_b]j 2³ and therefore [Fa;Qb] is centralized by O²(Gb). Consequently Za6[Fa;Qb] and so Za\[Fa;Qb]ˆZb. Next we consider

[F_a;Q_a\Q_b][F_a;Q_a]\[F_a;Q_b]ˆZ_a\[F_a;Q_b]ˆZ_b:

Therefore [VbFa;Qa\Qb]Zb. So Qa\QbCQb(VbFa=Zb). Recalling that [FaVb;Qb]ˆ[Fa;Qb] has order at least 2² we deduce thatQa\Qb ˆ

ˆC_Qb(V_bFa=Z_b). But thenQa\Q_b/G_b, contrary to Lemma 12.4(i). This completes the verification of (ii).

The groupsF_aandH_b will be important in later arguments in, for ex- ample, Section 13 and Part VI. Now we return to examine further the si- tuation in Lemma 12.3 in our next lemma, where Fa makes a brief ap- pearance.

LEMMA 12.6. Let (a;a⁰)2C, and assume that core_GaV_bˆV_{a 1}\

\V_b>Z_a.

(i) If V_b=Z_b4, then V_{a 1}\V_bˆ[V_b;G_ab] or V_{a 1}\V_b ˆ

ˆ[V_b;G_ab;G_ab].

(ii) Assume that Vb=Zb  4

1

, and set Yb ˆCVb(O²(Gb)). Then Y_b6V_{a 1}\V_band Y_b(V_{a 1}\V_b)6ˆV_b. Hence one of the following holds: