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

(S

, S

)-Amalgams VII.

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

Introduction.

This paper, which is the last of the series of papers [LPR1], completes the proof of the MAINTHEOREMstated in Section 1 of [LPR1]. More speci- fically, here we shall be investigating Cases 4 and 5 (as given in Section 12;

just as before we continue the section numbering of [LPR1]). In both of these cases we have that for each critical pair(a;a⁰)inG,[Z_a;Z_a⁰]ˆ1and a2O(S₃). (For notation see Section 1.) Sections 17 and 18 deal withCase 4 and Section 19 is devoted to Case 5. The short Section 20 reviews the main results of [LPR1] and this paper, and shows that together they es- tablishour MAINTHEOREM.

Section 14 concentrates upon the non-central chief factors ofGbwithin W_b(b2O(S₆)), the main conclusions being contained in Theorem 17.3. Note the crucial obstructing role the quadratic fours group(W_a⁰\G_a‡2a‡3)Q_b=Q_b (Gba‡2-conjugate tohs1;tias given in Proposition 2.5(ii)) in the theorem as highlighted in Lemma 17.4(ii). Also, the fact that ZbWa⁰ 2 when bˆ7 (so (17.3.2) doesn't hold) leads us to a lengthy tussle with the situationbˆ7 before we complete the proof of Theorem 17.3. As ever our old friend the central transvection is never far from the centre of the action; see particu- larly Section 18 where we build upon the results of the previous section and establishthat, for Case 4,b2 f3;5g. Case 4 proves to be a slippery customer.

For example, we are unable till quite late in the day to establish, in full generality, the following symmetry statement that for(a;a⁰)2 C we have V_a⁰6Q_b(proved in Lemma 18.7, in fact).

(*) 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 Kingdom.

(***) Indirizzo dell'A.: School of Mathematics, University of Manchester, Oxford Road, Manchester M13 9JL, United Kingdom.

That Case 5 is not so elusive as Case 4 is very much due to our having Yb>Zb (b2O(S6)). This gives us extra leverage in the form of the sub- groupsFaˆ hYbGai andHbˆ hFaGbi(where a2D(b)). These subgroups first appeared in Section 12 and were also of use in analyzing Case 2.

Finally, we point out that this paper may be read independently of the other parts with the exception of notation and module data given in Sec- tions 1 and 2, and a small amount of material relating to the case divisions in Section 12.

17. Case 4 - the non-central chief factors inW_b.

Sections 17 and 18 are concerned withCase 4. So, in these sections, we shall be assuming

HYPOTHESIS17.0. Vb=Zb4andcoreG_aVbˆVa 1\Vbˆ[Vb;Gab;Gab]

E(2³).

LEMMA17.1. h(Ga;Ua)ˆ3

PROOF. Sinceb>1 andGabˆQaQbby Lemmas 11.1(iii) and 12.2(ii), h(G_a;U_a=[U_a;Q_a])ˆ1. Because core_GaV_bˆ[V_b;G_ab;G_ab] wealso obtain h(G_a;[U_a;Q_a]=V_{a 1}\V_b)ˆ1, so givingh(G_a;U_a)ˆ3.

LEMMA17.2. Ifb>3, thenh(Gb;Wb)3.

PROOF. By combining Lemmas 12.2(ii) and 12.4(i) with [Theorem 1;

LPR2] weobtain thelemma.

The main result of this section which gives us a foothold in analysing Case 4 whenb>5is the following theorem.

THEOREM 17.3. Assume Hypothesis 17.0 holds, and b>5, and let d2O(S₆). Then h(G_d;W_d)ˆ3 with the non-central chief factors in W_d being isomorphic natural modules.

We will present the proof of Theorem 17.3 in a sequence of results.

Since we shall suppose the theorem is false and seek a contradiction, we shall assume in Lemma 17.4 and Theorems 17.5 and 17.6 that Hypoth- esis 17.0 andb>5holds but not the conclusions of the theorem. An im- portant intermediate step, achieved in Theorem 17.6, is that of finding a

critical pair(d;d⁰)for which[Zd‡1;W_d⁰]6ˆ1. Lemma 17.4 contains various observations needed for the proof of Theorem 17.6.

LEMMA 17.4. Let (a;a⁰)2 C and suppose that [Z_b;W_a⁰]ˆ1. Put XˆW_a⁰\G_a‡2a‡3. Then

(i) [W_a⁰ :X]ˆ2 and hence there exists a⁰‡32V(G) such that d(a⁰;a⁰‡3)ˆ3and(a⁰‡3;a‡3)2 C

(ii) XG_band, in the notation of Proposition 2.5(ii),XQ_b=Q_bis Gba‡2=Qb-conjugate tohs1;ti;

(iii) Za‡2[X;Vb]E(2³);

(iv) Z_a‡2W_a⁰; and

(v) [V_b;G_ba‡2]ˆ[X;V_b](V_b\V_a‡3).

PROOF. Since[Zb;Wa⁰]ˆ1, [Za‡2;Wa⁰]ˆ1. Therefore hGa‡2a‡3;Wa⁰i normalizesZ_a‡2 and so is a proper parabolic subgroup ofG_a‡3. Hence, by Lemma 3.10, [W_a⁰:X]2 andXQ_a‡2G_b. Notethat, asW_a⁰is abelian, X acts quadratically on V_b, so j[X;V_b]j 2³ by Proposition 2.5. If either XˆWa⁰ or j[X;Vb]j 2² hold, then Theorem 17.3 follows. Thus [Wa⁰ :X]ˆ2 andj[X;Vb]j ˆ2³. Weclaim that [X;Vb]6ˆ[Vb;Gba‡2;2]. For if not then [Vb;Gba‡2;2]ˆ[X;Vb]Wa⁰and Proposition 2.5(ii) yields that W_a⁰C_Ga‡3([V_a‡3;G_a‡2a‡3;2])ˆG_a‡2a‡3, contradicting [W_a⁰:X]ˆ2. So, since[X;V_b](V_b\V_a‡3)[V_b;G_ba‡2] weobtain (v). IfjXQ_b=Q_bj 2, then [Wa⁰ :CW_a0(Za)]2³and again we have Theorem 17.3. Bearing in mind that Xacts quadratically onVb, consulting Proposition 2.5(ii) gives thatXQb=Qb

isGba‡2=Qb-conjugatetohs1;tiandZa‡2[X;Vb]. This proves (i), and (ii) and (iii), and (iv) follows from (iii).

THEOREM17.5. Assume that for all(d;d⁰)2 C,[Z_d‡1;W_d⁰]ˆ1. Then for each(a;a⁰)2 Cwe haveVa⁰ 6Qb.

PROOF. Le t (a;a⁰)2 Cbesuch thatV_a⁰Q_b. ThenV_a⁰ G_a,V_a⁰ 6Q_a and V_a⁰ interchanges l and m where D(a)ˆ fl;m;bg. Further, we have Z_bˆ[V_b;V_a⁰].

(17.5.1) (i)UaQa⁰ 2Ga⁰ 1. (ii) Ifb>7, thenUaGa⁰.

If there exists (a 2;a⁰ 2)2 C, then Lemma 17.4(iv) applied to (a 2;a⁰ 2) gives Z_aW_a⁰ ₂Q_a⁰, as b>5. Therefore U_aQ_a⁰ ₂

Ga⁰ 1. Next we verify part (ii). Lemma 17.4(i) yields the existence of (a⁰‡3;a‡3)2 Cwithd(a⁰;a⁰‡3)ˆ3. Now using Lemma 17.4(iv) on this critical pair gives Za⁰‡1Wa‡3. Because b>7 by assumption [W_a‡3;U_a]ˆ1, and so [Z_a⁰_‡1;U_a]ˆ1. Hence U_a centralizes Z_a⁰ ₂Z_a⁰ˆ

ˆZ_a⁰ ₁and thusU_aQ_a⁰ ₁G_a⁰.

(17.5.2) If U_aG_a⁰, then [U_a;V_a⁰]E(2³) and Z_a⁰[U_a;V_a⁰]ˆ

ˆC_Va0(U_a)U_a.

SinceUaacts quadratically uponVa⁰,j[Ua;Va⁰]j 2³by Proposition 2.5.

On the other hand, Va⁰Ga, Va⁰ 6Qa and Lemma 17.1 imply that 2³ j[U_a;V_a⁰]j. HenceC_Va0(U_a)ˆ[U_a;V_a⁰]E(2³)withZ_a⁰[U_a;V_a⁰]U_a.

SetX_lˆW_l\G_a⁰ andX_mˆW_m\G_a⁰. (17.5.3) Ifb>7, th en[Wl:Xl]2.

First we show that W_lG_a⁰ ₃. If W_l6Q_a⁰ ₄, then we may find (a 4;a⁰ 4)2 C. Applying Lemma 17.4(iv) to (a 4;a⁰ 4) and using b>7givesZa 2Wa⁰ 4Qa⁰. But thenZlZa 2Qa⁰forcesZaQa⁰. SoWlQa⁰ 4Ga⁰ 3holds.

We now consider two following cases:- [U_a;V_a⁰ ₂]ˆ1 and [U_a;V_a⁰ ₂]6ˆ1. Suppose [U_a;V_a⁰ ₂]ˆ1. Then orders, (17.5.1)(ii) and (17.5.2) give Va⁰ 2\Va⁰ˆCV_a0(Ua)Ua. Hence, as b>5, Wl\Qa⁰ 3

commutes with Va⁰ 2\Va⁰ ˆ[Va⁰ 2;Ga⁰ 2a⁰ 1;2] whence we obtain Wl\Qa⁰ 3Ga⁰. So Wl\Qa⁰ 3Xl. Since a⁰ 32O(S3), we have shown that [W_l :X_l]2 when[U_a;V_a⁰ ₂]ˆ1. So now we examine the case[U_a;V_a⁰ ₂]6ˆ1. SinceV_a⁰centralizesV_a⁰ ₂and interchangeslandm, we have[Vl;Va⁰ 2]6ˆ1. IfVa⁰ 2Ql, th enZlˆ[Vl;Va⁰ 2]Va⁰ 2Qa⁰, against (a;a⁰)2 C. So Va⁰ 26Ql and therefore (x;l)2 C for some x2D(a⁰ 2). By Lemma 17.4(iv)Za⁰ 3Wl. BecauseWl in abelian this gives W_lQ_a⁰ ₃ and then using Z_a⁰ U_a (by (17.5.1)(ii) and (17.5.2)) together with the parabolic argument we conclude that [W_l:X_l]2.

This verifies (17.5.3).

Our next objective is to establisha weaker version of (17.5.3) when bˆ7.

(17.5.4) Supposebˆ7andZ_b6ˆZ_a‡5. Then[W_l :X_l]2.

By Proposition 2.5(ii) Zbˆ

Vb;Va⁰

Vb\Va‡3\Va‡5\Va⁰():

We divide the proof of (17.5.4) into the two following cases:-ZbZa‡4and Zb6Za‡4. Beginning with the former, the assumptionZb6ˆZa‡5and() imply that

Z_a‡4ˆZ_bZ_a‡5V_a‡3\V_a‡5\V_a⁰:

Also, Z_bZ_a‡4 yields Z_a‡2ˆZ_bZ_a‡3ˆZ_a‡4 and hence hG_a‡2a‡3;G_a‡3a‡4i 6ˆG_a‡3. This, when combined with Lemma 17.4(i), yields the existence of (a⁰‡3;a‡3)2 Cfor whichVa⁰‡2Qa‡3=Qa‡3 is not a central transvection of Ga‡3a‡4=Qa‡3 on Va‡3=Za‡3. Therefore [Va⁰‡2;Va‡3]6Za‡4. Since [Va⁰‡2;Va‡3]Va‡3\Va‡5\Va⁰ (again by Proposition 2.5(ii)), we deduce that

E 2³

Za‡4

Va⁰‡2;Va‡3

Va‡3\Va‡5\Va⁰:

ConsequentlyV_a‡3\V_a‡5ˆV_a‡5\V_a⁰. Since [U_a;V_a‡3]ˆ1,[U_a;Z_a‡6]ˆ1 and soU_aG_a⁰by (17.5.1)(i). Hence, using (17.5.2),

Va‡5\Va⁰ ˆCV_a0(Ua)Ua:

Now [W_l;U_a]ˆ1 and Proposition 2.5(ii) (applied twice) force WlGa⁰, so proving (17.5.4) in this case. Now we assume thatZb6Za‡4. So, by()

Va‡3\Va‡5 ˆZa‡4ZbˆZa‡4Za‡2:

If Za‡5Za‡2, then we obtain Za‡2ˆZa‡4 whence Va‡3\Va‡5 ˆ

ˆZ_a‡4Z_a‡2ˆZ_a‡2, which is impossible. ThusZ_a‡56Z_a‡2and hence V_a‡3\V_a‡5ˆZ_a‡4Z_a‡2ˆZ_a‡5Z_a‡2():

We claim that [Ua;Va‡5]ˆ1. For suppose [Ua;Va‡5]6ˆ1. Then, by Proposition 2.5(ii) and (17.5.1)(i),

Za‡5ˆ

Ua;Va‡5

Vb\Va‡3\Va‡5:

Now Z_a‡2ˆZ_bZ_a‡3V_b\V_a‡3\V_a‡5 together with our earlier deduc- tion Z_a‡56Z_a‡2 forces V_b\V_a‡3\V_a‡5E(2³). But then V_b\V_a‡3 ˆ

ˆVa‡3\Va‡5 whence Proposition 2.5(ii) gives Wa⁰ CGa‡3(Vb\Va‡3) Ga‡2a‡3, contradicting Lemma 17.4(i). Hence[Ua;Va‡5]ˆ1, as claimed.

SoUaGa⁰ and therefore

Z_a‡6C_Va0(U_a)U_a

by (17.5.2). Since[Wl;Ua]ˆ1, we get[Wl;Za‡6]ˆ1which, by(), im- plies that W_l centralizes V_a‡3\V_a‡5. Consequently W_lG_a‡5. Since [W_l;Z_a⁰]ˆ1, employing the parabolic argument gives [W_l :X_l]2 and completes the proof of (17.5.4).

We now show that we do in fact have some critical pairs to which we may apply (17.5.4).

(17.5.5) If bˆ7, then there exists (d;d⁰)2 C suchthat V_d⁰ Qd‡1 and Zd‡16ˆZd‡5.

If Z_b6ˆZ_a‡5, then we may take (d;d⁰)ˆ(a;a⁰). So we may assume ZbˆZa‡5. Thus Za‡2ˆZa‡4 and therefore hGa‡2a‡3;Ga‡3a‡4i 6ˆGa‡3. Then, by Lemma 17.4(i), we may find (a⁰‡3;a‡3)2 C suchthat V_a⁰_‡2Q_a‡3=Q_a‡3is not a central transvection ofG_a‡3a‡4=Q_a‡3onV_a‡3=Z_a‡3. IfV_a‡36Q_a⁰_‡2, the there existsx2D(a‡3) suchthat(x;a⁰‡2)2 C. Ap- plying Lemma 17.4(v) to(x;a⁰‡2)we see that

V_a‡3;G_a‡3a‡4

ˆ

V_a‡3;W_a⁰_‡2\G_a‡4a‡5

V_a‡3\V_a‡5 :

But thenVa⁰‡2centralizes[Va‡3;Ga‡3a‡4], a contradiction. SoVa‡3Qa⁰‡2

and[V_a⁰_‡2;V_a‡3]ˆZ_a⁰_‡2. FurtherZ_a⁰_‡26ˆZ_a‡5, sinceV_a⁰_‡2Q_a‡3=Q_a‡3is not a central transvection onV_a‡3=Z_a‡3. Taking(d;d⁰)ˆ(a⁰‡3;a‡3)we see that (17.5.5) holds

X_l;V_a⁰ X_m;V_a⁰

U_a\G_a⁰: (17:5:6)

Observe thatb>5impliesUaZ(Ga[4]). So, since[Xl;Va⁰]Ga[4], Xl;Va⁰

CV_a0 Ua\Ga⁰ : SupposeU_aG_a⁰ holds. Then, using (17.5.2),

X_l;V_a⁰

C_Va0 U_a

U_a\G_a⁰:

Now we consider the possibility Ua6Ga⁰. From (17.5.1)(i) and Lem- ma 17.1, we get [U_a;Z_a⁰]6ˆ1 and j(U_a\G_a⁰)Q_a⁰=Q_a⁰j 2². Hence, as U_a\G_a⁰ acts quadratically onV_a⁰=Z_a⁰, Proposition 2.5(ii) gives

Ua\Ga⁰;Va⁰

Za⁰C_Va0 Ua\Ga⁰

Xl;Va⁰ : From[Ua;Za⁰]6ˆ1,Za⁰ 6G^[4]_a and so we deduce that

X_l;V_a⁰

ˆ

U_a\G_a⁰;V_a⁰

U_a\G_a⁰:

A similar argument proves that[X_m;V_a⁰]U_a\G_a⁰ and so (17.5.6) holds.

Wl:Wl\Wm 2 (17:5:7)

SinceVa⁰ interchangeslandm,Va⁰acts uponXlXmandXl\Xm. (Note thatXlandXmnormalize eachother sinceb5.) So, by (17.5.6),

X_lX_m;V_a⁰

ˆ

X_l;V_a⁰ X_m;V_a⁰

U_a\G_a⁰ X_l\X_m:

Letx2Va⁰be suchthatlxˆm. IfXl\Xm<

cXl, then[XlXm;x]6Xl\Xm

contrary to [XlXm;Va⁰]Xl\Xm. Therefore Xl\XmˆXl, and so XlˆXm. Ifb>7, then (17.5.7) follows from (17.5.3). Forbˆ7, the above argument in conjunction with(17.5.4), (17.5.5) and Lemma 11.1(vii) also yields[W_l:W_l\W_m]2.

Appealing to Lemma 11.1(vii), (17.5.7) implies that[W_a⁰:W_a⁰\W_a⁰ ₂] 2[Wa⁰ 2:Wa⁰ 2\Wa⁰ 4], whence [Wa⁰ :Wa⁰\Wa⁰ 4]2². Since [Za;Wa⁰ 4]ˆ1, this forcesh(Ga⁰;Wa⁰)2, against Lemma 17.2. From this contradictory state of affairs we conclude thatV_a⁰ 6Q_b.

We are now ready to begin the proof of

THEOREM17.6. There exists(d;d⁰)2 Csuch that[Z_d‡1;W_d⁰]6ˆ1.

PROOF. Arguing for a contradiction wesupposethat [Z_d‡1;W_d⁰]ˆ1 for all (d;d⁰)2 C.

(17.6.1) For(a;a⁰)2 C,[Vb;Va⁰]Za‡2\Za⁰ 1.

Lemma 17.4(v) gives [Vb;Gba‡2]ˆ[X;Vb](Vb\Va‡3) (where XˆWa⁰\Ga‡2a‡3). SinceWa⁰ is abelian and[X;Vb]Wa⁰, together with [Va⁰;Vb\Va‡3]ˆ1, we get that Va⁰ centralizes [Vb;Gba‡2]. Thus [V_b;V_a⁰]Z_a‡2. By Theorem 17.5 V_a⁰ 6Q_b and so there exists (a⁰‡1;b)2 C. Then a similar argument gives[V_b;V_a⁰]Z_a⁰ ₁.

(17.6.2) For(a;a⁰)2 Cwe have[W_a⁰;V_a‡3]ˆZ_a‡4.

It follows directly from (17.6.1) that [W_a⁰;V_a‡3]Z_a‡4. Hence jW_a⁰Q_a‡3=Q_a‡3j 2 and so, by Lemma 17.4(i), W_a⁰\Q_a‡3ˆX (ˆWa⁰\Ga‡2a‡3). If[X;Va‡3]ˆ1, thenXcentralizes[Vb;Gba‡2]by Lem- ma 17.4(v) which impliesjXQb=Qbj 2, contradicting Lemma 17.4(ii). So Z_a‡3ˆ[X;V_a‡3][W_a⁰;V_a‡3]. Since W_a⁰ 6Q_a‡3, [W_a⁰;V_a‡3]6ˆZ_a‡3 and we have (17.6.2).

Za‡4Va⁰: (17:6:3)

For l with l6ˆa⁰ 2 and d(l;a⁰)ˆ2 we have, as V_a‡3 centralizes V_a⁰\V_l ˆ[V_l;G_{l 1l};2], that [V_a‡3;V_l]V_a⁰\V_l. So, since W_a⁰ ˆ

ˆVa⁰ Q

d(l;a⁰)ˆ2Vl, we get, using (17.6.2), that Z_a‡4ˆ

W_a⁰;V_a‡3

ˆ Y

d(l;a0)ˆ2 l6ˆa0 2

V_l;V_a‡3 V_a⁰:

(17.6.4) Let (a;a⁰)2 C and set XˆWa⁰\Ga‡2a‡3. Then we have Za‡3ˆ[Vb;Va⁰]ˆZa⁰ 2ˆVa⁰\[X;Vb].

By Lemma 17.4(iii) and (17.6.3), Za‡3Va⁰\[X;Vb]. Also we clearly have[Va⁰;Vb]Va⁰\[X;Vb]. Now suppose thatjVa⁰\[X;Vb]j>2, and put Wa⁰ ˆWa⁰=Va⁰. Then, by Lemma 17.4(i), (iii), j[Vb;X]j 2[Wa⁰ : X].

Hence h(G_a⁰;W_a⁰)2 and thus h(G_a⁰;W_a⁰)ˆ3 by Lemma 17.2, with V_b acting as a transvection upon each of the non-central chief factors within W_a⁰. By (17.6.1)V_balso acts as a transvection onV_a⁰=Z_a⁰and thus the non- central chief factors in Wa⁰ are isomorphic natural modules, a contra- diction. Therefore we deduce that jVa⁰\[X;Vb]j ˆ2 and so Za‡3ˆ

ˆV_a⁰\[X;V_b]ˆ[V_b;V_a⁰]. SinceV_a⁰ 6Q_b by Theorem 17.5, a symmetric argument gives[V_a⁰;V_b]ˆZ_a⁰ ₂. This completes the proof of (17.6.4).

(17.6.5) For(a;a⁰)2 Cwe haveZa‡4ˆZa⁰ 1.

By (17.6.4) applied to (a;a⁰) we obtain Z_a‡3ˆZ_a⁰ ₂. From Lem- ma 17.4(i) there exists(a‡3;a⁰‡3)2 Cand using (17.6.4) on this critical pair givesZ_a⁰ ˆZ_a‡5. Therefore

Za‡4ˆZa‡3Za‡5ˆZa⁰ 2Za⁰ˆZa⁰ 1; as required.

b>9:

(17:6:6)

Suppose (17.6.6) is false. Thenbˆ7or 9. Let(a;a⁰)2 C. Just as in (17.6.5) we have Za‡5 ˆZa⁰ and this rules out bˆ7. Since Va⁰ 6Qb by Theo- rem 17.5, we have(l;b)2 Cfor somel2D(a⁰)and likewise we deduce that Z_bˆZ_a⁰ ₄. So, asbˆ9,Z_bˆZ_a‡5ˆZ_a⁰. Therefore

Za⁰ ˆZbVa⁰\ X;Vb

ˆZa‡3

by Lemma 17.4(iii) and (17.6.4), whereasZb6ˆZa‡3. Thus (17.6.6) holds.

Now we fix(a;a⁰)2 Cand letx2D(a⁰)n fa⁰ 1gbe suchthat (i) Z_x6ˆZ_a‡6; and

(ii) hV_b;G_a⁰_xi ˆG_a⁰.

By Proposition 2.8(viii) we may choose r2D(a⁰)n fa⁰ 1g suchthat hV_b;G_a⁰_ri ˆG_a⁰. SinceZ_r6/G_a⁰and[V_b;Z_a‡6]ˆ1clearlyZ_r6ˆZ_a‡6, so we may takexˆr.

PutRxˆGx[4]andXˆWa⁰\Ga‡2a‡3. R_xQ_a‡5:

(17:6:7)

IfRx6Qa‡5, then there exists r suchthat d(x;r)ˆ4,d(r;a‡5)ˆb and Zr6Qa‡5. So (r;a‡5)2 C. Applying (17.6.5) to (r;a‡5) yields Zx ˆZa‡6contrary to the choice ofx. ThusRxQa‡5.

(17.6.8) (i) Rxˆ(Rx\Ga‡2a‡3)Wa⁰; and (ii) (R_x\G_a‡2a‡3)Q_bˆXQ_b.

By (17.6.6) G_a^0[5] is abelian. SinceZ_a‡2W_a⁰ andZ_a‡4ˆZ_a⁰ ₁W_a⁰ by Lemma 17.4(iv) and (17.6.5), R_x commutes withbothZ_a‡2 and Z_a‡4. Hence, using (17.6.7),

R_xQ_a‡4G_a‡3 and

G_a‡2a‡3;R_x

6ˆG_a‡3:

Therefore [R_x:R_x\G_a‡2a‡3]2. Since W_a⁰ R_x and [W_a⁰:X]ˆ2 by Lemma 17.4(ii), we have (i). Further, note that R_x\G_a‡2a‡3 G_b. From [V_b;R_x\G_a‡2a‡3]G_a^0[5]we have thatR_x\G_a‡2a‡3acts quadratically on Vb. BecauseXRx\Ga‡2a‡3, Lemma 17.4(ii) implies(Rx\Ga‡2a‡3)Qbˆ

ˆXQb, which completes the proof of (17.6.8).

Since, by Lemma 17.4(iii),Zb[X;Vb], (17.6.8)(ii) implies that Vb;Rx\Ga‡2a‡3

ˆ Vb;X

Wa⁰: Consequently

V_b;R_x

ˆ

V_b; R_x\G_a‡2a‡3 W_a⁰

ˆ

Vb;Rx\Ga‡2a‡3

Vb;Wa⁰

Wa⁰ Rx; by (17.6.8)(i). Hence

R_x/

V_b;G_a⁰_x

ˆG_a⁰;

a contradiction which completes the proof of Theorem 17.6.

We bring the following two groups into the fray:- F_d:ˆ

U_d;Q_d Hl:ˆ

FdGl

whered2O(S3)andl2D(d). These groups will play a somewhat similar role to theFaandHbdefined in Section 12; note that our presentFd,Hland their counterparts in Section 12 are entirely different groups.

LEMMA17.7. Let(a;a⁰)2 C.

(i) h(Ga;Fa)ˆ2.

(ii) FaVb6/Gb. (iii) Hb[Wb;Qb]Vb.

PROOF. Sinceh(Ga;Ua=[UaQa])ˆ1, (i) follows from Lemma 17.1.

Suppose (ii) is false. From[Qb;FaVb]ˆ[Qb;Fa][Qb;Vb]ˆ[Qb;Fa]Zbˆ

ˆ[Qb;Fa], we then get [Qb;Fa]/ Gb. Since j[Qb;Fa]j 2⁴ and Zb[Qb;Fa], we deduce thatO²(Gb)centralizes[Qb;Fa]. If(Va 1\Vb)\

\[Q_b;F_a]>Z_b, then the uniseriality ofG_ab onV_b=Z_bgivesZ_a [Q_b;F_a], against Lemma 1.1(ii). Therefore (V_{a 1}\V_b)\[Q_b;F_a]ˆZ_b. Also, [F_a;Q_a]V_{a 1}\V_band thus

Q_a\Q_b;F_aV_b

ˆ

Q_a\Q_b;F_a

Q_a\Q_b;V_b

Qb;Fa

\ Va 1\Vb

ˆZb:

Now we may obtain a contradiction as in Lemma 12.5(ii). This proves (ii).

Turning to (iii) we first show that Va 1\[Wb;Qb]Vbc>Vb\Va 1. By [Proof of Theorem 1; LPR2] (Qa\Q_b)Q_{a 1}=Q_{a 1} is not contained in the quadratic E(2³)-subgroup of G_{a 1a}=Q_{a 1} (on V_{a 1}=Z_{a 1}), and so [V_{a 1};Q_a\Q_b]6V_{a 1}\V_b. Since [V_{a 1};Q_a\Q_b]V_{a 1}\[W_b;Q_b], we see thatVa 1\[Wb;Qb]Vb>

cVb\Va 1. Hence V_{a 1}

W_b;Q_b V_b=

W_b;Q_b V_bˆ2

and therefore[V_{a 1};Q_a][W_b;Q_b]V_b. Since this holds for alla 12D(a), F_aˆ[U_a;Q_a][W_b;Q_b]V_band thusH_b[W_b;Q_b]V_b.

With Theorem 17.6 and Lemma 17.7 to hand we now start the

PROOF OFTHEOREM17.3. We suppose the theorem is false and seek a contradiction. By Theorem 17.6 we may choose (a;a⁰)2 C suchthat [Z_b;W_a⁰]6ˆ1. In particular,Z_b6W_a⁰ and henceV_a⁰ 6Q_b. So there exists r2D(a⁰)suchthat(r;b)2 C.

Wb6Qa⁰ 2 andWa⁰ 6Qa‡3: (17:3:1)

Suppose WbQa⁰ 2 holds. By Lemma 17.4(i) applied to(r;b)we de- duce that [Za⁰;Wb]6ˆ1. In particular, Za⁰ 6Wb. So, since Wb\Ga⁰ acts quadratically upon Va⁰, j[Wb\Ga⁰;Va⁰]j 2². But [Wb:Wb\Ga⁰]2 whence h(G_b;W_b)ˆ3 with all non-central chief factors withinW_b being isomorphic natural modules. Thus we must have W_b6Q_a⁰ ₂. A similar argument shows thatW_a⁰ 6Q_a‡3.

From now until (17.3.9) we assume, in addition, thatb>7.

By (17.3.1) there exists(b 3;a⁰ 2)2 Cwithd(b;b 3)ˆ3. We dis- play the part ofGthat will be of interest to us (lhas yet to be introduced).

Zb6Wa⁰ 2: (17:3:2)

Because b>7,ZbWa⁰ 2 would imply [Zb;Wa⁰]ˆ1 contrary to the choice of(a;a⁰).

(17.3.3) (i) Wa⁰ 26Gbb 1.

(ii) [W_a⁰ ₂:W_a⁰ ₂\Q_b]2².

(i) Suppose Wa⁰ 2Gbb 1 holds. Since [Wa⁰ 2;Zb 1]Zb and, by (17.3.2), Z_b6W_a⁰ ₂, [W_a⁰ ₂;Z_{b 1}]ˆ1 and so W_a⁰ ₂Q_{b 1}G_{b 2}. Thus j[W_a⁰ ₂;V_{b 2}]j 2³ which implies h(G_a⁰ ₂;W_a⁰ ₂)ˆ3 withall non- central chief factors within Wa⁰ 2 being isomorphic natural modules, a contradiction. ThereforeWa⁰ 26Gbb 1.

(ii) Assume that [Wa⁰ 2:Wa⁰\Qb]2 holds. Since Zb6Wa⁰ 2, [W_a⁰ ₂\Q_b;V_b]ˆ1. So [W_a⁰\Q_b;Z_{b 1}]ˆ1 and thus W_a⁰ ₂\Q_b Q_{b 1}G_{b 2}. Since the theorem is supposed false we must have j[Wa⁰ 2\Qb;Vb 2]j ˆ2³. Also from[Wa⁰ 2\Qb;Vb]ˆ1we have

V_b\V_{b 2}C_{Vb 2} W_a⁰ ₂\Q_b :

Consequently, as W_a⁰ ₂\Q_b acts quadratically on V_{b 2} and not as a transvection onVb 2=Zb 2,

V_b\V_{b 2}ˆC_{Vb 2} W_a⁰ ₂\Q_b

ˆ

W_a⁰ ₂\Q_b;V_{b 2}

W_a⁰ ₂: Therefore [V_b\V_{b 2};W_a⁰ ₂]ˆ1 whence W_a⁰ ₂G_{bb 1} by Proposition 2.5(ii), contrary to part (i). So we have proved (ii).

(17.3.4) (i) V_b\V_a‡3ˆZ_b[W_a⁰ ₂;V_b]ˆC_Vb(W_a⁰ ₂); and (ii) j[W_a⁰ ₂;V_b]j ˆ2².

By (17.3.3)(ii)W_a⁰ ₂acts as at least a quadratic fours group onV_b=Z_b.

ThereforeVb\Va‡3ˆCVb(Wa⁰ 2)andVb\Va‡3ˆZb[Wa⁰ 2;Vb]. Because [Wa⁰ 2;Vb]Wa⁰ 2 (17.3.2) implies that j[Wa⁰ 2;Vb]j ˆ2², and we have (17.3.4).

Zb 2;Wa⁰ 2 6ˆ1:

(17:3:5)

If we have[Z_{b 2};W_a⁰ ₂]ˆ1, then Lemma 17.4(iv) applied to the critical pair(b 3;a⁰ 2) givesZb 1Wa⁰ 2. HenceZbZb 1Wa⁰ 2, contra- dicting (17.3.2). Therefore[Zb 2;Wa⁰ 2]6ˆ1.

By (17.3.5) we have (b 3;a⁰ 2)2 C with [Z_{b 2};W_a⁰ ₂]6ˆ1. So we may repeat the procedure that produced (b 3;a⁰ 2) from (a;a⁰), this time starting with (b 3;a⁰ 2) to obtain (l;a⁰ 4)2 C with d(l;b 2)ˆ3 and [Zb 2;Wa⁰ 2]6ˆ1. As a consequence all the results obtained for(b 3;a⁰ 2)also hold for (l;a⁰ 4). In particular,

(17.3.6) (i) Z_{b 2}6W_a⁰ ₄(analogue of (17.3.2)); and (ii) j[W_a⁰ ₄;V_{b 2}]j ˆ2² (analogue of (17.3.4)(ii)).

(17.3.7) [W_a⁰ ₄;V_{b 2}]V_a⁰ ₄.

This follows from the fact thatWa⁰ 4ˆVa⁰ 4 Q

d(m;a⁰ 4)ˆ2Vmand, for each m,[Vm;Vb 2]Va⁰ 4\Vm.

Wa⁰ 4;Vb 2

ˆ

Wa⁰ 2;Vb

Vb\Vb 2: (17:3:8)

From [Vb;Wa⁰ 4]ˆ1, [Vb\Vb 2;Wa⁰ 4]ˆ1 and so Wa⁰ 4Gb 2b 1

and [Wa⁰ 4;Vb 2]Vb\Vb 2Vb. Now b>7 gives [Wa⁰ 2;Wa⁰ 4]ˆ1

and hence

Wa⁰ 4;Vb 2

CVb Wa⁰ 2

ˆVb\Va‡3;

using (17.3.4)(i). If [Wa⁰ 4;Vb 2]6[Wa⁰ 2;Vb], then, as [Vb\Va‡3: [W_a⁰ ₂;V_b]]ˆ2by (17.3.4)(ii),

Vb\Va‡3ˆ

Wa⁰ 4;Vb 2

Wa⁰ 2;Vb : Then, using (17.3.7), we deduce that

Z_bZ_a‡2V_b\V_a‡3ˆ

W_a⁰ ₄;V_{b 2}

W_a⁰ ₂;V_b Va⁰ 4

Wa⁰ 2;Vb

Wa⁰ 2;

against (17.3.2). Therefore [Wa⁰ 4;Vb 2] [Wa⁰ 2;Vb] and hence [Wa⁰ 4;Vb 2]ˆ[Wa⁰ 2;Vb]by (17.3.4)(ii) and (17.3.6)(ii). This establishes (17.3.8).

We now unveil the desired contradiction. Combining (17.3.4)(i) and (17.3.8) gives

V_b\V_a‡3ˆZ_b

W_a⁰ ₂;V_b

Z_b V_{b 2}\V_b

V_{b 2}\V_b:

Thus V_b\V_a‡3ˆV_{b 2}\V_b by orders. But then W_a⁰ ₂ centralizes V_{b 2}\V_b whence W_a⁰ ₂G_{bb 1} by Proposition 2.5(ii), contradicting (17.3.3)(i). From this we conclude that

bˆ7:

(17:3:9)

Before making essential use of (17.3.9) we observe the following gen- eration result forW_b. Fort2D(b), putL(b;t)ˆ fg2D(b)jZ_g6[V_b;G_bt]g.

(17.3.10) Lett2D(b)andx2GbtnQb. If[Wb=Vb:C_Wb=Vb(x)]2³, then W_bˆ

U_gjg2L(b;t) U_t:

From Lemma 17.7H_b[W_b;Q_b]V_b. Also we note that [U_a;Q_b;Q_b] [U_a;Q_a]H_b and [U_a;Q_a;Q_b;Q_b]V_b. Thus W_b=[W_b;Q_b]V_b, [Wb;Qb]Vb=Hb, Hb=[Hb;Qb] and [Hb;Qb]Vb=Vb are all GF(2)(Gb=Qb)- modules. NowHb=[Hb;Qb]Vbis generated by an involution centralized by Gab and[Hb;Qb]Vb=Vbis either trivial or generated by an involution cen- tralized byG_ab. Sinceh(G_b;W_b=V_b)2, our assumption implies

Hb=Vb:C_Hb=Vb(x) 2²

and so Proposition 2.15 applies to both these sections. ThusH_b=[H_b;Q_b]V_b and[H_b;Q_b]V_b=V_bare bothquotients of 4

1 1. Proceeding as in Lem- ma 5.17 gives

H_bˆ Ug;Qg

jg2L(b;t) Ut;Qt The same arguments apply toWb=Hb, and so (17.3.10) holds.

We shall make repeated use, often without reference, of the following facts.

(17.3.11) Let(d;d⁰)2 C.

(i) [Vd‡1;V_d⁰]Vd‡1\Vd‡3\Vd‡5\V_d⁰. (ii) [V_d‡3;W_d⁰]V_d‡3\V_d‡5\V_d⁰.

(iii) V_d‡1\V_d‡36ˆV_d‡3\V_d‡5. Further, if V_d⁰ 6Q_d‡1, then V_d‡3\

\V_d‡56ˆV_d‡5\V_d⁰.

Part (i) is a consequence of[V_d‡1;V_d‡5]ˆ[V_d⁰;V_d‡3]ˆ1, and (ii) fol- lows from (i). IfV_d‡1\V_d‡3ˆV_d‡3\V_d‡5, then, using Proposition 2.5(ii),

W_d⁰ Gd‡1. Hence j[W_d⁰;Vd‡1]j 2³, a contradiction. Therefore Vd‡1\

\Vd‡36ˆVd‡3\Vd‡5 and (iii) holds.

Our next goal is (17.3.25), which asserts that jWa⁰Qa‡3=Qa‡3j ˆ2 for any(a;a⁰)2 Cfor which[Z_b;W_a⁰]6ˆ1.

The results (17.3.12), (17.3.13) and (17.3.14) prepare the ground for the proof of (17 3.25). For the duration of these results(d;d⁰)is assumed to be a critical pair for which [Zd‡1;W_d⁰]6ˆ1 and jW_d⁰Qd‡3=Qd‡3j 6ˆ2. (Recall from (17.3.1) thatW_d⁰ 6Qd‡3and sojW_d⁰Qd‡3=Qd‡3j>2.) Also recall that V_d⁰ 6Q_d‡1.

(17.3.12) (i) [W_d⁰;Vd‡3]ˆVd‡3\Vd‡5\V_d⁰ E(2²).

(ii)Z_d‡36V_d⁰.

(iii)[W_d⁰\Qd‡3;Vd‡3]ˆ1.

By assumptionjW_d⁰Q_d‡3=Q_d‡3j>2and thereforej[W_d⁰;V_d‡3=Z_d‡3]j 2². Thus parts (i) and (ii) follow from (17.3.11)(ii), (iii). Because

W_d⁰\Q_d‡3;V_d‡3

Z_d‡3\V_d⁰; (ii) implies (iii).

(17.3.13) V_d⁰Q_d‡1=Q_d‡1 is a non-central transvection of G_d‡1d‡2=Q_d‡1 (acting onV_d‡1=Z_d‡1) andj[V_d‡1;V_d⁰]j ˆ2.

First we demonstrate that j[V_d‡1;V_d⁰]j ˆ2. Put V ˆV_d‡1\V_d‡3\

\V_d‡5\V_d⁰. If j[V_d‡1;V_d⁰]j>2, then, by (17.3.11) (i), (iii), [V_d‡1;V_d⁰]ˆV. Th us, by (17.13.12)(ii), Z_d‡3V ˆV_d‡1\V_d‡3 and con- sequently W_d⁰ centralizes [Vd‡3=Zd‡3;Gd‡2d‡3;2]. Th is forces W_d⁰ Gd‡2d‡3, and so [W_d⁰ :W_d⁰\Gd‡1]2. Since [W_d⁰;Zd‡1]6ˆ1 and W_d⁰ acts quadratically on Vd‡1, we must h ave [W_d⁰\Gd‡1;Vd‡1]ˆ

ˆ[V_d‡1;V_d⁰]V_d⁰. But this gives the impossible h(G_d⁰;W_d⁰=V_d⁰)1. So j[V_d‡1;V_d⁰]j ˆ2. Observe thatV_d⁰Q_d‡1=Q_d‡1 being a central transvection of G_d‡1d‡2=Q_d‡1 on V_d‡1=Z_d‡1 gives [V_d‡1;V_d⁰]Z_d‡2. By (17.3.12)(ii) [Vd‡1;V_d⁰]6ˆZd‡3 and hence Zd‡2ˆ[Vd‡1;V_d⁰]Zd‡3, wh ich yields [Zd‡1;W_d⁰]ˆ1, whereas[Zd‡1;W_d⁰]6ˆ1. This proves (17.3.13).

From [V_d‡1;V_d⁰]6ˆZ_d‡3, [V_d‡1;V_d⁰]V_d‡1\V_d‡3\V_d‡5 and V_d‡1\

\V_d‡36ˆV_d‡3\V_d‡5, we have that(V_d‡1\V_d‡3)(V_d‡3\V_d‡5)E(2⁴). Let E_d‡3 be suchthat G_d‡3d‡4E_d‡3>Q_d‡3 and E_d‡3 centralizes (Vd‡1\Vd‡3)(Vd‡3\Vd‡5)=Zd‡3; note thatEd‡3 induces a transvection on Vd‡3=Zd‡3.

(17.3.14) (i) W_d⁰\E_d‡3G_d‡1and[W_d⁰\E_d‡3;V_d‡1]V_d‡1\V_d‡3. (ii) [W_d⁰ :W_d⁰\Ed‡3]2².