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

(S

, S

)-Amalgams VI.

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

Introduction.

In this, the penultimate part of [LPR1], we analyse Case 3 (as de- scribed in [Section 12; [LPR1]). Our main result, Theorem 14.1, asserts that Case 3 does not occur. Since core_GaVbis so small, the core argument is, at times, especially deadly. Consequently we often end up in situa- tions, where the subgroups we are interested in have trivial action on all the non-central G_d-chief factors inQ_d for many vertices d2O(S₃). This is especially unfortunate since, as Lemma 15.1 shows,U_d (ford2O(S₃)) possesses at least four non-central Gd-chief factors. In fact the proof of Theorem 14.1 hinges upon overcoming just this type of situation. A significant step in dealing with this problem is made in Lemmas 16.3 and 16.4. It follows from these lemmas that there exists a critical pair (a;a⁰) andr2D(a⁰)n fa⁰ 1g for whichF_a\Q_a⁰6Q_r. The groupF_g and its accomplice Hd (where g2O(S3) and d2O(S6)) are defined in Sec- tion 14. These groups are ``small'' enough so as they fix many vertices yet are ``large'' from the point of view of the non-central chief factors they contain. Before the groupsF_gandH_dcan be successfully deployed we need to restrict the structure of the critical pairs inG. This we do in Section 15. Section 14, apart from definingFg andHb, is concerned with eliminating the possibilitybˆ3.

Finally, as before we continue the section numbering in [LPR1] and note that this paper only refers to results and notation contained in sec- tions 1, 2 and 12.

(*) 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, Alan Turing Building, Manchester M13 9PL, United Kingdom.

14. bˆ3 and the groupsTdg.

Throughout this paper the following hypothesis is assumed to hold.

HYPOTHESIS 14.0. (i) For each (a;a⁰)2 C we have a2O(S₃) and [Za;Za⁰]ˆ1; and

(ii) core_GaV_bˆV_b\V_{a 1}ˆZ_a. Our objective is to show that

THEOREM14.1. Hypothesis 14.0cannot hold.

Before tackling the case bˆ3 we give some notation and define the groupsT_dg.

Ford2O(S₆) we set

L_d:ˆO² G_d

; Yd:ˆCVd Ld

and C_d:ˆC_Qd V_d

:

We recall from Theorem 12.1 that ford2O(S6),h(Gd;Vd)ˆ1 andVd=Zdis a quotient of 4

1 1. Letg2D(d). IfV_d=Z_d4, then we define Tdg :ˆ

Vd;Qg;Qg

;

otherwise we defineTdgto be a normal fours subgroup ofGdg with Z_d<T_dgY_d:

Next we describe two groups which play a crucial role later in this section and in Section 16.

F_g:ˆ

T_dg^Gg Hd:ˆ

FgGd

The groupsFg andHd are similar to theFaandHbdefined in Section 12.

(Indeed ifjYdj ˆ2², then they are the same.) We will need a result analo- gous to Lemma 12.5.

LEMMA14.2. Let(a;a⁰)2 C. Then

(i) h(G_a;F_a)ˆ2with Z_aˆ[F_a;Q_a]andjF_a=Z_aj 2 f2²;2³g; and (ii) V_bc^<V_bF_ac^<H_bandh(G_b;H_b=V_b)1.

PROOF. First we observe that if Vb=Zb4, thenZa Tba. While, if Vb=Zb64, thenZa\TbaˆZbby Lemma 1.1(ii). HenceTbaZaE(2³) in either case. Since V₁(Z(Q_a))ˆZ_a by Lemma 12.2(i) and V_bQ_a, 16ˆ[T_ba;Q_a]Z_a and hence [F_a;Q_a]ˆZ_a. Because Z_a ˆcore_GaV_b we clearly haveF_ac^>T_baZ_aand now (i) follows.

Since [V_b;Q_b]ˆZ_b, (i) implies thatV_bc^<V_bF_aH_b. Also, from (i) and Lemma 12.2(ii), 2² j[Fa;Qb]j 2³. Now suppose that h(Gb;Hb=Vb)ˆ0.

ThenHbˆVbFa. HenceKb:ˆ[Hb;Qb]ˆ[Vb;Qb][Fa;Qb]ˆ[Fa;Qb]/Gb. SincejKbj 2³, [Kb;Lb]ˆ1 and soZa\KbˆZb. Therefore [Fa;Qa\Qb] [F_a;Q_a]\K_bˆZ_a\K_bˆZ_b. Hence

[H_b;Q_a\Q_b]ˆ[V_b;Q_a\Q_b][F_a;Q_a\Q_b]Z_b

and so Q_a\Q_bC_Qb(H_b=Z_b). SincejK_bj 2², we deduce thatQ_a\Q_bˆ

ˆC_Qb(H_b=Z_b)/G_b, contradicting Lemma 12.4(i). This proves (ii).

And now to the main business of this section.

THEOREM14.3. b5.

We suppose the theorem is false and, in the following lemmas, seek a contradiction. So, by Lemma 11.1(iii),bˆ3.

For (a;a⁰)2 Cwe label vertices as indicated.

SoD(b⁰)ˆ fb;g;a⁰g.

LEMMA14.4. Let(a;a⁰)2 C.

(i) V_b=Z_b4or 4 1 . (ii) R:ˆ[Vb;Va⁰]ˆZg.

(iii) There existsr2D(a⁰)n fb⁰gsuch that(r;b)2 C.

(iv) ZaQa⁰=Qa⁰ ˆVbQa⁰=Qa⁰is the central transvection of G_b⁰_a⁰=Qa⁰on Va⁰=Za⁰and ZrQb=QbˆVa⁰Qb=Qbis the central transvection of G_bb⁰=Qbon V_b=Z_b.

(v) C_Vb(V_a⁰)ˆ[V_b;Q_b⁰]ˆV_b\Q_a⁰<₂V_band CV_a0(Vb)ˆ[Va⁰;Q_b⁰]ˆVa⁰\Qb<2 Va⁰:

(vi) There existsd2D(b)such that(d;a⁰)2 Cand G_db;V_a⁰

ˆG_b: PROOF. First we note that

16ˆ

Za;Va⁰

R:ˆ

Vb;Va⁰

Vb\Va⁰ ˆZ_b⁰:

Since R is G_b⁰_g-invariant, Z_gR whence V_a⁰ 6Q_b and so there exists r2D(a⁰)n fb⁰gsuch that (r;b)2 C. Also we haveZ_b⁰ ˆZ_bR[V_b;G_b] and therefore V_bˆ[V_b;G_b]. So (i) holds. Further, RY_a⁰=Y_a⁰ˆZ_b⁰Y_a⁰=Y_a⁰ˆ

ˆC_Va0=Ya0(G_b⁰_a⁰) together with Proposition 2.5(iii) yields that ZaQa⁰=Qa⁰ˆ

ˆVbQa⁰=Qa⁰ is the central transvection ofG_b⁰_a⁰=Qa⁰ onVa⁰=Za⁰proving the first part of (iv). The second part of (iv) follows similarly.

Because Q_b⁰Q_bˆG_bb⁰ and V_b\Q_a⁰ is Q_b⁰-invariant with, by (iv), Z_bV_b\Q_a⁰<₂ V_b we infer that V_b\Q_a⁰ ˆ[V_b;G_bb⁰]. Then, by (iv), [Vb\Qa⁰;Va⁰]Zb\Za⁰ ˆ1 and thusCVb(Va⁰)ˆVb\Qa⁰ ˆ[Vb;Q_b⁰] with a symmetric statement withbanda⁰interchanged, so we have (v). From (v) V_a⁰ acts upon V_b as an involution and a transvection and sojRj ˆ2. Thus RˆZ_g.

Since V_a⁰6Q_b, by Proposition 2.8(viii) we may choose d2D(b) such thathGdb;Va⁰i ˆGb. IfZd Qa⁰, thenZdVb\Qa⁰and so [Zd;Va⁰]ˆ1 by (iv). But thenZd/Gb, a contradiction. Hence (d;a⁰)2 C, as required.

In view of Lemma 14.4(vi) we may, and shall, assume that (a;a⁰) is a fixed critical pair for whichhGab;Va⁰i ˆGb. Also we setJbˆ h[Ua;Qa]^Gbi.

LEMMA14.5.

(i) [Ua;Qa]Z(Ua).

(ii) VbHbJbCbGa⁰ and Ha⁰ Ca⁰ Gb. (iii) [Fa;Va⁰\Qb]ˆZb.

PROOF. From Lemma 14.4(v) V_b\Q_a⁰ˆ[V_b;Q_b⁰]ˆC_Vb(V_a⁰), and Qb6Q_b⁰ by Lemma 12.2(ii). Therefore, since [Vb;Q_b⁰] is Qb-invariant we get Vb\Qa⁰ˆCVb(Vg) and thus Vb\Qa⁰ Z(U_b⁰). This in turn

implies that

Z(U_b⁰)

Vb;Q_b⁰_Gb0

ˆ

U_b⁰;Q_b⁰

; from which (i) follows.

Appealing to Lemma 14.4(i) and Proposition 2.5(i) givesT_ba[V_b;Q_a] and hence Fa[Ua;Qa]. Consequently Hbˆ hFaGbi h[Ua;Qa]^Gbi ˆJb. By part (i) [Ua;Qa]CbwhenceJbCb. EvidentlyCbQ_b⁰ Ga⁰ and it is also clear that we haveHa⁰ Ca⁰Gb.

Now we prove (iii). By Lemma 14.4(v) V_a⁰\Q_b centralizes V_b and so V_a⁰\Q_bQ_a. Hence [F_a;V_a⁰\Q_b][F_a;Q_a]ˆZ_a by Lemma 14.2(i).

Since FaGa⁰ by (ii), we then get [Fa;Va⁰\Qb]Za\Va⁰ ˆZb. If [Fa;Va⁰\Qb]ˆ1, then Fa centralizes a hyperplane of Va⁰ and so FaQa⁰ˆZaQa⁰ ˆVbQa⁰. Hence, using Lemma 14.4(ii),

Fa;Va⁰ Za⁰

Vb;Va⁰

ˆZa⁰Zg ˆZ_b⁰ Vb:

But thenVbFais normalized byhGab;Va⁰i ˆGbby the choice of the critical pair, against Lemma 14.2(ii). This shows that [Fa;Va⁰\Q_b]ˆZ_b.

LEMMA 14.6. V_b=Z_b4. In particular, T_baˆ[V_b;Q_a;Q_a] and Faˆ[Ua;Qa;Qa].

PROOF. Assume thatVb=Zb 4

1 , and setD(a)ˆ fl;m;bg. Then by definition T_baˆY_bE(2²) and F_aˆ hY_l;Y_m;Y_bi. Put XˆV_a⁰\Q_b. By Lemma 14.4(v) X centralizes Vb and so XQaGl. Therefore [X;Yl]Zl. Since [X;Fa]ˆZbby Lemma 14.5(iii), [X;Yl]Zl\Zbˆ1.

Likewise [X;Ym]ˆ1 and thus [X;Fa]ˆ1, contrary to [X;Fa]ˆZb. Hence Vb=Zb6 4

1 and now the lemma follows from Lemma 14.4(i) and the definition ofTbaandFa.

LEMMA14.7. PutGa⁰ ˆGa⁰=Qa⁰. Then (i) [Jb;Jb]Zb; and

(ii) JbˆCbis the non-quadratic E(2³)-subgroup ofG_b⁰_a⁰on Va⁰=Za⁰. PROOF. Put D(a)ˆ fl;m;bg, Ga⁰ˆGa⁰=Qa⁰ and XˆVa⁰\Qb. From Lemma 14.5(iii) [F_a;X]ˆZ_b. NowT_baV_band [V_b;X]ˆ1 means we can assume, without loss of generality, that

T_la;X

ˆZ_b6ˆZ_l: (14:7:1)

From this and Lemma 14.6 we immediately deduce

(14.7.2) X6Q_l and X does not centralize T_la=Z_lˆ[V_l;Q_a;Q_a]=Z_l; in particular,XQl=Qlis not contained in theE(2³)-subgroup ofGla=Qlacting quadratically onVl=Zl.

By Lemmas 14.4(v) and 14.5(ii) we have JbCbQaGl and J_b[U_b⁰;Q_b⁰][V_a⁰;Q_b⁰]ˆX. Combining these observations with (14.7.2) and Lemma 11.1(vii) yields

(14.7.3) J_b is not contained in the E(2³)-subgroup of G_b⁰_a0 acting quad- ratically onVa⁰=Za⁰.

Since G_b⁰_a0 is a 2-group, F(C_b)ˆF(C_b)F(G_b⁰_a0). By Lemma 14.4(v) F(G_b⁰_a⁰)\Vbˆ1 and so F(Cb)\VbQa⁰. Because Vb=Zb is a Gb-chief factor we deduce that

F Cb

\VbZb

(14:7:4)

ClearlyF(C_b)\F(C_a⁰) is normalized byG_b⁰_g. IfF(C_b)\F(C_a⁰)6ˆ1, then F(Cb)\F(Ca⁰)V1(Z(G_b⁰_g))ˆZg and henceZg F(Cb)\Vb which con- tradicts (14.7.4). Therefore we have

F C_b

;F C_a⁰

ˆF C_b

\F C_a⁰

ˆ1:

(14:7:5)

Next, we assume thath(G_b;F(C_b))6ˆ0. Thenh(G_a⁰;F(C_a⁰))1 and thus F(C_b)Q_a⁰ by (14.7.5). Hence [F(C_b);V_a⁰]Z_a⁰. SinceV_a⁰ 6Q_b by Lem- ma 14.4(iii), we then get [F(Cb);Va⁰]ˆZa⁰ whenceZa⁰ F(Cb)\Vb, again contradicting (14.7.4). So we have shown that

h G_b;F C_b

ˆ0:

(14:7:6)

Now we suppose that F(Cb)6Qa⁰. Then F(Cb)ˆZ(G_b⁰_a⁰)\O²(Ga⁰)

E(2). Thus

Za⁰ F Cb

;Va⁰

ˆ

Va⁰;Q_b⁰;Q_b⁰ :

Clearly,Kb:ˆF(Cb)Vb/Gbwith, by (14.7.6),h(Gb;Kb=Vb)ˆ0. Therefore Vb[F(Cb);Va⁰]ˆVb[Va⁰;Q_b⁰;Q_b⁰] is normalized by Lb and hence by Q:ˆ[L_b;Q_b]. Appealing to Lemma 12.4(i) we conclude that

Vb

Va⁰;Q_b⁰;Q_b⁰

ˆVb

Vg;Q_b⁰;Q_b⁰

ˆVb

U_b⁰;Q_b⁰;Q_b⁰

ˆV_bF_b⁰/

L_b;G_bb⁰

ˆG_b:

ThusH_bˆV_bF_b⁰ ˆV_bF_awhich is impossible by Lemma 14.2(ii), and so F C_b

Q_a⁰: (14:7:7)

Now part (ii) is a direct consequence of Lemma 14.5(ii), (14.7.3), (14.7.7) and the fact thatJbandCbare normal subgroups ofQ_b⁰ˆG_b⁰_a⁰containing the central transvectionZa.

From part (ii) we infer that [J_b;[V_a⁰;Q_b⁰]]Z_b⁰ and hence [J_b;[U_b⁰;Q_b⁰]]Z_b⁰. Since J_bˆ h[U_b⁰;Q_b⁰]^Gbi, we then obtain

Jb;Jb

ˆ Jb;

U_b⁰Q_b⁰_Gb

Z_b^0Gb

ˆVb:

Noting that [Jb;Jb][Cb;Cb]F(Cb) an application of (14.7.4) yields that [J_b;J_b]Z_b, so completing the proof of the lemma.

LEMMA 14.8. Put V_bˆV_b[H_b;Q_b]. Then the following statements hold:

(i) h(G_b;C_b=V_b)ˆh(G_b;H_b=V_b)ˆ1and the only non-central G_b- chief factor within C_b=V_bis not isomorphic to V_b=Z_b, as a G_b=Q_b-module;

(ii) J_bˆH_bˆ[H_b;G_b]and H_b=V_b4or 4 1 ; and

(iii) V_bˆV_b[F_b⁰;Q_b] and V_b\C_a⁰ ˆ[V_b;Q_b⁰][F_b⁰;Q_b] with [Vb:Vb]2and[Vb :Vb\Ca⁰]ˆ2.

PROOF. BecauseC_a⁰G_b Ca⁰;Jb;Jb

Jb;Jb

ZbVa⁰; by Lemma 14.7(i). Therefore

(14.8.1) J_bacts quadratically onC_a⁰=V_a⁰and so (by Lemma 14.7(ii)) the non-centralG_a⁰-chief factors withinC_a⁰=V_a⁰are not isomorphic toV_a⁰=Z_a⁰as G_a⁰=Q_a⁰-modules.

Using Lemma 14.7(ii) we see that E 2³

Cb;Va⁰]

Va⁰;Q_b⁰

ˆVa⁰\QbE 2⁴ :

Now E(2²)Z_b⁰ V_a⁰\Q_b and so j[C_b=V_b;V_a⁰]j 2². Then Lemmas 14.2(ii), 14.4(iv) and (14.8.1) force

h G_b;C_b=V_b

ˆ1:

(14:8:2)

NowsetH_bˆ[H_b;G_b],and note thath(G_b;H_b)ˆ2 andH_b V_b.Also, by (14.8.1),j[H_b=V_b;V_a⁰]j ˆ2² and thus [V_a⁰;Q_b⁰]ˆV_a⁰\Q_bH_b. Em- ploying Lemma 11.1(vii) gives H_b[U_b⁰;Q_b⁰]. Since J_bˆ h[U_b⁰;Q_b⁰]^Gbi HbHbwe obtain

J_bˆH_bˆ H_b;G_b

: (14:8:3)

Ifh(Gb;Vb=Vb)6ˆ0, thenh(Gb;Hb=Vb)ˆ0 and soHbˆVb[Hb;Qb]Fa

which, commutating sufficiently often with Qb, forces the untenable H_bˆV_bFa. Thus h(G_b;V_b)ˆh(G_b;V_b)ˆh(G_b;H_b=V_b)ˆ1. Now [F_a:(F_a\V_b)[F_a;Q_b]]2 means that we have jF_aV_b=V_bj ˆ2 with H_b=V_bˆ h(F_aV_b=V_b)^Gbi. SinceH_bˆ[H_b;G_b] andh(G_b;H_b=V_b)ˆ1 we easily see thatH_b=V_b4 or 4

1 .

Fromh(G_b;V_b=V_b)ˆ0, we clearly haveV_bˆV_b[F_a;Q_b]ˆV_b[F_b⁰;Q_b] with [Vb:Vb]2,

V_b\C_a⁰ˆ V_b\C_a⁰ F_b⁰;Q_b

ˆ

V_b;Q_b⁰ F_b⁰;Q_b and [V_b:V_b\C_a⁰]ˆ2. This completes the proof of the lemma.

We are now in a position to complete the proof of Theorem 14.3. Using Lemma 14.6 we see that

Qb=Cb Vb=Zb

Vb=Zb: (14:3:1)

So we have (Q_b⁰\Qa⁰)=Ca⁰ E(2³). Since, by Lemmas 14.7(ii) and 14.8(ii), HbQa⁰=Qa⁰ is the non-quadratic E(2³)-subgroup of G_b⁰_a⁰=Qa⁰ on Va⁰=Za⁰, (14.3.1) implies

(14.3.2) [(Q_b⁰\Qa⁰)=Ca⁰;Hb]E(2); in particular [Hb\Qa⁰:Hb\Ca⁰]2 and therefore [H_b:H_b\C_a⁰]2⁴.

We now observe from Lemma 14.8(iii) thatVb\Ha⁰ˆVb\Ca⁰ with [Vb :Vb\Ha⁰]ˆ2. Combining Lemmas 14.4(iii), (vi), 14.7(ii) and 14.8(ii) with Proposition 2.5(ii) gives

Hb=Vb:

Hb=Vb;Ha⁰

ˆ2²; and consequently

H_b:H_b\H_a⁰

H_b:

H_b;H_a⁰

V_b\H_a⁰ 2³:

Since Hb\Ha⁰Hb\Ca⁰, this clearly contradicts (14.3.2), so proving Theorem 14.3.

15. The structure of critical pairs.

The main result in this section is Theorem 15.7.

From Theorem 14.3 we have b5, and soUa is elementary abelian.

Our first result shows that there is an abundance of non-central chief factors inUa.

LEMMA15.1. Let(a;a⁰)2 C.

(i) h(G_a;U_a)4.

(ii) If h(Ga;Ua)ˆ4, then [Ua;Qa;3]ˆ[Vb;Qa;3]ˆZa and Vb=Zb4 or 4

1 .

PROOF. PutVb(i) ˆ[Vb;Qa;i] andU_a⁽ⁱ⁾ˆ[Ua;Qa;i] fori2N[ f0g; let n2N be such that Vb(n)6ˆ1 and Vb(n‡1) ˆ1. By Lemma 12.2(i) V₁(Z(Q_a))ˆZ_a and sinceV_b⁽ⁿ⁾ is a G_ab-invariant subgroup ofV₁(Z(Q_a)), Theorem 12.1 and Proposition 2.5(i) imply

n3 andZbVb(n)ZaˆUa(n): (15:1:1)

Now suppose thath(G_a;U_a^(j)=U_a^(j‡1))ˆ0 for somej2 f0;1;. . .;n 1g.

Then Ua(j)ˆUa(j‡1)Vb(j) and hence Ua(j‡1)ˆUa(j‡2)Vb(j‡1). So Ua(j) ˆ

ˆUa(j‡2)Vb(j)and consequentlyUa(j)ˆVb(j)core_GaVbˆZa. This implies U_a^{(n 1)} Z_a whenceU_a⁽ⁿ⁾ ˆ1, a contradiction. Thus we have

h G_a;U_a^(j)=U_a^(j‡1)

1 forj2 f0;1;. . .;n 1g:

(15:1:2)

Clearly (15.1.1) and (15.1.2) give (i). Now assume thath(G_a;U_a)ˆ4. Then, by (15.1.1) and (15.1.2), Z_bV_b⁽³⁾U_a⁽³⁾ˆZ_a. Since V_b⁽³⁾6Z_b, V_b⁽³⁾ˆU_a⁽³⁾ˆZ_awhich establishes (ii).

LEMMA15.2. For(a;a⁰)2 C, Va⁰ 6Ga; in particular Va⁰6Qb.

PROOF. Let (a;a⁰)2 C, putD(a)ˆ fl;m;bgand assume by way of con- tradiction thatV_a⁰ G_a. SinceZ_aG_a⁰andZ_a 6Q_a⁰, we have

V_a⁰G_ab with [Z_a;V_a⁰]ˆZ_b6ˆZ_a⁰: (15:2:1)

Note that there existsr2D(a⁰) such thatZ_r6Q_a; moreoverZ_racts as an involution onUa.

Ua6Ga⁰; in particular Ua6Qa⁰ 1: (15:2:2)

Suppose thatU_aG_a⁰holds. Then, sinceU_ais abelian andV_a⁰ G_a,U_a acts quadratically on V_a⁰. Since h(G_a;U_a)4 by Lemma 15.1(i) and [Ua\Qa⁰:Ua\Qa⁰\Qr]2, we see that UaQa⁰=Qa⁰E(2³) with h(Ga;Ua)ˆ4. Also, from j[Ua;Va⁰]j 2⁴, Lemma 15.1(ii) and Proposition 2.5(ii) forceVa⁰=Za⁰  4

1 with [Ua;Va⁰]E(2⁴).

Since Z_r interchanges l and m, Z_r normalizes V :ˆV_lV_m. Since Vl\VmˆZaandVa⁰ E(2⁶),V :ˆV=ZaˆVlVmwithVlE(2⁴)Vm

and hence [V;Zr]E(2⁴). Now [Za;Zr]ˆZbandV Uayields E 2⁵

V;Z_r

U_a;V_a⁰

; contrary to [Ua;Va⁰]E(2⁴). So we have (15.2.2).

Becauseb5 andV_l

ZrV_m, [V_l;V_a⁰ ₂]ˆ[V_m;V_a⁰ ₂]V_l\V_mˆZ_aand so [Ua;V_a⁰ ₂]Za. From (15.2.2) and [Ua;V_a⁰ ₂]C_Za(Va⁰)ˆZ_bwe obtain

U_a;V_a⁰ ₂

ˆZ_b: (15:2:3)

U_a6Q_a⁰ ₂ (15:2:4)

Assume U_aQ_a⁰ ₂ holds. Since U_a6Q_a⁰ ₁ by (15.2.2) and Z_a⁰ ₁ˆ

ˆZ_a⁰ ₂Z_a⁰, [U_a;Z_a⁰]6ˆ1. Hence [U_a\Q_a⁰ ₁\Q_a⁰;V_a⁰]ˆ1 and so U_a\Q_a⁰ ₁\Q_a⁰C_Ua(Z_r). Now [U_a:U_a\Q_a⁰ ₁]2 and h(G_a;U_a)4 forces (Ua\Qa⁰ 1)Qa⁰=Qa⁰to be the quadraticE(2³)-subgroup ofGa⁰ 1a⁰=Qa⁰

onVa⁰=Za⁰and [Ua :Ua\Qa⁰ 1]ˆ2. Also we note that [Ua\Qa⁰ 1;Va⁰]Za⁰

has index 2² in Va⁰ and that [Ua\Qa⁰ 1;Va⁰]Ua. So Ua centralizes [U_a\Q_a⁰ ₁;V_a⁰] and U_a6Q_a⁰ ₁ whence, using the core argument, [U_a\Q_a⁰ ₁;V_a⁰]core_Ga0 1V_a⁰ ˆZ_a⁰ ₁. But this is impossible since (Ua\Qa⁰ 1)Qa⁰=Qa⁰ in the quadratic E(2³)-subgroup of Ga⁰ 1a⁰=Qa⁰ on Va⁰=Za⁰. ThereforeUa 6Qa⁰ 2.

(15.2.5) [Z_a;V_a⁰]ˆ[U_a;V_a⁰ ₂]ˆZ_bV_a⁰ ₂\V_a⁰ˆZ_a⁰ ₁andZ_a⁰ ₁ˆZ_a⁰ ₂Z_b. By (15.2.3) and (15.2.4) Z_b6ˆZ_a⁰ ₂. Now (15.2.5) follows from (15.2.1) and (15.2.3).

Put Ve_a⁰ ₂ˆV_a⁰ ₂=Y_a⁰ ₂ and P_a⁰ ₂ˆ hU_a;G_a⁰ _2a⁰ ₁i. Since Z_a⁰ ₁ˆ

ˆZ_a⁰ ₂Z_b, P_a⁰ ₂ centralizes Z_a⁰ ₁ and, as U_a6G_a⁰ _2a⁰ ₁ by (15.2.2), P_a⁰ ₂:ˆP_a⁰ ₂=Q_a⁰ ₂S₄Z₂ withVe_a⁰ ₂jPa0 2 1

21 0

@ 1

A. However, by (15.2.5), [U_a;Ve_a⁰ ₂]ˆZe_bˆZe_a⁰ ₁ˆCe_Va0 2(P_a⁰ ₂) which implies U_aO₂(P_a⁰ ₂) Ga⁰ 2a⁰ 1. This contradiction completes the proof of Lemma 15.2.

LEMMA 15.3. Let (a;a⁰)2 C and Rˆ[Vb;Va⁰]. If h(Gb;Wb)ˆ2 or h(Gb;[Wb;Qb]Vb=Vb)ˆ0, then the following hold:

(i) RZa‡2\Za⁰ 1 with RZbˆZa‡2 and RZa⁰ˆZa⁰ 1. If, more- over, bˆ5, then RˆZa‡3ˆZa⁰ 2;

(ii) Vbˆ[Vb;Gb]and so Vb=Zb4or 4 1 ;

(iii) VbQa⁰=Qa⁰ is the central transvection of Ga⁰ 1a⁰=Qa⁰ acting on

Va⁰=Za⁰with a symmetric statement holding for Va⁰with the roles ofband a⁰interchanged; and

(iv) either W_bQ_a⁰ ₂ or W_bQ_a⁰ ₂=Q_a⁰ ₂is the central transvection of G_a⁰ _3a⁰ ₂=Q_a⁰ ₂on V_a⁰ ₂=Z_a⁰ ₂, and so[W_b:W_b\Q_a⁰ ₂]2.

PROOF. Put W_b⁽¹⁾ˆ[W_b;Q_b] and W_bˆW_b⁽¹⁾V_b. Note that h(Gb;Wb=Wb)6ˆ0 else WbˆWb(1)Ua which produces the absurd WbˆUa. So we haveh(Gb;Wb=Vb)ˆ0. Appealing to Lemma 12.7 gives RVa⁰ 2\Va⁰ ˆZa⁰ 1. Since Va⁰ 6Qb by Lemma 15.2, there exists r2D(a⁰) such that (r;b)2 C, and likewise we deduceRV_b\V_a‡3ˆZ_a‡2. This proves (i)

Since Z_a‡2ˆRZ_b[V_b;G_b], the statements in (ii) and (iii) follow readily while (iv) is an easy consequence of (iii).

LEMMA15.4. Let(a;a⁰)2 C. If Z_a⁰V_b, thenjV_a⁰Q_b=Q_bj ˆ2.

PROOF. Let (a;a⁰)2 C be such that Z_a⁰ V_b and assume that jVa⁰Qb=Qbj 6ˆ2. So, by Lemma 15.2, jVa⁰Qb=Qbj 2². Also Lemma 15.3 implies that

h Gb;Wb 3:

(15:4:1)

Clearly we haveW_bG_a⁰ ₂. Sinceb5,Z_a⁰V_bZ(W_b) and soW_b centralizes Za⁰ 1ˆZa⁰ 2Za⁰. Thus WbCG_a0 2(Za⁰ 1Ya⁰ 2=Ya⁰ 2) and thereforeX:ˆWb\Ga⁰ 2a⁰ 1 has index at most 2 inWb by the parabolic argument (Lemma 3.10). Note that XC_Ga0 1(Za⁰ 1)ˆQa⁰ 1Ga⁰, and setXˆ(X\Q_a⁰)V_b. Then [X:X]2³and

X;Va⁰

ˆ

X\Qa⁰;Va⁰

RZa⁰RVb:

Hence, puttingWebˆWb=Vb, we haveXeCW~b(Va⁰). Using (15.4.1) we then get

2⁴ eWb:CW~b(Va⁰)

eWb:Xe

ˆ

Wb:X

ˆ

W_b:X X:X

2⁴ and consequently [W_b:X]ˆ2 and [X:X]ˆ2³. In particular

G_a⁰ _1a⁰ˆQ_a⁰X:

(15:4:2)

Since [X;Va⁰]X\Va⁰ Qb\Va⁰ and [Va⁰ :Qb\Va⁰]2², (15.4.2) together with Proposition 2.5(i) implies that

(15:4:3) X;V_a⁰

ˆX\V_a⁰ˆQ_b\V_a⁰<₄V_a⁰ withV_a⁰=Z_a⁰ 14 or 1 4 1

It now follows from (15.4.3) that Za⁰ 1ˆZa⁰ 2Za⁰ Qb\Va⁰ˆ

ˆ[X;Va⁰][Va⁰;Ga⁰] whenceVa⁰ˆ[Va⁰;Ga⁰], which contradicts the struc- ture ofVa⁰ given in (15.4.3). Thus we infer thatjVa⁰Q_b=Q_bj ˆ2 must hold.

COROLLARY15.5. If(a;a⁰)2CandjV_bQ_a⁰=Q_a⁰j2², thenjV_a⁰Q_b=Q_bj 2². PROOF. Suppose we have (a;a⁰)2 C with jVbQa⁰=Qa⁰j 2². Then Zb6Va⁰ by Lemmas 15.2 and 15.4. Hence [Va⁰\Qb;Vb]ˆ1. Since Vb

cannot centralize a hyperplane ofV_a⁰, [V_a⁰:V_a⁰\Q_b]2²which proves the result.

LEMMA 15.6. Let (a;a⁰)2 C and suppose that jVbQa⁰=Qa⁰j 2², and hence by Corollary15.5 jVa⁰Qb=Qbj 2². Then the following statements hold.

(i) VbQa⁰=Qa⁰Va⁰Qb=QbE(2²);

(ii) W_b\G_a⁰ ₁G_a⁰;

(iii) P_a⁰ ₂:ˆ hW_b;Q_a⁰ ₁i G_a⁰ ₂ ˆQ_a⁰ ₂P_a⁰ ₂ and P_a⁰ ₂L_a⁰ ₂; (iv) bˆ5; and

(v) if Gba⁰ˆGa⁰=Qa⁰, then Wba⁰ 2E(2³) and Wa⁰ 2 acts quad- ratically on Va⁰=Za⁰ with

4

1 V_a⁰=Z_a⁰ 4

1 1:

PROOF. PutRˆ[Vb;Va⁰] andGba⁰ˆGa⁰=Qa⁰. By Lemmas 15.2 and 15.4 Za⁰ 6VbandZb6Va⁰. Hence [Vb;Ya⁰]ˆ[Va⁰;Yb]ˆ1. Since [Va⁰:RYa⁰]ˆ

ˆ[V_b:RY_b]ˆ2²we obtain (i). Noting thatR6Z_a⁰ ₁and [W_b;V_b]ˆ1, the core argument yields (ii).

Lemma 15.3 (iii) gives h Gb;Wb

3:

(15:6:1)

VbQa⁰=Qa⁰ Wb\Ga⁰

Qa⁰=Qa⁰E 2³ : (15:6:2)

SinceW_b\G_a⁰centralizesRY_a⁰=Y_a⁰ E(2²), Proposition 2.5(vi) implies that (Wb\Ga⁰)Qa⁰=Qa⁰ is elementary abelian, so giving (15.6.2).

We now establish part (iii). Suppose that Q_a⁰ ₂P_a⁰ ₂6ˆG_a⁰ ₂. Then hWb;Ga⁰ 2a⁰ 1i 6ˆGa⁰ 2 and so, by the parabolic argument, [Wb:W]2 where W:ˆWb\Ga⁰ 2a⁰ 1. By (ii) WGa⁰, and by (15.6.2) [W:V_b(W\Q_a⁰)]2. Clearly [V_b(W\Q_a⁰);V_a⁰]RZ_a⁰. Putting W_bˆ

ˆW_b=V_b we have [W_b:V_b(W\Q_a⁰)]2² and j[V_b(W\Q_a⁰);V_a⁰]j 2.

Since Va⁰Qb=QbE(2²), this gives h(Gb;Wb)1, contradicting (15.6.1).

This proves part (iii).

As a direct consequence of (iii), Proposition 2.5(ii) and Lemmas 12.7 and 15.2 we have

(15.6.3) (i)Za⁰Ris not normal inGa⁰ 1a⁰andZbRis not normal inGba‡2. (ii) Either

(a) 4

V_a⁰=Z_a⁰

41 andVb_b hs₁;tior (b) 4

1 V_a⁰=Z_a⁰ 4

1 1 and Vb_b hs₁;ti or hs₂;ti (using the notation of Proposition 2.5).

Moving onto part (iv), we now assumeb>5 and derive a contradiction.

SetW₀ˆW_b\Q_a⁰ ₂. By (ii)W₀G_a⁰. SinceW_bis abelian we observe that [W_b:W₀]2³[W₀:W₀\Q_a⁰]. NowW_bbeing abelian and (iii) imply that Z_a⁰6W_b. Hence [W₀\Q_a⁰;V_a⁰]ˆ1, and so [W_b:C_Wb(V_a⁰)]2⁶. Therefore, by (15.6.1),h(Gb;Wb)ˆ3, [Wb:CWb(Va⁰)]ˆ2⁶andW0Qa⁰=Qa⁰E(2³).

(15.6.4) The non central G_b chief factors inW_b are isomorphic natural S₆-modules.

Observe thatVa⁰acts quadratically uponWbsince [Wb;Va⁰]Wa⁰ 2. By (15.6.3)(i)Va⁰Qb=Qb6Z(Gba‡2=Qb) from which (15.6.4) follows.

Since W0Qa⁰=Qa⁰E(2³) withW0 acting quadratically on Va⁰, we may chooset2W0 so thatVbhtiQa⁰ ˆW0Qa⁰ withtacting as a transvection on Va⁰=Ya⁰. Hence t acts as a transvection on Va⁰=Za⁰ and, recalling that Z_a⁰6W_b, we see thatC:ˆC_Va0(t) has index 2 in V_a⁰. ThereforeC6Q_b. Now

[W₀;C]ˆ

V_bhti W₀\Q_a⁰

;C

ˆ V_b;C

V_b

and consequently [Wb=Vb:C_Wb=Vb(C)]2³. Since h(Gb;Wb=Vb)ˆ2, C must induce a transvection on at least one non-central Gb-chief factor withinW_b=V_b. Appealing to (15.6.4) gives thatCinduces a transvection on V_b=Y_band hence onV_b=Z_b. BecauseZ_b6V_a⁰,Cinduces a transvection on Vb. But then [W0;C]ˆ[Vb;C]Z2 whereas [W0;C=Ya⁰]E(2²). From this untenable situation we deduce thatbˆ5.

Finally we consider part (v). By (15.6.3)Vb_b6/Gb_a⁰ _1a⁰ˆQb_a⁰ ₁. So, since V_bW_a⁰ ₂ (as bˆ5) and Wb_a⁰ ₂/Qb_a⁰ ₁, Ve_bc^<Wb_a⁰ ₂. Evidently we have [V_b;W_a⁰ ₂]V_b and consulting (15.6.3)(ii) we see thatVb_b\(Gb_a⁰ _1a⁰)⁰ˆ1.