On Classifying All Full Factorisations and Multiple-Factorisations of the Finite Almost Simple Groups

2 6 : A7 ,

A8 ,

A7 .

FACTORISATIONS OF ALMOST SIMPLE GROUPS

159

Clearly if L is isomorphic to one of 2 6 : A 8 , 2 6 : A 7 , then L is determined uniquely by its isomorphism type up to 2 6 : A 8-conjugacy; if L is isomorphic to A 7 or A 8 , then L is a complement to 2 6 in 2 6 : A 7 , 2 6 : A 8 , respectively, and a standard argument Žsee Ex. 6 of Chapter 6 of w1x. shows that there is a unique 2 6 : A 8-conjugacy class of such subgroups if L ( A 7 , but that there are two such if L ( A 8 . Observe that all possibilities have orders divisible by all prime divisors of < T < and so the condition p ŽT . : p Ž A. l p Ž B . is automatically satisfied by any overgroups A, B. Observe also that all of the above listed possibilities for A l T and B l T are contained in a unique maximal subgroup of T ; thus given the above remarks, each such possibility is uniquely determined up to T-conjugacy by its isomorphism type and the isomorphism type of a maximal subgroup of T containing it, with the exception of A 8 contained in 2 6 : A 8 , for which there are two T-conjugacy classes. This information must be taken into account when calculating the entries in the Ž l, m.-column of Table I, and when verifying the final sentence of note Žii. for Table I. Rather than now considering what overgroups exist, we instead reduce to the case G s T by the following argument. The information on maximal subgroups of almost simple groups with socle T given by w8, p. 85x shows that for any choice of H, if L is a subgroup of Aut T satisfying X : p Ž L.

and

L l T F H,

then L F NAut T Ž H . ( H.2. Hence A F NG Ž H . and by a symmetrical argument, B F NG Ž K .. Now G s AB implies that GrT s Ž ATrT . Ž BTrT . F Out T s S3 . As ATrT, BTrT are either trivial or of order two, we see that GrT is also either trivial or of order two and that, on interchanging A and B if necessary, we have ATrT F BTrT. Recall that the normalizers NAut T Ž H . and NAut T Ž K . induce distinct subgroups of Out T. It follows that ATrT is trivial and A F T. Thus T s T l Ž AB . s A Ž T l B . whence T s Ž A l T .Ž B l T . as required, and more importantly, the tuple ŽT, G s T, A l T, B l T . satisfies the hypotheses of Theorem 1.1. ŽNote that we have already verified in the previous paragraph that p ŽT . : p Ž A l T . l p Ž B l T .. We deduce that the tuples ŽT, G, A, B . satisfying the hypotheses of Theorem 1.1 with G ( T.2 can be obtained from those with G s T by replacing precisely one of A and B by its normalizer in Aut T and G by AB. Thus it remains only to verify that one of lines 6]10 hold for G s T.

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We assume from now on that G s T whence A F H and B F K. Hence A, B actually appear in the above lists of subgroups of H and K. We consider four cases distinguished by H and K corresponding to lines 5]8 of Table VI. In the following the notation W Ž H ., W Ž K . denote subgroups of H, K, respectively, isomorphic to W. Case 6Ža.. H ( K ( PSp6 Ž2.. Here H and K are non-conjugate subgroups of T both isomorphic to PSp6 Ž2.. By inspecting the above list of suitable subgroups of PSp6 Ž2. we see that, on interchanging A and B if necessary either A F S8Ž H . and B F S8Ž K ., or A s H and B G AŽ7K ., for suitable choices of S8Ž H ., S8Ž K ., and AŽ7K .. In the former case the 3-part of < AB < is at most the 3-part of Ž8!. 2 , namely 3 4 , whereas < T < 3 s 3 5. Hence A s H and B G AŽ7K .. It is sufficient to show that T s HAŽ7K .. As T s HK, we have T s HAŽ7K . if and only if K s Ž H l K . AŽ7K .. Now by w16, 3.1.1Žvi.x H l K ( G 2 Ž2. and is a maximal subgroup of H. Now L2 of TAB. 2 with q s 2 and « s q shows that K factorises as K s S8Ž K . Ž H l K ., whence K s Ž H l K . AŽ7K . if and only if S8Ž K . s Ž H l S8Ž K . . AŽ7K ., which in turn holds if and only if H l S8Ž K . is a transitive subgroup of S8Ž K . in its natural action on 8 points and is not contained in AŽ8K .. Furthermore L2 of TAB. 2 shows that H l S8Ž K . is a maximal subgroup of S8Ž K . ; by orders and by inspecting a list of maximal subgroups of S8Ž K . we see that it is L2 Ž7.:2 acting on one-dimensional subspaces of a 2-dimensional F 7-vector space. In particular, it is transitive and not contained in AŽ8K .. Case 6Žb.. H ( PSp6 Ž2. and K ( 2 6 : A 8 . In this case we need to perform some easy calculations involving the natural module for T. To this end, we view T as the group of all 8 = 8 invertible matrices with entries in F 2 acting on the F 2-vector space V s F 28 by right multiplication and preserving the quadratic form Q: V ª F 2 with associated symmetric bilinear form Ž , .: V = V ª F 2 given by Q Ž ei . s Q Ž fi . s Ž ei , e j . s Ž fi , f j . s 0

and Ž e i , f j . s d i j for all i , j s 1, . . . , 4,

where  e1 , . . . , e4 , f 1 , . . . , f 4 4 is the standard basis Ž1, 0, . . . , 0., . . . ,Ž0, . . . , 0, 1.4 for V. Up to Aut T-conjugacy we may assume that H, K are the set-wise stabilizers T² e1qf 1 : , T² e1 , e 2 , e 3 , e 4 : in T of the subspaces ² e1 q f 1 :, ² e1 , e2 , e3 , e4 :, respectively. For any square matrix a we use aT to denote the transpose of a. Explicitly we find that Ks

1 b

0 1

½ž / ž

( 2 6 : A8 .

a 0

0 y1 T

Ža .

/

:

a g GL4 Ž 2 . , b g M4 Ž 2 . , bT s b, and bii s 0 for i , . . . , 4

5

161

FACTORISATIONS OF ALMOST SIMPLE GROUPS

By inspecting the above lists of suitable subgroups of H ( PSp6 Ž2. and K ( 2 6 : A 8 we see that either A F S8Ž H ., or A s H and B G AŽ7K ., for suitable choices of S8Ž H . and AŽ7K .. In fact by conjugating if necessary by some element of K we may assume that AŽ7K . is contained in Ls

½ž

a

0 y1 T

Ža .

0

/

5

: a g GL4 Ž 2 . ( L4 Ž 2 . ( A 8 .

Ž 3.E.

To verify line 7 of Table I it is thus sufficient to show that T / S8Ž H . K and that T s HAŽ7K .. The former inequality holds if and only if H / S8Ž H . Ž H l K .; as T s HK we have < H l K < s 2 6 .168, and it follows from the information given by w8, p. 46x that H l K is a maximal subgroup of H isomorphic to 2 6 : L3 Ž2. and is uniquely determined up to H-conjugacy}moreover, the permutation characters given there show that H does not factorise as S8Ž H . Ž H l K .. To see that T s HAŽ7K . we start by noting that H l L s L² e1qf 1 : . An easy calculation shows that for a g GL4Ž2. we have

ž

a 0

0 y1 T

Ža .

/

gHlL

as

if and only if

ž

1 0

0 w

/

for some w g GL3 Ž2.. Thus H l L ( L3 Ž2. and is invariant under the outer automorphism of L ( GL4 Ž2. induced by x ¬ Ž xy1 .T. In A 8 there are three A 8-conjugacy classes of subgroups isomorphic to L3 Ž2.; namely two intransitive classes interchanged by an outer automorphism of A 8 , and one transitive class invariant under an outer automorphism. It follows that H l L lies in the transitive class, whence L s AŽ7K . Ž H l L. and we have < HAŽ7K . < s H Ž H l L . AŽ7K . s < HL < s

< H < < L< < H l L<

s
by direct calculation. Thus T s HAŽ7K . as required. Case 6Žc.. H ( PSp6 Ž2. and K ( A 9 . We start by noting that T s HK and that the information given on L8 of TAB. 4 on this factorisation gives that H l K is a maximal subgroup of both H and K. Observe that as T s HK we have < H: H l K < s 960 and < K: H l K < s 120. We split into three cases: Ži. A s H, Žii. B s K, and Žiii. A - H and B - K. Firstly we suppose that A s H. To verify lines 8 and 9 of Table I in this instance we must show that T s HAŽ7K .. By w8, p. 36x we see that H l K is

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BADDELEY AND PRAEGER

uniquely determined up to K-conjugacy by its index; in fact, H l K ( L2 Ž8.:3 and the embedding H l K ¨ A 9 ( K is best understood as that induced by the action on one-dimensional subspaces of a 2-dimensional F 8-vector space. It is straightforward to verify that H l K is 2-transitive in the natural action of K on 9 points, and so K factorises as Ž H l K . AŽ7K . whence HAŽ7K . s H Ž H l K . AŽ7K . s HK s T as required. Now suppose that B s K. Here we must show that T s AK if and only if A G AŽ8H . ; equivalently, that T s AŽ8H . K but that T / S7Ž H . K. By w8, p. 46x we see that H l K is uniquely determined up to H-conjugacy by its index; moreover, the permutation characters given there show that H s S8Ž H . Ž H l K .. It follows that < K l S8Ž H . < s 42 and by Sylow’s theorems K l S8Ž H . can be identified as the normalizer in S8Ž K . of a cyclic subgroup of order 7. Hence K l S8Ž H . contains an odd permutation but is not transitive in the natural action of S8Ž H . on 8 points. We deduce that S8Ž H . factorises as Ž K l S8Ž H . . AŽ8H ., but not as Ž K l S8Ž H . . S7Ž H .. This completes the analysis of this case since for any subgroup L F S8Ž H . of H we have T s LK if and only if S8Ž H . s LŽ K l S8Ž H . .. Finally suppose that A - H and B - K. We must show that T / AB. Now both < A < and < B < divide 8! whence < AB < has 3-part at most 3 4 . As < T < 3 s 3 5 we have T / AB as required. Case 6Žd.. H ( A 9 and K ( 2 6 : A 8 . As T s HK we have H l K s 2 .3.7 and by inspecting the lists of maximal subgroups of A 8 and A 9 given in w8x we see that H l K, in the natural action of H ( A 9 on 9 points, is intransitive Žand comprises all affine transformations of a 3-dimensional F 2-vector space.. Now if A is AŽ8H ., S7Ž H ., or AŽ7H . then A is also intransitive and by w21, Theorem Dx we have AŽ H l K . / H, whence AK / T. So A s H. It remains only to show that T s H Ž2 6 : A 7 .Ž K . s HAŽ8K . and that T / HAŽ7K .. We view T as a group of matrices in the same way as in Case 6Žb. above we also assume once again that K s T² e1 , e 2 , e 3 , e 4 : . Let M be the normal subgroup of K isomorphic to 2 6 . The information given by w8, p. 85x shows that all non-trivial elements of M belong to T-conjugacy classes labelled either 2A, 2C, or 2D. ŽIn fact we cannot have elements from both 2C and 2D but in this situation we cannot distinguish which of these classes we have.. By analysing the permutation character given in w8x for the action of T on the coset space of H we find that H contains no elements from T-conjugacy classes 2A, 2C, or 2D. ŽRecall that T s HK whence there are 6

FACTORISATIONS OF ALMOST SIMPLE GROUPS

163

only two choices up to T-conjugacy for H given K.. Thus H l M is trivial and the quotient map K ª KrM ( A 8 restricts to give a monomorphism H l K ª A 8 . In the previous paragraph we noted that H l K ( 2 3 : L3 Ž2. and on inspecting a list of maximal subgroups of A 8 we see that the image of H l K in A 8 is transitive of degree 8. Hence K s Ž H l K .Ž2 6 : A 7 .Ž K . and so H Ž2 6 : A7 .

ŽK.

s H Ž H l K . Ž2 6 : A7 .

ŽK.

s HK s T

as required. We now verify that T s HAŽ8K .. Observe that M is precisely the subgroup of K fixing the maximal isotropic subspace ² e1 , e2 , e3 , e4 : elementwise. Thus w8x not only tells us that non-trivial elements of M belong to T-conjugacy classes 2A, 2C, or 2D, but moreover, that if x g T _  14 fixes any maximal isotropic subspace element-wise, then x belongs to one of the classes 2A, 2C, or 2D. Up to K-conjugacy we may assume that AŽ8K . s L where L is as defined by Ž3.E.. Let x, y g L be given by xs

ž

a o

0

ys

Ža .

/

0 1 0 0

0 0 , 0 1

y1 T

,

ž

b o

0

Žb .

/

0 1 0 0

0 0 . 0 1

y1 T

where 1 1 as 0 0



0 0 1 0

1 1 bs 0 0

0 

0 0 1 1

0

An easy calculation shows that the subspaces Fix V Ž x ., Fix V Ž y . of V fixed element-wise by x, y, respectively, are given by Fix V Ž x . s ² e1 , e3 , e4 , f 2 , f 3 , f 4 :

and

Fix V Ž y . s ² e1 , e3 , f 2 , f 4 : .

Hence both x and y fix element-wise the maximal isotropic subspace ² e1 , e3 , f 2 , f 4 :, and so belong to T-conjugacy classes 2A, 2C, or 2D. Note also that x and y cannot be T-conjugate as their fixed-point spaces have different dimensions. Since L ( A 8 has precisely two conjugacy classes of order 2 we conclude that all elements of order 2 in L belong to one of the classes 2A, 2C, 2D, which implies that H l L has odd order since, as noted above, H contains no elements from such classes. Recall that H l K ( 2 6 : L3 Ž2. whence < H l L < F 21. Conversely, Sylow’s theorem implies that any subgroup of K ( 2 6 : A 8 of order 7 is conjugate to one of the form ² z : with z g L of order 7. It is straightforward to see that NK Ž² z :.

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BADDELEY AND PRAEGER

is a Frobenius group of order 21 and is contained in L. We again apply Sylow’s theorem to deduce that any group of order 21 has a normal subgroup of order 7, whence any subgroup of K of order 21 is K-conjugate to NK Ž² z :. F L. So, by replacing H if necessary by some K-conjugate of H, we have < H l L < G 21. Equality follows, and we see that < HL < s

< H < < L< < H l L<

s

9!8! 4.21

s
which gives the required factorisation T s HAŽ8K . of T. Finally we show that HAŽ7K . / T. Now given that T s HAŽ8K ., the latter holds if and only if AŽ8K . / Ž H l AŽ8K . . AŽ7K . where AŽ7K . is chosen to be a subgroup of AŽ8K .. In the previous paragraph we saw that < H l AŽ8K . < s 21 which means that H l AŽ8K . is an intransitive subgroup of AŽ8K . in its action on 8 points, and so AŽ8K . / Ž H l AŽ8K . . AŽ7K .. The proof of Theorem 1.1 is now finished.

4. PROOF OF THEOREM 1.2 Throughout this section we suppose that G is almost simple with socle T, and that M is a multiple-factorisation of G. We split the proof of Theorem 1.2 into the following five cases: Ž1. Ž2. Ž3. Ž4. Ž5.

M M M M M

maximal and < M < s 3; maximal and < M < s 4; maximal and < M < G 5; strong and < M < s 3; strong and < M < G 4.

Case Ž1.. M maximal and < M < s 3. We write M s  A, B, C 4 ; thus A, B, C are maximal subgroups of G not containing T and satisfying G s AB s BC s AC. We must show that the pair Ž G, M . is Aut T-equivalent to a pair listed either by Table II or by Table III. ŽRecall from note Ži. for Tables II]V that Ž G 0 ,  A 0 , B0 , C0 4. is Aut T-equivalent to Ž G, M . if and only if there exist s g Aut T and x, y, z g G such that G s G 0s

and

 A, B, C 4 s  As0 x , B0s y , C0s z 4 ..

Conversely, we must also show that the implications of each line of Tables II and III are valid, i.e., that the information so given is correct and that

FACTORISATIONS OF ALMOST SIMPLE GROUPS

165

each pair Ž G, M . so specified does indeed correspond to a maximal multiple-factorisation of G. If T ( A n and G s A n or Sn for some n G 5, then we may assume that B and C are either both transitive or both intransitive Žin the natural action of G on n points.. Given that G s BC, Corollary 5 on p. 9 w21x implies that n s 6, that both B and C are transitive, and that, up to T-conjugacy and interchanging B and C if necessary, B s PGL2 Ž 5 . l G

C s Ž S3 X S2 . l G.

and

ŽHere PGL2 Ž5. is viewed as a subgroup of S6 via its action on one-dimensional subspaces of a 2-dimensional F5-vector space, and S3 X S2 is viewed as a subgroup of S6 by identification with the stabilizer in S6 of a partition of  1, . . . , 64 into 2 equal parts.. If A is also a transitive subgroup, then by applying w21, Corollary 5x to the factorisation G s AB we have that A and C are G-conjugate, whence G / AC}a contradiction. Hence A is intransitive. Choose 1 F k F 3 such that A F Ž Sk = S6yk . l G. It is clear that C is not 2-homogenous and so given that G s AC, w21, Corollary 5x implies that k s 1. Thus A is unique up to T-conjugacy. It follows that line 1 of Table II holds. Conversely, it is clear that all implications of line 1 of Table II are valid. If T s A 6 and G g S6 , then by w8, p. 4x there are precisely three G-conjugacy classes of maximal subgroups of G. As no pair of A, B, C are G-conjugate, we may assume that A, B, C have orders dividing 2 3 .5,

2 4 .3 2 ,

25,

respectively. But then < T < does not divide < B < < C < }a contradiction. So we may assume that T is not an alternating group. All maximal factorisations of T are listed in TABS. 1]6 of w21x. We proceed by considering in turn T exceptional of Lie type, T sporadic, T classical and qŽ . Ž . not isomorphic to PVq 8 q , and lastly T ( PV 8 q . T Exceptional of Lie Type All factorisations of G are listed in TAB. 5 of w21x. By inspection it is clear that given T, there are at most two G-conjugacy classes of maximal factorisations of G and so no examples arise. T Sporadic All factorisations of G are listed in TAB. 6 of w21x. As G must have at least three G-conjugacy classes of maximal factorisations for which there is no factor common to all of them, we see that T is isomorphic to one of M12 ,

M24 ,

or

Co1 .

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Suppose first that T s M12 . As all maximal factorisations not involving M11 , share a common factor, namely L2 Ž11., we may assume that A l T s M11 , whence G s T and A s M11 by the maximality of A in G. Note that A here is uniquely determined only up to Aut T-conjugacy, and not up to T-conjugacy. Suppose that B is also isomorphic to M11 . As A, B are not G-conjugate, we have B s A a for some a g Aut T _ Inn T, in which case  A, B4 is unique up to T-conjugacy. Now the factorisation G s AB corresponds to a L3 of TAB. 6, and C satisfies G s AC s A C. Inspection of the permutaw x tion characters given by 8, p. 33 together with L4]9 of TAB. 6, shows that one of lines 2]5 of Table II holds. Conversely, it is clear that the implications of these lines are valid. We may now assume that neither B nor C is isomorphic to M11 . Hence the factorisation G s BC corresponds to L10 or to L11 of TAB. 6. Both involve L2 Ž11. and so we may assume that B s L2 Ž11., which is uniquely determined up to T-conjugacy. This forces C ( M10 :2 or M9 :S3 . By inspecting the permutation characters given by w8, p. 33x we see that G s AB and that for either choice of isomorphism type of C, we have G s BC but that G s AC if and only if C is chosen from one of two possible T-conjugacy classes Žthat is unique given A.. We deduce that lines 6 and 7 of Table II hold, but as there are two choices for A up to T-conjugacy, we see that each line corresponds to two T-equivalence classes of examples. This observation is borne out by direct calculation of X; in each case X s T. Conversely, it is clear that the implications of both of these lines are valid. We now suppose that T s M24 . Inspection of w8, p. 96x shows that any maximal subgroup of G is uniquely determined up to G-conjugacy by its isomorphism type. Now the maximal factorisations of G are given by L16]23 of TAB. 6. As all involve either M23 or L 2 Ž23. we may assume that A s M23 and B s L2 Ž23.. But then C satisfies G s AC s BC and by inspection no such C exists. Finally we suppose that T s Co1. Inspection of w8, p. 183x shows that any maximal subgroup of G is uniquely determined up to T-conjugacy by its isomorphism type. Now the maximal factorisations of G are given by L35]39 of TAB. 6. As all involve either Co 2 or Co 3 we may assume that A s Co 2 and B s Co 3 . But then G / AB and so no examples arise. Ž . T Classical and T \ PVq 8 q All factorisations of G are listed in TABS. 1]3 of w21x. We deal first with the case T s Sp4 Ž q . with q even. As T alternating has already been considered we may suppose that q G 4. Now Out T involves a graph automorphism and this means that care must be taken in interpreting the

FACTORISATIONS OF ALMOST SIMPLE GROUPS

167

results of w21x. We start by observing that G F P GSp4 Ž q ., as if not, then G possesses no maximal factorisations Žcf. w21, 5.1.7Žb.x. Now by Theorem 5.6 of w15x Žparts of which are originally due to Mitchell w22x and Flesner w10x. if H is a maximal subgroup of G then H l T is one of the following, where each possibility corresponds to a unique T-conjugacy class, but where the first six become Aut T-conjugate in pairs as indicated Žand Pi is the stabilizer of an isotropic i-dimensional subspace, for i s 1, 2.:

½

P1 P2

½

O4y Ž q . PSp 2 Ž q 2 . .2

Sp4 Ž q0 . q s q0c

½

O4q Ž q . Sp 2 Ž q . X S2

Sz Ž q . q non-square.

Recall the concept of primitive prime divisors as introduced in Section 3. We see that the primitive prime divisor q4 divides < G < and so must divide the orders of at least two of A, B, C since < G < divides each of < A < < B <, < B < < C <, < A < < C <. As the only possibilities with order divisible by q4 correspond to O4yŽ q ., PSp 2 Ž q 2 ..2, or Sz Ž q ., and as the first two of these are AutT-conjugate, we may assume up to Aut T-conjugacy that A l T s PSp 2 Ž q 2 ..2 and that B l T s O4yŽ q . or Sz Ž q .. If the latter then G / AB and so the former holds. Observe that the factorisation G s AB is given by L6 of TAB. 1 and that the pair  A, B4 is in fact uniquely determined up to T-conjugacy. We now consider the possibilities for C in turn. Although the factorisation G s AP1 is possible ŽL5 of TAB. 1., on inspecting w21, 5.1.7b.x, we see that G s BP1 is not; hence C l T / P1. By applying a graph automorphism we also see that C l T / P2 . Now C l T / O4yŽ q . nor PSp 2 Ž q 2 ..2 as C is G-conjugate to neither A nor B. If C l T s O4qŽ q . or Sp 2 Ž q . X S2 then by L6 and 11 of TAB. 1, G s AC s BC if and only if q s 4 and G s P GSp4 Ž q . contains all field automorphisms. In this case C is unique up to Aut T-conjugacy, but there are two choices up to G-conjugacy, and line 9 of Table II holds Žfor m s 2.. If C l T s Sp4Ž q0 . then C is unique up to T-conjugacy, and by L13]14 of TAB. 1, G s AC s BC if and only if q s q02 s 4 or 16, and G s P GSp4Ž q . contains all field automorphisms. In each case we have a single T-equivalence class of examples corresponding to lines 10 and 11 of Table II. This leaves only C l T s Sz Ž q . to consider: here G / AC and so we have no examples. Conversely, it is clear from the above that the implications of lines 9]11 of Table II Žfor m s 2. are all valid. We now assume that T / PSp4 Ž q . with q even, and start by supposing that all of the maximal factorisations G s AB s BC s AC are listed in TAB. 1 of w21x. It is straightforward to see that no such examples arise if T

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is linear, unitary, or orthogonal, and so we suppose that T s PSp 2 mŽ q . with m G 2 and either q odd or m / 2. If q is odd then only L5 of TAB. 1 applies and no examples arise; thus q is even and m ) 2. In this case all subgroups of T appearing in TAB. 1 correspond to a unique T-conjugacy class. Thus for fixed q, each of the subgroups P1 , Pm , N2 , Spm Ž q . X S2 , Sp 2 m Ž q 1r2 . is involved in at most one maximal factorisation listed in TAB. 1; it follows that none of the factorisations given by L5, 7]9, 12]14 of TAB. 1 correspond to any of G s AB s BC s AC. We are left with only L6, 10, and 11 of TAB. 1 to consider. Inspection shows that all possible examples correspond to either line 8 or line 9 of Table II, and that all implications of these lines are valid. Secondly we suppose that all of the maximal factorisations G s AB s BC s AC are listed in TABS. 1]2, with at least one from TAB. 2. In fact, as the factorisations of TAB. 1 involve only geometric subgroups, while those in TAB. 2 involve at least one non-geometric subgroup, we may assume that A is not geometric and that both G s AB and G s AC are factorisations listed in TAB. 2. Given that by assumption T / PSp4Ž q . with q even, we see that either T s PSp6 Ž q . with q even,

A l T s G2 Ž q . ,

and

 B l T , C l T 4 :  O6y Ž q . , O6q Ž q . , P1 , N2 4 , or T s V 7 Ž q . with q odd,

A l T s G2 Ž q . ,

 B l T , C l T 4 :  P1 ,

N1q,

N1y,

N2q,

N2y

and

4.

The factorisation G s BC must then be listed in TAB. 1. Inspection shows that examples only arise when the former holds, and that such examples correspond to either line 12 or line 13 of Table II; moreover we see that all implications of lines 12 and 13 of Table II are valid. Finally we suppose that at least one of the factorisations G s AB s BC s AC is listed in Table 3 of w21x. Thus T is highly restricted and inspection shows that no examples arise unless T s PV 7 Ž3.. Here the information given by w8, p. 109x can be used to show both that any such example is Aut T-equivalent to one given either by line 14 or by line 15 of Table II; and also that the implications of these lines are valid. Ž . T ( PVq 8 q All factorisations of G are listed in TAB. 4 of w21x. Recall that in note Žiii. for Tables II and III, we defined various elements and subgroups of

FACTORISATIONS OF ALMOST SIMPLE GROUPS

169

Aut T for the sake of precision. We use the notation set up there in what follows. In particular, we gave a set of generators  r, rw , d , a , f 4 for Out T and noted that Out T is isomorphic to S3 = e if q is even, and to S4 = e if q is odd. To assist in some of the calculations which follow, particularly in those which determine X Žwhere X is defined as in Note Živ. on Tables II]V., we would like to make this isomorphism explicit by noting that f can be taken as the generator of the cyclic subgroup of order e, and that the other generators can be identified as rw ¬ Ž 12 . ,

a ¬ Ž 123 . ,

r¬

½

Ž 34 . Ž 12 .

d¬

½

Ž 13 . Ž 24 . id

if q odd; if q even.

We now start by assuming that both A l T and B l T are Aut T-conjugate to N1 , which we may identify as the stabilizer in T of the subspace ² ¨ : where ¨ s e4 q f 4 . Thus up to Aut T-conjugacy we can assume A l T s Ž N1 . a. Given that G s AB, L1 of TAB. 4 implies that B s At for some triality t . As t is Out T-conjugate to a , we can take t s a whence 2 B l T s Ž N1 . a . The maximality of A and B in G implies that G F NAut T Ž A l T . l NAut T Ž B l T . . Now NAut T Ž N1 . s ² r , rw , f : and it follows that G F F s ²T, f :. To see what possible subgroups C can arise we consider which factorisations listed in w21x involve any Aut Tconjugate of N1. Such factorisations are precisely those given by L1]6, 8, 11, 12, and 16 of TAB. 4; from these we obtain the examples given by lines 1]10 of Table III. In all but one such case the information given in w21x together with w8x is sufficient to make the determination of the possible G and the calculation of X straightforward. The exceptional case is that corresponding to L12 of TAB. 4. Here q s 3. Lemma C of 5.1.15 of w21x Ž . demonstrates that T s HK for some Aut T-conjugates H, K of N1 , Vq 8 2 , respectively; we need to decide precisely which conjugates. Let u be any non-singular vector in V, and let ru be the reflection in u, i.e., ru : x ¬ x y

Ž x, u . u QŽ u.

for all x g V .

Let H s NAut T ŽT² u: . and let K be the normalizer in Aut T of any Ž . Aut T-conjugate of Vq 8 2 .

170

BADDELEY AND PRAEGER

Our claim is that T s Ž H l T . Ž K l T . if and only if T F HK , and that this in turn holds if and only if ru f TK , or equivalently, if and Ž 4.A. only if ru f K . To see this we start by noting that up to T-conjugacy there are eight Ž . choices for the pair  H, K 4 . Under the conjugation action of PGOq 8 q , these eight choices fall into two orbits of size 4, distinguished by whether ru g TK or not. Since Lemma C of w21, 5.1.15x shows that T s Ž H l T .Ž K l T . F HK for at least one choice of H and K, it is sufficient to show that T g HK for at least one choice of H, K satisfying ru g TK. We view Ž . Ž . 2.Oq 8 2 as the Weyl group W s W E8 of type E8 acting naturally on a Euclidean space U spanned by a root system F, i.e., W is the group of isometries of U generated by the reflections in the hyperplanes in U orthogonal to each of the roots in F. We further assume that F and U are as described by Section 3.6 of w7x, except that we change notation so that x 1 , . . . , x 8 is an orthonormal basis for U and x1 y x 2 , x 2 y x 3 , x 3 y x 4 , x 4 y x5 , x5 y x6 , x6 y x 7 , x6 q x 7 , y

1 2

8

Ý xi is1

is a fundamental system F. Let ² , :U be the inner product on U, and let P be the above set of fundamental roots. Observe that ² a , b :U g Z for all roots a , b g F. ŽA complete listing of roots is given in w7, 3.6x.. Now the lattice ZF s ZP of all integral linear combinations of roots is preserved by the action of W. We let Y be the tensor product ŽZF . mZ F 3 of the additive group of ZF with the additive group of F 3 , and we view Y as a 8-dimensional F 3-vector space with basis  a m 1: a g P4 Žcf. Section 4.4 of w7x.. Let Ž , . Y be the symmetric bilinear form Y = Y ª F 3 induced by ² , :U and reduction modulo 3, and let Q Y be the quadratic form associated with Y, i.e., Q Y Ž y . s yŽ y, y . for all y g Y. A straightforward calculation shows that Y contains an orthonormal basis, namely x 1 m 1, . . . , x 8 m 1, whence Ž , . Y is non-degenerate of type q by 2.5.10 and 2.5.12 of w18x. Observe that the action of W on U gives rise to an action of W on Y preserving Q Y and Ž , . Y . By identifying Y and the vector space V on which T acts, we obtain Ž . an embedding of W in Oq 8 3 ; by factoring out scalars we obtain an qŽ . embedding V 8 2 ¨ T. Let K 0 be the image of this embedding, so that Ž . Ž . K 0 is Aut T-conjugate to Vq 8 2 ; set K s NAut T K 0 . Let a 0 be the highest root in F with respect to the fundamental system P, that is, a 0 s x 1 y x 8 . Then Q Y Ž a 0 m 1. s 1, whence a 0 m 1 is a non-singular vector in Y. Set u s a 0 m 1 and H s NAut T ŽT² u: .; we verify Ž4.A. by showing that ruf K and that T g HK. Using w7, 2.5.4x we see that NW Ž ² u: . ( W Ž A1 . = W Ž E7 . ( 2 = Sp6 Ž 2 . = 2.

FACTORISATIONS OF ALMOST SIMPLE GROUPS

171

Thus Ž K 0 .² u: involves Sp6 Ž2., and by inspecting a list of maximal subgroups Ž . w x Ž . Ž . < < of Vq 8 2 as given by 8, p. 84 we see that K 0 ² u: ( Sp6 2 . Now HK divides < T² u: K 0 < NAut T Ž T² u: . < K < s < T² u: <

< K0 <

Ž K 0 . ² u:

NAut T Ž T² u: . < K <

s < T² u: < .120.4.6, which is not divisible by < T < since < T : T² u: < s 1066. Hence T g HK. Also W clearly contains the reflection ru in u as u is a root, whence rug K and Ž4.A. holds. We now use Ž4.A. to verify that line 9 of Table III holds under the 2 assumption that A l T s N1a , B l T s N1a , and C l T is Aut T-conqŽ . jugate to V 8 2 . Certainly we have G s T, since we have already noted that T F G F F, but for q s 3, T s F. Since both T s N1a C

and

T s N1a C, 2

by applying a 2 , a , respectively, we deduce that T s N1C a

2

and

T s N1C a .

Then by Ž4.A. we have r s r¨ f NAut T Ž C a

2

. l NAut T Ž C a . .

On recalling the identification between Out T and S4 in which r ¬ Ž34. Žsince q s 3 is odd., a ¬ Ž123., and NAut T Ž Vq 8 Ž 2 . . s ² Ž 12 . , Ž 123 .: , Ž . we deduce that C s Vq 8 2 whence line 9 of Table IV does indeed hold. Ž . Conversely, the claim shows that T s N1 Vq 8 2 whence a q T s N1a Vq 8 Ž 2 . s N1 V 8 Ž 2 . , 2

Ž . since a normalises Vq 8 2 up to T-conjugacy. Also X can be calculated by noting firstly that ² : X F TNAut T Ž Vq 8 Ž 2 . . s T , rw , a , secondly that a f X since  N1a , N1a 4 a is not, up to T-conjugacy, equal to 2  N1a , N1a 4 , and finally that rw g X since 2

N1a r w s Ž N1r w .

a rw

and

N1a

2

rw

s Ž N1r w .

Ž a 2 .rw

172

BADDELEY AND PRAEGER

are T-conjugate to N1a and N1a , respectively, as N1r w s N1 and a r w s a 2. Hence all implications of line 9 of Table IV are valid. We now suppose that A l T s N1 and that N1 is not Aut T-conjugate to either B l T or C l T. If either B l T or C l T is Aut T-conjugate to P1 , then by L2 of TAB. 4 2 we may assume that B l T s P1a or P1a . These options are interchanged on conjugating by rw g NAut T Ž N1 . and so up to Aut T-equivalence we may assume that B l T s P1a. Direct calculation shows that the maximality of A and B in G implies that G F ² F, rrw :. Given that G s BC, TAB. 4 Ž . shows that C l T is Aut T-conjugate to one of N2y, A 9 with q s 2, Vq 8 2 with q s 3, or 2 6 : A 8 with q s 3. The latter does not occur since there is no corresponding N1Ž2 6 : A 8 . factorisation; the former three give rise to lines 11]13 of Table IV and we leave the verification of the fine detail to the reader. We now suppose that neither B l T nor C l T is Aut T-conjugate to P1; if one is Aut T-conjugate to N2y, then by L3 of TAB. 4 we may assume 2 that B l T s Ž N2y. a or Ž N2y. a . Again these options are interchanged on conjugating by rw g NAut T Ž N1 . and so up to Aut T-equivalence we may assume that B l T s Ž N2y. a. Given that G s BC, TAB. 4 shows that C l T is Aut T-conjugate to one of A 9 with q s 2 or N2q with q s 4. These possibilities give rise to lines 14 and 15 of Table II, respectively; we again leave the verification of the fine detail to the reader. We now suppose that neither B l T nor C l T is Aut T-conjugate to P1 or N2y. As the factorisation G s BC does not involve N1 , P1 , or N2y it must be given by L14 of TAB. 4; but then one of B, C does not allow a factorisation with A. Finally it remains to consider the case in which none of A l T, B l T, C l T is Aut T-conjugate to N1. Thus the factorisations G s AB s AC s BC are given by three of L7, 9, 10, 13]15, and 17 of TAB. 4; it is clear that the only possibility is that given by line 16 of Table II. Again we leave the verification of the fine detail to the reader. This completes Case Ž1. of the proof of Theorem 1.2. 2

Case Ž2.. M maximal and < M < s 4. We write M s  A, B, C, D4 ; thus A, B, C, D are maximal subgroups of G not containing T and satisfying G s AB s AC s AD s BC s BD s CD. We must show that the pair Ž G, M . is Aut T-equivalent to a pair listed by Table IV. ŽRecall from note Ži. for Tables II]V that Ž G 0 ,  A 0 , B0 , C0 , D 0 4. is Aut T-equivalent to Ž G, M . if and only if there exist s g Aut T and w, x, y, z g G such that G s G 0s

and

 A, B, C, D 4 s  As0 w , B0s x , C0s y , D 0s z 4 ..

FACTORISATIONS OF ALMOST SIMPLE GROUPS

173

Conversely, we must also show that the implications of each line of Table IV are valid, i.e., that the information so given is correct and that each pair Ž G, M . so specified does indeed correspond to a maximal multiplefactorisation of G. Observe that each of

 A, B, C 4 ,  A, B, D 4 ,  A, C, D 4 ,  B, C, D 4

Ž 4.B.

is a maximal multiple-factorisation of G. By Case Ž1. the possibilities for each are listed in Tables II and III. Observe also that no two of A, B, C, D are G-conjugate to each other. It is easy to see that none of the Ž . possibilities in Table II applies, and so T s PVq 8 q . In order to treat this case in as uniform a fashion as possible we first prove the following lemma. Ž . LEMMA 4.1. Let G be almost simple with socle T s PVq 8 q and let N be a maximal multiple-factorisation of G with < N < s 3. Note that Aut T has a normal subgroup Y containing all field automorphisms and such that Aut TrY ( S3 , and moreo¨ er, that such a normal subgroup is unique. Ž In the notation of note Žiii. for Tables II]V we ha¨ e Y s ²T, f , d :.. Let j : Aut T ª S3 be any map with ker j s Y. Then either Syl 2 Ž S3 . s  j Ž NAut T Ž L l T . . : L g N 4 , where Syl 2 Ž S3 . is the set of Sylow 2-subgroups of S3 , or q s 3, N is Aut T-equi¨ alent to a multiple-factorisation as specified either by line 9 or by Ž . line 13 of Table III, and there exists L g N with L l T ( Vq 8 2 . Proof. Given that Case Ž1. above has been proved, this follows by inspection of Table III. The pigeon-hole argument shows, with j as in Lemma 4.1, that

j Ž NAut T Ž A l T . . , j Ž NAut T Ž B l T . . , j Ž NAut T Ž C l T . . , j Ž NAut T Ž D l T . . are not pair-wise distinct elements of Syl 2 Ž S3 .. Thus at least one of the maximal multiple-factorisations listed by Ž4.B. is Aut T-equivalent to a multiple-factorisation as specified either by line 9 or by line 13 of Table III. Moreover, up to Aut T-conjugacy and relabelling A, B, C, D if necŽ . essary, we may assume that A l T s Vq 8 2 . It is now straightforward, and we leave it to the reader, to show that M is Aut T-equivalent to a multiple-factorisation as specified by Table IV, and conversely that all implications of Table IV are valid. This completes Case Ž2. of the proof of Theorem 1.2.

174

BADDELEY AND PRAEGER

Case Ž3.. M maximal and < M < G 5. By the results of Case Ž2. we have Ž . T s PVq 8 3 and may assume that M contains the groups A, B, C of Table IV, together with D s N1 and E s P1. Then, since DE / G, we see that there are no examples with < M < G 5. Case Ž4.. M strong and < M < s 3. We write M s  A, B, C 4 ; thus A, B, C are subgroups of G not containing T and satisfying G s AŽ B l C . s B Ž A l C . s C Ž A l B . .

Ž 4.C.

We must show that the pair Ž G, M . is Aut T-equivalent to a pair listed by Table V. ŽAs before Ž G 0 ,  A 0 , B0 , C0 4. is Aut T-equivalent to Ž G, M . if and only if there exist s g Aut T and x, y, z g G such that G s G 0s

and

 A, B, C 4 s  As0 x , B0s y , C0s z 4 ..

Conversely, we must also show that the implications of each line of Table V hold. We use a reduction step based upon the following lemma. LEMMA 4.2. Let G be almost simple with socle T, and let  A, B, C 4 be a strong multiple-factorisation of G. Then the following all hold: Ži. let G 0 be a subgroup of G containing T and at least one of A, B, C, and set A 0 s A l G 0 , B0 s B l G 0 , and C0 s C l G 0 ; then  A 0 , B0 , C0 4 is a strong multiple-factorisation of G 0 ; Žii. set G1 s AT l BT l CT and A1 s A l G1 , B1 s B l G1 , and C1 s C l G1; then  A1 , B1 , C14 is a strong multiple-factorisation of G1 , and G1 s A1T s B1T s C1T ; Žiii. let G1 , A1 , B1 , C1 be as in Žii., and let A 2 , B2 , C2 be maximal subgroups of G1 containing A1 , B1 , C1 , respecti¨ ely; then  A 2 , B2 , C2 4 is a multiple-factorisation of G1 that is both maximal and strong. Proof. We start by noting that G 0 Žin Ži.., and G1 Žin Žii.., are both almost simple with socle T, and so it is sensible to consider multiplefactorisations of G 0 and G1. We turn to part Ži.. Certainly A 0 , B0 , C0 are subgroups of G 0 not containing T and so it is enough to show that G 0 s A 0 Ž B0 l C0 . s B0 Ž A 0 l C0 . s C0 Ž A 0 l B0 . .

Ž 4.D.

Without loss of generality we may assume that A F G 0 , and so Ž4.D. is equivalent to G0 s AŽ B l C l G0 . s Ž B l G0 . Ž A l C . s Ž C l G0 . Ž A l B . . This follows by intersecting Ž4.C. with G 0 and noting that A, A l C, and A l B are all subgroups of G 0 .

FACTORISATIONS OF ALMOST SIMPLE GROUPS

175

We now consider part Žii.. It is easy to see that G1 s AT l BT l CT s Ž Ž AT l B . T l C . T . By applying part Ži., firstly to the strong multiple-factorisation  A, B, C 4 of G with G 0 s AT, secondly to the strong multiple-factorisation  A, AT l B, AT l C 4 of AT with G 0 s Ž AT l B .T, and thirdly to the strong multiple-factorisation  A l Ž AT l B .T, AT l B, C l Ž AT l B .T 4 of Ž AT l B .T with G 0 s Ž C l Ž AT l B .T .T, we deduce that  A1 , B1 , C14 is a strong multiple-factorisation of G1 as required. Now C1 s C l G1 s C l Ž Ž AT l B . T l C . T s Ž Ž AT l B . T l C . Ž C l T . , whence C1T s Ž Ž AT l B . T l C . Ž C l T . T s Ž Ž AT l B . T l C . T s G1 . The equalities G1 s A1T s B1T follow by symmetry and so Žii. holds. Finally we consider part Žiii.. Since G1 s A1T s B1T s C1T, we see that A 2 , B2 , C2 are maximal subgroups of G1 not containing T. Given that  A1 , B1 , C14 is a strong multiple-factorisation of G1 the remainder of Žiii. follows easily. In the following we consider three exhaustive and mutually exclusive subcases distinguished by the following conditions on the strong multiplefactorisation  A, B, C 4 of G: ŽA.  A, B, C 4 is also a maximal multiple-factorisation of G, or equivalently, A, B, C are all maximal subgroups of G Žwhence G s AT s BT s CT .; ŽB. G s AT s BT s CT but A, B, C are not all maximal subgroups of G; ŽC. one of AT, BT, CT is a proper subgroup of G. The key significance of the above lemma is that if Subcase ŽC. applies, then using Lemma 4.2Žii. we can reduce to a strong multiple-factorisation for which either Subcase ŽA. or Subcase ŽB. applies, while if Subcase ŽB. applies, then using Lemma 4.2Žiii. we can reduce to a strong multiplefactorisation for which Subcase ŽA. applies, namely to a strong multiplefactorisation that is also maximal. We shall also need the following lemma. LEMMA 4.3. Let K, L, M be subgroups of a group H. Then the following are equi¨ alent: Ži. H s K Ž L l M . s LŽ K l M . s M Ž K l L.; Žii. H s KL s LM s KM and K s Ž K l M .Ž K l L.; Žiii. H s KL s LM s KM and H s M Ž K l L..

176

BADDELEY AND PRAEGER

Proof. That Ži. implies Žiii. is obvious. We assume Žiii.; as H s M Ž K l L. we have K s K l H s K l Ž M Ž K l L. . s Ž K l M . Ž K l L. , and Žii. holds. We assume Žii.. Then H s MK s M Ž K l M .Ž K l L. s M Ž K l L.. A similar argument shows that H s Ž K l M . L. Finally, as H s LM, H s M Ž K l L., and H s KL we have KŽ L l M. s s


s
< L< < M <



< M < < K l L<

s < H <,

whence H s K Ž L l M . and Ži. holds. Subcase ŽA.. M s  A, B, C 4 is a maximal multiple-factorisation of G. Given Case Ž1. of the proof of Theorem 1.2 the pair Ž G, M . is Aut Tequivalent to one listed either by Table II or by Table III. Thus it is sufficient to consider each line of Tables II and III in turn, and to show that the maximal multiple-factorisation so listed is a strong multiplefactorisation if and only if one of the following holds: line 1 of Table II with G s S6 , line 8 of Table II with b s 2, line 12 of Table II, or line 13 of Table III. ŽThe first possibility corresponds to line 1 of Table V, the second to line 2, the third to line 3, and the fourth to line 7.. We start by using an elementary divisibility argument to discount lines 2]7, 10, 11, 14, and 15 of Table II and lines 11, 12, 14]16 of Table III. If  A, B, C 4 is a strong multiple-factorisation of G then by Lemma 4.3 we have C s Ž A l C .Ž B l C . which implies that < C : A l C < < C : B l C < s < C : A l B l C <. Now G s AC s AT implies that < G: A < s < C: A l C < s < T : A l T <; likewise, G s BC s BT implies that < G: B < s < C: B l C < s < T : B l T <. It follows that < T : A l T < < T : B l T < divides < C <. Now if one of lines 2]7, 14, 15 of Table II or one of lines 12, 14, 16 of Table III applies then G s T, in which case A s A l T, B s B l T, and Ž . C s C l T, and T is one of M12 , PV 7 Ž3., or PVq 8 2 . For such T, lists of maximal subgroups appear in w8, pp. 33, 109, 85x, respectively and we use the information there to tabulate in Table IX the values of < T : A <, < T : B <, and < C <. In each case, we find that < T : A l T < < T : B l T < does not divide < C < and so  A, B, C 4 is not a strong multiple-factorisation of G. This leaves lines 11 and 15 of Table III and lines 10 and 11 of Table II to be eliminated using the same divisibility argument a little more carefully.

177

FACTORISATIONS OF ALMOST SIMPLE GROUPS

TABLE IX Elimination by Divisibility Argument T

Table

Line




III

2 3 4 5 6 7

2 2 .3

2 2 .3

2 2 .3.5.11 2 4 .3.5 2 6 .3 2 3.3 2 2 5 .3 2 .5 2 4 .3 3

M12

2 4 .3 2

PV 7 Ž3.

III

12 13

2.3 3.7

2 3.3 3.5

2 9.3 4 .5.7 2 7.3 4 .5.7

Ž . PVq 8 2

IV

12 14 16

2 3.3.5

3 3.5 2 5 .5.7 3 3.5

2 6 .3 4 .5.7

2 5 .5.7

Ž . If line 11 of Table III holds then T s PVq 8 q and inspection shows that 1
2

Ž 2, q y 1 .

2

q 6 Ž q 3 q 1 .Ž q 4 y 1 .Ž q 2 y 1 . Ž q q 1 . .

As G F ² F, rrw : we see that < C < divides Ž2, q y 1. e < C l T < where q s p e. However, 2 eq 3 Ž q 2 y 1 . Ž 2, q y 1 . e < C l T < s
178

BADDELEY AND PRAEGER

We now eliminate all remaining lines of Table III except line 13, i.e., lines 1]10. To do this we assume that ŽT, G, A, B, C . is so listed and  A, B, C 4 is a strong multiple-factorisation of G. In all cases T s PVq Ž . 8 q , 2 A l T s N1a , and B l T s N1a . By w16, 3.1.1Žvi.x N1 l N1a s N1 l N1a s N1a l N1a s G 2 Ž q . 2

2

and is normalised by a . Note also that by w16, 3.1.1Žv.x, G 2 Ž q . is normalised by the generating field automorphism f and so, since G F F s ²T, f :, we have NG Ž N1 . l NG Ž N1a . s NG Ž N1 . l NG Ž N1a

2

s NG Ž N1a . l NG Ž N1a

. 2

. s NG Ž G2 Ž q . . .

Since in each case G is also normalised by a and since A s NG Ž N1a . is a maximal subgroup of G, the factorisation G s Ž A l B .C gives rise to the maximal factorisations G s NG Ž N1 . C s NG Ž N1a . C s NG Ž N1a . C 2

of G. The only possibility for C for which this is true is that given by line 9 Ž .. is also normalised by a , and from w16x of Table III; here C s NG Ž Vq 8 2 we deduce that G.² a : s Ž Ž A l B . .² a : . Ž C.² a : . is a maximal factorisation of G.² a : }this is a contradiction as by w21x no such factorisation exists. We now turn to considering maximal multiple-factorisations that do give rise to examples of strong multiple-factorisations. We assume that line 1 of Table II holds. Thus T s A 6 and G s A 6 or S6 . If G s S6 then up to G-conjugacy we may assume that A s  p g S 6 : 6p s 6 4 , B s Aa , and C s  p g S6 :  1, 2, 3 4 p s  1, 2, 3 4 or  4, 5, 6 4 4 , where a is an outer automorphism of S6 normalizing C. Clearly A l C s  p g S6 : 6p s 6 and  1, 2, 3 4 p s  1, 2, 3 4 4 ( S3 = S2 . Since S6 s AB we have < A l B < s 20 and by Sylow’s theorems A l B is the normalizer in A of a Sylow 5-subgroup of A. Hence up to A-conjugacy we have A l B s NA² Ž 12345.: ( 5:4.

FACTORISATIONS OF ALMOST SIMPLE GROUPS

179

By direct calculation we find A l B l C s  Ž 13 . Ž 45 . , id 4 . ŽFor later usage we note that A l B l C is contained in the derived subgroups A9, B9, C9 of A, B, C, respectively.. It follows that

Ž A l B. Ž A l C. s

20.12 2

s 120 s < A < .

Thus A s Ž A l B .Ž A l C . and by Lemma 4.3,  A, B, C 4 is a strong multiple-factorisation of G as required. On the other hand, if G s A 6 then a similar calculation again shows that < A l B l C < s 2 and by orders we have A / Ž A l B .Ž A l C .. We assume now that either line 8 or line 9 of Table II holds. So G s P GSp 2 mŽ q . with 2 e s q s 2 or 4. Now A s Sp 2 aŽ q b ..be and the argument of w21, 3.2.1Žd.x shows that b A l B s Oy 2 a Ž q . .be

and

b A l C s Oq 2 a Ž q . .be

are both maximal subgroups of A. Now A is almost simple since q b G 4 and w21x shows that A s Ž A l B .Ž A l C . if and only if q s b s 2. ŽTo see this for a G 2 we consult the sections of TABS. 1, 2, 3 for symplectic groups, and for a s 1 the sections for linear groups.. By Lemma 4.3, A s Ž A l B .Ž A l C . if and only if  A, B, C 4 is a strong multiple-factorisation of G, and as the condition q s b s 2 corresponds to those assumed in line 2 of Table V we have finished this subcase. We next assume that either line 12 or line 13 of Table II holds. So G s P GSp6 Ž q . with q s 2 or 4. We must show that we have a strong multiple-factorisation if and only if q s 2. Note that the results of w21x show that both A l B and A l C are maximal in A. If q s 4, then A s G 2 Ž4..2, B s O6yŽ 4..2, and C s O6qŽ 4..2 Žwhere the 2 on top is a field automorphism in each case.. Calculation shows that < A: A l B < s < G: B < s 2016

and

< A: A l C < s < G: C < s 2080.

From the information on p. 97 of w8x we deduce that A l B ( U3 Ž4.:4 and A l C ( 3. L3 Ž4.:2 2 . All factorisations of A are listed in TAB. 5 and inspection shows that A / Ž A l B .Ž A l C .. On the other hand, if q s 2 then A s G 2 Ž2. ( U3 Ž3..2. Again calculation shows that < A: A l B < s < G: B < s 28

and

< A: A l C < s < G: C < s 36

and the information given by p. 14 of w8x shows that A s Ž A l B .Ž A l C .. By Lemma 4.3 we see that  A, B, C 4 is a strong multiple-factorisation of G if and only if q s 2 as required.

180

BADDELEY AND PRAEGER

Ž . Finally we assume that line 13 of Table III holds. So G s T s PVq 8 3 . The results of w21x show that both A l B and A l C are maximal in A. Now G s AB s AC implies that < A: A l B < s < G: B < s 1120

and

< A: A l C < s < G: C < s 28431.

We have A ( PV 7 Ž3. and by w8, p. 109x, A l B ( 3 3q 3 : L3 Ž3., which is P3 in the notation of w21x, and A l C ( 2 6 : A 7 are both unique up to A-conjugacy. Now L28 of TAB. 3 implies that A s Ž A l B .Ž A l C . whence by Lemma 4.3,  A, B, C 4 is a strong multiple-factorisation of G as required. This finishes the analysis of Subcase ŽA.. Before moving on we remark that by Lemma 4.2 we have now shown that if  A, B, C 4 is a strong multiple-factorisation of G, then T is one of A 6 , PSp4 mŽ2. with m G 2, Ž . Sp6 Ž2., or PVq 8 3 . This will be used in the following. Subcase ŽB.. We assume that G s AT s BT s CT but that not all of A, B, C are maximal subgroups of G. We shall show that all examples in this subcase correspond to one of lines 4]6 of Table V. Choose A 2 , B2 , C2 to be maximal subgroups of G containing A, B, C, respectively. By Lemma 4.2Žiii.,  A 2 , B2 , C2 4 is a strong multiple-factorisation of G satisfying the assumptions of Subcase ŽA., and so by the above is listed up to Aut T-equivalence by one of lines 1]3 and 7 of Table V. Before considering each such line in turn we make some general observations. Now G s AŽ B l C . implies G s AŽ B2 l C2 . whence A 2 s A 2 l G s A Ž A 2 l B2 l C2 . .

Ž 4.E.

From this it follows that A 2 s AŽ A 2 l B2 . s AŽ A 2 l C2 ., while by Lemma 4.3 we have A 2 s Ž A 2 l B2 .Ž A 2 l C2 .. So if A 2 is almost simple and A, A 2 l B2 , A 2 l C2 do not contain the socle Soc A 2 of A 2 , then  A, A 2 l B2 , A 2 l B2 4 is a strong multiple-factorisation of the almost simple group A 2 ; by the remark made at the end of Subcase ŽA., this can only Ž . happen if Soc A 2 is one of A 6 , PSp4 m Ž2. with m G 2, Sp6 Ž2., or PVq 8 3 .

Note that, in each case, if A 2 is almost simple, then it is clear that A 2 l B2 and A 2 l C2 do not contain the socle of A 2 and so the only difficulty in applying this observation is in deciding whether A G Soc A 2 or not. Note also that symmetric observations hold for B2 and C2 . Assume first that the strong multiple-factorisation  A 2 , B2 , C2 4 is as given by line 1 of Table V. Recall that while considering Subcase ŽA. we saw that A 2 l B2 l C2 has order two and is contained in the derived subgroup of each of A 2 , B2 , C2 . By considering group orders the condition Ž4.E. implies that A has index one or two in A 2 . Hence A G AX2 and we

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see that A 2 s AŽ A 2 l B2 l C2 . if and only if A s A 2 . A similar argument works for B and C, and we deduce that A s A 2 , B s B2 , and C s C2 , which gives the required contradiction Žsince we assumed that not all of A, B, C were maximal.. We next assume that the strong multiple-factorisation  A 2 , B2 , C2 4 is as given by line 2 of Table V, i.e., that G s Sp4 aŽ2. with a G 2, A 2 s Sp 2 aŽ4..2, B2 s O4yaŽ2., and C2 s O4qaŽ2.. Note that A 2 , B2 , C2 are all almost simple. For convenience we set Y s A 2 l B2 l C2 and we claim that Y is contained in the socle of each of A 2 , B2 , C2 . In the final paragraph of Ž . 3.2.1Žd. of w21x it is observed that A 2 l C2 is contained in Vq 4 m 2 s Soc C 2 . Now by viewing Y as the intersection of A 2 l B2 and A 2 l C2 , we see that Y is determined by 3.2.4Že. of w21x; it is clear from the discussion there that Y F Soc A 2 . However, it is not immediately apparent that Y F Soc B2 , and so we expand on the argument of 3.2.4Že.. We let  a i , bi : 1 F i F m q 14 Ž . be a standard basis for an orthogonal geometry of type Oq 2 mq2 4 with q Ž . associated quadratic form P. Writing R s V 2 mq2 4 we may identify Soc A 2 s Sp 2 mŽ4. with the stabilizer R ² ¨ : of the subspace generated by the non-singular vector ¨ s a1 q b1 Žso that P Ž ¨ . s 1.. Let 1 / v g F4 Ž . be a cube root of unity. As noted in w21x we can identify Oy 2 m 4 with the H stabilizer in R ² ¨ : of the hyperplane V of ¨ spanned by  x, y, a i , bi : 4 Ž 3 F i F m q 1 where x s a2 q b 2 and y s v a1 q b1 . q a2 . Now the embedding y Oy 2 m Ž 2 . ¨ O4 m Ž 2 . s B2

is obtained by viewing V as an F 2-vector space in the natural way with associated quadratic form Q equal to the composition of P with the trace map F4 ª F 2 . Now 3.2.4Že. identifies Y as Y s ² h : = Y0 , where Y0 ( Sp 2 my2 Ž4. is simple and h is such that h: a1 ¬ b1 ¬ a1 , ai ¬ ai ,

a2 ¬ b 2 ¬ a2 ,

bi ¬ bi

for all i G 3.

So Y F Soc B2 if and only if h g Soc B2 ; furthermore, h g Soc B2 if and only if h, viewed as a map on the anisotropic subspace W s ² x, y : of V, Ž . lies in Soc O4yŽ 2. s Vy 4 2 when W is viewed as an F 2 -vector space in the natural way. Now W as an F 2-vector space has basis  x, y, v x, v y4 and contains 5 singular non-zero vectors, namely x, v y, v Ž x q y . , x q y q v y, v Ž x q y . q y.

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Since h: x ¬ x, y ¬ x q y we see that h fixes x and interchanges the pairs  v y, v Ž x q y .4 and  x q y q v y, v Ž x q y . q y4 ; thus h induces an even permutation on the singular non-zero vectors. However, O4yŽ 2. ( S5 Ž . acts transitively on such vectors and Vy 4 2 is precisely the subgroup that Ž . Ž induces even permutations, whence h g Vy 4 2 as required. The reader will note that we have essentially repeated a calculation performed first when considering line 1 of Table V in Subcase ŽA., but in a more general setting.. Thus Y F Soc B2 . So Y : Soc A 2 l Soc B2 l Soc C2 . This, together with Ž4.E. and its analogues, implies that if any of A, B, C is a proper subgroup of A 2 , B2 , C2 , respectively, then it does not contain Soc A 2 , Soc B2 , Soc C2 , respectively. From the observations made at the beginning of Subcase ŽB., we deduce that one of A 2 , B2 , C2 has socle equal to one of A 6 , PSp4 mŽ2. Ž . with m G 2, Sp6 Ž2., or PVq 8 3 which is not the case. We now assume that the strong multiple-factorisation  A 2 , B2 , C2 4 is as given by line 3 of Table V, i.e., that G s Sp6 Ž2., A 2 s G 2 Ž2. s U3 Ž3..2, B2 s O6yŽ 2. s U4 Ž2..2, and C2 s O6qŽ 2. s S8 . Note that A 2 , B2 , C2 are all almost simple and that information about these groups is given in w8, pp. 14, 26, 22x, respectively. Now  A 2 , B2 , C2 4 is a strong multiple-factorisation of G. It follows that < A 2 l B2 l C2 < s


,

and by calculation we find that < A 2 l B2 l C2 < s 12. Similarly we have < A l B l C< s


,

whence < A l B l C< s

< A 2 l B2 l C2 < < A 2 : A < < B2 : B < < C2 : C <

.

Ž 4.F.

In particular, < A 2 : A <, < B2 : B <, < C2 : C < are all divisors of < A 2 l B2 l C2 < s 12. From the information given in w8x on maximal subgroups of A 2 , B2 , and C2 we deduce that Soc A 2 F A,

Soc B2 F B,

and

Soc C2 F C.

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In fact, as < A 2 : Soc A 2 < s < B2 : Soc B2 < s < C2 : Soc C2 < s 2 we see that there are precisely 7 possibilities to consider for  A, B, C 4 . ŽBy assumption we do not have A s A 2 , B s B2 , and C s C2 .. We make two claims: Ži. ŽSoc A 2 . l C2 s A 2 l ŽSoc C2 . s ŽSoc A 2 . l ŽSoc C2 .; Žii. ŽSoc B2 . l C2 s B2 l ŽSoc C2 . s ŽSoc B2 . l ŽSoc C2 .. To see Ži. we observe that A 2 l C2 is a subgroup of index 36 in A 2 . By p. 14 of w8x we have A 2 l C2 s L2 Ž7..2 and ŽSoc A 2 . l C2 s L2 Ž7. s SocŽ A 2 l C2 .. As ŽSoc A 2 . l ŽSoc C2 . is a normal subgroup of ŽSoc A 2 . l C2 of index at most 2, we further deduce that ŽSoc A 2 . l ŽSoc C2 . s ŽSoc A 2 . l C2 . Moreover, from p. 22 of w8x we see that A 2 l ŽSoc C2 . s L2 Ž7. s L3 Ž2., whence Ži. does indeed hold. To see Žii. we consider G as the group of automorphisms of a 6-dimensional F 2-vector space V that preserve a non-degenerate sympletic form Ž , . on V. We fix a basis  e1 , e2 , e3 , f 1 , f 2 , f 34 for V such that for all i, j

Ž ei , e j . s Ž fi , f j . s 0

and

Ž ei , fi . s di j ,

and let Qy, Qq be the quadratic forms on V such that for all ¨ , w g V and for all i, j

Ž ¨ , w . s Qy Ž ¨ q w . y Qy Ž ¨ . y Qy Ž w . s Qq Ž ¨ q w . y Qq Ž ¨ . y Qq Ž w . , Qq Ž e i . s Qq Ž f j . s Qy Ž e1 . s Qy Ž e2 . s Qy Ž f 1 . s Qy Ž f 2 . s 0, and Qy Ž e3 . s Qy Ž f 3 . s 1. Then B2 , C2 can be identified with the subgroups of G preserving the forms Qy, Qq, respectively. Set u s e3 q f 3 so that Qy Ž u. s QqŽ u. s 1, and let W be the largest subset of V on which Qy and Qq agree. Routine calculations show that W is a subspace of V with basis  e1 , e2 , f 1 , f 2 , u4 , that W H s ² u:, and moreover that B2 l C2 s Stab B 2Ž u . s StabC 2Ž u . . Now Proposition 4.1.7ŽII. of w18x determines both Stab Soc B 2Ž u. and Stab Soc C 2Ž u.: the calculations given are independent of the quadratic form Qy or Qq. It follows that

Ž Soc B2 . l C2 s Stab Soc B 2Ž u . s Stab Soc C 2Ž u . s B2 l Ž Soc C2 . , whence Žii. holds as required.

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From Ži. and Žii. we see that

Ž Soc A 2 . l B2 l C2 s Ž Soc A 2 . l B2 l Ž Soc C2 . s A 2 l B2 l Ž Soc C2 . s Ž Soc A 2 . l Ž Soc B2 . l C2 s Ž Soc A 2 . l Ž Soc B2 . l Ž Soc C2 . s A 2 l Ž Soc B2 . l C2 s A 2 l Ž Soc B2 . l Ž Soc C2 . . We use Y to denote any of these subgroups. From p. 14 of w8x we see that Soc A 2 s ŽŽSoc A 2 . l B2 .ŽŽSoc A 2 . l C2 ., whence < Y < s 6. Using condition Ž4.F. we see that < A 2 : A < < B2 : B < < C2 : C < s 2, whence  A, B, C 4 is one of  Soc A 2 , B2 , C2 4 ,  A 2 , Soc B2 , C2 4 , and Ž A 2 , B2 , Soc C2 4 . We now show that each of these three possibilities is a strong multiple-factorisation Žcorresponding to lines 4]6 of Table V, respectively.. Given that Y / A 2 l B2 l C2 we see that A 2 l B2 l C2 is not contained in any of Soc A 2 , Soc B2 , Soc C2 , whence A 2 l B2 l C2 supplements Soc A 2 , Soc B2 , Soc C2 in A 2 , B2 , C2 , respectively. Using condition 4.3Žiii. and the fact that  A 2 , B2 , C2 4 is a strong multiple-factorisation, the required result now follows easily. Finally to complete our analysis of Subcase ŽB. we assume that line 7 of a Ž . Ž . Table V holds. So G s PVq 8 3 , A 2 s N1 ( PV 7 3 , B2 s P1 , and C 2 s qŽ . V 8 2 . As both A 2 and C2 are almost simple and moreover are equal to their socles, we may use the observations made at the beginning of Subcase ŽB. to deduce that A s A 2 and C s C2 . To show that there are no examples in Subcase ŽB., it is enough to show that B s B2 . Now B is isomorphic to the semi-direct product of 3 6 by L4 Ž3. in which L4 Ž3. ( 6 Ž . PVq 6 3 acts on 3 as a six-dimensional orthogonal geometry of type q over F 3 ; we note that this action is irreducible. By orders we have < A 2 l B2 l C2 < s 144. An analogue of Ž4.E. implies that < B2 : B < divides 144. The maximal subgroups of L4Ž3. are listed on p. 69 of w8x }none have index dividing 144. Thus B must induce L4Ž3. on the base group 3 6 and so acts irreducibly. As 3 6 does not divide 144, we conclude that B s B2 . Subcase ŽC.. We assume that one of AT, BT, CT is a proper subgroup of G. It is easy to see that the lines of Table V satisfying this assumption, namely lines 8]11, do indeed give rise to strong multiple-factorisations; so to complete the proof of Theorem 1.2 it is enough to show that one of these lines holds under the assumption that  A, B, C 4 is a strong multiplefactorisation of G. Set G1 s AT l BT l CT which is a proper subgroup of G containing T ; also set A1 s A l G1 , B1 s B l G1 , and C1 s C l

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185

G1. By Lemma 4.2Žii.  A1 , B1 , C14 is strong multiple-factorisation of G satisfying the assumptions of either Subcase ŽA. or Subcase ŽB.. The above shows that  A1 , B1 , C1 4 is listed by one of lines 1]7 of Table V. We consider each possibility for ŽT, G1 , A1 , B1 , C1 . in turn. If line 1 of Table V holds, then T s A 6 and A1 ( B1 ( S5 . As both A1 and B1 , as subgroups of Aut T, are maximal subject to not containing T, we have A s A1 and B s B1 , whence G s AB s A1 B1 s G1 }a contradiction. If line 2 of Table V holds, then T s Sp4 aŽ2. with a G 2. As Aut T s T we have G s G1 }again a contradiction. If one of lines 3]6 of Table V holds, then T s Sp6 Ž2.. As Aut T s T we have G s G1 }again a contradiction. Ž . Thus line 7 of Table V holds, so T s PVq denote the 8 3 . As usual let quotient map Aut T ª Out T ; note that G /  id4 since we are in Subcase ŽC.. As  A, B, C 4 is a strong multiple-factorisation, we see that G s AB s AC s BC s A Ž B l C . . Now inspection shows that A F ² r , rw : s ² Ž 12 . , Ž 34 .: ( S2 = S2 , B F ² r , rw , d :a s ² Ž 23 . , Ž 12 . Ž 34 .: ( S2 X S2 , C F ²rw , a : s ² Ž 12 . , Ž 123 .: ( S3 , where we have identified Out T with S4 via the explicit isomorphism given Ž . Ž . while considering T ( PVq 8 q during Case 1 above. As G s AB, we see that G is a 2-group. If A is non-trivial and either Ž23. or Ž13. g C, then it is straightforward to see that ² A, C :, and hence G, is not a 2-group. Hence either C F ²Ž12.: or A is trivial. As G s AC we see that either G s ²Ž12., Ž34.: or G has order two. If the former holds, then C F ²Ž12.: and B F ²Ž12.Ž34.: ( Z2 ; as G s BC we must have equality in both, whence the condition G s AŽ B l C . forces A s G, and line 11 of Table V holds. If < G < s 2, then at least two of A, B, C equal G, and it is an easy matter to see that the only possibilities are those given by lines 8]10 of Table V. This completes Case Ž4. of the proof of Theorem 1.2. Case Ž5.. M strong and < M < G 4. Here we must reach a contradiction. To do this we use Cases Ž1. ] Ž4. above, but must do so with a little care as it does not appear to be possible, by using only elementary arguments, to reduce directly to the case in which M is also maximal.

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We suppose that M is a strong multiple-factorisation of G with < M < G 4. Let M0 be any subset of M of size 4. Then M0 is also a strong multiple-factorisation of G. We write M0 s  A, B, C, D4 and note that each of

 A, B, C 4 ,  A, B, D 4 ,  A, C, D 4 ,  B, C, D 4 is a strong multiple-factorisation of G. Given Case Ž4., the possibilities for each of these strong multiple-factorisations of size 3 are given by Table V. Ž . In particular, T is one of A 6 , PS4 mŽ2. with m G 2, Sp6 Ž2., or PVq 8 3 . Set G 0 s S6 if T s A 6 , and G 0 s T otherwise; in both cases set A 0 s NG 0Ž A l G 0 . and define B0 , C0 , D 0 analogously. By inspecting Tables II and V we find that each of

 A 0 , B0 , C0 4 ,  A 0 , B0 , D 0 4 ,  A 0 , C0 , D 0 4 ,  B0 , C0 , D 0 4 is both a strong and a maximal multiple-factorisation of G 0 , whence  A 0 , B0 , C0 , D 0 4 is both a strong and a maximal multiple-factorisation of G 0 . The required contradiction now follows easily from Cases Ž2. and Ž4.. This completes our analysis of Case Ž5. and so finishes the proof of Theorem 1.2.

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