e J. London Math. Soc. (2) 69 (2004) 128–140 C 2004 London Mathematical Society DOI: 10.1112/S0024610703004848
A CHARACTERISTIC 5 IDENTIFICATION OF THE LYONS GROUP
CHRISTOPHER PARKER and PETER ROWLEY
1. Introduction Downloaded from https://academic.oup.com/jlms/article/69/1/128/917235 by guest on 27 September 2021 In characteristic 5 there is an exceptional symplectic amalgam [9] which is related to the sporadic simple group whose existence was predicted by Lyons [7]. This amalgam of groups also appears in the ongoing investigation of groups of local characteristic p instigated by Meierfrankenfeld, Stellmacher and Stroth [8]. A significant part of this programme will involve the identification of various simple groups from certain local data. The aim of this paper and [10]istoprovide results of this kind. The exceptional configuration considered here appears as an anomalous situation in [10] where we consider weak BN-pairs related to G2(5). In fact the subgroups appearing in this amalgam closely resemble those in the 2+1+2 . weak BN-pair related to G2(5). In this amalgam we have Lα ∼ 5 .2 Alt(5), ∼ 1+4 . ∩ ∈ ∩ Lβ 5+ .2 Alt(6), and Lα Lβ Syl5(Lα) Syl5(Lβ), and, of course, no non- trivial subgroup of Lα ∩ Lβ is normalized by both Lα and Lβ. The abstract amalgam of these groups will be denoted by L = A(Lα,Lβ,Lα ∩ Lβ). Notice that L may not be unique up to isomorphism of amalgams. That there is at least one such amalgam follows from the existence of the Lyons group. Further evidence, if any is needed, of the Lyons group’s aspirations to be a characteristic 5 object is its possessing a rank 3 minimal parabolic geometry of characteristic 5 and being the only sporadic finite simple group to have such a geometry. The universal cover of this geometry is an affine building of type G2 which does not arise from an algebraic group over a locally compact local field. As a consequence, and as a point particularly relevant to this paper, the Lyons group cannot be identified solely by its action on its characteristic 5 geometry. For further information on such exceptional phenomena the reader may consult [5, 11]. A faithful completion of L isjustagroupG which is a quotient of the universal completion of L by some homomorphism which is an isomorphism when restricted to Lα and Lβ.IfG is a faithful completion of L we may, and indeed do, identify Lα and Lβ with their images in G. Suppose that p is a prime; then a finite group G is said to be of local characteristic p so long as for every non-trivial p-subgroup P of G we have CG(Op(NG(P ))) Op(NG(P )) (that is, NG(P )isp-constrained). A K-group is a group in which every non-abelian composition factor is isomorphic to either a group of Lie type (including the Tits group), an alternating group of degree at least 5 or one of the 26 sporadic simple groups. Following Meierfrankenfeld, Stellmacher and Stroth [8], we say that a finite group G is a K2-group provided
Received 18 September 2002. 2000 Mathematics Subject Classification 20D08. characteristic 5 identification of the lyons group 129 that G is a group in which the normalizer of every non-trivial 2-subgroup of G is a K-group. Our main theorem is as follows.
Theorem 1.1. Suppose that G is a K2-group of local characteristic 5 which is L ∩ ∈ a finite faithful completion of .IfLα Lβ Syl5(G), then there is an involution t ∼ . in G such that CG(t) = 2 Alt(11) and G is isomorphic to the Lyons group.
Because of the work of Lyons [7] and Sims [12] to prove Theorem 1.1, we only ∼ . need to show that G has an involution t with CG(t) = 2 Alt(11). To do this we Downloaded from https://academic.oup.com/jlms/article/69/1/128/917235 by guest on 27 September 2021 study centralizers of elements of order 5 in G to rule out the possibility that CG(t) ∗ has a large normal subgroup of order coprime to 5 and to discover that F (CG(t)) is either 2 .Alt(11) or a central product of two copies of 2.Alt(6). The elimination of this latter possibility, though not stated explicitly, forms a substantial part of the proof of Theorem 1.1 and hinges on examining the action of G upon a certain coset graph Γ. This analysis takes up most of Section 3, where our attention is focused on the set of vertices of Γ fixed by a subgroup T of G which is isomorphic to Z4 × Z4. Our notation is mostly standard and is taken from [1] and [3] with a few additions from [9].
2. The 4-dimensional 2 .Alt(6)-module and a characterization of Alt(11) In this section we discuss the 4-dimensional GF(5)2 .Alt(6)-modules and prove a K-group statement which characterizes Alt(11). We begin with the 4-dimensional GF(5)2 .Alt(6)-modules. For this we need to recall that the irreducible modules for . ∼ A =2 Alt(5) = SL2(5) in characteristic 5 are of dimensions 1, 2, 3, 4 and 5 [2]. Only the 2- and 4-dimensional modules are faithful. The 2-dimensional module is the one which we call the natural module and the 4-dimensional module can be considered as the vector space over GF(5) which consists of homogeneous polynomials of degree 3 in two commuting variables. Thus we may easily calculate in both these modules. Suppose that t ∈ A has order 3. Then for V the natural module we have CV (t)=0 and for V the irreducible 4-dimensional module we have dim CV (t)=2. For any vector space W we let P(W ) denote the set of 1-dimensional subspaces of W .
∼ . Lemma 2.1. Suppose that H = 2 Alt(6) and let H1 and H2 be representatives ∼ . of the two conjugacy classes of subgroups of H isomorphic to SL2(5) = 2 Alt(5). Assume that V is a faithful irreducible 4-dimensional GF(5)H-module. For S ∈ C { | ∈ } Syl5(H),setVS = CV (S) and = VS S Syl5(H) . Then the following hold. | | (i) Up to changing notation, we have that V H1 is irreducible and V H2 is an indecomposable module with two composition factors each isomorphic to the natural GF(5)H2-module. ∈ | C⊆P (ii) For S Syl5(H) we have V S is uniserial, dim VS =1 and (V ). ∈ (iii) If R, S Syl5(H) with R = S, then VR = VS. ∈ P ⊆C (iv) If R, S Syl5(H) with R = S and (VR + VS) , then R, S is conjugate to H2.
| Proof. Since V is a faithful module, all the composition factors of V Hi,for ∈ | i =1, 2, have dimension either 2 or 4. Let D Syl3(H). If, for i =1, 2, V Hi is 130 christopher parker and peter rowley irreducible, then each element of order 3 in D centralizes a 2-dimensional subspace of V . Since, by [9, Lemma 2.42], V/CV (D)= 1= E
When we deal with Lie type groups in the following proposition we shall encounter ∼ ∼ the many classical group incarnations of Alt(6). These are Alt(6) = PSL2(9) = ∼ ∼ ∼ − ∼ + ∼ PSp2(9) = PSU2(9) = Ω3(9) = Ω4 (3) = Sp4(2) . Additionally, we recall that O2 (q) = − − ∼ Dih(2(q 1)) and O2 (q) = Dih(2(q + 1)). We also recall that, for p a prime, a finite ∈ group X is p-closed provided that Op(X) Sylp(X).
Proposition 2.2. K ∈ Suppose that H is a finite simple -group, that S Syl5(H) and that the following hold. (i) S is elementary abelian of order 25. # ∼ (ii) For x ∈ S , either CH (x) is 5-closed or CH (x) = 5 × Alt(6), and both possibilities occur. ∼ Then H = Alt(11).
Proof. First we observe that C (S)=C (S)=S, where y ∈ S# is such that ∼ H CH (y) CH (y) = 5 × Alt(6). Therefore CH (S)=S. We next elucidate the structure of CH (x) # for x ∈ S in the case when CH (x) is 5-closed. Since S = O5(CH (x)) is elementary abelian of order 25, from the structure of GL2(5) we infer that CH (x)/CH (S)is isomorphic to a subgroup of Z4.Thus,asCH (S)=S, it follows that CH (x)is isomorphic to a subgroup of 5 × Frob(20), where Frob(20) is the Frobenius group characteristic 5 identification of the lyons group 131 of order 20. Hence we have the following. Suppose that z ∈ S# and that N is a non-trivial normal subgroup ∼ ∼ of CH (z) with N ∩z = 1. Then N = Alt(6) and CH (z) = 5 × Alt(6) or N is a normal subgroup of the Frobenius group of order 20. (∗) We now use the fact that H is a simple K-group and consider each of the possibilities for H. Assume that H is a Lie type group defined over a field of order m ∼ # q = r . Since CH (y) = 5 × Alt(6) is not 5-constrained for some y ∈ S , we see that Downloaded from https://academic.oup.com/jlms/article/69/1/128/917235 by guest on 27 September 2021 r =5. We first deal with the classical groups. Thus assume that H is a normal sub- group of a quotient of a classical group H defined over a field of characteristic not 5. Let V be the natural module for H and let f be the associated form defining H (so f is trivial if H is SLn(q)). Thus H = K/Z(K), where K is a normal ∼ subgroup of H. If 5 divides |Z(K)|, then (see [3]) we must have H = An(q), with 5 ∼ 2 dividing both n +1andq − 1orH = An(q) with 5 dividing both n +1andq +1.In 2 particular, n 4. Scrutinizing the orders of An(q) (respectively An(q)) we see that (q − 1)4 (respectively (q +1)4) must then be a divisor of |H|, against assumption | | ∈ ∼ (i). Consequently 5 does not divide Z(K) .LetS Syl5(K). Of course S = S. # Let z ∈ S . Then V = CV (z) ⊥ [V,z]. We claim the following. # dim CV (z) 1 for all z ∈ S . (†) ∈ # | Suppose that dim CV (z) 2 for some z S and define L =O(CV (z),f CV (z)). By Witt’s lemma [1, Section 20, p. 81], L H and so L CH(z). In the case when f is a symplectic form, we note that H = K. Because of the limitations on the normal subgroups of CH (z) as given in (∗), we infer that one of the following holds. + ∼ Z (a) q =2, L =O2 (2) = 2 and dim CV (z)=2. ∼ − ∼ (b) q =4, L = O (4) = Dih(10) and dim C (z)=2. ∼ 2 V (c) q =9, L = O3(9) and dim CV (z)=3. ∼ − (d) q =3, L = O4 (3) and dim CV (z)=4. In case (a), since CV (z) admits S, we see that S must centralize CV (x)and # so CV (x)=CV (S)=CV (z) for all x ∈ S , which is a contradiction. For case (b) we select l ∈ L of order 5 and note that 2 dim CV (l) dim [V,z] = dim V − 2 ∼ + so that dim V = 4. Since V is an orthogonal space, we either have H = Ω4 (4) ∼ − ∼ or H = Ω4 (4) = SL2(16). In the first case we have a contradiction to H being a simple group and in the second case we contradict the Sylow 5-subgroups of H being elementary abelian. Thus case (b) cannot hold. Suppose now that case (c) holds. Then as in case (b) we infer that dim V 6 − 3 ∼ + and dim V 5. Therefore, as O6 (9) has Sylow 5-subgroups of order 5 , H = O6 (9) + − × − or O5(9). Now O6 (9) contains a subgroup isomorphic to J =O2 (9) O4 (9), and so − the centralizer of the element of order 5 in the left-hand factor of J contains O4 (9), ∗ ∼ + × which contradicts ( ). For H = O5(9), we see that the subgroup O2 (9) O3(9) contains elements of order 10 centralizing O3(9) and this is also against (∗). Finally ∼ we consider case (d). Then as before, we have dim V 8andsoH = On(3), where ∼ + n 8. Through examining the orders of these groups we have H = O8 (3). Now in ∼ # H we have CH (z) = 5 × Alt(6) for all z ∈ S (see the atlas [3]) and so this group fails to satisfy the full force of condition (ii). This contradiction shows that (d) cannot hold, so completing the verification of (†). 132 christopher parker and peter rowley
From V = CV (z), 1 = z
# (†) implies that dim V 6 and that CV (z) = 0 for some z ∈ S . Since CV (z)isa non-degenerate space, we immediately have H is not a symplectic group. Let # x ∈ S be such that CV (x) = 0. Then, by (†) and the decomposition of V as centralizers from elements of S,wehaveCV (x) = CV (S). In particular, S/x ∈ # operates faithfully on the one-dimensional space C (x). Furthermore, if y S Downloaded from https://academic.oup.com/jlms/article/69/1/128/917235 by guest on 27 September 2021 ∼ V is such that CH (y) = 5 × Alt(6), then as CV (y) admits CH (y), we see that CV (y)=CV (S), a contradiction. Thus CV (y) = 0 and we infer that dim V 5. Assume that H is a linear group. Then as S/x operates faithfully on CV (x), we have 5 divides q − 1. Noting that (q − 1)2 divides |H| whenever dim V 3, we have a contradiction to (∗) in this case. Since the Sylow 5-subgroups of SL2(q) are cyclic we now know that H is not a linear group. Simple orthogonal groups of dimension at most 5 are isomorphic to linear or symplectic groups, and so we do not need to consider these possibilities. Therefore we finally assume that H is a unitary group. Then 5 divides either q − 1orq + 1. However, for n 2, J =SU2(q) SUn(q)and so, for z ∈ J of order 5, dim CV (z) 1 implies that n = 2 or 3. Plainly n cannot be 2, so n = 3. Then, through considering the order of SU3(q), we infer that 5 divides 2 q + 1. Since U3(q) contains an abelian subgroup of order (q +1) /(3,q+ 1), we have q = 4. Looking in the atlas [3], we see that U3(4) has no 5-elements with centralizer 5 × Alt(6). This completes the investigation of the classical groups. ∼ We now consider the cases where H is an exceptional Lie type group. If H = E6(q), 21 then H contains a parabolic subgroup of shape [q ].Z(q−1,2).PSL6(q). Thus there is an element of order 5 whose centralizer involves SL2(q) and so we infer that q =2 3 or 4. Furthermore, 5 divides |E6(4)|, so giving q = 2, and then the maximal rank subgroup SL6(2) × SL2(2) reveals a contradiction. That H is not of type E7 or E8 ∼ 2 follows from the result for E6. Assume that H = E6(q). Then the Levi complement − × ∼ 2 × involving O4 (q) SL3(q) = SL2(q ) SL3(q) contains elements of order 5 in the first factor and so it is impossible for the centralizer of that element to have structure as ∼ indicated in (∗). Next assume that H = F4(q). In this case H contains a maximal ∼ − × rank subgroup of type B4(q) = Ω9(q) which in turn contains a subgroup O4 (q) | − | ∗ O5(q). Once again, as 5 divides O4 (q) , we have a contradiction to ( ). Assume ∼ 2 m that H = F4(2 ) with m>1. Then H contains a subgroup isomorphic to the Tits 2 group T = F4(2) . Now a Sylow 5-subgroup X of T is elementary abelian of order 2 . 25 and NT (X)=5 :4 Alt(4) (see [3]) . Hence T has exactly one conjugacy class of elements of order 5 and so it is impossible for H to satisfy condition (ii). ∼ 6 6 2 6 2 2 2 Suppose that H = G2(q). Then |H| = q (q − 1)(q − 1) = q (q − 1) (q + q +1) (q2 − q + 1). It is easy to see that the factors in this decomposition are coprime. Furthermore, there are cyclic subgroups of order (q2 + q + 1) and (q2 − q +1)inH (as they are the orders of maximal tori). Thus we see that if there is an elementary abelian subgroup of H of order 25, then 25 divides (q2 − 1)2.ThusS can be found in a subgroup of H isomorphic to SL (q) ∗ SL (q). Since S is self-centralizing we infer 2 2 ∼ that q = 4 and in this case we consult the atlas [3] to obtain a contradiction. If H = 3 12 2 2 2 2 2 D4(q), then we may factorize the order of H as q (q + q +1) (q − q +1) (q − 1)2(q4 − q2 + 1). Note that 5 does not divide any pair of terms in this factorization. Thus the order of the Sylow 5-subgroup divides (q2 +q +1)2,(q2 −q +1)2,(q2 −1)2 4 2 3 or q − q + 1. From the list of maximal subgroups of D4(q) given in [6], we know characteristic 5 identification of the lyons group 133 that if 5 divides q4 − q2 + 1, then the Sylow 5-subgroups are cyclic, whereas in the other cases they are homocyclic with two generators. If 5 divides (q2 ± q +1)2, then all the cyclic subgroups of order 5 in S are conjugate and so condition (ii) is not satisfied. Thus we may assume that 5 divides (q2 − 1)2. However, then the Sylow 3 5-subgroup of H can be seen in SL2(q )∗SL2(q) and, in particular, the centralizer of 3 ∼ 3 some element of order 5 contains SL2(q ), a contradiction to (∗). Thus H = D4(q). 2 n 2 m Finally we note that the Ree groups B2(2 )and G2(3 ) have cyclic Sylow 5-subgroups [4, Theorem 6.5.4, Theorem 6.5.5] and with this we have completed the investigation of the Lie type groups. Downloaded from https://academic.oup.com/jlms/article/69/1/128/917235 by guest on 27 September 2021 Suppose now that H is an alternating group Alt(n). Then, as |S| = 25, we have # ∼ 10 n 14. Let y ∈ S be such that CH (y) = 5 × Alt(6). Since the centralizer of elements of cycle type 52 does not contain Alt(6), we infer that y is a 5-cycle and then that n = 11. This is our example. Finally we examine the sporadic simple groups using [4, Tables 5.3.1–5.3.26], and we see that there are no possibilities for H. This completes the proof of the proposition.
3. The amalgam L From here on we suppose that G satisfies the hypothesis of Theorem 1.1. For γ ∈{α, β},wesetQγ = O5(Lγ ), Zγ = Z(Qγ )andSαβ = Lα ∩ Lβ. Notice that, by ∈ hypothesis, Sαβ Syl5(G). We now begin the proof of Theorem 1.1 with some observations about the structure of L. Since G is of local characteristic 5, both Lα and Lβ are 5-constrained. Further, it is easy to check that L is a symplectic amalgam [9].
Lemma 3.1. The following hold. (i) Sαβ = QαQβ, Zβ = Z(Sαβ)=Z(Lβ) and |Zβ| =5. ∼ . (ii) Zα =Ω1(Z(Qα)) is a natural SL2(5)( = 2 Alt(5))-module. (iii) Qβ is extraspecial and Lβ operates irreducibly on Qβ/Zβ. ∩ | | (iv) Qα Qβ =[Qβ,Sαβ] and CQβ /Zβ (Sαβ) =5.
(v) NLβ (Zα)=NLβ (Sαβ). (vi) Qα and Qβ are both characteristic subgroups of Sαβ.
Proof. Since L is a symplectic amalgam, parts (i) and (ii) follow from [9, Lemma 4.15] and (iii) holds as Lβ is 5-constrained (and the minimal dimension of a GF(5)2 .Alt(6) module is 4). Part (iv) now follows from (iii) and Lemma 2.1(ii). From (iv) it follows that Zα is the second centre of Sαβ. Hence NLβ (Sαβ) NLβ (Zα) and then, by the structure of 2 .Alt(6), we get equality. Finally we come to (vi), since Zα is a characteristic subgroup of Sαβ and Qα = CSαβ (Zα), so also is Qα. Using (iv) again gives Qβ/Zβ is the unique abelian subgroup of Sαβ/Zβ of order 4 5 ,andsoQβ is also characteristic in Sαβ.
Lemma 3.2. We have the following. ∼ (i) NG(Zβ)/CG(Zβ) = Z4. (ii) CG(Zβ)=Lβ. (iii) NG(Qβ)=NG(Zβ). (iv) NG(Qβ)/Qβ ∼ 4.Sym(6). 134 christopher parker and peter rowley
Proof. Let N = NG(Zβ)andC = CG(Zβ). Then, as |Zβ| =5, N/C is cyclic of order at most 4. However, in L , we see, using Lemma 3.1(ii), that N (Z )/ ∼ α Lα β Sαβ = Z4. Hence (i) holds. ∈ By Lemma 3.1(i), Lβ C.LetQ = O5(C). Then, as Lβ C and Sαβ Syl5(G), Q Qβ. Furthermore, by Lemma 3.1(iii), Q = Qβ or Q = Zβ. The latter possibility is impossible because C is 5-constrained. Therefore O5(C)=Qβ.NowQβ is 5 extraspecial of order 5 , and thus C/Qβ embeds into Sp4(5). A glance at the atlas ∼ ∈ [3] shows that C = Lβ or C/Qβ = Sp4(5). Because we have Sαβ Syl5(G), the latter possibility fails. Thus (ii) is proven and, as Zβ is a characteristic subgroup of Qβ, Downloaded from https://academic.oup.com/jlms/article/69/1/128/917235 by guest on 27 September 2021 (iii) also follows. The structure of NG(Qβ)/Qβ can now be read from the atlas [3].
Lemma 3.3. The following hold. (i) N (Q )=N (Z )=L N (Z ). G α G∼ α α Lβ α (ii) NG(Qα)/Qα = GL2(5).
Proof. From Lemma 3.1(ii), we have NG(Qα) NG(Zα). Furthermore, since 2 Zα is elementary abelian of order 5 ,wehaveNG(Zα)/CG(Zα) is isomorphic to a subgroup of GL2(5) which contains SL2(5). In particular, because Lα operates transitively on the elements of order 5 in Zα,wehave
NG(Zα)=LαNCG (Zβ )(Zα)=LαNLβ (Zα). Since, by Lemma 3.1(v) and (vi), NLβ (Zα)=NLβ (Sαβ) NG(Qα), part (i) holds. We have NG(Zα)/CG(Zα) is isomorphic to a subgroup of GL2(5) which contains SL2(5) and from (i), NG(Zα)=LαNLβ (Zα). Also NLβ (Zα)=NLβ (Sαβ) and NLβ (Sαβ)/Sαβ is cyclic of order 4. Since O5,2(Lβ)invertsQβ/Zβ, we conclude that N (S )/S acts faithfully on Z /Z and, of course, Lβ αβ αβ ∼ α β centralizes Zβ.ThusNG(Zα)/CG(Zα) = GL2(5). Now, using Lemma 3.2(ii), we get C (Z ) C (Z ) S which gives Q C (Z ) core (S )=Q .Thus G α L∼β α αβ α G α NG (Zα ) αβ α NG(Qα)/Qα = GL2(5), completing the proof of the lemma.
For γ ∈{α, β} define Nγ = NG(Qγ ).
Lemma 3.4. We have Nα ∩ Nβ = NG(Sαβ).
Proof. By Lemma 3.1(v), both Qα and Qβ are characteristic subgroups of Sαβ, whence Nα ∩ Nβ NG(Sαβ). On the other hand, as Qα = Qβ,wehaveSαβ = QαQβ and so Nα ∩ Nβ NG(Sαβ).
From Lemma 3.4 we see that Nα ∩ Nβ = NG(Sαβ) has index 36 in Nβ and 6 in Nα. Let Γ = Γ(G, Nα,Nβ,Nα ∩ Nβ) be the coset graph of the amalgam A(Nα,Nβ,Nα ∩ Nβ)inG. Therefore Γ is a bipartite graph which admits G acting edge transitively.
4. A T -apartment in Γ ∼ Let T be a fixed complement to S in N ∩ N . Then, as N /Q = GL (5), we ∼ αβ α β α α 2 have T = Z4 × Z4. Let Θ be the connected component of the fixed graph of T which contains the edge {α, β} (so if Γ were a building, then Θ would be an apartment, characteristic 5 identification of the lyons group 135
σ1r rρ
αr τ r rσ3
β r σ2r Figure 1. Downloaded from https://academic.oup.com/jlms/article/69/1/128/917235 by guest on 27 September 2021 which in this rank 2 case would be an ordinary polygon). For γ ∈ Γ, Γ(γ) denotes the neighbours of γ in Γ and, if γ ∈ Θ, we set Θ(γ)=Θ∩ Γ(γ). We easily see that T normalizes exactly two Sylow 5-subgroups of Nα and precisely four Sylow 5-subgroups of Nβ.
Lemma 4.1. The following hold.
(i) NNβ (T )/T is elementary abelian of order 4 and acts regularly on Θ(β).
(ii) NNα (T )/T is of order 2 and acts regularly on Θ(α).
Proof. Once we observe that NQα (T )=NQβ (T ) = 1, this boils down to an elementary calculation in Sym(6) in the first case and GL2(5) in the second.
Set NT = NNα (T ),NNβ (T ) . Then, by Lemma 4.1, NT has exactly two orbits on Θ. Furthermore, we see that Θ is a subgraph of Γ which has vertices which are in the same NT -orbit as β of valency 4 and otherwise the vertices have valency 2. We abuse notation slightly and, for γ ∈ Θ, set Nγ = Nγ /Qγ . Denote the three ∈ involutions of Ω1(T )byt1, t2 and t3.Forγ Θ, we define Jγ = CNγ (s), where ∈{ } ∈ ∞ s t1,t2,t3 is such that s Z(Nγ )andKγ = Jγ . We have the following elementary result about the structure of Jγ and Kγ .
Lemma 4.2. Assume that γ ∈ Θ. Then one of the following holds. ∼ . (i) γ is NT -conjugate to β and Jγ ∼ 5.4.Sym(6) with Kγ = 2 Alt(6). ∼ . (ii) γ is NT -conjugate to α and Jγ ∼ 5.GL2(5) with Kγ = 2 Alt(5).
Assume that t1 ∈ Z(Nβ)andt2 ∈ Z(Nα). Let τ ∈ Θ(α)\{β} (so τ is uniquely determined). As t1 centralizes Zβ and inverts Zτ and t2 inverts Zα ( Zτ ), we see that t3 centralizes Zτ and so t3 ∈ Z(Nτ ) with t3 inverting Qτ /Zτ . Let us denote the \{ } three vertices in Θ(τ) α by σ1, σ2 and σ3. Then, as NNτ (T )swapst1 and t2 and ∈ centralizes t3, we may assume without loss of generality that t2 Z(Nσ1 ) and that, ∈ \{ } for j =2, 3, t1 Z(Nσj ). Now let ρ be the unique member of Θ(σ1) τ and note that t1 ∈ Z(Nρ). We depict these vertices of Θ in Figure 1.
Lemma 4.3. Up to interchanging the roles of σ2 and σ3, the following hold. (i) If γ,δ ∈ Θ(τ) and Zγ = Zδ, then γ = δ. ∩ ∩ (ii) Qσ2 Qσ3 = Zσ2 Zσ3 and Qα Qσ1 = ZαZσ1 . (iii) Zρ Kβ and Zβ Kρ. (iv) Sατ ,Sτσ1 acts irreducibly on Qτ /Zτ . (v) Zβ Qσ2 , Zβ Qσ3 , Zρ Qσ3 and Zρ Qσ2 . (vi) Kσ2 Kβ and Kσ3 Kρ. 136 christopher parker and peter rowley Proof. Part (i) is an immediate consequence of Lemma 2.1(iii). By the selection of σ2 and σ3,forj =2, 3, we have t1 Z(Nσj ); therefore ∩ [Qτ ,t1] Qσj and Zσj is inverted by t1. In particular, Zσ2 Zσ3 Qσ2 Qσ3 and, 3 as Qτ is extraspecial and, by (i), Zσ2 Zσ3 has order 5 , we conclude that the first part of (ii) holds. Applying a similar argument with t2, α and σ1 in place of t1, σ2 and σ3, we obtain the second part of (ii). By (ii), Zρ Zσ1 Qα Sαβ and Zβ Zα Qσ1 Sσ1ρ.Now,Zρ is centralized by t1 and so, if Zρ Qβ, we then have Zρ = CQβ (t1)=Zβ, which contradicts (i).
Therefore Zρ Qβ. However, Zρ centralizes t1 and so Zρ Jβ.LetY = ZβZρ. Then, Downloaded from https://academic.oup.com/jlms/article/69/1/128/917235 by guest on 27 September 2021 ∈ by Lemma 4.2(i), Y Syl5(Jβ). Assuming that Zρ Kβ, we see that T normalizes three cyclic subgroups of Y of order 5. Since T/t1 operates faithfully on Y ,we have a contradiction. Thus Zρ Kβ. A similar argument shows that Zβ Kρ. Hence (iii) holds. Put X = Sατ ,Sτσ and assume that X does not operate irreducibly on Qτ /Zτ . 1 ∼ . ∼ Since σ1 = α, Lemma 2.1(i) implies that X = 2 Alt(5) = SL2(5) and Qτ /Zτ is an indecomposable GF(5)X-module whose composition factors consist of two ∩ natural X-modules. Now Rσ1 = Qτ Qρ is a normal subgroup of Sτσ1 of order 3 5 . Hence, by Lemma 2.1(i) and (ii), Rσ1 is invariant under X. However, then Zβ Zα Rσ1 Qρ, contrary to part (iii). Let F = CT (Zβ)andE = CT (Zρ). Then, by (iii), [E,Kβ] = 1 and [F, Kρ]=1. Hence E = F . In particular, ZαZσ1 /Zτ is a direct sum of two non-isomorphic T - ∩ modules both of which are centralized by t1. Since Qσ2 Qσ3 is inverted by t1, we have Qτ = ZαZσ1 Zσ2 Zσ3 and Qτ /Zσ2 Zσ3 is a direct sum of non-isomorphic T - ∩ ∩ modules. It follows that Qτ Qσ2 and Qτ Qσ3 are the only proper T -invariant 4 subgroups of order 5 in Qτ which contain Zσ2 Zσ3 . However, then, after changing ∩ ∩ notation if necessary, Qτ Qσ2 = ZαZσ2 Zσ3 and Qτ Qσ3 = Zσ1 Zσ2 Zσ3 . Since
Qτ = ZαZσ1 Zσ2 Zσ3 , this proves (v). ∩ By (v), Zβ Qσ2 .ThusZβ = Jσ2 Qσ2 and so Jσ2 NG(Zβ)=Nβ. However, then Jσ2 Jβ and the first part of (vi) follows as Kσ2 is perfect. A similar argument proves the second part of (vi) and completes the verification of the lemma.
In passing, we note that Lemma 4.3(v) shows that β = ρ. Now assume that γ ∈ Θ is NT -conjugate to β and assume that s ∈{t1,t2,t3} with s ∈ Z(Nγ ). Define Φγ to be the set of vertices ψ ∈ Θ which have distance 4 from γ and have s ∈ Z(Nψ).
Lemma 4.4. Assume that µ ∈ Θ is NT -conjugate to β. Then the set of Sylow 5-subgroups of Kµ which are normalized by T coincides with {Zψ|ψ ∈ Φµ}.In particular, |Φµ| =4.
Proof. It suffices to prove the result for β. Since NNβ (T ) is transitive on the set of Sylow 5-subgroups of Kβ which are normalized by T , the first part of the statement follows from Lemma 4.3(iii). Thus we have |Φβ| 4. Following the argument in the development of the vertices in Figure 1, we see that | | | | Φβ NNβ (T )/T =4.
We now develop a larger portion of Θ. From Lemma 4.1(ii), N (T ) operates Jσ 2 transitively on Θ(σ2) and, by Lemma 4.3(vi), N (T ) N . Jσ 2 β characteristic 5 identification of the lyons group 137
σ1r rρ
αr τ r rσ3
β r σ2 r
τ Downloaded from https://academic.oup.com/jlms/article/69/1/128/917235 by guest on 27 September 2021 α rrr σ3
σ1 r r ρ Figure 2.
σ1r rρ r
αr τ r rσ3 r ξ r
β r σ2 r r r
τ rrr σ3 r r
r r r
Figure 3.
Thus, conjugating the subgraph of Θ in Figure 1 by N (T ), we obtain the sub- Jσ 2 graph shown in Figure 2. Our main result in this section is the following lemma.
Lemma 4.5. [Kσ2 ,Kσ3 ] =1.
Proof. Aiming for a contradiction, we assume that Kσ2 and Kσ3 do in fact commute. Then, by the construction of the subgraph in Figure 2, we have Kσ3 = Kσ Kρ . We now conjugate the graph in Figure 2 by NKσ (T ). This fixes the 3 3 vertices ρ, σ3, σ3 and ρ . Thus we obtain Figure 3. Notice that t2 ∈ Z(Nτ )andt2 ∈ Z(Nξ). Therefore ξ ∈ Φτ . However, then, as there are two paths of length 4 in Θ from τ to ξ,wehave|Φτ | < 4 in contradiction to Lemma 4.4.
Obviously Lemmas 4.3(vi) and 4.5 together imply the following corollary.
Corollary 4.6. [Kρ,Kβ] =1 . 138 christopher parker and peter rowley
5. Centralizers of 5-elements
We continue the notation of the previous section. Set E = ZβZρ. Notice ∈ ∩ that by Lemma 4.3(iii), E Syl5(Jβ) Syl5(Jρ). Also Lemma 4.3(v) shows that ∈ ∩ E Syl5(Jσ2 ) Syl5(Jσ3 ). We first record a fact which follows from Lemma 2.1(i) and (iv) and Lemma 4.3(iv). Lemma 5.1. ∈ ∈ There exists x E such that x ξ∈Γ(τ) Zξ. Downloaded from https://academic.oup.com/jlms/article/69/1/128/917235 by guest on 27 September 2021
Recall that for a prime p,ap-element x of a finite group X is p-central if CX (x) contains a Sylow p-subgroup of X.
Lemma 5.2. For x ∈ E, one of the following holds. ∈ (i) x ξ∈Γ(τ) Zξ, CG(x) is 5-closed and O5(CG(x)) = CQτ (x). ∈ (ii) x ξ∈Γ(τ) Zξ and x is 5-central.
∈ ∈ # Proof. If x Zξ for some ξ Γ(τ), then, as Lξ operates transitively on Zξ , x is ∈ 5-central and we are done. Thus we may suppose that x ξ ∈ Γ(τ) Zξ. Since E Qτ | | 4 and Qτ is extraspecial, we have CQτ (x) =5 . Set C = CQτ (x)andF = NG(C). Since C is not abelian, we have [C, C]=Zτ . Therefore Lemma 3.2(iii) implies that F NG(Qτ ). Furthermore, we obviously have Qτ F . Since Z(C)=Zτ ,x,the | | ∈ choice of x implies that F/Qτ has order coprime to 5. However, then Qτ Syl5(F ) ∈ and, as Qτ does not centralize x, we conclude that CQτ (x) Syl5(CG(x)). Let X = O5(NG(x)). Then X C. Assume that X O5(CG(x)) = CQτ (x), as claimed in (i). # Lemma 5.3. Assume that x ∈ E is 5-central. Then t = t1 inverts O5(CG(x))/ x . In particular, E O5(CG(x)) and E = CEO5(CG (x))(t). ∼ . Proof. Clearly t1 ∈ CG(x) and, by Lemma 3.2(ii), CG(x)/O5(CG(x)) = 2 Alt(6). . ∼ Since 2 Alt(6) = SL2(9) has a unique involution, we have tO5(CG(x)) ∈Z(CG(x)/ O5(CG(x))) and so the result follows. 6. The centralizer of an involution We now set t = t1 and K = CG(t). Our objective is to determine the structure of K. We first deal with the normal 5-subgroups of K. characteristic 5 identification of the lyons group 139 Lemma 6.1. O5 (K)=t. Proof. Set U = O5 (K). Because E is elementary abelian of order 25, we have # U = CU (x) | x ∈ E . If x ∈ E is not 5-central, then CU (x) CG(x), which is 5-closed by Lemma 5.2. Thus [C (x),O (C (x)) ∩ K] U ∩ O (C (x)) = 1, U 5 G 5 G Downloaded from https://academic.oup.com/jlms/article/69/1/128/917235 by guest on 27 September 2021 so, as E O5(CG(x)) ∩ K,wehave[CU (x),E] = 1. In particular, as E contains 5-central elements, we may as well assume that x is 5-central. Hence CG(x)/ ∼ . O5(CG(x)) = 2 Alt(6) and CU (x) O5(CG(x)). Since, by Lemma 5.3, E . O5(CG(x)), we get CU (x) t by considering the structure of 2 Alt(6). Thus U = t, as claimed. Lemma 6.2. ∈ E Syl5(K). Proof. Aiming for a contradiction, we suppose that E0 is a 5-subgroup of K # which normalizes E with E0 >E. Then E0 centralizes some x ∈ E .Ifx is not 5- ∩ central, then E0 CQτ (x) by Lemma 5.2. Since t inverts Zτ , E0 Zτ = 1 and hence E0 is abelian. However, then CQτ (x)=E0Zτ must also be abelian, a contradiction. Therefore, x is 5-central. Then E0 CG(x) and we obtain a contradiction from Lemma 5.3. Hence no such E0 exists and the result is valid. We now let F = F ∗(K). By Lemma 6.1, 5 divides the order of F and consequently, by Lemma 6.2, E1 = F ∩ E = 1. Assume that E1 = O5(Jβ)=Zβ. Then, ∈ by Lemma 4.3(iii), Kρ F . Moreover, since O5 (K)= t and E1 Syl5(F ), we see that there is a unique component contained in F . Since Kρ does not split over t, F is quasi-simple. Because G is K2-group, F satisfies the Schreier 2 property whence Kβ FCK (F )=F . However, then ZβKβ F implies that 5 divides |F |, a contradiction. Hence E1 = Zβ and similarly we can prove that E1 = Zρ. ∈{ } Jγ Since Lemma 4.2 implies that, for γ β,ρ , Kγ E1 ,wehave Kβ,Kρ F . Furthermore, Kβ,Kρ E(K), and so, by Lemma 4.3(iii), E E(K). If F has more than two components, then, as the Sylow 5-subgroups of K have order 25, Lemma 6.1 implies that F contains exactly two components, and it is easy to see that Kβ is contained in one of them and Kρ is contained in the other. However, then [Kβ,Kρ] = 1, contrary to Corollary 4.6. Thus F has a unique component and is consequently quasi-simple with centre t. ∼ Employing Proposition 2.2, we see that F = 2 .Alt(11), and so either K = F or K ∼ 2 . Sym(11). In the latter case, however, we note that the centralizer of an elementoforder5projectingtoa5cycleinSym(11) has centralizer 5 × 2 . Sym(6), ∼ . which is not the case by Lemma 3.2(ii). Thus CG(t)=K = 2 Alt(11), and this completes the proof of Theorem 1.1. References 1. M. 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Stroth, ‘Finite groups of local Downloaded from https://academic.oup.com/jlms/article/69/1/128/917235 by guest on 27 September 2021 characteristic p: an overview’, Proceedings of Durham Conference on Finite Groups, 2001, to appear. 9. C. W. Parker and P. J. Rowley, Symplectic amalgams (Springer, 2002). 10. C. W. Parker and P. J. Rowley, ‘Local characteristic p completions of weak BN-pairs for p odd’, in preparation. 11. M. A. Ronan and G. Stroth, ‘Minimal parabolic geometries for the sporadic groups’, European J. Combin. 5 (1984) 59–91. 12. C. C. Sims, ‘The existence and uniqueness of Lyons’ group’, Finite groups ’72 (ed. T. Gagen, M. P. Hale and E. E. Shult, North-Holland, 1973) 138–141. Christopher Parker Peter Rowley School of Mathematics and Department of Mathematics Statistics University of Manchester University of Birmingham Institute of Science and Technology Edgbaston PO Box 88 Birmingham B15 2TT Manchester M60 1QD [email protected] [email protected]