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AN IMPROVED 3-LOCAL CHARACTERISATION OF McL AND ITS AUTOMORPHISM GROUP 2 CHRIS PARKERANDGERNOT STROTH 1 0 2 n a Abstract. This article presents a 3-local characterisation of the spo- J radic simple group McL and its automorphism group. The theorem is 5 underpinnedby two further identification theorems theproofs of which arecharactertheoretic.Themaintheoremisappliedinourinvestigation ] R of groups with a large 3-subgroup [10]. G . h 1. Introduction t a This article extends earlier work of Parker and Rowley [8] in which the m McLaughlin sporadic simple group McL and its automorphism group are [ characterised by certain 3-local information. Suppose that p is a prime and 1 G is a finite group. Then the normalizer of a non-trivial p-subgroup of G is v called a p-local subgroup of G. AsubgroupM of G is said to beof character- 7 istic p provided F∗(M) = O (M) where F∗(M) is the generalized Fitting 7 p 0 subgroup of M. See [1] for the fundamental properties of the generalized 1 Fitting subgroup. The group G is of local characteristic p if every p-local . 1 subgroup of G has characteristic p and G is of parabolic characteristic p 0 if every p-local subgroup of G which contains a Sylow p-subgroup of G has 2 characteristic p.Thedifferencebetween thesetwogrouptheoretic properties 1 : is the difference between the characterisation theorem presented in [8] and v thetheorempresentedinthisarticle. Themaintheoremoftheformerarticle i X essentially assumes that the group under investigation is of local character- r istic 3. The theorem we prove here in essence only assumes that the group a G has parabolic characteristic 3 though it is not necessary to articulate this explicitly in the statement of the theorem. Theorem 1.1. Suppose that G is a finite group, S Syl (G), Z = Z(S) ∈ 3 and J is an elementary abelian subgroup of S of order 34. Further assume that (i) O3′(N (Z)) 31+4.2.Alt(5); (ii) O3′(NG(J)) ≈34+.Alt(6); and G ≈ (iii) C (O (C (Z))) O (C (Z)). G 3 G 3 G ≤ Then G= McL or Aut(McL). ∼ Date: January 6, 2012. 1 2 ChrisParker andGernotStroth The main theorem of [8] carries the additional hypothesis C (O (C (x))) O (C (x)) G 3 G 3 G ≤ for all x J#. Hence to prove Theorem 1.1, we just need to show that this ∈ statement is a consequence of the assumptions in Theorem 1.1. Theorem1.1isappliedinourinvestigationofexceptionalcaseswhicharise in the determination of groups with a large p-subgroup [10]. A p-subgroup Q of a group G is large if and only if (L1) F∗(N (Q)) = Q; and G (L2) for all non-trivial subgroups U of Z(Q), then N (U) N (Q). G G ≤ Itisanelementary observationthatmostofthegroupsofLietypeincharac- teristic phave alarge p-subgroup.TheonlyLie typegroupsin characteristic p and rank at least 2 which do not contain such a subgroup are PSp (2a), 2n F (2a) and G (3a). It is not difficult to show that groups G which contain 4 2 a large p-subgroup are of parabolic characteristic p (see [10, Lemma 2.1]). Thework in[7]beginsthedetermination of thestructureof the p-local over- groups of S which are not contained in N (Q). The idea is to collect data G about the p-local subgroups of G which contain a fixed Sylow p-subgroup and then using this information show that the subgroup generated by them is a group of Lie type. However sometimes one is confronted with the following situation: some (but perhaps not all) of the p-local subgroups of G containing a given Sylow p-subgroup S of G generate a subgroup H and F∗(H) is known to be isomorphic to a Lie type group in characteristic p. Usually G = H. To show this, we assume that H is a proper subgroup of G, andfirstestablish that H contains all the p-local subgroupsof G which contain S. The next step then demonstrates that H is strongly p-embedded in Gatwhichstage [9]isapplicable anddelivers G= H.Thelast twosteps are reasonably well understood,at least for groups with mild extra assumptions imposed. However it might be that the first step cannot be made. Typically this occurs only when N (Q) is not contained in H. The main theorem of G this article is applied in just this type of situation. Specifically, it is applied inthecasethatprimep = 3andF∗(H)isthegroupPSU (3). Inthiscase in 4 F∗(H) the large 3-subgroup Q is extraspecial of order 35 and is the largest normal 3-subgroup in the normalizer of a root group in F∗(H). In this con- figuration we are not able to show that N (Q) H, as is demonstrated by G ≤ noting that PSU (3) is a subgroup of McL. In fact in [10] we show that, if 4 F∗(H) = PSU (3) and N (Q) H, then F∗(G) = McL,Co , or PSU (2). ∼ 4 G 6≤ ∼ 2 6 It is precisely for the identification of McL that we need the result of this paper. The route to prove Theorem 1.1 is paved by two lesser results. These theorems state that under certain hypotheses a subgroup H of a group G is actually equal to G. The two theorems are as follows: 3 Theorem 1.2. Suppose G is a finite group and H is a subgroup of G with H = (1,2,3),(4,5,6),(7,8,9),(1,2)(4,5),(1,2)(7,8) 33 : 22. ∼ h i ≈ Let T Syl (H) and assume N (T) = H and C (t) H for all t T#. ∈ 3 G G ≤ ∈ Then G= H. The hypothesis about the embedding of H in G in Theorem 1.2 is equiv- alent to saying that H is strongly 3-embedded in G. We denote by K the subgroup of Alt(9) which normalizes J = (1,2,3),(4,5,6),(7,8,9) . h i Thus K = (1,2,3),(1,4,7)(2,5,8)(3,6,9),(1,2)(4,5),(1,4)(2,5)(3,6)(7,8) . h i Theorem 1.3. Suppose G is a finite group, H G with H = K and ≤ ∼ J = O (H). If C (j) H for all j J# and J is strongly closed in H with 3 G ≤ ∈ respect to G, then G = H. The proofs of both of these theorem exploit the methods introduced by Suzuki which permit parts of the character table of G to be constructed. Details of the theory behind the Suzuki method are very well presented in [4]. The embedding properties in both Theorems 1.2 and 1.3 which assert that centralizers of certain elements of H are contained in H are precisely the requirements needed to make the Suzuki theory of special classes work. The result of the Suzuki method are fragments of possible character tables for G. These fragmentary tables provide all the character values on certain elements of order3 inH.Thiscalculation is performedbyhandand ingreat detail for the proof of Theorem 1.2. For the proof of Theorem 1.3, however, as we start to build up the required decompositions there is an overwhelm- ing number of possibilities and so we have performed this calculation using Magma [2]. The result of this computation (which takes a few hours) is four fragments of possible character tables for G. However this statement is rather disingenuous as in fact each fragment represents many possible character tables as the entries of the partial tables are only known up to sign choices. Recall that for x,y,z G the G-structure constant ∈ a = (a,b) xG yG ab = z xyz |{ ∈ × | } is determined by the character table of G by the following equation G χ(x)χ(y)χ(z−1) a = | | . xyz C (x) C (y) X χ(1) G G | || |χ∈Irr(G) Noticethatifweselectx,y H suchthatC (x)andC (y)arecontained in G G ∈ H, then we know C (x) = C (x) and C (y) = C (y). Furthermore, G H G H | | | | | | | | for certain choices of x,y and z, we know all the character values of x, y and z. Thus the only unknown quantity on the left hand side of the structure constant equation is G. The proofs of both the above theorems pivot on | | 4 ChrisParker andGernotStroth the following fundamental fact [5] about groups generated by two elements of order 3 which have product of order 3: Suppose that X = x,y x3 = y3 = (xy)3 = 1 . Then G h | i has an abelian normal subgroup of index 3. This fact allows us to show that for certain z H, and for certain pairs x,y ∈ of elements of order 3 in G, if xy = z then x,y H. This means that we ∈ can calculate a in H and so we know the left hand side of the structure xyz constant formula. Hence in principle we can determine G. What in fact | | happens is that we can discover enough information about G to decide | | that the possiblepartial character tables are invalid or to show that G= H. OnceTheorems1.2and1.3areproven,inSection3weproveTheorem1.1. Now suppose G and J be as in Theorem 1.1 and set Q = O (N (Z)) and 3 G M = N (J). The initial part of the proof recalls some pertinent facts from G [8]. In particular, we recall that M/J = Mat(10) or 2 Mat(10). For y ∼ × ∈ Q Z, we also show that O3(C (y))/ y is isomorphic to the group H in M \ h i Theorem 1.2 if N (J)/J = Mat(10) and to the group H in Theorem 1.3 if G ∼ N (J)/J =2 Mat(10). Themaintechnical resultinthissection proves,for G ∼ × x O (C (y)/ y ), C (x) C (y)/ y and exploits the theorem ∈ 3 M h i CG(y)/hyi ≤ M h i of Smith and Tyrer [11]. Once this is proved we quickly finish the proof of Theorem 1.1 with the help of Theorems 1.2 and 1.3. Ournotationfollowsthatof[1]and[6].WeuseAtlas[3]notationforgroup extensions.For oddp,theextraspecialgroupsofexponentpandorderp2n+1 are denoted by p1+2n. The quaternion group of order 8 is Q and Mat(10) is + 8 the Mathieu group of degree 10. A non-trivial central product of groups H and K will be denoted H K. For a subset X of a group G, XG is the set of ◦ G-conjugatesofX.Fromtimetotimeweshallgivesuggestivedescriptionsof groups which indicate the isomorphism type of certain composition factors. We refer to such descriptions as the shape of a group. Groups of the same shape have normal series with isomorphic sections. We use the symbol to ≈ indicate the shape of a group. All the groups in this paper are finite groups. Acknowledgement. The first author is grateful to the DFG for their sup- port and thanks the mathematics department in Halle for their hospitality. 2. Proof of Theorems 1.2 and 1.3 Inthissection weusetheSuzukimethodwhichexploits virtualcharacters to prove Theorems 1.2 and 1.3. We recall the following definition from [4, Definition 14.5]. Definition 2.1 (Suzuki Special Classes). Let G be a group and H be a subgroup of G. Suppose that = n is a union of conjugacy classes of C Si=1Ci H. Then is called a set of special classes in H provided the following three C conditions hold. (i) C (h) H for all h ; G ≤ ∈ C (ii) G = for 1 i n; and Ci ∩C Ci ≤ ≤ 5 (iii) if h and h = f , then f . ∈ C h i h i ∈C As mentioned in the introduction, the Suzuki method and the theory behind it are well explain in [4] and we refer the reader explicitly to Section 14B in [4]. 2.1. Proof of Theorem 1.2. Suppose that G, H and T are as in the statement of Theorem 1.2. Thus H = (1,2,3),(4,5,6),(7,8,9),(1,2)(4,5),(1,2)(7,8) 33 : 22, ∼ h i ≈ T Syl (H), H = N (T) and C (t) = C (t) for all t T#. Note that ∈ 3 G G H ∈ T = (1,2,3),(4,5,6),(7,8,9) is elementary abelian of order 27. h i Wefixthefollowingnotationfortherepresentativesofthenon-trivialcon- jugacy classes of H: s = (1,2)(7,8), s = (4,5)(7,8), s = (1,2)(4,5), t = 1 2 3 1 (1,2,3),t = (4,5,6),t = (7,8,9),t = (1,2,3)(4,5,6), t = (1,2,3)(7,8,9), 2 3 4 5 t = (4,5,6)(7,8,9), t = (1,2,3)(4,5,6)(7,8,9), t = (1,3,2)(4,6,5)(7,9,8), 6 7 8 f = (1,2)(4,5,6)(7,8), f = (1,2,3)(4,5)(7,8) and f = (1,2)(4,5)(7,8,9). 1 2 3 For 1 i 8, define ≤ ≤ = tH T#. Ci i ⊆ Lemma 2.2. The following hold: (i) T Syl (G); ∈ 3 (ii) H controls G-fusion of elements of order 3 in T; (iii) = 8 = T# is set of special classes in H; and C Si=1Ci (iv) if t T# and x,y G satisfy xy = t, then either x , y and t ∈ ∈ C h i h i h i are all G-conjugate or x,y H. ∈ Proof. As H = N (T) and T Syl (H), T Syl (G). Thus (i) holds and, G ∈ 3 ∈ 3 as T is abelian, (ii) follows from a theorem of Burnside. That is a set of C special classes follows from (i), (ii), the hypothesis that C (t) H for all G ≤ t T# and the fact that = T#. So (iii) holds. ∈ C Suppose that x,y G, t T# and that xy = t. Assume that x , y ∈ C ∈ h i h i and t are not all G-conjugate and set X = x,y . Then x Syl (X). h i h i h i 6∈ 3 Since x,y and t have order 3, X has an abelian normal subgroup Y of index 3 and, as x Syl (X), Y has order divisible by 3. In particular, C (t) is h i 6∈ 3 | Y | divisible by 3. By hypothesis C (t) C (t) H. Hence Y T = 1. Since Y G ≤ ≤ ∩ 6 Y is abelian, Y C (Y T) H. Thus Y H and T Y Syl (Y). ≤ G ∩ ≤ ≤ ∩ ∈ 3 But then 1 = C (x) and consequently x C (C (x)) H. Similarly, T∩Y G T∩Y y H. Henc6 e X H and this proves (iv). ∈ ≤ (cid:3) ∈ ≤ We obtain the character table of H. This can of course be done by hand, but we have used Magma [2]. The vector space of class functions on H which vanish off the special classes of H has dimension 8 (see [4, Lemma 14.6]) and is easily seen to 6 ChrisParker andGernotStroth Class 1 s s s t t t t t t t t f f f 1 2 3 1 2 3 4 5 6 7 8 1 2 3 χ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 χ 1 1 -1 -1 1 1 1 1 1 1 1 1 1 -1 -1 2 χ 1 -1 -1 1 1 1 1 1 1 1 1 1 -1 -1 1 3 χ 1 -1 1 -1 1 1 1 1 1 1 1 1 -1 1 -1 4 χ 2 0 0 -2 2 2 -1 2 -1 -1 -1 -1 0 0 1 5 χ 2 0 0 2 2 2 -1 2 -1 -1 -1 -1 0 0 -1 6 χ 2 0 2 0 -1 2 2 -1 -1 2 -1 -1 0 -1 0 7 χ 2 0 -2 0 -1 2 2 -1 -1 2 -1 -1 0 1 0 8 χ 2 -2 0 0 2 -1 2 -1 2 -1 -1 -1 1 0 0 9 χ 2 2 0 0 2 -1 2 -1 2 -1 -1 -1 -1 0 0 10 χ 4 0 0 0 4 -2 -2 -2 -2 1 1 1 0 0 0 11 χ 4 0 0 0 -2 -2 4 1 -2 -2 1 1 0 0 0 12 χ 4 0 0 0 -2 4 -2 -2 1 -2 1 1 0 0 0 13 χ 4 0 0 0 -2 -2 -2 1 1 1 α β 0 0 0 14 χ 4 0 0 0 -2 -2 -2 1 1 1 β α 0 0 0 15 Table 1. The character table of H. Here α = 2 3ω and − − β = 1+3ω where ω = e2πi/3. have the following basis: λ = χ +χ +χ +χ χ , 1 1 2 3 4 11 − λ = χ +χ χ , 2 5 6 11 − λ = χ +χ χ , 3 7 8 11 − λ = χ +χ χ , 4 9 10 11 − λ = χ χ , 5 11 12 − λ = χ χ , 6 11 13 − λ = χ χ and 7 11 14 − λ = χ χ . 8 11 15 − For 1 i 8, define ≤ ≤ 15 γ = χ (t )χ . i X j i j j=1 For s−1 H we have ∈ \C 15 γ (s−1)= χ (t )χ (s−1), i X j i j j=1 which is zero by the second orthogonality relation. Hence, for 1 i 8, ≤ ≤ γ λ 1 j 8 . i j ∈ h | ≤ ≤ i 7 In the matrix C below, row i describes the decomposition of γ in terms of i the λ . j 1 2 −1 2 2 2 2 2 1 2 2 −1 2 −4 2 2   1 −1 2 2 −4 2 2 2 1 2 −1 −1−1 2 −1 −1 C = . 1 −1−1 2 2 −1 −1 −1 1 −1 2 −1 2 2 −1 −1 1 −1−1 −1−1 −1−α −β   1 −1−1 −1−1 −1 −β −α The matrix of inner products (λ ,λ ), 1 i,j 8 is i j ≤ ≤ 5 1 1 1 −1−1 −1−1  1 3 1 1 −1−1 −1−1 1 1 3 1 −1−1 −1−1 D = 1 1 1 3 −1−1 −1−1 . −1 −1−1 −1 2 1 1 1  −1 −1−1 −1 1 2 1 1    −1 −1−1 −1 1 1 2 1   −1 −1−1 −1 1 1 1 2 For 1 i 8 set µ = λG. We use the fact that (µ ,µ ) = (λ ,λ ) (see [4, ≤ ≤ i i i j i j Lemma 14.9]) to determine the possibilities for the decomposition of µ as i a linear combination of irreducible characters of G. Let Irr(G) = θ ,...,θ be the irreducible characters of G with the 1 m { } principal character being θ . We know by Frobenius reciprocity that θ has 1 1 multiplicity 1 in µ and multiplicity 0 in µ ,...,µ . Furthermore, for 1 1 2 8 ≤ i 8, we also have µ (1) = 0 again by [4, Lemma 14.9]. i ≤ We have (µ ,µ )= (λ ,λ )= 5. Suppose µ = θ 2θ . Then 1 1 1 1 1 1 2 ± 1 = (λ ,λ )= (µ ,µ ) = 2(θ ,µ ), 1 2 1 2 2 2 ± which is a contradiction. Hence we have, up to reordering the irreducible characters of G, µ = θ +ε θ +ε θ +ε θ +ε θ 1 1 2 2 3 3 4 4 5 5 where ε 1, 1 for 2 j 5. j ∈ { − } ≤ ≤ Now as (µ ,µ ) = (λ ,λ ) = 2 and µ (1) = 0 for 5 i 8, µ ,µ ,µ and i i i i i 5 6 7 ≤ ≤ µ must be the difference of two characters of G. Therefore 8 µ = σ τ +σ τ , 5 1 1 2 2 µ = σ τ +σ τ , 6 3 3 2 4 µ = σ τ +σ τ and 7 5 5 6 6 µ = σ τ +σ τ 8 7 7 8 8 where for 1 i 8, τ Irr(G) and σ 1, 1 . Using the fact that i i ≤ ≤ ∈ ∈ { − } (µ ,µ ) = 1 for 5 i < j 8, we deduce that, again up to changing i j ≤ ≤ notation, σ = σ = σ = σ and τ = τ = τ = τ and that the other 1 3 5 7 1 3 5 7 characters arepairwisedistinctandthesignsarenegative σ .Thusinreality 1 8 ChrisParker andGernotStroth we have µ = σ τ σ τ , 5 1 1 1 2 − µ = σ τ σ τ , 6 1 1 1 4 − µ = σ τ σ τ and 7 1 1 1 6 − µ = σ τ σ τ 8 1 1 1 8 − where τ ,τ ,τ ,τ = 4. Now we decompose µ ,µ and µ . These excep- 2 4 6 8 2 3 4 |{ }| tional characters involve three irreducible constituents. µ = δ η +δ η +δ η , 2 1 1 2 2 3 3 µ = δ η +δ η +δ η and 3 4 4 5 5 6 6 µ = δ η +δ η +δ η 4 7 7 8 8 9 9 where, for 1 i 9, η Irr(G) and δ 1, 1 . We have that (µ ,µ ) = i i 2 j ≤ ≤ ∈ ∈ { − } 1 for 5 j 8. If (µ ,τ ) = 0, then τ ,τ ,τ and τ must all appear 2 1 2 4 6 8 − ≤ ≤ in the expression for µ and this is absurd as there are only 3 irreducible 2 constituents. Hence (µ ,τ ) = σ . Similarly (µ ,τ ) = σ and (µ ,τ ) = 2 1 1 3 1 1 4 1 − − σ . Since (µ ,µ )= (µ ,µ )= (µ ,µ ) = 1, we now have 1 2 3 2 4 3 4 − µ = σ τ +δ η +δ η , 2 1 1 2 2 3 3 − µ = σ τ +δ η +δ η and 3 1 1 5 5 6 6 − µ = σ τ +δ η +η γ 4 1 1 8 8 9 9 − whereagainthecharactersη arepairwisedistinctandarenotin τ ,τ ,τ ,τ . i 2 4 6 8 { } Now suppose that (µ ,τ ) = 0. Then η ,η ,η ,τ ,τ ,τ ,τ all appear in µ , 1 1 2 5 8 2 4 6 8 1 as (µ ,µ ) = 1 for all i, which is not possible. Hence 1 i ± µ = θ σ τ +ε θ +ε θ +ε θ . 1 1 1 1 3 3 4 4 5 5 − Finally, again after adjusting notation yet again, we have proved that µ = θ +ε θ +ε θ +ε θ +ε θ , 1 1 2 2 3 3 4 4 5 5 µ = ε θ +ε θ +ε θ , 2 2 2 6 6 7 7 µ = ε θ +ε θ +ε θ , 3 2 2 8 8 9 9 µ = ε θ +ε θ +ε θ , 4 2 2 10 10 11 11 µ = ε θ +ε θ , 5 2 2 2 12 − µ = ε θ +ε θ , 6 2 2 2 13 − µ = ε θ +ε θ and 7 2 2 2 14 − µ = ε θ +ε θ 8 2 2 2 15 − where ε 1, 1 and θ ,...,θ Irr(G) are pairwise different. This j 1 15 ∈ { − } ∈ provides a matrix B with Irr(G) columns (those columns indexed by θ i | | with i 16 consist only of zero entries and have been suppressed) which ≥ 9 describes this decomposition 1 ε2 ε3 ε4 ε5 0 0 0 0 0 0 0 0 0 0 ... 0 ε2 0 0 0 ε6 ε7 0 0 0 0 0 0 0 0 ... 0 ε2 0 0 0 0 0 ε8 ε9 0 0 0 0 0 0 ... B = 0 ε2 0 0 0 0 0 0 0 ε10 ε11 0 0 0 0 ... . 0−ε2 0 0 0 0 0 0 0 0 0 ε2 0 0 0 ... 0−ε2 0 0 0 0 0 0 0 0 0 0 ε2 0 0 ... 0−ε2 0 0 0 0 0 0 0 0 0 0 0 ε2 0 ... 0−ε2 0 0 0 0 0 0 0 0 0 0 0 0 ε2 ... We now deploy Suzuki’s Theorem on special classes [4, Theorem 14.11] to obtain a fragment of the character table for G by forming the product (CB)t. We let the degree of θ be denoted by d . Notice that as µ (1 ) = 0, i i i G we know that d = d = d = d = d . It is important to note here that 2 12 13 14 15 every character θ , i 16, of G takes value zero when evaluated on t for i j ≥ 1 j 8. That is, the part of the character table of G below the columns ≤ ≤ that are presented consists entirely of zeros. Class 1 t t t t t t t t 1 2 3 4 5 6 7 8 θ 1 1 1 1 1 1 1 1 1 1 θ d 4ε 2ε 2ε 2ε 2ε ε ε ε 2 2 2 2 2 2 2 2 2 2 − − − − θ d ε ε ε ε ε ε ε ε 3 3 3 3 3 3 3 3 3 3 θ d ε ε ε ε ε ε ε ε 4 4 4 4 4 4 4 4 4 4 θ d ε ε ε ε ε ε ε ε 5 5 5 5 5 5 5 5 5 5 θ d 2ε 2ε ε 2ε ε ε ε ε 6 6 6 6 6 6 6 6 6 6 − − − − − θ d 2ε 2ε ε 2ε ε ε ε ε 7 7 7 7 7 7 7 7 7 7 − − − − − θ d ε 2ε 2ε ε ε 2ε ε ε 8 8 8 8 8 8 8 8 8 8 − − − − − θ d ε 2ε 2ε ε ε 2ε ε ε 9 9 9 9 9 9 9 9 9 9 − − − − − θ d 2ε ε 2ε ε ε ε ε ε 10 10 10 10 10 10 10 10 10 10 − − − − − − θ d 2ε ε 2ε ε ε ε ε ε 11 11 11 11 11 11 11 11 11 11 − − − − − − θ d 2ε 2ε 4ε ε 2ε 2ε ε ε 12 2 2 2 2 2 2 2 2 2 − − − − θ d 2ε 4ε 2ε 2ε ε 2ε ε ε 13 2 2 2 2 2 2 2 2 2 − − − − θ d 2ε 2ε 2ε ε ε ε ε α ε β 14 2 2 2 2 2 2 2 2 2 − − − − − θ d 2ε 2ε 2ε ε ε ε ε β ε α 15 2 2 2 2 2 2 2 2 2 − − − − − . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Table 2. A few columns of the character table of G WecannowdeterminethestructureconstantsforG.Weusethestructure constants formula G m θ (x)θ (y)θ (z−1) i i i a = | | . xyz C (x) C (y) X θ (1) G G i | || | i=1 Because of Lemma 2.2 (iv), so long as x, y and z do not generate conjugate cyclic subgroups, we can determine a by calculating in H (we can either xyz use the character table or calculate by hand in H). However we may also use our partial character table of G given in Table 2 to obtain expressions for the structure constants. 10 ChrisParker andGernotStroth We have G ε ε ε 8ε 8ε 2ε 2ε 2ε 3 4 5 6 7 8 9 10 1 = a = | | (1+ + + + + + + + t1t2t4 (54)2 d d d d d d d d 3 4 5 6 7 8 9 10 2ε 44ε 11 2 + ), d − d 11 2 and, as α2+β2 = 13, − G ε ε ε 2ε 2ε ε ε 2ε 3 4 5 6 7 8 9 10 2 = a = | | (1+ + + + + + t7t7t1 (27)2 d d d d d − d − d d 3 4 5 6 7 8 9 10 2ε 26ε 11 2 + ). d − d 11 2 Subtracting a at7t7t1 and multiplying by 2, we get t1t2t4 − 4 G 12ε 12ε 6ε 6ε 36ε 6 7 8 9 2 (1) 1 = | | ( + + + ). (54)2 d d d d − d 6 7 8 9 2 Now we determine the structure constants a = a = 0 and obtain t1t2t3 t1t1t2 ε ε ε 4ε 4ε 4ε 4ε 4ε 3 4 5 6 7 8 9 10 0 = a = (1+ + + t1t2t3 d d d − d − d − d − d − d 3 4 5 6 7 8 9 10 4ε 32ε 11 2 ) − d − d 11 2 and ε ε ε 8ε 8ε 2ε 2ε 4ε 3 4 5 6 7 8 9 10 0 = a = (1+ + + + + + + t1t1t2 d d d d d d d − d 3 4 5 6 7 8 9 10 4ε 40ε 11 2 + ). − d d 11 2 Subtracting a a gives t1t2t3 − t1t1t2 72ε 12ε 12ε 6ε 6ε 2 6 7 8 9 (2) = + + + . − d d d d d 2 6 7 8 9 Finally, we substitute Eqn. 2 into Eqn. 1 to get G ε 2 1 = −| | 108 . (54)2 · · d 2 Thus G = 27ε d . In particular, ε = 1. Now there are at least 5 2 2 2 | | − − characters of G which have degree d . Hence 5d2 G = 27d and so 2 2 ≤ | | 2 d 5 and G 27 5. Since 108 divides G, we have G = H and this 2 com≤pletes th|e d|e≤monst·ration of Theorem 1.2|. | (cid:3) 2.2. Proof of Theorem 1.3. InthissubsectionK isthesubgroupofAlt(9) which normalizes J = (1,2,3),(4,5,6),(7,8,9) . Thus h i K = (1,2,3),(1,4,7)(2,5,8)(3,6,9),(1,2)(4,5),(1,4)(2,5)(3,6)(7,8) . h i We remark that K has shape 33:Sym(4) but is not the unique group of this shape which has characteristic 3. We assume that G, H and J are as in the statement of Theorem 1.3. Hence we may identify H with K and for all