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$\Text {M} $, $\Text {B} $ and $\Text {Co} 1 $ Are Recognisable by Their M, B AND Co1 ARE RECOGNISABLE BY THEIR PRIME GRAPHS MELISSA LEE AND TOMASZ POPIEL Abstract. The prime graph, or Gruenberg–Kegel graph, of a finite group G is the graph Γ(G) whose vertices are the prime divisors of |G|, and whose edges are the pairs {p, q} for which G contains an element of order pq. A finite group G is recognisable by its prime graph if every finite group H with Γ(H) = Γ(G) is isomorphic to G. By a result of Cameron and Maslova, every such group must be almost simple, so one natural case to investigate is that in which G is one of the 26 sporadic simple groups. Existing work of various authors answers the question of recognisability by prime graph for all but three of these groups, namely the Monster, M, the Baby Monster, B, and the first Conway group, Co1. We prove that these three groups are recognisable by their prime graphs. 1. Introduction The Gruenberg–Kegel graph of a finite group G, introduced in an unpublished 1975 manuscript of Karl W. Gruenberg and Otto H. Kegel, is the (labelled) graph Γ(G) whose vertices are the prime divisors of G , and whose edges are the pairs p,q for which G contains an element of order pq. It| is| also called the prime graph of G{, and} we use this name here for brevity. A (finite) group G is recognisable by its prime graph if every finite group H with Γ(H)=Γ(G) is isomorphic to G. More generally, G is k-recognisable by its prime graph if there are precisely k pairwise non-isomorphic groups with the same prime graph as G. If no such k exists, then G is unrecognisable by its prime graph. The question of recognisability of various groups by their prime graphs has attracted significant interest. We refer the reader to the recent article of Cameron and Maslova [2] for an up-to-date review of the literature (and several new results). Much work has focused on simple groups, and this is justified succinctly by [2, Theorem 1.2], which states, in particular, that if G is k-recognisable by its prime graph for some k, then G is almost simple (i.e. G0 6 G 6 Aut(G0) for some simple group G0). It is therefore natural to ask, in particular, which of the 26 sporadic simple groups are recognisable by their prime graphs. As summarised in Table 1, this question has been answered for all but three of these groups: the Monster, M, the Baby Monster, B, and the first Conway group, Co1. The purpose of this note is to settle these remaining cases. We prove the following theorem. Theorem 1.1. M, B and Co1 are recognisable by their prime graphs. arXiv:2107.12755v1 [math.GR] 27 Jul 2021 2. Proof of Theorem 1.1 — Co1 21 9 4 2 Recall that Co1 = 2 3 5 7 11 13 23. The vertex set of Γ(Co1) is therefore | | · · · · · · 2, 3, 5, 7, 11, 13, 23 , and we see from the ATLAS [3] that Γ(Co1) has two connected components,{ one of} which is the isolated vertex 23 . { } Our argument begins along the lines of Kondrat’ev’s proofs [6, 7]. Suppose that G is a finite group with Γ(G) = Γ(Co1). Then [4, Theorem 3] implies that G/F (G) ∼= Co1 and that the prime divisors of the Fitting subgroup F (G) belong to the set 2, 3 . Assume { } towards a contradiction that the maximal normal p-subgroup Op(G) of G is nontrivial for at least one p 2, 3 . Then it follows that every g G of order 23 must act fixed-point ∈ { } ∈ Date: July 28, 2021. 1 2 MELISSA LEE AND TOMASZ POPIEL G Recognisable? Γ(H)=Γ(G)? Reference Co2, J1, M22, M23, M24 Yes Hagie [4] M11 2-recognisable H =L2(11) Hagie [4] J2, M12 Unrecognisable Hagie [4] J4 Yes Zavarnitsine [10] Ru Yes Kondrat’ev [6] HN 2-recognisable H = HN.2 Kondrat’ev [6] Fi22 3-recognisable H Fi22.2, Suz.2 Kondrat’ev [6] ∈{ } He, McL, Co3 Unrecognisable Kondrat’ev [6] ′ Fi23, Fi24, J3, Ly, O’N, Suz, Th Yes Kondrat’ev [7] HS 2-recognisable H = U6(2) Kondrat’ev [7] M, B, Co1 Yes Theorem1.1 Table 1. Recognisability of the sporadic simple groups by their prime graphs. freely on Op(G), and hence on a minimal normal subgroup of Op(G), which is necessarily elementary abelian and can therefore be regarded as an FCo1-module for some field F of characteristic p. (Otherwise, the product of g with a commuting h Op(G) would have ∈ order 23p, so Γ(G) would contain the edge 23,p , contradicting Γ(G) = Γ(Co1).) { } In particular, every element of order 23 in a maximal subgroup Co2 of G/F (G) ∼= Co1 must act fixed-point freely on O3(G). However, according to the 3-modular Brauer character table of Co2, which is available in GAP [1, 5], the elements of order 23 in Co2 do not act fixed-point freely on any irreducible FG-module defined over a field F of characteristic 3. Therefore, O3(G) must be trivial, and so O2(G) must be nontrivial. Let V be a minimal normal subgroup of O2(G), and let χ be the corresponding Brauer character for Co1. Consider again a maximal subgroup Co2. According to the 2-modular Brauer character table of Co2, the only irreducible module for Co2 in characteristic 2 on which elements of order 23 act fixed-point freely is the module, call it W , of dimension 22. Therefore, each of the k composition factors of the restriction V Co2 of V to Co2 ↓ is isomorphic to W , and so the restriction of χ to Co2 is k times the Brauer character of W . Now consider a maximal subgroup 3.Suz.2 of Co1. There are exactly 6 non-trivial conjugacy classes of elements of odd order in Co1 that intersect both 3.Suz.2 and Co2. These classes are listed in Table 2, together with the corresponding values of the hypoth- esised character χ, and the corresponding classes in 3.Suz.2 and Co2. The restriction of χ to 3.Suz.2 must be a linear combination of the irreducible 2-modular Brauer characters for 3.Suz.2, with non-negative integer coefficients. Using the 2-modular Brauer character table for 3.Suz.2, we thereby obtain a system of equations for the multiplicities of the characters appearing in this restriction. There are 9 equations (one for each conjugacy class of 3.Suz.2 in Table 2, plus one for the identity element) in 27 variables (one for each irreducible 2-modular Brauer character). Using Mathematica [9], we check that this system has no (non-trivial) non-negative integer solutions, so we have a contradiction. Therefore, O2(G) is also trivial, so F (G) is trivial and hence G ∼= Co1, as claimed. 3. Proof of Theorem 1.1 — M Recall that M = 246 320 59 76 112 133 17 19 23 29 31 41 47 59 71. The prime graph Γ(M)| | has three· isolated· · · vertices· 41· , · 59· and· 71· ,· and· all other· · vertices { } { } { } lie in a connected component. Our argument is similar to the argument given for Co1. Suppose that G is a finite group with Γ(G) = Γ(M). Then [4, Theorem 3] implies that G/F (G) ∼= M and that F (G) is a 3-group. If O3(G) is nontrivial, then every element of order 71 in G must act fixed-point freely on O3(G). In particular, every element of order M, B AND Co1 ARERECOGNISABLEBYTHEIRPRIMEGRAPHS 3 Co1 Co2 3.Suz.2 χ 3B 3B 3C, 3D 4k 3C 3A 3E 5k 5B 5B 5B− 2k 9C 9A 9B k 11A 11A 11A 0 15D 15A 15E k − Table 2. The 2-regular conjugacy classes of Co1 that intersect both max- imal subgroups Co2 and 3.Suz.2, and the corresponding conjugacy classes in these subgroups. Class names are as in the ATLAS [3]. The final column lists the corresponding values of the 2-modular Brauer character χ on the hypothesised Co1-module V in the proof of Theorem 1.1. 71 in a maximal subgroup L2(71) of G/F (G) ∼= M must act fixed-point freely on O3(G). Let V be a minimal normal subgroup of O3(G), and let χ be the corresponding Brauer character for M. By the 3-modular Brauer character table of L2(71), the only irreducible modules for L2(71) in characteristic 3 on which elements of order 71 act fixed-point freely are the two modules of dimension 70 and the two modules of dimension 35. Hence, every composition factor of V L2(71) must be isomorphic to one of these four modules. The ↓ conjugacy classes 2B and 7B of M intersect L2(71) in its conjugacy classes 2A and 7A–C. The elements of the L2(71)-classes 7A–C all have vanishing Brauer character on each of the four aforementioned modules, so we infer that χ(7B) = 0. We now use the fact that every element of order 41 in G must act fixed-point freely on V . In particular, every such element in a maximal subgroup L2(41) must act fixed-point freely on V . (Recall that such a subgroup exists [8].) The only irreducible modules for L2(41) in characteristic 3 on which elements of order 41 act fixed-point freely are the four modules of dimension 40.
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