Integral group ring of the McLaughlin simple group

We consider the Zassenhaus conjecture for the normalized unit group of the integral group ring of the McLaughlin sporadic group McL. As a consequence, we confirm for this group the Kimmerle’s conjecture on prime graphs.

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Datum:	2007
Hauptverfasser:	Bovdi, V.A., Konovalov, A.B.
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Veröffentlicht:	Інститут прикладної математики і механіки НАН України 2007
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spelling	irk-123456789-1573532019-06-21T01:28:39Z Integral group ring of the McLaughlin simple group Bovdi, V.A. Konovalov, A.B. We consider the Zassenhaus conjecture for the normalized unit group of the integral group ring of the McLaughlin sporadic group McL. As a consequence, we confirm for this group the Kimmerle’s conjecture on prime graphs. 2007 Article Integral group ring of the McLaughlin simple group / V.A. Bovdi, A.B. Konovalov // Algebra and Discrete Mathematics. — 2007. — Vol. 6, № 2. — С. 43–53. — Бібліогр.: 26 назв. — англ. 1726-3255 2000 Mathematics Subject Classification: 16S34, 20C05; 20D08. http://dspace.nbuv.gov.ua/handle/123456789/157353 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection	DSpace DC
language	English
description	We consider the Zassenhaus conjecture for the normalized unit group of the integral group ring of the McLaughlin sporadic group McL. As a consequence, we confirm for this group the Kimmerle’s conjecture on prime graphs.
format	Article
author	Bovdi, V.A. Konovalov, A.B.
spellingShingle	Bovdi, V.A. Konovalov, A.B. Integral group ring of the McLaughlin simple group Algebra and Discrete Mathematics
author_facet	Bovdi, V.A. Konovalov, A.B.
author_sort	Bovdi, V.A.
title	Integral group ring of the McLaughlin simple group
title_short	Integral group ring of the McLaughlin simple group
title_full	Integral group ring of the McLaughlin simple group
title_fullStr	Integral group ring of the McLaughlin simple group
title_full_unstemmed	Integral group ring of the McLaughlin simple group
title_sort	integral group ring of the mclaughlin simple group
publisher	Інститут прикладної математики і механіки НАН України
publishDate	2007
url	http://dspace.nbuv.gov.ua/handle/123456789/157353
citation_txt	Integral group ring of the McLaughlin simple group / V.A. Bovdi, A.B. Konovalov // Algebra and Discrete Mathematics. — 2007. — Vol. 6, № 2. — С. 43–53. — Бібліогр.: 26 назв. — англ.
series	Algebra and Discrete Mathematics
work_keys_str_mv	AT bovdiva integralgroupringofthemclaughlinsimplegroup AT konovalovab integralgroupringofthemclaughlinsimplegroup
first_indexed	2025-07-14T09:47:50Z
last_indexed	2025-07-14T09:47:50Z
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fulltext	Jo u rn al A lg eb ra D is cr et e M at h . Algebra and Discrete Mathematics RESEARCH ARTICLE Number 2. (2007). pp. 43 – 53 c© Journal “Algebra and Discrete Mathematics” Integral group ring of the McLaughlin simple group V. A. Bovdi, A. B. Konovalov Dedicated to V.I. Sushchansky on the occasion of his 60th birthday Abstract. We consider the Zassenhaus conjecture for the normalized unit group of the integral group ring of the McLaughlin sporadic group McL. As a consequence, we confirm for this group the Kimmerle’s conjecture on prime graphs. 1. Introduction, conjectures and main results Let V (ZG) be the normalized unit group of the integral group ring ZG of a finite group G. A long-standing conjecture of H. Zassenhaus (ZC) says that every torsion unit u ∈ V (ZG) is conjugate within the rational group algebra QG to an element in G (see [26]). For finite simple groups the main tool for the investigation of the Zassenhaus conjecture is the Luthar–Passi method, introduced in [22] to solve it for A5. Later M. Hertweck in [17] extended the Luthar–Passi method and applied it for the investigation of the Zassenhaus conjecture for PSL(2, pn). The Luthar–Passi method proved to be useful for groups containing non-trivial normal subgroups as well. For some recent results we refer to [5, 7, 16, 17, 18, 19]. Also, some related properties and some weakened variations of the Zassenhaus conjecture can be found in [1, 23] and [3, 21]. First of all, we need to introduce some notation. By #(G) we denote the set of all primes dividing the order of G. The Gruenberg–Kegel graph (or the prime graph) of G is the graph π(G) with vertices labeled by the The research was supported by OTKA grant No. K 68383 2000 Mathematics Subject Classification: 16S34, 20C05; 20D08. Key words and phrases: Zassenhaus conjecture, Kimmerle conjecture, torsion unit, partial augmentation, integral group ring. Jo u rn al A lg eb ra D is cr et e M at h .44 Integral group ring primes in #(G) and with an edge from p to q if there is an element of order pq in the group G. In [21] W. Kimmerle proposed the following weakened variation of the Zassenhaus conjecture: (KC) If G is a finite group then π(G) = π(V (ZG)). In particular, in the same paper W. Kimmerle verified that (KC) holds for finite Frobenius and solvable groups. We remark that with respect to the so-called p-version of the Zassenhaus conjecture the inves- tigation of Frobenius groups was completed by M. Hertweck and the first author in [4]. In [6, 7, 8, 9, 10, 12] (KC) was confirmed for the Mathieu simple groups M11, M12, M22, M23, M24 and the sporadic Janko simple groups J1, J2 and J3. Here we continue these investigations for the McLaughlin simple group McL. Although using the Luthar–Passi method we cannot prove the ra- tional conjugacy for torsion units of V (ZMcL), our main result gives a lot of information on partial augmentations of these units. In particular, we confirm the Kimmerle’s conjecture for this group. Let G = McL. It is well known (see [15]) that \|G\| = 27 · 36 · 53 · 7 · 11 and exp(G) = 23 · 32 · 5 · 7 · 11. Let C = {C1, C2a, C3a,C3b, C4a, C5a, C5b, C6a, C6b, C7a, C7b, C8a, C9a, C9b, C10a, C11a, C11b, C12a, C14a, C14b, C15a, C15b, C30a, C30b} be the collection of all conjugacy classes of McL, where the first index denotes the order of the elements of this conjugacy class and C1 = {1}. Suppose u = ∑ αgg ∈ V (ZG) has finite order k. Denote by νnt = νnt(u) = εCnt (u) = ∑ g∈Cnt αg the partial augmentation of u with respect to Cnt. From the Berman–Higman Theorem (see [2] and [25], Ch.5, p.102) one knows that ν1 = α1 = 0 and ∑ Cnt∈C νnt = 1. (1) Hence, for any character χ of G, we get that χ(u) = ∑ νntχ(hnt), where hnt is a representative of the conjugacy class Cnt. Our main result is the following Theorem 1. Let G denote the McLaughlin simple group McL. Let u be a torsion unit of V (ZG) of order \|u\|. Denote by P(u) the tuple (ν2a, ν3a, ν3b,ν4a, ν5a, ν5b, ν6a, ν6b, ν7a, ν7b, ν8a, ν9a, ν9b, ν10a, ν11a, ν11b, ν12a, ν14a, ν14b, ν15a, ν15b, ν30a, ν30b) ∈ Z23 of partial augmentations of u in V (ZG). The following properties hold. Jo u rn al A lg eb ra D is cr et e M at h .V. A. Bovdi, A. B. Konovalov 45 (i) There is no elements of orders 21, 22, 33, 35, 55, 77 in V (ZG). Equivalently, if \|u\| 6∈ {18, 20, 24, 28, 36, 40, 45, 56, 60, 72, 90, 120, 180, 360}, then \|u\| coincides with the order of some element g ∈ G. (ii) If \|u\| = 2, then u is rationally conjugate to some g ∈ G. (iii) If \|u\| = 3, then all components of P(u) are zero except possibly ν3a and ν3b, and the pair (ν3a, ν3b) is one of { (−2, 3), (−1, 2), (0, 1), (1, 0) }. (iv) If \|u\| = 5, then all components of P(u) are zero except possibly ν5a and ν5b, and the pair (ν5a, ν5b) is one of { (−4, 5), (−3, 4), (−2, 3), (−1, 2), (0, 1), (1, 0) }. (v) If \|u\| = 7, then all components of P(u) are zero except possibly ν7a and ν7b, and the pair (ν7a, ν7b) is one of { (ν7a, ν7b) \| −86 ≤ ν7a ≤ 87, ν7a + ν7b = 1 }. (vi) If \|u\| = 11, then all components of P(u) are zero except possibly ν11a and ν11b, and the pair (ν11a, ν11b) is one of { (ν11a, ν11b) \| −9 ≤ ν11a ≤ 10, ν11a + ν11b = 1 }. As an immediate consequence of part (i) of the Theorem we obtain Corollary 1. If G = McL then π(G) = π(V (ZG)). 2. Preliminaries The following result is a reformulation of the Zassenhaus conjecture in terms of vanishing of partial augmentations of torsion units. Proposition 1. (see [22] and Theorem 2.5 in [24]) Let u ∈ V (ZG) be of order k. Then u is conjugate in QG to an element g ∈ G if and only if for each d dividing k there is precisely one conjugacy class C with partial augmentation εC(ud) 6= 0. The next result now yield that several partial augmentations are zero. Proposition 2. (see [16], Proposition 3.1; [17], Proposition 2.2) Let G be a finite group and let u be a torsion unit in V (ZG). If x is an element of G whose p-part, for some prime p, has order strictly greater than the order of the p-part of u, then εx(u) = 0. Jo u rn al A lg eb ra D is cr et e M at h .46 Integral group ring The key restriction on partial augmentations is given by the following result that is the cornerstone of the Luthar–Passi method. Proposition 3. (see [22, 17]) Let either p = 0 or p a prime divisor of \|G\|. Suppose that u ∈ V (ZG) has finite order k and assume k and p are coprime in case p 6= 0. If z is a complex primitive k-th root of unity and χ is either a classical character or a p-Brauer character of G, then for every integer l the number µl(u, χ, p) = 1 k ∑ d\|k TrQ(zd)/Q{χ(ud)z−dl} (2) is a non-negative integer. Note that if p = 0, we will use the notation µl(u, χ, ∗) for µl(u, χ, 0). Finally, we shall use the well-known bound for orders of torsion units. Proposition 4. (see [13]) The order of a torsion element u ∈ V (ZG) is a divisor of the exponent of G. 3. Proof of the Theorem Throughout this section we denote McL by G. The character table of G, as well as the p-Brauer character tables, which will be denoted by BCT(p) where p ∈ {2, 3, 5, 7, 11}, can be found using the computational algebra system GAP [15], which derives these data from [14, 20]. For the characters and conjugacy classes we will use throughout the paper the same notation, indexation inclusive, as used in the GAP Character Table Library. First of all we will investigate units of orders 2, 3, 5, 7 and 11, since the group G possesses elements of these orders. After this, by Proposition 4, the order of each torsion unit divides the exponent of G, so to prove the Kimmerle’s conjecture, it remains to consider units of orders 21, 22, 33, 35, 55 and 77. We prove that no units of all these orders do appear in V (ZG). Now we consider each case separately. • Let u be an involution. Using Proposition 2 we obtain that all partial augmentations except one are zero. Thus by Proposition 1 the proof of part (ii) of Theorem 1 is done. • Let u be a unit of order 3. By (1) and Proposition 2 we get ν3a+ν3b = 1. Put t1 = 5ν3a − 4ν3b. By (2) we obtain the system of inequalities µ0(u, χ2, ∗) = 1 3(−2t1 + 22) ≥ 0; µ1(u, χ2, ∗) = 1 3(t1 + 22) ≥ 0, Jo u rn al A lg eb ra D is cr et e M at h .V. A. Bovdi, A. B. Konovalov 47 from which t1 ∈ {−22,−19,−16,−13,−10,−7,−4,−1, 2, 5, 8, 11}. Now for each possible value of t1 consider the system of linear equations ν3a + ν3b = 1, 5ν3a − 4ν3b = t1. Since ∣ ∣ 1 1 5 −4 ∣ ∣ 6= 0, this system always has the unique solution. First we select only integer solutions, and then using the condition that all µi(u, χj , ∗) are non-negative integers, we obtain only four pairs (ν3a, ν3b) listed in part (iii) of Theorem 1. • Let u be a unit of order 5. By (1) and Proposition 2 we get ν5a+ν5b = 1. Put t1 = 3ν5a − 2ν5b. By (2) we obtain the system of inequalities µ0(u, χ2, ∗) = 1 5(−4t1 + 22) ≥ 0; µ1(u, χ2, ∗) = 1 5(t1 + 22) ≥ 0, so t1 ∈ {−22,−17,−12,−7,−2, 3}. Using the same arguments as in the previous case, we obtain only six pairs (ν5a, ν5b) listed in part (iv) of Theorem 1. • Let u be a unit of order 7. By (1) and Proposition 2 we get ν7a+ν7b = 1. Put t1 = 4ν7a − 3ν7b. Using BCT(3) and BCT(5), by (2) we have µ3(u, χ7, 3) = 1 7(t1 + 605) ≥ 0; µ1(u, χ12, 5) = 1 7(−t1 + 3245) ≥ 0; µ1(u, χ7, 3) = 1 7(−3ν7a + 4ν7b + 605) ≥ 0. It follows that we have only 174 pairs (ν7a, ν7b), given in part (v) of Theo- rem 1. Note that using our implementation of the Luthar–Passi method, which we intended to make available in the GAP package LAGUNA [11], we checked that it is not possible to further reduce the number of so- lutions, and the same remark also applies for the remaining part of the paper. • Let u be a unit of order 11. By (1) and Proposition 2 we have ν11a + ν11b = 1. Using BCT(3), by (2) we obtain the system of inequalities µ1(u, χ3, 3) = 1 11(6ν11a − 5ν11b + 104) ≥ 0; µ2(u, χ3, 3) = 1 11(−5ν11a + 6ν11b + 104) ≥ 0, that has only that twenty pairs (ν11a, ν11b) listed in part (vi) of the The- orem 1. • Let u be a unit of order 21. By (1) and Proposition 2 we have ν3a + ν3b + ν7a + ν7b = 1. Put t1 = 5ν3a−4ν3b−ν7a−ν7b, t2 = 5ν3a+2ν3b and t3 = 3ν7a−4ν7b. Since \|u7\| = 3, for any character χ of G we need to consider four cases, defined by part (iii) of the Theorem. Now we consider each case separately: Jo u rn al A lg eb ra D is cr et e M at h .48 Integral group ring Case 1. Let χ(u7) = χ(3a). Using (2), we obtain the system of inequalities µ3(u, χ2, ∗) = 1 21(2t1 + 11) ≥ 0; µ0(u, χ2, ∗) = 1 21(−12t1 + 18) ≥ 0, which has no integral solution. Case 2. Let χ(u7) = χ(3b). Again, using (2), we obtain the system of inequalities µ0(u, χ2, ∗) = 1 21(−12t1 + 36) ≥ 0; µ7(u, χ2, ∗) = 1 21(6t1 + 24) ≥ 0; µ0(u, χ3, ∗) = 1 21(36t2 + 243) ≥ 0; µ7(u, χ3, ∗) = 1 21(−18t2 + 225) ≥ 0; µ1(u, χ16, ∗) = 1 21(−t3 + 8386) ≥ 0; µ9(u, χ16, ∗) = 1 21(2t3 + 8386) ≥ 0; µ1(u, χ5, ∗) = 1 21(−13ν3a + 5ν3b + 765) ≥ 0. This yields t1 ∈ {−4, 3}, t2 ∈ {−5, 2, 9} and t3 ∈ {7 + 21k \| −200 ≤ k ≤ 399}, but none of possible combinations of ti’s gives us any solution. Case 3. Let χ(u7) = −2χ(3a) + 3χ(3b). Then using (2), we obtain the system µ1(u, χ2, ∗) = 1 21(−t1 − 1) ≥ 0; µ7(u, χ2, ∗) = 1 21(6t1 + 6) ≥ 0; µ0(u, χ3, ∗) = 1 21(36t2 + 207) ≥ 0; µ7(u, χ3, ∗) = 1 21(−18t2 + 243) ≥ 0; µ1(u, χ16, ∗) = 1 21(−t3 + 8218) ≥ 0; µ9(u, χ16, ∗) = 1 21(2t3 + 8218) ≥ 0, from which t1 = −1, t2 ∈ {−4, 3, 10} and t3 ∈ {7 + 21k \| −196 ≤ k ≤ 391}, and again we have no solution for every combination of ti’s. Case 4. Let χ(u7) = −χ(3a) + 2χ(3b). Using (2), we obtain the system µ7(u, χ2, ∗) = 1 21(6t1 + 15) ≥ 0; µ0(u, χ2, ∗) = 1 21(−12t1 + 54) ≥ 0; µ0(u, χ3, ∗) = 1 21(36t2 + 225) ≥ 0; µ7(u, χ3, ∗) = 1 21(−18t2 + 234) ≥ 0; µ9(u, χ16, ∗) = 1 21(2t3 + 8015) ≥ 0; µ1(u, χ16, ∗) = 1 21(−t3 + 8015) ≥ 0, so t1 = 1, t2 ∈ {−1, 6, 13} and t3 ∈ {14 + 21k \| −191 ≤ k ≤ 381}, that also gives us no solutions. • Let u be a unit of order 22. By (1) and Proposition 2 we have ν2a + ν11a + ν11b = 1. Now by (2) we obtain the system of inequalities µ0(u, χ2, ∗) = 1 22(60ν2a+28) ≥ 0; µ11(u, χ2, ∗) = 1 22(−60ν2a+16) ≥ 0, which has no integral solution. Jo u rn al A lg eb ra D is cr et e M at h .V. A. Bovdi, A. B. Konovalov 49 • Let u be a unit of order 33. By (1) and Proposition 2 we have ν3a + ν3b + ν11a + ν11b = 1. Put t1 = 5ν3a−4ν3b, t2 = 5ν3a+2ν3b and t3 = 32ν3a−4ν3b−6ν11a+5ν11b. Since \|u11\| = 3, for any character χ of G we need to consider four cases, defined by part (iii) of the Theorem. Case 1. Let χ(u11) = χ(3a). Then by (2) we obtain the system of inequalities µ11(u, χ2, ∗) = 1 33(10t1 + 27) ≥ 0; µ0(u, χ2, ∗) = 1 33(−20t1 + 12) ≥ 0, that has no integral solution. Case 2. Let χ(u11) = χ(3b). Now (2) gives us the system µ11(u, χ2, ∗) = 1 33(10t1 + 18) ≥ 0; µ0(u, χ2, ∗) = 1 33(−20t1 + 30) ≥ 0, which also has no integral solution. Case 3. Let χ(u11) = −2χ(3a) + 3χ(3b). By (2) we obtain that µ1(u, χ2, ∗) = 1 33(−5ν3a + 4ν3b) ≥ 0; µ11(u, χ2, ∗) = 1 33(50ν3a − 40ν3b) ≥ 0; µ0(u, χ3, ∗) = 1 33(60t2 + 207) ≥ 0; µ11(u, χ3, ∗) = 1 33(−30t2 + 243) ≥ 0; µ1(u, χ7, ∗) = 1 33(t3 + 978) ≥ 0; µ3(u, χ7, ∗) = 1 33(−2t3 + 750) ≥ 0. It follows that t1 = 0, t2 = 7 and t3 ∈ {12 + 33k \| −30 ≤ k ≤ 11}, and we have no solutions again. Case 4. Let χ(u11) = −χ(3a) + 2χ(3b). By (2) we obtain the system µ11(u, χ2, ∗) = 1 33(10t1 + 9) ≥ 0; µ0(u, χ2, ∗) = 1 33(−20t1 + 48) ≥ 0, which has no integral solution. • Let u be a unit of order 35. By (1) and Proposition 2 we have ν5a + ν5b + ν7a + ν7b = 1. Put t1 = 3ν5a − 2ν5b − ν7a − ν7b, t2 = 6ν5a + ν5b and t3 = 6ν5a + ν5b + 3ν7a − 4ν7b. Since \|u7\| = 5, for any character χ of G we need to consider six cases, defined by part (iv) of the Theorem. Case 1. Let χ(u7) = χ(5a). By (2) we obtain the system µ5(u, χ2, ∗) = 1 35(4t1 + 9) ≥ 0; µ0(u, χ2, ∗) = 1 35(−24t1 + 16) ≥ 0, which has no integral solutions. Case 2. Let χ(u7) = χ(5b). Now the non-compatible inequalities are µ7(u, χ2, ∗) = 1 35(6t1 + 26) ≥ 0; µ0(u, χ2, ∗) = 1 35(−24t1 + 36) ≥ 0. Jo u rn al A lg eb ra D is cr et e M at h .50 Integral group ring Case 3. Let χ(u7) = −2χ(5a) + 3χ(5b). By (2) we obtain the system µ7(u, χ2, ∗) = 1 35(6t1 + 16) ≥ 0; µ0(u, χ2, ∗) = 1 35(−24t1 + 76) ≥ 0, which has no integral solution. Case 4. Let χ(u7) = −3χ(5a) + 4χ(5b). By (2) we obtain the system µ0(u, χ2, ∗) = 1 35(−24t1 + 96) ≥ 0; µ7(u, χ2, ∗) = 1 35(6t1 + 11) ≥ 0; µ0(u, χ3, ∗) = 1 35(24t2 + 175) ≥ 0; µ7(u, χ3, ∗) = 1 35(−6t2 + 245) ≥ 0; µ15(u, χ16, ∗) = 1 35(4t3 + 8071) ≥ 0; µ1(u, χ16, ∗) = 1 35(−t3 + 8001) ≥ 0; µ0(u, χ2, 3) = 1 35(−96ν5a + 24ν5b + 85) ≥ 0, so t1 = 4, t2 ∈ {0, 35} and t3 ∈ {21+35k \| −58 ≤ k ≤ 228}, and we have no solutions again. Case 5. Let χ(u7) = −4χ(5a) + 5χ(5b). Using (2), we obtain µ7(u, χ2, ∗) = 1 35(6t1 + 6) ≥ 0; µ1(u, χ2, ∗) = 1 35(−t1 − 1) ≥ 0; µ0(u, χ3, ∗) = 1 35(24t2 + 155) ≥ 0; µ5(u, χ3, ∗) = 1 35(−4t2 + 155) ≥ 0; µ15(u, χ16, ∗) = 1 35(4t3 + 8091) ≥ 0; µ1(u, χ16, ∗) = 1 35(−t3 + 7996) ≥ 0. Then t1 = 1, t2 ∈ {−5, 30} and t3 ∈ {16 + 35k \| −58 ≤ k ≤ 228}, and we have no solutions as before. Case 6. Let χ(u7) = −χ(5a) + 2χ(5b). By (2) we have incompatible inequalities µ7(u, χ2, ∗) = 1 35(6t1 + 21) ≥ 0; µ0(u, χ2, ∗) = 1 35(−24t1 + 56) ≥ 0. • Let u be a unit of order 55. By (1) and Proposition 2 we have ν5a + ν5b + ν11a + ν11b = 1. Put t1 = 3ν5a − 2ν5b, t2 = 6ν5a + ν5b and t3 = 4ν5a − ν5b + 6ν11a − 5ν11b. Since \|u11\| = 5, for any character χ of G we need to consider six cases, defined by part (iv) of the Theorem. Case 1. Let χ(u11) = χ(5a). Then by (2) we obtain incompatible inequalities µ5(u, χ2, ∗) = 1 55(4t1 + 10) ≥ 0; µ0(u, χ2, ∗) = 1 55(−40t1 + 10) ≥ 0. Case 2. Let χ(u11) = χ(5b). Using (2) we obtain the system µ11(u, χ2, ∗) = 1 55(10t1 + 20) ≥ 0; µ0(u, χ2, ∗) = 1 55(−40t1 + 30) ≥ 0; µ0(u, χ3, ∗) = 1 55(40t2 + 235) ≥ 0; µ11(u, χ3, ∗) = 1 55(−10t2 + 230) ≥ 0; µ5(u, χ7, ∗) = 1 55(4t3 + 939) ≥ 0; µ1(u, χ7, ∗) = 1 55(−t3 + 934) ≥ 0, Jo u rn al A lg eb ra D is cr et e M at h .V. A. Bovdi, A. B. Konovalov 51 so t1 = −2, t2 ∈ {1, 12, 23} and t3 ∈ {−1 + 55k \| −4 ≤ k ≤ 17}, and in every case we have no integer solution. Case 3. Let χ(u11) = −2χ(5a) + 3χ(5b). By (2) we obtain that µ11(u, χ2, ∗) = 1 55(10t1 + 10) ≥ 0; µ0(u, χ2, ∗) = 1 55(−40t1 + 70) ≥ 0; µ0(u, χ3, ∗) = 1 55(40t2 + 195) ≥ 0; µ11(u, χ3, ∗) = 1 55(−10t2 + 240) ≥ 0; µ5(u, χ7, ∗) = 1 55(4t3 + 946) ≥ 0; µ1(u, χ7, ∗) = 1 55(−t3 + 891) ≥ 0. From this follows that t1 = −1, t2 ∈ {2, 13, 24} and t3 ∈ {11+55k \| −4 ≤ k ≤ 16}, and for every combination of ti’s we have no solution. Case 4. Let χ(u11) = −3χ(5a)+4χ(5b). By (2) we obtain the system µ11(u, χ2, ∗) = 1 55(10t1 + 5) ≥ 0; µ0(u, χ2, ∗) = 1 55(−40t1 + 90) ≥ 0, which has no integral solution. Case 5. Let χ(u11) = −4χ(5a) + 5χ(5b). By (2) we have that µ11(u, χ2, ∗) = 1 55(10t1) ≥ 0; µ1(u, χ2, ∗) = 1 55(−t1) ≥ 0; µ0(u, χ3, ∗) = 1 55(40t2 + 155) ≥ 0; µ11(u, χ3, ∗) = 1 55(−10t2 + 250) ≥ 0; µ5(u, χ7, ∗) = 1 55(4t3 + 986) ≥ 0; µ1(u, χ7, ∗) = 1 55(−t3 + 881) ≥ 0, so t1 = 0, t2 ∈ {3, 14, 25} and t3 ∈ {1 + 55k \| −4 ≤ k ≤ 16}, and again we have no solutions in every combination of ti’s. Case 6. Let χ(u11) = −χ(5a) + 2χ(5b). Using (2) we obtain the system µ11(u, χ2, ∗) = 1 55(10t1 + 15) ≥ 0; µ0(u, χ2, ∗) = 1 55(−40t1 + 50) ≥ 0, which has no integral solution. • Let u be a unit of order 77. By (1) and Proposition 2 we have ν7a + ν7b + ν11a + ν11b = 1. Then using (2) we obtain the non-compatible system of inequalities µ0(u, χ2, ∗) = 1 77(60(ν7a + ν7b) + 28) ≥ 0; µ11(u, χ2, ∗) = 1 77(−10(ν7a + ν7b) + 21) ≥ 0. 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Konovalov School of Computer Science, University of St Andrews, Jack Cole Building, North Haugh, St Andrews, Fife, KY16 9SX, Scot- land E-Mail: konovalov@member.ams.org

Integral group ring of the McLaughlin simple group

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