Integral group ring of the McLaughlin simple group

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Journal Algebra Discrete Math.

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 algebraQGto an element inG (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 forP SL(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 ofG. The Gruenberg–Kegel graph (or the prime graph) ofGis the graphπ(G) with vertices labeled by the

The research was supported by OTKA grant No. K 68383

2000 Mathematics Subject Classification: 16S34, 20C05; 20D08.

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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) IfG is a finite group thenπ(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-calledp-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 ofV(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 = P

α_gg ∈ V(ZG) has finite order k. Denote by ν_nt =

νnt(u) =εCnt(u) = P

g∈Cntαg the partial augmentation ofuwith respect toCnt. From the Berman–Higman Theorem (see [2] and [25], Ch.5, p.102)

one knows thatν₁ =α₁ = 0 and

X

Cnt∈C

νnt = 1. (1)

Hence, for any characterχ ofG, we get thatχ(u) =P

νntχ(hnt), where

hnt is a representative of the conjugacy classCnt.

Our main result is the following

Theorem 1. Let Gdenote the McLaughlin simple group McL. Letu 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

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(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, thenu is rationally conjugate to some g∈G.

(iii) If|u|= 3, then all components ofP(u)are zero except possiblyν₃_a andν3b, and the pair(ν3a, ν3b) is one of

{(−2,3), (−1,2), (0,1), (1,0)}.

(iv) If|u|= 5, then all components ofP(u)are zero except possiblyν5a

andν₅_b, and the pair(ν₅_a, ν₅_b) is one of

{(−4,5), (−3,4), (−2,3), (−1,2), (0,1), (1,0)}.

(v) If|u|= 7, then all components ofP(u)are zero except possiblyν7a

andν7b, 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 ν₁₁_aand ν₁₁_b, and the pair(ν₁₁_a, ν₁₁_b) 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. IfG=McL thenπ(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]) Letu∈V(ZG)be of order k. Then u is conjugate in QG to an element g ∈ G if and only if for eachddividingkthere is precisely one conjugacy classC with partial augmentationεC(ud)6= 0.

The next result now yield that several partial augmentations are zero.

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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 kand assume kand p are coprime in case p6= 0. Ifz is a complex primitivek-th root of unity and χ is either a classical character or a p-Brauer character of G, then for every integer lthe number

µ_l(u, χ, p) = 1_kP

d|kT rQ(zd

)/Q{χ(ud)z−dl} (2)

is a non-negative integer.

Note that if p= 0, we will use the notation µ_l(u, χ,∗) for µ_l(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 elementu∈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) wherep∈ {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 and11, since the groupGpossesses 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 inV(ZG).

Now we consider each case separately.

• Letu 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.

•Letube a unit of order3. By (1) and Proposition 2 we getν3a+ν3b = 1.

Put t₁= 5ν₃_a−4ν₃_b. By (2) we obtain the system of inequalities

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from which t1 ∈ {−22,−19,−16,−13,−10,−7,−4,−1,2,5,8,11}. Now for each possible value oft1 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.

•Letube a unit of order5. By (1) and Proposition 2 we getν5a+ν5b= 1.

Put t1= 3ν5a−2ν5b. By (2) we obtain the system of inequalities

µ0(u, χ2,∗) = 1₅(−4t1+ 22)≥0; µ1(u, χ2,∗) = 1₅(t1+ 22)≥0,

sot1 ∈ {−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.

•Letube a unit of order7. By (1) and Proposition 2 we getν7a+ν7b= 1.

Put t1= 4ν7a−3ν7b. Using BCT(3) andBCT(5), by (2) we have

µ3(u, χ7,3) = 1₇(t1+ 605)≥0; µ1(u, χ12,5) = 1₇(−t1+ 3245)≥0;

µ1(u, χ7,3) = 1₇(−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.

• Letu be a unit of order 11. By (1) and Proposition 2 we haveν11a+

ν11b= 1. UsingBCT(3), by (2) we obtain the system of inequalities

µ1(u, χ3,3) = ₁₁1 (6ν11a−5ν11b+ 104)≥0;

µ2(u, χ3,3) = ₁₁1 (−5ν11a+ 6ν11b+ 104)≥0,

that has only that twenty pairs(ν11a, ν11b) listed in part (vi) of the

The-orem 1.

•Let ube a unit of order21. By (1) and Proposition 2 we have

ν3a+ν3b+ν7a+ν7b = 1.

Putt1 = 5ν3a−4ν3b−ν7a−ν7b,t2= 5ν3a+2ν3bandt3 = 3ν7a−4ν7b. Since

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Case 1. Let χ(u7) = χ(3a). Using (2), we obtain the system of inequalities

µ₃(u, χ₂,∗) = ₂₁1(2t₁+ 11)≥0; µ₀(u, χ₂,∗) = ₂₁1 (−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,∗) = ₂₁1 (−12t₁+ 36)≥0; µ₇(u, χ₂,∗) = ₂₁1(6t₁+ 24)≥0;

µ0(u, χ3,∗) = ₂₁1 (36t2+ 243)≥0; µ7(u, χ3,∗) = ₂₁1(−18t2+ 225)≥0;

µ1(u, χ16,∗) = ₂₁1 (−t3+ 8386)≥0; µ9(u, χ16,∗) = ₂₁1(2t3+ 8386)≥0;

µ1(u, χ5,∗) = ₂₁1(−13ν3a+ 5ν3b+ 765)≥0.

This yieldst1 ∈ {−4,3},t2∈ {−5,2,9} and t3 ∈ {7 + 21k| −200≤k≤ 399}, but none of possible combinations of t_i’s gives us any solution.

Case 3. Let χ(u7) = −2χ(3a) + 3χ(3b). Then using (2), we obtain the system

µ1(u, χ2,∗) = ₂₁1 (−t1−1)≥0; µ7(u, χ2,∗) = ₂₁1 (6t1+ 6)≥0;

µ₀(u, χ₃,∗) = ₂₁1 (36t₂+ 207)≥0; µ₇(u, χ₃,∗) = ₂₁1(−18t₂+ 243)≥0;

µ1(u, χ16,∗) = ₂₁1 (−t3+ 8218)≥0; µ9(u, χ16,∗) = ₂₁1(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 ofti’s.

Case 4. Let χ(u7) = −χ(3a) + 2χ(3b). Using (2), we obtain the system

µ7(u, χ2,∗) = ₂₁1 (6t1+ 15)≥0; µ0(u, χ2,∗) = ₂₁1(−12t1+ 54)≥0;

µ0(u, χ3,∗) = ₂₁1 (36t2+ 225)≥0; µ7(u, χ3,∗) = ₂₁1(−18t2+ 234)≥0;

µ9(u, χ16,∗) = ₂₁1 (2t3+ 8015)≥0; µ1(u, χ16,∗) = ₂₁1 (−t3+ 8015)≥0,

so t1 = 1,t2 ∈ {−1,6,13} and t3 ∈ {14 + 21k| −191≤k ≤381}, that also gives us no solutions.

• Letu be a unit of order22. By (1) and Proposition 2 we have

ν2a+ν11a+ν11b = 1.

Now by (2) we obtain the system of inequalities

µ₀(u, χ₂,∗) = ₂₂1(60ν₂_a+28)≥0; µ₁₁(u, χ₂,∗) = ₂₂1 (−60ν₂_a+16)≥0,

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ν3a+ν3b+ν11a+ν11b = 1.

Putt1 = 5ν3a−4ν3b,t2 = 5ν3a+2ν3bandt3 = 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,∗) = ₃₃1(10t1+ 27)≥0; µ0(u, χ2,∗) = ₃₃1(−20t1+ 12)≥0,

that has no integral solution.

Case 2. Letχ(u11) =χ(3b). Now (2) gives us the system

µ11(u, χ2,∗) = ₃₃1(10t1+ 18)≥0; µ0(u, χ2,∗) = ₃₃1(−20t1+ 30)≥0,

which also has no integral solution.

Case 3. Letχ(u11) =−2χ(3a) + 3χ(3b). By (2) we obtain that

µ1(u, χ2,∗) = ₃₃1(−5ν3a+ 4ν3b)≥0; µ11(u, χ2,∗) = ₃₃1 (50ν3a−40ν3b)≥0;

µ0(u, χ3,∗) = ₃₃1(60t2+ 207)≥0; µ11(u, χ3,∗) = ₃₃1(−30t2+ 243)≥0;

µ₁(u, χ₇,∗) = ₃₃1(t₃+ 978)≥0; µ₃(u, χ₇,∗) = ₃₃1 (−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. Letχ(u11) =−χ(3a) + 2χ(3b). By (2) we obtain the system

µ11(u, χ2,∗) = ₃₃1(10t1+ 9)≥0; µ0(u, χ2,∗) = ₃₃1(−20t1+ 48)≥0,

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 characterχ ofGwe need to consider

six cases, defined by part (iv) of the Theorem.

Case 1. Letχ(u7) =χ(5a). By (2) we obtain the system

µ₅(u, χ₂,∗) = ₃₅1(4t₁+ 9)≥0; µ₀(u, χ₂,∗) = ₃₅1(−24t₁+ 16)≥0,

which has no integral solutions.

Case 2. Letχ(u7) =χ(5b). Now the non-compatible inequalities are

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Case 3. Let χ(u7) =−2χ(5a) + 3χ(5b). By (2) we obtain the system

µ7(u, χ2,∗) = ₃₅1(6t1+ 16)≥0; µ0(u, χ2,∗) = ₃₅1(−24t1+ 76)≥0,

Case 4. Let χ(u7) =−3χ(5a) + 4χ(5b). By (2) we obtain the system

µ0(u, χ2,∗) =₃₅1(−24t1+ 96)≥0; µ7(u, χ2,∗) = ₃₅1(6t1+ 11)≥0;

µ0(u, χ3,∗) =₃₅1(24t2+ 175)≥0; µ7(u, χ3,∗) = ₃₅1 (−6t2+ 245)≥0;

µ15(u, χ16,∗) =₃₅1(4t3+ 8071)≥0; µ1(u, χ16,∗) = ₃₅1(−t3+ 8001)≥0;

µ₀(u, χ₂,3) = ₃₅1(−96ν₅_a+ 24ν₅_b+ 85)≥0,

sot1 = 4,t2 ∈ {0,35}andt3 ∈ {21 + 35k| −58≤k≤228}, and we have no solutions again.

Case 5. Letχ(u7) =−4χ(5a) + 5χ(5b). Using (2), we obtain

µ7(u, χ2,∗) =₃₅1(6t1+ 6)≥0; µ1(u, χ2,∗) = ₃₅1(−t1−1)≥0;

µ₀(u, χ₃,∗) =₃₅1(24t₂+ 155)≥0; µ₅(u, χ₃,∗) = ₃₅1 (−4t₂+ 155)≥0;

µ15(u, χ16,∗) =₃₅1(4t3+ 8091)≥0; µ1(u, χ16,∗) = ₃₅1(−t3+ 7996)≥0.

Thent1 = 1,t2∈ {−5,30}andt3 ∈ {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,∗) = ₃₅1(6t₁+ 21)≥0; µ₀(u, χ₂,∗) = ₃₅1 (−24t₁+ 56)≥0.

• Letu be a unit of order55. By (1) and Proposition 2 we have

ν5a+ν5b+ν11a+ν11b= 1.

Put t₁= 3ν₅_a−2ν₅_b,t₂ = 6ν₅_a+ν₅_b and t₃ = 4ν₅_a−ν₅_b+ 6ν₁₁_a−5ν₁₁_b. 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,∗) = ₅₅1(4t1+ 10)≥0; µ0(u, χ2,∗) = ₅₅1 (−40t1+ 10)≥0.

Case 2. Letχ(u11) =χ(5b). Using (2) we obtain the system

µ11(u, χ2,∗) = ₅₅1 (10t1+ 20)≥0; µ0(u, χ2,∗) = ₅₅1(−40t1+ 30)≥0;

µ₀(u, χ₃,∗) = ₅₅1 (40t₂+ 235)≥0; µ₁₁(u, χ₃,∗) = ₅₅1(−10t₂+ 230)≥0;

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sot1 =−2, t2 ∈ {1,12,23} and t3 ∈ {−1 + 55k| −4≤k ≤17}, and in every case we have no integer solution.

Case 3. Letχ(u11) =−2χ(5a) + 3χ(5b). By (2) we obtain that

µ11(u, χ2,∗) = ₅₅1(10t1+ 10)≥0; µ0(u, χ2,∗) = ₅₅1(−40t1+ 70)≥0;

µ0(u, χ3,∗) = ₅₅1(40t2+ 195)≥0; µ11(u, χ3,∗) = ₅₅1(−10t2+ 240)≥0;

µ₅(u, χ₇,∗) = ₅₅1(4t₃+ 946)≥0; µ₁(u, χ₇,∗) = ₅₅1(−t3+ 891)≥0.

From this follows thatt1=−1,t2∈ {2,13,24}andt3∈ {11+55k| −4≤

k≤16}, and for every combination of ti’s we have no solution.

Case 4. Letχ(u11) =−3χ(5a) + 4χ(5b). By (2) we obtain the system

µ11(u, χ2,∗) = ₅₅1(10t1+ 5)≥0; µ0(u, χ2,∗) = ₅₅1 (−40t1+ 90)≥0,

Case 5. Letχ(u11) =−4χ(5a) + 5χ(5b). By (2) we have that

µ11(u, χ2,∗) = ₅₅1(10t1)≥0; µ1(u, χ2,∗) = ₅₅1 (−t1)≥0;

µ0(u, χ3,∗) = ₅₅1(40t2+ 155)≥0; µ11(u, χ3,∗) = ₅₅1(−10t2+ 250)≥0;

µ5(u, χ7,∗) = ₅₅1(4t3+ 986)≥0; µ1(u, χ7,∗) = ₅₅1(−t3+ 881)≥0,

sot1 = 0,t2 ∈ {3,14,25} and t3 ∈ {1 + 55k| −4 ≤k≤16}, and again we have no solutions in every combination oft_i’s.

Case 6. Let χ(u11) = −χ(5a) + 2χ(5b). Using (2) we obtain the system

µ11(u, χ2,∗) = ₅₅1(10t1+ 15)≥0; µ0(u, χ2,∗) = ₅₅1(−40t1+ 50)≥0,

ν7a+ν7b+ν11a+ν11b = 1.

Then using (2) we obtain the non-compatible system of inequalities

µ0(u, χ2,∗) = ₇₇1(60(ν7a+ν7b) + 28)≥0;

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Contact information

V. A. Bovdi Institute of Mathematics, University of De-brecen, P.O. Box 12, H-4010 DeDe-brecen, Hungary Institute of Mathematics and In-formatics, College of Ny´ıregyháza, Sóstói út 31/b, H-4410 Ny´ıregyháza, Hungary E-Mail: [email protected]

A. B. Konovalov School of Computer Science, University of St Andrews, Jack Cole Building, North Haugh, St Andrews, Fife, KY16 9SX, Scot-land