Torsion units for a Ree group, Tits group and a Steinberg triality group

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1 Rend. Circ. Mat. Palermo (2016) 65: DOI /s Torsion units for a Ree group, Tits group and a Steinberg triality group Joe Gildea 1 Received: 5 November 2015 / Accepted: 5 December 2015 / Published online: 28 December 2015 The Author(s) This article is published with open access at Springerlink.com Abstract We investigate the Zassenhaus conjecture for the Steinberg triality group 3 D 4 (2 3 ), Tits group 2 F 4 (2) and the Ree group 2 F 4 (2). Consequently, we prove that the Prime Graph question is true for all three groups. Keywords Zassenhaus conjecture Prime graph question Torsion unit Partial augmentation Integral group ring Mathematics Subject Classification 16S34 20C05 1 Introduction and main results Let U(ZG) be the unit group of the integral group ring of a finite group G.Itiswellknown that U(ZG) ={±1} V (ZG), where V (ZG) is the group of units of augmentation one. Throughout this paper, G will always represent a finite group and torsion units will always represent torsion units in V (ZG)\{1}. The next conjecture is very important in theory of integral group rings: Conjecture 1 If G is a finite group, then for each torsion unit u V (ZG) there exists g G, such that u = g. Hans Zassenhaus formulated a stronger version of this conjecture in [1], which states: Conjecture 2 A torsion unit in V (ZG) is said to be rationally conjugate to a group element if it is conjugate to an element of G by a unit of the rational group ring QG. B Joe Gildea j.gildea@chester.ac.uk 1 Department of Mathematics, Faculty of Science and Engineering, University of Chester, Chester, England

2 140 J. Gildea This conjecture was confirmed for Nilpotent Groups in [2,3] and cyclic-by-abelian groups in [4]. However, these techniques do not transfer to simple groups. The Luthar Passi Method (which was introduced in [5]) is the main investigative tool for simple groups in relation to the Zassenhaus conjecture. The Zassenhaus conjecture was confirmed true for all groups up to order 71 in [6]. This conjecture was also validated for A 5, S 5, central extensions of S 5 and other simple finite groups in [5,7 9]. In [10] partial results were given for A 6,and the remaining cases were dealt with in [11]. Alternating groups of higher order were also considered in [12,13]. Additionally, it was proved for PSL(2, p) when p ={7, 11, 13} in [14] andp ={8, 17} in [15] andpsl(2, p) when p ={19, 23} in [16]. See [17,18] for further results regarding PSL(2, p). Let H be a group with a torsion part t(h) of finite exponent and #H be the set of primes dividing the order of elements from the set t(h). The prime graph of H (denoted by π(h)) is a graph with vertices labeled by primes from #H, such that vertices p and q are adjacent if and only if there is an element of order pq in the group H. The following, was composed as a problem in [19] (Problem 37): Question 1 (Prime Graph Question) If G is a finite group, then π(g) = π(v (ZG)). This question is upheld for Frobenius and Solvable groups in [20]. It was also confirmed for some Sporadic Simple groups in [21 32]. Additionally, it was confirmed for the sympletic simple group S 4 (4) in [33]. The prime graph question is an intermediate step towards the Zassenhaus conjecture. The family of Steinberg Triality groups (denoted by 3 D 4 (q 3 )) were introduced by Steinberg in [34]. The Ree groups 2 F 4 (2 n+1 ) were initially constructed in [35]. These groups are simple if n 1 and Tits constructed a new simple group from the derived subgroup of 2 F 4 (2) (denoted by 2 F 4 (2) )in[36]. We combine the Luthar Passi Method together with techniques developed in [14,37] to obtain our results. Our main results are as follows: Theorem 1 Let G = 3 D 4 (2 3 ) and u be a torsion unit of V (ZG). Subsequently the following conditions hold: (i) If u {7}, then u is rationally conjugate to some g in G. (ii) There are no elements of orders 26, 39 or 91 in V (ZG). (iii) If u =2, thenν rx = 0 rx / {ν 2a,ν 2b } and (ν 2a,ν 2b ) {(2, 1), (1, 0), (0, 1), ( 1, 2), ( 2, 3)}. (iv) If u =3, thenν rx = 0 rx / {ν 3a,ν 3b } and (ν 3a,ν 3b ) {(3, 2), (2, 1), (1, 0), (0, 1), ( 1, 2)}. (v) If u =13, thenν rx = 0 rx / {ν 13a,ν 13b,ν 13c } and (ν 13a,ν 13b,ν 13c ) {(1, 1, 1), (1, 0, 0), (0, 0, 1), (1, 1, 1), (0, 1, 0), ( 1, 1, 1)}. Theorem 2 Let G = 2 F 4 (2) and u be a torsion unit of V (ZG). Subsequently the following conditions hold: (i) If u {3, 5}, then u is rationally conjugate to some g in G. (ii) There are no elements of orders 15, 26, 39 or 65 in V (ZG). (iii) If u =2, thenν rx = 0 rx / {ν 2a,ν 2b } and (ν 2a,ν 2b ) {(3, 2), (2, 1), (1, 0), (0, 1), ( 1, 2), ( 2, 3), ( 3, 4)}.

3 Torsion units for a Ree group (iv) If u =10, thenν rx = 0 rx / {ν 2a,ν 2b,ν 5a,ν 10a } and (ν 2a,ν 2b,ν 5a,ν 10a ) {( 2, 2, 0, 5), ( 1, 3, 0, 5), ( 1, 3, 5, 0), ( 1, 1, 0, 3), ( 1, 1, 5, 2), ( 1, 1, 0, 1), (0, 4, 5, 0), (0, 2, 0, 3), (0, 2, 5, 2), (0, 0, 0, 1), (0, 0, 5, 4), (0, 2, 0, 1), (0, 4, 5, 2), (0, 4, 0, 3), (1, 1, 0, 1), (1, 3, 5, 2), (1, 3, 0, 3), (2, 2, 0, 3)}. (v) If u =13, thenν rx = 0 rx / {ν 13a,ν 13b } and (ν 13a,ν 13b ) {(9, 8), (8, 7), (7, 6), (6, 5), (5, 4), (4, 3), (3, 2), (2, 1), (1, 0), (0, 1), ( 1, 2), ( 2, 3), ( 3, 4), ( 4, 5), ( 5, 6), ( 6, 7), ( 7, 8), ( 8, 9)}. Theorem 3 Let G = 2 F 4 (2) and u be a torsion unit of V (ZG). Subsequently the following conditions hold: (i) If u {3, 5, 13}, then u is rationally conjugate to some g in G. (ii) There are no elements of orders 15, 26, 39 or 65 in V (ZG). (iii) If u =2, thenν rx = 0 rx / {ν 2a,ν 2b } and (ν 2a,ν 2b ) {(3, 2), (2, 1), (1, 0), (0, 1), ( 1, 2), ( 2, 3), ( 3, 4)}. (iv) If u =10, thenν rx = 0 rx / {ν 2a,ν 2b,ν 5a,ν 10a } and (ν 2a,ν 2b,ν 5a,ν 10a ) {( 2, 2, 0, 5), ( 1, 3, 0, 5), ( 1, 3, 5, 0), ( 1, 1, 0, 3), ( 1, 1, 5, 2), ( 1, 1, 0, 1), (0, 4, 5, 0), (0, 2, 0, 3), (0, 2, 5, 2), (0, 0, 0, 1), (0, 0, 5, 4), (0, 2, 0, 1), (0, 4, 5, 2), (0, 4, 0, 3), (1, 1, 0, 1), (1, 3, 5, 2), (1, 3, 0, 3), (2, 2, 0, 3)}. Corollary 1 The Prime Graph question is true for the groups 3 D 4 (2), 2 F 4 (2) and 2 F 4 (2). Let u = a g g be a torsion unit of V (ZG). Then, the sum g X G a g Z is the partial augmentation (denoted by ε C (u)) ofu with respect to it s conjugacy classes X G in G. Let ν i = ε Ci (u) be the i-th partial augmentation of u. It was proved that ν 1 = 0andν j = 0if the conjugacy class C j consists of a central element by G. Higman and S. D. Berman [38]. Therefore, ν 2 + ν 3 + +ν l = 1, where l denotes the number of non-central conjugacy classes of G. Proposition 1 ([39]) Let u be torsion unit of V (ZG). The order of u divides the exponent of G. The following propositions provide relationships between the partial augmentations and the order of a torsion unit. Proposition 2 (Proposition 3.1 in [37]) Let u be a torsion unit of V (ZG). Let C be a conjugacy class of G. If p is a prime dividing the order of a representative of C but not the order of u then the partial augmentation ε C (u) = 0. Proposition 3 (Proposition 2.2 in [14]) Let G be a finite group and let u be a torsion unit in V (ZG).

4 142 J. Gildea (i) If u has order p n,thenε x (u) = 0 for every x of G whose p-part is of order strictly greater than p n. (ii) If x is an element of G whose p-part, for some prime, has order strictly greater than the order of the p-part of u, then ε x (u) = 0. Proposition 4 ([5]) Let u be a torsion unit of V (ZG) of order k. Then u is conjugate in QG to an element g G iff for each d dividing k there is precisely one conjugacy class νi d with partial augmentation ε νid (u d ) = 0. Proposition 5 ([5]) Let p be equal to zero or a prime divisor of G. Suppose that u is an element of V (ZG) of order k. Let z be a primitive kth root of unity. Then for every integer l and any character χ of G, the number {χ(u d )z dl} is a non-negative integer. μ l (u,χ,p) = 1 k d k Tr Q(z d )/Q We will use the notation μ l (u,χ, ) when p = 0. The LAGUNA package [40] forthe GAP system [41] is a very useful tool when calculating μ l (u,χ,p). 2 Proof of Theorem 1 Let G = 3 D 4 (2 3 ). Clearly G = = and exp(g) = 6552 = Initially, for any torsion unit of V (ZG) of order k: ν 2a + ν 2b + ν 3a + ν 3b + ν 4a + ν 4b + ν 4c + ν 6a + ν 6b + ν 7a + ν 7b + ν 7c + ν 7d + ν 8a + ν 8b + ν 9a + ν 9b + ν 9c + ν 12a + ν 13a + ν 13b + ν 13c + ν 14a + ν 14b + ν 14c + ν 18a + ν 18b + ν 18c + ν 21a + ν 21b + ν 21c + ν 28a + ν 28b + ν 28c = 1. In order to prove that the Zassenhaus Conjecture holds, we need to consider torsion units of V (ZG) of order 2, 3, 4, 6, 7, 8, 9, 12, 13, 14, 18, 21, 28, 24, 26, 36, 39, 42, 56, 63 and 91 (by Proposition 1). For the purpose of this paper and due to the complexity of certain orders, we shall consider elements of order 2, 3, 7, 13, 26, 39 and 91. We shall now consider each case separately. Case (i). Letu V (ZG) where u =2. Using Propositions 2 and 3, ν 2a + ν 2b = 1. Applying Proposition 5, we obtain the following system of inequalities: μ 0 (u,χ 2, ) = 2 1 ( 2γ ) 0; μ 1 (u,χ 2, ) = 2 1 (2γ ) 0; μ 0 (u,χ 3, ) = 2 1 (4γ ) 0; μ 1 (u,χ 3, ) = 2 1 ( 4γ ) 0 where γ 1 = 3ν 2a ν 2b and γ 2 = 5ν 2a ν 2b. It follows that the only possible integer solutions for (ν 2a,ν 2b ) are listed in part (iii) of Theorem 1. Case (ii). Letu V (ZG) where u =3. Using Propositions 2 and 3, ν 3a + ν 3b = 1.

5 Torsion units for a Ree group Applying Proposition 5, we obtain the following system of inequalities: μ 0 (u,χ 3, ) = 1 3 (2γ ) 0; μ 1 (u,χ 3, ) = 1 3 ( γ ) 0; μ 0 (u,χ 2, 2) = 1 3 (2γ 2 + 8) 0; μ 1 (u,χ 2, 2) = 1 3 ( γ 2 + 8) 0 where γ 1 = 7ν 3a 2ν 3b and γ 2 = 2ν 3a ν 3b. It follows that the only possible integer solutions for (ν 2a,ν 2b ) are listed in part (iv) of Theorem 1. Case (iii). Letu V (ZG) where u =7. Using Propositions 2 and 3, ν 7a + ν 7b + ν 7c + ν 7d = 1. μ 0 (u,χ 2, ) = 1 7 (6γ ) 0; μ 1 (u,χ 2, ) = 1 7 ( γ ) 0; μ 0 (u,χ 5, ) = 1 7 (42γ ) 0; μ 1 (u,χ 5, ) = 1 7 ( 7γ ) 0; μ 0 (u,χ 6, ) = 1 7 (6γ ) 0; μ 1 (u,χ 6, ) = 1 7 ( γ ) 0; μ 0 (u,χ 10, ) = 1 7 (6γ ) 0; μ 1 (u,χ 10, ) = 1 7 ( γ ) 0; μ 0 (u,χ 6, 13) = 1 7 (6γ ) 0; μ 1 (u,χ 6, 13) = 1 7 ( γ ) 0; μ 0 (u,χ 2, 3) = 1 7 (4γ ) 0; μ 1 (u,χ 2, 3) = 1 7 ( γ ) 0; μ 0 (u,χ 2, 2) = 1 7 ( 2γ 6 + 8) 0; μ 0 (u,χ 12, ) = 1 7 (6γ ) 0; μ 1 (u,χ 12, ) = 1 7 ( γ ) 0; μ 1 (u,χ 2, 2) = 1 7 (γ 8 + 8) 0; μ 3 (u,χ 2, 2) = 1 7 (γ 9 + 8) 0; μ 2 (u,χ 2, 2) = 1 7 (γ ) 0; μ 1 (u,χ 18, ) = 1 7 (γ ) 0; μ 2 (u,χ 18, ) = 1 7 (γ ) 0; μ 3 (u,χ 18, ) = 1 7 (γ ) 0; μ 0 (u,χ 25, ) = 1 7 (γ ) 0; μ 1 (u,χ 25, ) = 1 7 (γ ) 0; μ 2 (u,χ 25, ) = 1 7 (γ ) 0; μ 3 (u,χ 25, ) = 1 7 (γ ) 0; μ 1 (u,χ 6, 2) = 1 7 (γ ) 0; μ 2 (u,χ 6, 2) = 1 7 (γ ) 0; μ 3 (u,χ 6, 2) = 1 7 (γ ) 0; μ 0 (u,χ 9, 2) = 1 7 (γ ) 0; μ 1 (u,χ 9, 2) = 1 7 (γ ) 0; μ 2 (u,χ 9, 2) = 1 7 (γ ) 0; μ 3 (u,χ 9, 2) = 1 7 (γ ) 0; μ 1 (u,χ 13, 2) = 1 7 (γ ) 0; μ 2 (u,χ 13, 2) = 1 7 (γ ) 0; μ 3 (u,χ 13, 2) = 1 7 (γ ) 0; μ 1 (u,χ 22, ) = 1 7 (γ ) 0; μ 2 (u,χ 22, ) = 1 7 (γ ) 0; μ 3 (u,χ 22, ) = 1 7 (γ ) 0 where γ 1 = 5ν 7a + 5ν 7b + 5ν 7c 2ν 7d,γ 2 = ν 7a + ν 7b + ν 7c,γ 3 = 9ν 7a + 9ν 7b + 9ν 7c + 2ν 7d,γ 4 = 6ν 7a + 6ν 7b + 6ν 7c ν 7d,γ 5 = 8ν 7a + 8ν 7b + 8ν 7c + ν 7d,γ 6 = 4ν 7a + 4ν 7b + 4ν 7c 3ν 7d,γ 7 = 3ν 7a + 3ν 7b + 3ν 7c 4ν 7d,γ 8 = 8ν 7a ν 7b + 13ν 7c ν 7d,γ 9 = ν 7a + 13ν 7b 8ν 7c ν 7d,γ 10 = 13ν 7a 8ν 7b ν 7c ν 7d,γ 11 = 36ν 7a 27ν 7b 6ν 7c + ν 7d,γ 12 = 6ν 7a + 36ν 7b 27ν 7c + ν 7d,γ 13 = 27ν 7a 6ν 7b + 36ν 7c +ν 7d,γ 14 = 22ν 7a 22ν 7b 22ν 7c +6ν 7d,γ 15 = 34ν 7a ν 7b 22ν 7c ν 7d,γ 16 = 22ν 7a + 34ν 7b ν 7c ν 7d,γ 17 = ν 7a 22ν 7b + 34ν 7c ν 7d,γ 18 = 20ν 7a + 8ν 7b + 15ν 7c + ν 7d,γ 19 = 15ν 7a 20ν 7b + 8ν 7c + ν 7d,γ 20 = 8ν 7a + 15ν 7b 20ν 7c + ν 7d,γ 21 = 34ν 7a 34ν 7b 34ν 7c 6ν 7d,γ 22 = 20ν 7a 13ν 7b +50ν 7c +ν 7d,γ 23 = 50ν 7a 20ν 7b 13ν 7c + ν 7d,γ 24 = 13ν 7a + 50ν 7b 20ν 7c + ν 7d,γ 25 = 35ν 7a + 28ν 7b + 7ν 7c,γ 26 =

6 144 J. Gildea 7ν 7a 35ν 7b + 28ν 7c,γ 27 = 28ν 7a + 7ν 7b 35ν 7c,γ 28 = 35ν 7a 14ν 7b 14ν 7c,γ 29 = 14ν 7a + 35ν 7b 14ν 7c and γ 30 = 14ν 7a 14ν 7b + 35ν 7c. It follows that the only possible integer solutions for (ν 7a,ν 7b,ν 7c,ν 7d ) are (1, 0, 0, 0), (0, 0, 0, 1), (0, 0, 1, 0) and (0, 1, 0, 0). Therefore, u is rationally conjugated to some element g G by Proposition 4. Case (iv). Letu V (ZG) where u =13. Using Propositions 2 and 3, ν 13a + ν 13b + ν 13c = 1. μ 1 (u,χ 9, 2) = 13 1 (γ ) 0; μ 1 (u,χ 6, 2) = 13 1 ( γ ) 0; μ 2 (u,χ 6, 2) = 13 1 (γ ) 0; μ 2 (u,χ 9, 2) = 13 1 ( γ ) 0; μ 1 (u,χ 2, 2) = 13 1 (γ 3 + 8) 0; μ 2 (u,χ 2, 2) = 13 1 (γ 3 + 8) 0; μ 4 (u,χ 2, 2) = 13 1 (γ 4 + 8) 0 where γ 1 = 9ν 13a 4ν 13b 4ν 13c, γ 2 = 4ν 13a 9ν 13b + 4ν 13c, γ 3 = 5ν 13a 8ν 13b + 5ν 13c, γ 4 = 5ν 13a + 5ν 13b 8ν 13c and γ 5 = 8ν 13a + 5ν 13b + 5ν 13c. Clearly γ 1 {9 + 13k 13 k 3} and γ 2 {4 + 13k 4 k 12}. It follows that the only possible integer solutions for (ν 13a,ν 13b,ν 13c ) are listed in part (v) of Theorem 1. Case (v). Letu V (ZG) where u =26. Using Propositions 2 and 3, ν 2a + ν 2b + ν 13a + ν 13b + ν 13c = 1. Let γ 1 = 3ν 2a ν 2b, γ 2 = 5ν 2a ν 2b, γ 3 = 28ν 2a 4ν 2b + ν 13a + ν 13b + ν 13c, γ 4 = 63ν 2a +15ν 2b 4ν 13a 4ν 13b +9ν 13c and γ 5 = 63ν 2a +15ν 2b +9ν 13a 4ν 13b 4ν 13c. We shall now separately consider the following cases involving χ(u n ) for n {2, 13}: χ(u 13 ) = m 1 χ(2a)+m 2 χ(2b) and χ(u 2 ) = m 3 χ(13a)+m 4 χ(13b)+m 5 χ(13c) where (m 1, m 2, m 3, m 4, m 5 ) {(1, 0, 1, 0, 0), (1, 0, 0, 1, 0), (1, 0, 0, 0, 1), (1, 0, 1, 1, 1), (1, 0, 1, 1, 1), (1, 0, 1, 1, 1), (2, 1, 1, 0, 0), (2, 1, 0, 1, 0), (2, 1, 0, 0, 1), (2, 1, 1, 1, 1), (2, 1, 1, 1, 1), (2, 1, 1, 1, 1), ( 1, 2, 1, 0, 0), ( 1, 2, 0, 1, 0), ( 1, 2, 0, 0, 1), ( 1, 2, 1, 1, 1), ( 1, 2, 1, 1, 1), ( 1, 2, 1, 1, 1), ( 2, 3, 1, 0, 0), ( 2, 3, 0, 1, 0), ( 2, 3, 0, 0, 1), ( 2, 3, 1, 1, 1), ( 2, 3, 1, 1, 1), ( 2, 3, 1, 1, 1)} (24γ 1 + 8) 0; μ 0 (u,χ 2, ) = 26 1 ( 24γ ) 0. It follows that there are no possible integer solutions for (ν 2a,ν 2b,ν 13a,ν 13b,ν 13c ) in all cases. χ(u 13 ) = m 1 χ(2a)+m 2 χ(2b) and χ(u 2 ) = m 3 χ(13a)+m 4 χ(13b)+m 5 χ(13c) where (m 1, m 2, m 3, m 4, m 5 ) {(0, 1, 1, 0, 0), (0, 1, 0, 1, 0), (0, 1, 0, 0, 1), (0, 1, 1, 1, 1), (0, 1, 1, 1, 1), (0, 1, 1, 1, 1)}.

7 Torsion units for a Ree group (24γ ) 0; μ 0 (u,χ 2, ) = 26 1 ( 24γ ) 0; μ 0 (u,χ 3, ) = 26 1 (48γ ) 0; μ 13 (u,χ 3, ) = 26 1 ( 48γ ) 0; μ 1 (u,χ 4, ) = 26 1 (γ ) 0; μ 8 (u,χ 31, ) = 26 1 (γ 4 + k 1 ) 0; μ 1 (u,χ 31, ) = 26 1 ( γ 4 + k 2 ) 0; μ 2 (u,χ 31, ) = 26 1 (γ 5 + k 3 ) 0; μ 3 (u,χ 31, ) = 26 1 ( γ 5 + k 4 ) 0 where (m 1, m 2, m 3, m 4, m 5 ) = (0, 1, 1, 0, 0) when (k 1, k 2, k 3, k 4 ) = (3993, 3950, 3980, 3963), (m 1, m 2, m 3, m 4, m 5 ) = (0, 1, 0, 1, 0) when (k 1, k 2, k 3, k 4 ) = (3980, 3963, 3993, 3950), (m 1, m 2, m 3, m 4, m 5 ) = (0, 1, 0, 0, 1) when (k 1, k 2, k 3, k 4 ) = (3980, 3950, 3980, 3950), (m 1, m 2, m 3, m 4, m 5 ) = (0, 1, 1, 1, 1) when (k 1, k 2, k 3, k 4 ) = (3993, 3937, 3967, 3963), (m 1, m 2, m 3, m 4, m 5 ) = (0, 1, 1, 1, 1) when (k 1, k 2, k 3, k 4 ) = (3993, 3963, 3993, 3963) and (m 1, m 2, m 3, m 4, m 5 ) = (0, 1, 1, 1, 1) when (k 1, k 2, k 3, k 4 ) = (3967, 3963, 3993, 3937). It follows that there are no possible integer solutions for (ν 2a,ν 2b,ν 13a,ν 13b, ν 13c ) in all cases. Case (vi). Letu V (ZG) where u =39. Using Propositions 2 and 3, ν 3a + ν 3b + ν 13a + ν 13b + ν 13c = 1. Consider the cases χ(u 13 ) = m 1 χ(3a) + m 2 χ(3b) and χ(u 3 ) = m 3 χ(13a) + m 4 χ(13b) + m 5 χ(13c) where (m 1, m 2, m 3, m 4, m 5 ) {(1, 0, 1, 0, 0), (1, 0, 0, 1, 0), (1, 0, 0, 0, 1), (0, 1, 1, 0, 0), (0, 1, 0, 1, 0), (0, 1, 0, 0, 1), (1, 0, 1, 1, 1), (0, 1, 1, 1, 1), (1, 0, 1, 1, 1), (0, 1, 1, 1, 1), (1, 0, 1, 1, 1), (0, 1, 1, 1, 1), (3, 2, 1, 0, 0), (3, 2, 0, 1,),(3, 2, 0, 0, 1), (3, 2, 1, 1, 1), (3, 2, 1, 1, 1), (3, 2, 1, 1, 1), (2, 1, 1, 0, 0), (2, 1, 0, 1, 0), (2, 1, 0, 0, 1), (2, 1, 1, 1, 1), (2, 1, 1, 1, 1), (2, 1, 1, 1, 1), ( 1, 2, 1, 0, 0), ( 1, 2, 0, 1, 0), ( 1, 2, 0, 0, 1), ( 1, 2, 1, 1, 1), ( 1, 2, 1, 1, 1), ( 1, 2, 1, 1, 1)} (12γ + 27) 0; μ 0(u,χ 2, ) = 39 1 ( 24γ + 24) 0; μ 1 (u,χ 2, ) = 39 1 ( γ + 27) 0 where γ = ν 3a + ν 3b. It follows that there are no possible integer solutions for (ν 3a,ν 3b,ν 13a,ν 13b,ν 13c ). Case (vii). Letu V (ZG) where u =91. Using Propositions 2 and 3, ν 7a + ν 7b + ν 7c + ν 7d + ν 13a + ν 13b + ν 13c = 1. Let γ = 5ν 7a + 5ν 7b + 5ν 7c 2ν 7d. We shall now separately consider the following cases involving χ(u n ) for n {7, 13}: χ(u 13 ) = χ(7k) and χ(u 7 ) = m 1 χ(13a) + m 2 χ(13b) + m 3 χ(13c) where k {a, b, c} and (m 1, m 2, m 3 ) {(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1), (1, 1, 1), ( 1, 1, 1)}.

8 146 J. Gildea μ 0 (u,χ 2, ) = 91 1 (72γ + 56) 0; μ 13(u,χ 2, ) = 91 1 ( 12γ + 21) 0. It follows that there are no possible integer values for (ν 7a,ν 7b,ν 7c,ν 7d,ν 13a,ν 13b,ν 13c ). χ(u 13 ) = χ(7d) and χ(u 7 ) = m 1 χ(13a) + m 2 χ(13b) + m 3 χ(13c) where (m 1, m 2, m 3 ) {(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1), (1, 1, 1), ( 1, 1, 1)}. μ 0 (u,χ 2, ) = 91 1 (72γ + 14) 0; μ 7(u,χ 2, ) = 91 1 ( 6γ + 14) 0. It follows that there are no possible integer values for (ν 7a,ν 7b,ν 7c,ν 7d,ν 13a,ν 13b,ν 13c ). We shall now consider the prime graph of G = 3 D 4 (2 3 ). G contains elements of order 6, 14 and 21. Therefore [2, 3], [2, 7] and [3, 7] are adjacent in π(g) and consequently adjacent in π(v (ZG)). Clearly π(g) = π(v (ZG)), since there are no torsion units of order 26, 39 and 91 in V (ZG). This completes the proof. 3 Proof of Theorem 2 Let G = 2 F 4 (2). Clearly G = = and exp(g) = 3120 = Initially, for any torsion unit of V (ZG) of order k: ν 2a + ν 2b + ν 3a + ν 4a + ν 4b + ν 4c + ν 5a + ν 6a + ν 8a + ν 8b + ν 8c + ν 8d + ν 10a + ν 12a + ν 12b + ν 13a + ν 13b + ν 16a + ν 16b + ν 16c + ν 16d = 1 For the purpose of this paper, we shall consider torsion units of V (ZG) of order 2, 3, 5, 10, 15, 26, 39 and 65. We shall now consider each case separately. Case (i). Letu V (ZG) where u =2. Using Propositions 2 and 3, ν 2a + ν 2b = 1. Applying Proposition 5, we obtain the following system of inequalities: μ 0 (u,χ 2, ) = 2 1 ( 2γ + 26) 0; μ 1(u,χ 2, ) = 2 1 (2γ + 26) 0 where γ = 3ν 2a ν 2b. It follows that the only possible integer solutions for (ν 2a,ν 2b ) are listed in part (iii) of Theorem 2. Case (ii). Letu V (ZG) where u =3. By Proposition 2, ν kx = 0forall kx {2a, 2b, 4a, 4b, 4c, 5a, 6a, 8a, 8b, 8c, 8d, 10a, 12a, 12b, 13a, 13b, 16a, 16b, 16c, 16d}. Therefore, u is rationally conjugated to some element g G by Proposition 4. Case (iii). Letu V (ZG) where u =5. By Proposition 2, ν kx = 0forall kx {2a, 2b, 3a, 4a, 4b, 4c, 6a, 8a, 8b, 8c, 8d, 10a, 12a, 12b, 13a, 13b, 16a, 16b, 16c, 16d}. Therefore, u is rationally conjugated to some element g G by Proposition 4. Case (iv). Letu V (ZG) where u =10. Using Propositions 2 and 3, ν 2a + ν 2b + ν 5a + ν 10a = 1. Let γ 1 = 6ν 2a 2ν 2b ν 5a + ν 10a, γ 2 = 14ν 2a 2ν 2b + 3ν 5a ν 10a, γ 3 = 5ν 2a + ν 2b, γ 4 = 5ν 2a 11ν 2b and γ 5 = 32ν 2a + 15ν 2b + ν 5a + ν 10a. We shall now separately consider the following cases involving χ(u n ) for n {2, 5}:

9 Torsion units for a Ree group χ(u 5 ) = m 1 χ(2a) + m 2 χ(2b) and χ(u 2 ) = χ(5a) where (m 1, m 2 ) {(1, 0), (0, 1), ( 1, 2)}. μ 0 (u,χ 2, ) = k 1 ) 0; μ 1 (u,χ 2, ) = k 2 ) 0; μ 5 (u,χ 2, ) = k 3 ) 0; μ 0 (u,χ 6, ) = k 4 ) 0; μ 1 (u,χ 6, ) = k 5 ) 0; μ 5 (u,χ 6, ) = k 6 ) 0 μ 0 (u,χ 7, ) = k 7 ) 0; μ 5 (u,χ 7, ) = k 8 ) 0; μ 0 (u,χ 9, ) = k 9 ) 0; μ 5 (u,χ 9, ) = k 10 ) 0 where the values for k i s and corresponding (m 1, m 2 ) values are as follows: (m 1, m 2 ) (k 1, k 2, k 3, k 4, k 5, k 6, k 7, k 8, k 9, k 10 ) (1, 0) (24, 31, 36, 104, 61, 76, 280, 320, 386, 324) (0, 1) (32, 23, 28, 88, 77, 92, 296, 304, 370, 340) ( 1, 2) (40, 15, 20, 72, 93, 108, 312, 288, 354, 356) It follows that the only possible integer solutions for (ν 2a,ν 2b,ν 13a ) are (0, 0, 5, 4), (0, 0, 0, 1), (0, 4, 5, 0), (1, 1, 0, 1), ( 1, 1, 0, 1), ( 1, 3, 0, 5), ( 1, 3, 5, 0), (2, 2, 0, 3) and (0, 2, 0, 1) in all cases. χ(u 5 ) = 3χ(2a) 2χ(2b) and χ(u 2 ) = χ(5a). μ 2 (u,χ 2, ) = ) 0; μ 0 (u,χ 2, ) = ) 0; μ 1 (u,χ 6, ) = ) 0; μ 5 (u,χ 6, ) = ) 0; μ 5 (u,χ 7, ) = ) 0; μ 0 (u,χ 7, ) = ) 0; μ 0 (u,χ 9, ) = ) 0. Clearly γ 1 = 3, γ 2 { 29, 19, 9, 1, 11} and γ 3 {3 + 5k 5 k 2}. It follows that the only possible integer solutions for (ν 2a,ν 2b,ν 13a ) are ( 2, 2, 0, 5) and (1, 3, 0, 3). χ(u 5 ) = 2χ(2a) χ(2b) and χ(u 2 ) = χ(5a). μ 2 (u,χ 2, ) = 10 1 (γ ) 0; μ 0 (u,χ 2, ) = 10 1 ( 4γ ) 0; μ 0 (u,χ 6, ) = 10 1 (4γ ) 0; μ 5 (u,χ 6, ) = 10 1 ( 4γ ) 0; μ 1 (u,χ 6, ) = 10 1 (γ ) 0; μ 5 (u,χ 7, ) = 10 1 (16γ ) 0; μ 0 (u,χ 7, ) = 10 1 ( 16γ ) 0; μ 0 (u,χ 8, ) = 10 1 (4γ ) 0; μ 0 (u,χ 9, ) = 10 1 (4γ ) 0. Clearly γ 1 { 1, 11}, γ 2 { 25, 15, 5, 5, 15} and γ 3 {4 + 5k 5 k 2}. It follows that the only possible integer solutions for (ν 2a,ν 2b,ν 13a ) are ( 1, 1, 0, 3), ( 1, 1, 5, 2), (0, 4, 5, 2) and (0, 4, 0, 3).

10 148 J. Gildea χ(u 5 ) = 2χ(2a) + 3χ(2b) and χ(u 2 ) = χ(5a). μ 5 (u,χ 2, ) = 10 1 (4γ ) 0; μ 1 (u,χ 2, ) = 10 1 ( γ 1 + 7) 0; μ 0 (u,χ 6, ) = 10 1 (4γ ) 0; μ 5 (u,χ 6, ) = 10 1 ( 4γ ) 0; μ 1 (u,χ 6, ) = 10 1 (γ ) 0; μ 5 (u,χ 7, ) = 10 1 (16γ ) 0; μ 0 (u,χ 7, ) = 10 1 ( 16γ ) 0; μ 0 (u,χ 8, ) = 10 1 (γ ) 0; μ 0 (u,χ 9, ) = 10 1 (γ ) 0. Clearly γ 1 { 3, 7} γ 2 { 9, 1, 11, 21, 3} and γ 3 {3 + 5k 4 k }. It follows that the only possible integer solutions for (ν 2a,ν 2b,ν 13a ) are (1, 3, 0, 3), (0, 2, 5, 2), (1, 3, 5, 2) and (0, 2, 0, 3). χ(u 5 ) = 3χ(2a) + 4χ(2b) and χ(u 2 ). μ 5 (u,χ 2, ) = 10 1 (4γ 1 + 4) 0; μ 1 (u,χ 2, ) = 10 1 ( γ 1 1) 0; μ 0 (u,χ 6, ) = 10 1 (4γ ) 0; μ 2 (u,χ 6, ) = 10 1 ( γ ) 0; μ 5 (u,χ 7, ) = 10 1 (16γ ) 0; μ 0 (u,χ 7, ) = 10 1 ( 16γ ) 0; μ 0 (u,χ 9, ) = 10 1 (4γ ) 0; μ 5 (u,χ 9, ) = 10 1 ( 4γ ) 0. Clearly γ 1 = 1, γ 2 { 5, 5, 15, 25} and γ 3 {4 + 5k 4 k 3}. It follows that there are no possible integer solutions for (ν 2a,ν 2b,ν 13a ) in this case. Case (v). Letu V (ZG) where u =13. Using Propositions 2 and 3, ν 13a + ν 13b = 1. Applying Proposition 5, we obtain the following system of inequalities: μ 1 (u,χ 7, 5) = 13 1 (γ ) 0; μ 2 (u,χ 15, 5) = 13 1 ( γ ) 0; μ 2 (u,χ 7, 5) = 13 1 (γ ) 0 where γ 1 = 8ν 13a 5ν 13b and γ 2 = 5ν 13a + 8ν 13b. It follows that the only possible integer solutions for (ν 3a,ν 3b ) are listed in part (v) of Theorem 2. Case (vi). Letu V (ZG) where u =15. Using Propositions 2 and 3, ν 3a + ν 5a = 1. Applying Proposition 5, we obtain the following system of inequalities: μ 0 (u,χ 2, ) = 15 1 ( 8γ + 28) 0; μ 5(u,χ 2, ) = 15 1 (4γ + 31) 0 where γ = ν 3a ν 5a. Clearly γ = 4 and there are no possible integer solutions for (ν 3a,ν 5a ). Case (vii). Letu V (ZG) where u =26. Using Propositions 2 and 3, ν 2a + ν 2b + ν 13a + ν 13b = 1. Let γ 1 = 3ν 2a ν 2b, γ 2 = 7ν 2a ν 2b, γ 3 = 20ν 2a 4ν 2b + ν 13a + ν 13b and γ 4 = 6ν 13a 7ν 13b. We shall now separately consider the following cases involving χ(u n ) for n {2, 13}: χ(u 13 ) = m 1 χ(2a) + m 2 χ(2b) and χ(u 2 ) = m 3 χ(5a) + m 4 χ(5b) where (m 1, m 2, m 3, m 4 ) {(1, 0, 1, 0), (1, 0, 0, 1), (1, 0, 9, 8), (1, 0, 8, 7), (1, 0, 7, 6), (1, 0, 6, 5), (1, 0, 5, 4), (1, 0, 4, 3), (1, 0, 3, 2), (1, 0, 2, 1), (1, 0, 1, 2), (1, 0, 2, 3), (1,

11 Torsion units for a Ree group , 3, 4), (1, 0, 4, 5), (1, 0, 5, 6), (1, 0, 6, 7), (1, 0, 7, 8), (1, 0, 8, 9)}. Applying Proposition 5, we obtain: 26 1 (24γ ) 0; μ 0 (u,χ 2, ) = 26 1 ( 24γ ) 0. It follows that there are no possible integer values for (ν 2a,ν 2b,ν 13a,ν 13b ). χ(u 13 ) = m 1 χ(2a) + m 2 χ(2b) and χ(u 2 ) = m 3 χ(5a) + m 4 χ(5b) where (m 1, m 2, m 3, m 4 ) {(2, 1, 1, 0), (2, 1, 0, 1), (2, 1, 9, 8), (2, 1, 8, 7), (2, 1, 7, 6), (2, 1, 6, 5), (2, 1, 5, 4), (2, 1, 4, 3), (2, 1, 3, 2), (2, 1, 2 1), (2, 1, 1, 2), (2, 1, 2, 3), (2, 1, 3, 4), (2, 1, 4, 5), (2, 1, 5, 6), (2, 1, 6, 7), (2, 1, 7, 8), (2, 1, 8, 9)} (24γ ) 0; μ 0 (u,χ 2, ) = 26 1 ( 24γ ) 0. It follows that there are no possible integer values for (ν 2a,ν 2b,ν 13a,ν 13b ). χ(u 13 ) = m 1 χ(2a) + m 2 χ(2b) and χ(u 2 ) = m 3 χ(5a) + m 4 χ(5b) where (m 1, m 2, m 3, m 4 ) {( 2, 3, 1, 0), ( 2, 3, 0, 1), ( 2, 3, 9, 8), ( 2, 3, 8, 7), ( 2, 3, 7, 6), ( 2, 3, 6, 5), ( 2, 3, 5, 4), ( 2, 3, 4, 3), ( 2, 3, 3, 2), ( 2, 3, 2, 1), ( 2, 3, 1, 2), ( 2, 3, 2, 3), ( 2, 3, 3, 4), ( 2, 3, 4, 5), ( 2, 3, 5, 6), ( 2, 3, 6, 7), ( 2, 3, 7, 8), ( 2, 3, 8, 9)} (24γ 1 + 8) 0; μ 0 (u,χ 2, ) = 26 1 ( 24γ ) 0. It follows that there are no possible integer values for (ν 2a,ν 2b,ν 13a,ν 13b ). χ(u 13 ) = m 1 χ(2a) + m 2 χ(2b) and χ(u 2 ) = m 3 χ(5a) + m 4 χ(5b) where (m 1, m 2, m 3, m 4 ) {(0, 1, 9, 8), (0, 1, 8, 9)} ) 0; μ 0 (u,χ 2, ) = 26 1 ( 24γ ) 0; μ 0 (u,χ 6, ) = ) 0; μ 13 (u,χ 6, ) = ) 0; μ 1 (u,χ 7, ) = ) 0. Clearly γ 1 = γ 2 = 1. It follows that there are no possible integer values for (ν 2a, ν 2b,ν 13a,ν 13b ). χ(u 13 ) = m 1 χ(2a) + m 2 χ(2b) and χ(u 2 ) = m 3 χ(5a) + m 4 χ(5b) where (m 1, m 2, m 3, m 4 ) {(3, 2, 9, 8), (3, 2, 8, 9)}. μ 2 (u,χ 2, ) = 26 1 (2γ 1 + 4) 0; μ 0 (u,χ 2, ) = 26 1 ( 24γ 1 + 4) 0; μ 0 (u,χ 6, ) = 26 1 (24γ ) 0; μ 13 (u,χ 6, ) = 26 1 ( 24γ ) 0. Clearly γ 1 = 2andγ 2 = 3. It follows that there are no possible integer values for (ν 2a,ν 2b,ν 13a,ν 13b ). χ(u 13 ) = m 1 χ(2a) + m 2 χ(2b) and χ(u 2 ) = m 3 χ(5a) + m 4 χ(5b) where (m 1, m 2, m 3, m 4 ) {( 3, 4, 9, 8), ( 3, 4, 8, 9)} (24γ 1) 0; μ 1 (u,χ 2, ) = 26 1 ( 2γ 1) 0; μ 0 (u,χ 6, ) = 26 1 (24γ ) 0; μ 13 (u,χ 6, ) = 26 1 ( 24γ ) 0. Clearly γ 1 = 0andγ 2 = 1. It follows that there are no possible integer values for (ν 2a,ν 2b,ν 13a,ν 13b ).

12 150 J. Gildea χ(u 13 ) = m 1 χ(2a) + m 2 χ(2b) and χ(u 2 ) = m 3 χ(5a) + m 4 χ(5b) where (m 1, m 2, m 3, m 4 ) {(0, 1, 1, 0), (0, 1, 0, 1), (0, 1, 8, 7), (0, 1, 7, 6), (0, 1, 6, 5), (0, 1, 5, 4), (0, 1, 4, 3), (0, 1, 3, 2), (0, 1, 2, 1), (0, 1, 1, 2), (0, 1, 2, 3), (0, 1, 3, 4), (0, 1, 4, 5), (0, 1, 5, 6), (0, 1, 6, 7), (0, 1, 7, 8)}. Applying Proposition 5, we obtain: 26 1 (24γ ) 0; μ 0 (u,χ 2, ) = 26 1 ( 24γ ) 0; μ 0 (u,χ 6, ) = 26 1 (24γ ) 0; μ 13 (u,χ 6, ) = 26 1 ( 24γ ) 0; μ 1 (u,χ 7, ) = 26 1 (γ ) 0; μ 4 (u,χ 21, ) = 26 1 (γ 4 + k) 0; μ 1 (u,χ 21, ) = 26 1 ( γ 4 + k) 0 where the corresponding k values to the above cases are: (m 1, m 2, m 3, m 4 ) k (m 1, m 2, m 3, m 4 ) k (0,1,1,0) 2041 (0,1,0,1) 2054 (0, 1, 8, 7) 1950 (0, 1, 7, 6) 1963 (0, 1, 6, 5) 1976 (0, 1, 5, 4) 1989 (0, 1, 4, 3) 2002 (0, 1, 3, 2) 2015 (0, 1, 2, 1) 2028 (0, 1, 1, 2) 2067 (0, 1, 2, 3) 2080 (0, 1, 3, 4) 2093 (0, 1, 4, 5) 2106 (0, 1, 5, 6) 2119 (0, 1, 6, 7) 2132 (0, 1, 7, 8) 2145 It follows that there are no possible integer values for (ν 2a,ν 2b,ν 13a,ν 13b ) in all cases. χ(u 13 ) = m 1 χ(2a) + m 2 χ(2b) and χ(u 2 ) = m 3 χ(5a) + m 4 χ(5b) where (m 1, m 2, m 3, m 4 ) {(3, 2, 1, 0), (3, 2, 0, 1), (3, 2, 8, 7), (3, 2, 7, 6), (3, 2, 6, 5), (3, 2, 5, 4), (3, 2, 4, 3), (3, 2, 3, 2), (3, 2, 2, 1), (3, 2, 1, 2), (3, 2, 2, 3), (3, 2, 3, 4), (3, 2, 4, 5), (3, 2, 5, 6), (3, 2, 6, 7), (3, 2, 7, 8)}. μ 2 (u,χ 2, ) = 26 1 (2γ 1 + 4) 0; μ 0 (u,χ 2, ) = 26 1 ( 24γ 1 + 4) 0; μ 0 (u,χ 6, ) = 26 1 (24γ ) 0; μ 13 (u,χ 6, ) = 26 1 ( 24γ ) 0; μ 4 (u,χ 21, ) = 26 1 (γ 4 + k) 0; μ 1 (u,χ 21, ) = 26 1 ( γ 4 + k) 0. where the corresponding k-values to the above cases are: (m 1, m 2, m 3, m 4 ) k (m 1, m 2, m 3, m 4 ) k (3, 2, 1, 0) 2041 (3, 2, 0, 1) 2054 (3, 2, 8, 7) 1950 (3, 2, 7, 6) 1963 (3, 2, 6, 5) 1976 (3, 2, 5, 4) 1989 (3, 2, 4, 3) 2002 (3, 2, 3, 2) 2015 (3, 2, 2, 1) 2028 (3, 2, 1, 2) 2067 (3, 2, 2, 3) 2080 (3, 2, 3, 4) 2093 (3, 2, 4, 5) 2106 (3, 2, 5, 6) 2119 (3, 2, 6, 7) 2132 (3, 2, 7, 8) 2145

13 Torsion units for a Ree group It follows that there are no possible integer values for (ν 2a,ν 2b,ν 13a,ν 13b ) in all cases. χ(u 13 ) = m 1 χ(2a) + m 2 χ(2b) and χ(u 2 ) = m 3 χ(5a) + m 4 χ(5b) where (m 1, m 2, m 3, m 4 ) {( 3, 4, 1, 0), ( 3, 4, 0, 1), ( 3, 4, 8, 7), ( 3, 4, 7, 6), ( 3, 4, 6, 5), ( 3, 4, 5, 4), ( 3, 4, 4, 3), ( 3, 4, 3, 2), ( 3, 4, 2, 1), ( 3, 4, 1, 2), ( 3, 4, 2, 3), ( 3, 4, 3, 4), ( 3, 4, 4, 5), ( 3, 4, 5, 6), ( 3, 4, 6, 7), ( 3, 4, 7, 8)}. Applying Proposition 5, we obtain: 26 1 (24γ 1) 0; μ 1 (u,χ 2, ) = 26 1 ( 2γ 1) 0; μ 0 (u,χ 6, ) = 26 1 (24γ ) 0; μ 13 (u,χ 6, ) = 26 1 ( 24γ ) 0; μ 4 (u,χ 21, ) = 26 1 (γ 4 + k) 0; μ 1 (u,χ 21, ) = 26 1 ( γ 4 + k) 0. where the corresponding k values to the above cases are: (m 1, m 2, m 3, m 4 ) k (m 1, m 2, m 3, m 4 ) k ( 3, 4, 1, 0) 2041 ( 3, 4, 0, 1) 2054 ( 3, 4, 8, 7) 1950 ( 3, 4, 7, 6) 1963 ( 3, 4, 6, 5) 1976 ( 3, 4, 5, 4) 1989 ( 3, 4, 4, 3) 2002 ( 3, 4, 3, 2) 2015 ( 3, 4, 2, 1) 2028 ( 3, 4, 1, 2) 2067 ( 3, 4, 2, 3) 2080 ( 3, 4, 3, 4) 2093 ( 3, 4, 4, 5) 2106 ( 3, 4, 5, 6) 2119 ( 3, 4, 6, 7) 2132 ( 3, 4, 7, 8) 2145 It follows that there are no possible integer values for (ν 2a,ν 2b,ν 13a,ν 13b ) in all cases. Case (viii). Letu V (ZG) where u =39. Using Propositions 2 and 3, ν 3a + ν 13a + ν 13b = 1. Consider the cases χ(u 13 ) = χ(3a) and χ(u 3 ) = m 1 χ(13a) + m 2 χ(13b) where (m 1, m 2 ) {(1, 0), (0, 1), (9, 8), (8, 7), (7, 6), (6, 5), (5, 4), (4, 3), (3, 2), (2, 1), ( 1, 2), ( 2, 3), ( 3, 4), ( 4, 5), ( 5, 6), ( 6, 7), ( 7, 8), ( 8, 9)} (12ν 3a + 27) 0; μ 0 (u,χ 2, ) = 39 1 ( 24ν 3a + 24) 0; μ 1 (u,χ 2, ) = 39 1 ( ν 3a + 27) 0. It follows that there are no possible integer solutions for (ν 3a,ν 13a,ν 13b ). Case (ix). Letu V (ZG) where u =65. Using Propositions 2 and 3, ν 5a + ν 13a + ν 13b = 1. Consider the cases χ(u 13 ) = χ(5a) and χ(u 5 ) = m 1 χ(13a) + m 2 χ(13b) where (m 1, m 2 ) {(1, 0), (0, 1), (9, 8), (8, 7), (7, 6), (6, 5), (5, 4), (4, 3), (3, 2), (2, 1), ( 1, 2), ( 2, 3), ( 3, 4), ( 4, 5), ( 5, 6), ( 6, 7), ( 7, 8), ( 8, 9)}.

14 152 J. Gildea μ 0 (u,χ 2, ) = 65 1 (48ν 5a + 30) 0; 65 1 ( 12ν 5a + 25) 0. It follows that there are no possible integer solutions for (ν 5a,ν 13a,ν 13b ). We shall now consider the prime graph of G = 2 F 4 (2). G contains elements of order 6 and 10. Therefore [2, 3] and [2, 5] are adjacent in π(g) and consequently adjacent in π(v (ZG)). Clearly π(g) = π(v (ZG)), since there are no torsion units of order 15, 26, 39 and 65 in V (ZG). This completes the proof. 4 Proof of Theorem 3 Let G = 2 F 4 (2). Clearly G = = and exp(g) = 3120 = Initially, for any torsion unit of V (ZG) of order k: ν 2a + ν 2b + ν 3a + ν 4a + ν 4b + ν 4c + ν 5a + ν 6a + ν 8a + ν 8b + ν 8c + ν 10a + ν 12a + ν 13a + ν 16a + ν 16b + ν 4d + ν 4e + ν 4 f + ν 4g + ν 8d + ν 8e + ν 12b + ν 12c + ν 16c + ν 16d + ν 20a + ν 20b = 1 For the purpose of this paper, we shall consider torsion units of V (ZG) of order 2, 3, 5, 10, 15, 26, 39 and 65. We shall now consider each case separately. Case (i). Letu V (ZG) where u =2. Using Propositions 2 and 3, ν 2a + ν 2b = 1. Applying Proposition 5, we obtain the following system of inequalities: μ 1 (u,χ 3, ) = 2 1 (4γ + 52) 0; μ 0(u,χ 3, ) = 2 1 ( 4γ + 52) 0 where γ = 3ν 2a ν 2b. Clearly γ {k 13 k 13}. It follows that the only possible integer solutions for (ν 2a,ν 2b ) are listed in part (iii) of Theorem 3. Case (ii). Letu V (ZG) where u =3. By Proposition 2, ν kx = 0forall kx {2a, 2b, 4a, 4b, 4c, 5a, 6a, 8a, 8b, 8c, 10a, 12a, 13a, 16a, 16b, 4d, 4e, 4 f, 4g, 8d, 8e, 12b, 12c, 16c, 16d, 20a, 20b}. Therefore, u is rationally conjugated to some element g G by Proposition 4. Case (iii). Letu V (ZG) where u =5. By Proposition 2, ν kx = 0forall kx {2a, 2b, 3a, 4a, 4b, 4c, 6a, 8a, 8b, 8c, 10a, 12a, 13a, 16a, 16b, 4d, 4e, 4 f, 4g, 8d, 8e, 12b, 12c, 16c, 16d, 20a, 20b}. Therefore, u is rationally conjugated to some element g G by Proposition 4. Case (iv). Letu V (ZG) where u =10. Using Propositions 2 and 3, ν 2a + ν 2b + ν 5a + ν 10a = 1. Let γ 1 = 6ν 2a 2ν 2b ν 5a + ν 10a, γ 2 = 14ν 2a 2ν 2b + 3ν 5a ν 10a, γ 3 = 5ν 2a + ν 2b, γ 4 = 5ν 2a 11ν 2b and γ 5 = 31ν 2a + 15ν 2b + ν 5a + ν 10a. We shall now separately consider the following cases involving χ(u n ) for n {2, 5}:

15 Torsion units for a Ree group χ(u 5 ) = m 1 χ(2a) + m 2 χ(2b) and χ(u 2 ) = χ(5a) where (m 1, m 2 ) {(1, 0), (0, 1), ( 1, 2)}. μ 0 (u,χ 3, ) = k 1 ) 0; μ 5 (u,χ 3, ) = k 2 ) 0; μ 0 (u,χ 8, ) = k 3 ) 0; μ 1 (u,χ 8, ) = k 4 ) 0; μ 5 (u,χ 8, ) = k 5 ) 0; μ 0 (u,χ 10, ) = k 6 ) 0; μ 5 (u,χ 10, ) = k 7 ) 0; μ 0 (u,χ 14, ) = k 8 ) 0; μ 5 (u,χ 14, ) = k 9 ) 0 where the values for k i s and corresponding (m 1, m 2 ) values are as follows: (m 1, m 2 ) (k 1, k 2, k 3, k 4, k 5, k 6, k 7, k 8, k 9 ) (1, 0) (48, 72, 104, 61, 76, 280, 320, 386, 324) (0, 1) (64, 56, 88, 77, 92, 296, 304, 370, 340) ( 1, 2) (80, 40, 72, 93, 108, 312, 288, 354, 356) It follows that the only possible integer solutions for (ν 2a,ν 2b,ν 13a ) are (0, 0, 5, 4), (0, 0, 0, 1), ( 1, 1, 0, 1), (1, 1, 0, 1), (0, 4, 5, 0), (0, 2, 0, 1), ( 1, 3, 5, 0), ( 1, 3, 0, 5) and (2, 2, 0, 3) in all cases. χ(u 5 ) = 2χ(2a) χ(2b) and χ(u 2 ) = χ(5a). μ 2 (u,χ 3, ) = 10 1 (2γ ) 0; μ 0 (u,χ 3, ) = 10 1 ( 8γ ) 0; μ 0 (u,χ 8, ) = 10 1 (4γ ) 0; μ 5 (u,χ 8, ) = 10 1 ( 4γ ) 0; μ 1 (u,χ 8, ) = 10 1 (γ ) 0; μ 5 (u,χ 10, ) = 10 1 (16γ ) 0; μ 0 (u,χ 10, ) = 10 1 ( 16γ ) 0; μ 0 (u,χ 12, ) = 10 1 (4γ ) 0; μ 0 (u,χ 14, ) = 10 1 (4γ ) 0. Clearly γ 1 { 11, 6, 1, 4}, γ 2 {4+5k 5 k 2} and γ 3 {4+5k 5 k 2}. It follows that the only possible integer solutions for (ν 2a,ν 2b,ν 13a ) are ( 1, 1, 0, 3), ( 1, 1, 5, 2), (0, 4, 5, 2) and (0, 4, 0, 3). χ(u 5 ) = 2χ(2a) + 3χ(2b) and χ(u 2 ) = χ(5a). μ 5 (u,χ 3, ) = 10 1 (8γ ) 0; μ 1 (u,χ 3, ) = 10 1 ( 2γ ) 0; μ 0 (u,χ 8, ) = 10 1 (4γ ) 0; μ 5 (u,χ 8, ) = 10 1 ( 4γ ) 0; μ 1 (u,χ 8, ) = 10 1 (γ ) 0; μ 5 (u,χ 10, ) = 10 1 (16γ ) 0; μ 0 (u,χ 10, ) = 10 1 ( 16γ ) 0; μ 0 (u,χ 12, ) = 10 1 (4γ ) 0; μ 0 (u,χ 14, ) = 10 1 (4γ ) 0. Clearly γ 1 { 3, 2, 7}, γ 2 { 9, 1, 11, 21, 31} and γ 3 {3+5k 4 k 3}. It follows that the only possible integer solutions for (ν 2a,ν 2b,ν 13a ) are (0, 2, 0, 3), (0, 2, 5, 2), (1, 3, 5, 2) and (1, 3, 0, 3).

16 154 J. Gildea χ(u 5 ) = 3χ(2a) 2χ(2b) and χ(u 2 ) = χ(5a). μ 2 (u,χ 3, ) = 10 1 (2γ 1 + 6) 0; μ 0 (u,χ 3, ) = 10 1 ( 8γ ) 0; μ 1 (u,χ 8, ) = 10 1 (γ ) 0; μ 5 (u,χ 8, ) = 10 1 ( 4γ ) 0; μ 5 (u,χ 10, ) = 10 1 (16γ ) 0; μ 0 (u,χ 10, ) = 10 1 ( 16γ ) 0; μ 0 (u,χ 14, ) = 10 1 (4γ ) 0. Clearly γ 1 { 3, 2} γ 2 { 29, 19, 9, 1, 11} and γ 3 {3 + 5k 5 k 2}. It follows that the only possible integer solutions for (ν 2a,ν 2b,ν 13a ) are ( 2, 2, 0, 5) and (1, 3, 0, 3). χ(u 5 ) = 3χ(2a) + 4χ(2b) and χ(u 2 ) = χ(5a). μ 5 (u,χ 3, ) = 10 1 (8γ 1 + 8) 0; μ 1 (u,χ 3, ) = 10 1 ( 2γ 1 2) 0; μ 0 (u,χ 8, ) = 10 1 (4γ ) 0; μ 2 (u,χ 8, ) = 10 1 ( γ ) 0; μ 5 (u,χ 10, ) = 10 1 (16γ ) 0; μ 0 (u,χ 10, ) = 10 1 ( 16γ ) 0; μ 0 (u,χ 14, ) = 10 1 (4γ ) 0; μ 5 (u,χ 14, ) = 10 1 ( 4γ ) 0. Clearly γ 1 = 1, γ 2 { 5, 5, 15, 25} and γ 3 {4 + 5k 4 k 3}. It follows that there are no possible integer solutions for (ν 2a,ν 2b,ν 13a ) in this case. Case (v). Letu V (ZG) where u =13. By Proposition 2, ν kx = 0forall kx {2a, 2b, 3a, 4a, 4b, 4c, 5a, 6a, 8a, 8b, 8c, 10a, 12a, 16a, 16b, 4d, 4e, 4 f, 4g, 8d, 8e, 12b, 12c, 16c, 16d, 20a, 20b}. Therefore, u is rationally conjugated to some element g G by Proposition 4. Case (vi). Letu V (ZG) where u =15. Using Propositions 2 and 3, ν 3a + ν 5a = 1. Applying Proposition 5, we obtain the following system of inequalities: μ 5 (u,χ 3, ) = 15 1 (8γ + 62) 0; μ 0(u,χ 3, ) = 15 1 ( 16γ + 56) 0 where γ = ν 3a ν 5a. Clearly γ = 4 and there are no possible integer solutions for (ν 3a,ν 5a ). Case (vii). Letu V (ZG) where u =26. Using Propositions 2 and 3, ν 2a + ν 2b + ν 13a = 1. Let γ 1 = 3ν 2a ν 2b, γ 2 = 7ν 2a ν 2b and γ 3 = 20ν 2a 4ν 2b + ν 13a.Weshallnow separately consider the following cases involving χ(u n ) for n {2, 13}: χ(u 13 ) = m 1 χ(2a) + m 2 χ(2b) and χ(u 2 ) = χ(13a) where (m 1, m 2 ) {(1, 0), (2, 1), ( 1, 2), ( 2, 3)}. μ 13 (u,χ 3, ) = 26 1 (48γ 1 + k 1 ) 0; μ 0 (u,χ 3, ) = 26 1 ( 48γ 1 + k 2 ) 0 where (k 1, k 2 ) = (64, 40) when (m 1, m 2 ) = (1, 0), (k 1, k 2 ) = (80, 24) when (m 1, m 2 ) = (2, 1), (k 1, k 2 ) = (32, 72) when (m 1, m 2 ) = ( 1, 2) and (k 1, k 2 ) = (16, 80) when (m 1, m 2 ) = ( 2, 3). It follows that there are no possible integer solutions for (ν 2a,ν 2b,ν 13a ) in all cases.

17 Torsion units for a Ree group χ(u 13 ) = 3χ(2a)+4χ(2b) and χ(u 2 ) = χ(13a). μ 13 (u,χ 3, ) = 1 26 (48γ 1) 0; μ 1 (u,χ 3, ) = 1 26 ( 4γ 1) 0; μ 0 (u,χ 8, ) = 1 26 (24γ ) 0; μ 13 (u,χ 8, ) = 1 26 ( 24γ ) 0. It follows that there are no possible integer solutions for (ν 2a,ν 2b,ν 13a ). χ(u 13 ) = χ(2b) and χ(u 2 ) = χ(13a). μ 13 (u,χ 3, ) = ) 0; μ 0 (u,χ 3, ) = ) 0; μ 0 (u,χ 8, ) = ) 0; μ 13 (u,χ 8, ) = ) 0; μ 1 (u,χ 10, ) = 26 1 (γ ) 0. It follows that there are no possible integer solutions for (ν 2a,ν 2b,ν 13a ). χ(u 13 ) = 3χ(2a) 2χ(2b) and χ(u 2 ) = χ(13a). μ 2 (u,χ 3, ) = 1 26 (4γ 1 + 8) 0; μ 0 (u,χ 3, ) = 1 26 ( 48γ 1 + 8) 0; μ 0 (u,χ 8, ) = 1 26 (24γ ) 0; μ 13 (u,χ 8, ) = 1 26 ( 24γ ) 0. It follows that there are no possible integer solutions for (ν 2a,ν 2b,ν 13a ). Case (viii). Letu V (ZG) where u =39. Using Propositions 2 and 3, ν 3a + ν 13a = 1. Applying Proposition 5, we obtain the following system of inequalities: μ 13 (u,χ 3, ) = 1 39 (24ν 3a + 54) 0; μ 0 (u,χ 3, ) = 1 39 ( 48ν 3a + 48) 0; μ 1 (u,χ 3, ) = 1 39 ( 2ν 3a + 54) 0. Clearly ν 3a = 1 and there are no possible integer solutions for (ν 3a,ν 5a ). Case (ix). Letu V (ZG) where u =65. Using Propositions 2 and 3, ν 5a + ν 13a = 1. Applying Proposition 5, we obtain the following system of inequalities: μ 0 (u,χ 3, ) = 1 65 (96ν 5a + 60) 0; μ 13 (u,χ 3, ) = 1 65 ( 24ν 5a + 50) 0; Clearly there are no possible integer solutions for (ν 5a,ν 13a ). We shall now consider the prime graph of G = 2 F 4 (2). Again,G contains elements of order 6 and 10. Therefore [2, 3] and [2, 5] are adjacent in π(g) and consequently adjacent in π(v (ZG)). Clearly π(g) = π(v (ZG)), since there are no torsion units of order 15, 26, 39 and 65 in V (ZG). This completes the proof. Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

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