Journal of Combinatorial Theory, Series A 136 (2015) 198–200
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Journal of Combinatorial Theory, Series A www.elsevier.com/locate/jcta
On a relation between the rank and the proportion of derangements in finite transitive permutation groups R. Guralnick a,1 , I.M. Isaacs b , P. Spiga c a
Department of Mathematics, University of Southern California, Los Angeles, CA 90089-2532, USA b University of Wisconsin, Department of Mathematics, 480 Lincoln drive, Madison, WI 53706, USA c University of Milano-Bicocca, Dipartimento di Matematica Pura e Applicata, Via Cozzi 55, 20125 Milano, Italy
a r t i c l e
i n f o
Article history: Received 17 June 2015 Available online xxxx Keywords: Derangements Rank Permutation groups
a b s t r a c t Let G be a finite transitive group of rank r. We give a short proof that the proportion of derangements in G is at most 1 − 1/r and we classify the permutation groups attaining this bound. © 2015 Elsevier Inc. All rights reserved.
Let G be a finite transitive group on the set Ω. An element g ∈ G is called a derangement if g has no fixed points on Ω, that is, αg = α for each α ∈ Ω. Moreover, the rank of G is the number of orbits of G in its natural component-wise action on the Cartesian product Ω × Ω. The group action is said to be regular if the identity is the only element of G fixing an element of Ω.
E-mail addresses:
[email protected] (R. Guralnick),
[email protected] (I.M. Isaacs),
[email protected] (P. Spiga). 1 Guralnick was partially supported by NSF grant DMS-1302886. http://dx.doi.org/10.1016/j.jcta.2015.07.003 0097-3165/© 2015 Elsevier Inc. All rights reserved.
R. Guralnick et al. / Journal of Combinatorial Theory, Series A 136 (2015) 198–200
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The main result of this paper is the following. Theorem 1. If G is a finite transitive group on Ω of rank r, then the proportion of elements of G which are derangements on Ω is at most 1 − 1/r. Moreover, this upper bound is met if and only if G acts regularly. In light of Theorem 1, one might wonder whether, by excluding the family of regular permutation groups, the upper bound 1 − 1/r can be improved further. This is not the case. For example, in the group PSL2 (q) in its natural 2-transitive action, the proportion 1 1 of derangements is 12 − q+1 when q is odd and 12 − 2(q+1) when q is even, and this number tends to 1/2 as q tends to infinity. In the Suzuki group 2 B2 (q) and in the Ree group 2 G2 (q) in their natural 2-transitive actions on the points of a Steiner system, the proportion of q 2 +1 1 derangements are 21 − 2(qq+1 2 +1) and 2 − q 3 +1 (respectively), and both of these ratios tend to 1/2 as q tends to infinity. Similar examples can be obtained for larger ranks. The proof of Theorem 1 uses some character theory and is very much in the same spirit as the proof of the beautiful Cameron–Cohen lower bound (r − 1)/|Ω| for the proportion of derangements [2]. This lower bound has motivated a conjecture due independently to Shalev and Boston et al. [1], and in turn this conjecture has spurred some interesting research culminating with a complete solution by the first author and Fulman [4,7,6,5]. The first half of Theorem 1 was also proved in [3, Theorem 3.1] using the so-called second moment method, which is a standard result in probability theory. (The proof of [3, Theorem 3.1] is also very short.) Our proof only uses the Cauchy–Schwarz inequality and characterises the groups attaining the bound. Proof of Theorem 1. Let π be the permutation character of G and let D be the set of derangements of G. Observe that π, π = r and π(g) = 0 for every g ∈ D, and hence
|G|r =
π(g)2 =
g∈G
g∈G\D
⎛ ⎞2 1 ⎝ π(g)2 ≥ π(g)⎠ , |G| − |D| g∈G\D
where the last inequality is a direct application of the Cauchy–Schwarz inequality. As π(g) = 0 for every g ∈ D, this yields ⎛ |G|r ≥
1 ⎝ |G| − |D|
g∈G
⎞2 π(g)⎠ =
|G|2 |G| − |D|
and the first half of the theorem follows. Suppose now that |D|/|G| = 1 − 1/r. By chasing the previous series of inequalities, 2 we get g∈G\D π(g)2 = g∈G\D π(g) . In turn, the Cauchy–Schwarz inequality gives π(g) = π(h) for every g, h ∈ G \ D, that is, π(g) = π(1) = |Ω| for every g ∈ G \ D. Since
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R. Guralnick et al. / Journal of Combinatorial Theory, Series A 136 (2015) 198–200
G acts faithfully, we must have G \ D = {1} and hence the identity of G is the only permutation fixing a point, that is, G is regular. 2 References [1] N. Boston, W. Dabrowski, T. Foguel, et al., The proportion of fixed-point-free elements in a transitive group, Comm. Algebra 21 (1993) 3259–3275. [2] P.J. Cameron, A.M. Cohen, On the number of fixed-point-free elements in a permutation group, Discrete Math. 106 (1992) 135–138. [3] P. Diaconis, J. Fulman, R. Guralnick, On fixed points of permutations, J. Algebraic Combin. 28 (2008) 189–218. [4] J. Fulman, R. Guralnick, Derangements in simple and primitive groups, in: Groups, Combinatorics, and Geometry, Durham, 2001, World Sci. Publ., River Edge, NJ, 2003, pp. 99–121. [5] J. Fulman, R. Guralnick, Bounds on the number and sizes of conjugacy classes in finite Chevalley groups with applications to derangements, Trans. Amer. Math. Soc. 364 (2012) 3023–3070. [6] J. Fulman, R. Guralnick, Derangements in subspace actions of finite classical groups, Trans. Amer. Math. Soc. (2015), in press. [7] J. Fulman, R. Guralnick, Derangements in finite classical groups for actions related to extension field and imprimitive subgroups and the solution of the Boston–Shalev conjecture, preprint.