CrossMark

Timothy C. Burness ■ Hung P. Tong-Viet

Primitive permutation groups and derangements of prime power order

Received: 19 May 2015 / Accepted 16 October 2015 Published online: 6 November 2015

Abstract. Let G be a transitive permutation group on a finite set of size at least 2. By a well known theorem of Fein, Kantor and Schacher, G contains a derangement of prime power order. In this paper, we study the finite primitive permutation groups with the extremal property that the order of every derangement is an r-power, for some fixed prime r. First we show that these groups are either almost simple or affine, and we determine all the almost simple groups with this property. We also prove that an affine group G has this property if and only if every two-point stabilizer is an r-group. Here the structure of G has been extensively studied in work of Guralnick and Wiegand on the multiplicative structure of Galois field extensions, and in later work of Fleischmann, Lempken and Tiep on r'-semiregular pairs.

1. Introduction

Let G be a transitive permutation group on a finite set ^ of size at least 2. An element x e G is a derangement if it acts fixed-point-freely on An easy application of the orbit-counting lemma shows that G contains derangements (this is originally a classical theorem of Jordan [36]), and we will write A(G) for the set of derangements in G. Note that if H is a point stabilizer, then x is a derangement if and only if xG n H is empty, where xG denotes the conjugacy class of x in G, so we have

The existence of derangements in transitive permutation groups has interesting applications in number theory and topology (see Serre's article [48], for example).

Various extensions of Jordan's theorem on the existence of derangements have been studied in recent years. For example, if 8(G) = |A(G)|/|G| denotes the proportion of derangements in G, then a theorem of Cameron and Cohen [13] states that 8(G) > |^|-1, with equality if and only if G is sharply 2-transitive. More recently, Fulman and Guralnick have established the existence of an absolute

T. C. Burness (B): School of Mathematics, University of Bristol, Bristol BS8 1TW, UK. e-mail: t.burness@bristol.ac.uk

H. P. Tong-Viet: Department of Mathematical Sciences, Kent State University, Kent, OH 44242, USA. e-mail: htongvie@kent.edu

Mathematics Subject Classification: Primary 20B15 ■ Secondary 20D05 DOI: 10.1007/s00229-015-0795-x

constant e > 0 such that S(G) > e for any simple transitive group G (see [21-24]). This latter result confirms a conjecture of Boston et al. [4] and Shalev.

The study of derangements with special properties has been another major theme in recent years. By a theorem of Fein et al. [18], A(G) contains an element of prime power order (their proof requires the classification of finite simple groups), and this result has important number-theoretic applications. For instance, it implies that the relative Brauer group of any finite extension of global fields is infinite. In most cases, A(G) contains an element of prime order, but there are some exceptions, such as the 3-transitive action of the smallest Mathieu group M11 on 12 points. The transitive permutation groups with this property are called elusive groups, and they have been investigated by many authors; see [14,26,27], for example.

In this paper, we are interested in the permutation groups with the special property that every derangement is an r-element (that is, has order a power of r) for some fixed prime r. One of our main motivations stems from a theorem of Isaacs et al. [35], which describes the finite transitive groups in which every derangement is an involution; by [35, Theorem A], such a group is either an elementary abelian 2-group, or a Frobenius group with kernel an elementary abelian 2-group. In [9], this result is used to classify the finite groups whose irreducible characters vanish only on involutions. It is natural to consider the analogous problem for odd primes, and more generally for prime powers. As noted in [35], it is easy to see that such a generalization will involve a wider range of examples. For instance, if p is an odd prime then every derangement in the affine group ASL2(p) = SL2(p):p2 (of degree p2) has order p (if p = 2, the derangements have order 2 or 4). Our first result is a reduction theorem.

Theorem 1. Let G be a finite primitive permutation group such that every derangement in G is an r-element for some fixed prime r. Then G is either almost simple or affine.

Our next result, Theorem 2 below, describes all the almost simple primitive groups that arise in Theorem 1. Notice that in Table 1, we write Pi for a maximal parabolic subgroup of L2(q) or L3 (q), which can be defined as the stabilizer of a

1-dimensional subspace of the natural module (similarly, P2 is the stabilizer of a

2-dimensional subspace). In addition, we define

E(G) = {|x| : x e A(G)}.

Theorem 2. Let G be a finite almost simple primitive permutation group with point stabilizer H. Then every derangement in G is an r-element for some fixed prime r if and only if (G, H, r) is one of the cases in Table 1. In particular, every derangement hasorderr ifandonlyif IE(G)| = 1.

Remark 3. Let us make a couple of comments on the cases arising in Table 1.

(i) Firstly, notice that the group G is recorded up to isomorphism. For example, the case (G, H) = (A6, (S3 ? S2) n A6) is listed as (L2(9), P1), (G, H) = (A5, A4) appears as (L2(4), P1), and we record (G, H) = (L2(7), S4) as (L3(2), P1), etc.

Table 1. The cases (G, H, r) in Theorem 2

r £ (G)

Conditions

L3 (q ) Pi, P2

r r, r r r

q2 + q + 1 = (3, q — 1)r q2 + q + 1 = 3r2

rL2(q) NG (D2(q+i)) rL2(8) Ng (Pi), NG (D14)

2 2i, 1 < i < e + 1 r r1, 1 < i < e

r = q — 1 Mersenne prime

PGL2 (q) Ng (P1) L2(q) P1

P1, D2(q—1)

D2(q+1)

r r r r

q = 2e+1

— 1 Mersenne prime q = 2re — 1 r = q + 1 Fermât prime r = q — 1 Mersenne prime

L2(8) P1, D14 M11 L2(11)

3 3,9 2 4,8

(ii) In the first two rows of the table we have G = L3 (q) and H = P1 or P2. Here q2 + q + 1 e {r, 3r, 3r2}, which implies that either q = 4, or q = pf for a prime p and f is a 3-power (see Lemma 2.9).

Now let us turn our attention to the affine groups that arise in Theorem 1. In order to state Theorem 4 below, we need to introduce some additional terminology. Let F be a field and let V be a finite dimensional vector space over F. Let H < GL(V) be a finite group and let r be a prime. Recall that x e H is an r'-element if the order of x is indivisible by r. Following Fleischmann et al. [19], the pair (H, V) is said to be r'-semiregular if every nontrivial r'-element of H has no fixed points on V \{0} (equivalently, no nontrivial r'-element of H has eigenvalue 1 on V).

Theorem 4. Let G = HV < AGL(V) be a finite affine primitive permutation group with point stabilizer H = G0 and socle V = (Zp )k, where p is a prime and k > 1. Then every derangement in G is an r-element for some fixed prime r if and only ifr = p and the pair (H, V) is r'-semiregular.

Let G = HV be an affine group as in Theorem 4 and notice that (H, V) is r'-semiregular if and only if every two-point stabilizer in G is an r-group. As a special case, observe that if G is a Frobenius group then every two-point stabilizer is trivial and it is clear that every derangement in G has order r. Therefore, it is natural to focus our attention on the non-Frobenius affine groups arising in Theorem 4, which correspond to r'-semiregular pairs (H, V) such that r divides |H|. In this situation, Guralnick and Wiegand [33, Section 4] obtain detailed information on the structure of H, which they use to investigate the multiplicative structure of finite Galois field extensions. Similar results were established in later work of Fleischmann et al. [19]. We refer the reader to the end of Sect. 5 for further details (see Propositions 5.4 and 5.5).

Transitive groups with the property in Theorem 1 arise naturally in several different contexts. For instance, let us recall that the existence of a derangement of prime power order in any finite transitive permutation group implies that the relative Brauer group B(L/K) of any finite extension L /K of global fields is infinite. More precisely, let L = K(a) be a separable extension of K, let E be a Galois closure of

L over K, and let ^ be the set of roots in E of the minimal polynomial of a over K. Then the r-primary component B(L/K)r is infinite if and only if the Galois group Gal(E/K) contains a derangement of r-power order on ^ (see [18, Corollary 3]). In this situation, it follows that the relative Brauer group B(L/K) has a unique infinite primary component if and only if every derangement in Gal(E/K) is an r -element for some fixed prime r.

In a different direction, our property arises in the study of permutation groups with bounded movement. To see the connection, let G < Sym(^) be a transitive permutation group of degree n and set

m = max{|Tx\r| : V c x e G}eN,

where Tx = {yx : y e T}. Following Praeger [47], we say that G has movement m. If G is not a 2-group and n = \2mp/(p — 1)J, where p > 5 is the least odd prime dividing | G|, then p divides n and every derangement in G has order p (see [34, Proposition 4.4]). Moreover, the structure of these groups is described in [34, Theorem 1.2].

Some additional related results are established by Mann and Praeger in [43]. For instance, [43, Proposition 2] states that if G is a transitive p -group, where p = 2 or 3, then every derangement in G has order p only if G has exponent p. It is still not known whether or not the same conclusion holds for any prime p (see [43, p. 905]), although [34, Proposition 6.1] does show that the exponent of such a group is bounded in terms of p only.

Remark 5. Let G = HV < AGL(V) be a finite affine primitive permutation group as above, and assume that every derangement in G is an r-element for some fixed prime r. Let P be a Sylow r-subgroup of G and set K = H n P. As explained in Proposition 5.6, P is a transitive permutation group on P/K with E(G) = E(P), so E(G) = {r} if and only if E(P) = {r}, and we will show that E(P) = {r} if and only if P has exponent r (see Theorem 5.7).

There is also a connection between our property and 2-coverings of abstract groups. First notice that Jordan's theorem on the existence of derangements is equivalent to the well known fact that no finite group G can be expressed as the union of G-conjugates of a proper subgroup (see (1)). However, it may be possible to express G as the union of the G-conjugates of two proper subgroups; if H and K are proper subgroups such that

G = (J Hg U J Kg,

geG geG

then G is said to be 2-coverable and the pair (H, K) is a 2-covering for G. This notion has been widely studied in the context of finite simple groups. For instance, Bubboloni [8] proves that An is 2-coverable if and only if 5 < n < 8, and similarly Ln (q) is 2-coverable if and only if 2 < n < 4 (see [10]). We refer the reader to [11] and [46] for further results in this direction. The connection between 2-coverable groups and the property in Theorem 1 is transparent. Indeed, if G is a transitive

permutation group with point stabilizer H, then every derangement in G is an r-element (for some fixed prime r) if and only if (H, K) is a 2-covering for G, where K is a Sylow r-subgroup of G.

Finally, some words on the organisation of this paper. In Sect. 2 we record several preliminary results that we will need in the proofs of our main theorems. The proof of Theorem 1 is given in Sect. 3, and the almost simple groups arehandled in Sect. 4, where we prove Theorem 2. Finally, in Sect. 5 we turn to affine groups and we establish Theorem 4.

Notation Our group-theoretic notation is standard, and we adopt the notation of Kleidman and Liebeck [38] for simple groups. For instance,

PSLn(q) = L+(q) = L„(q), PSU„(q) = L—(q) = U„(q).

If G is a simple orthogonal group, then we write G = PUfn (q), where e = + (respectively —) if n is even and G has Witt defect 0 (respectively 1), and e = o if n is odd (in the latter case, we also write G = Qn(q)). Following [38], we will sometimes refer to the type of a subgroup H, which provides an approximate description of the group-theoretic structure of H.

For integers a and b, we use (a, b) to denote the greatest common divisor of a and b. If p is a prime number, then we write a = ap ■ ap, where ap is the largest power of p dividing a. Finally, if X is a finite set, then n(X) denotes the set of prime divisors of | X |.

2. Preliminaries

In this section we record several preliminary results that will be useful in the proofs of our main theorems. Let H be a proper subgroup of a finite group G and set

AH(G) = G\ Hg.

Notice that if G is a transitive permutation group with point stabilizer H, then A(G) = AH(G) is the set of derangements in G (see (1)). It will be convenient to define the following property:

Every element in A H (G) is an r - element for some fixed prime r. (*)

Lemma 2.1. Let H be a proper subgroup of a finite group G. If (*) holds, then

(i) n(G) = n(H) U {r}; and

(ii) CG (x) is an r-group for every x e AH (G).

Proof. If s e n(G)\n(H), then AH(G) contains an s-element, so (i) follows. Now consider (ii). Let x e AH(G) and assume s = r is a prime divisor of |CG(x)|. Let y e CG(x) with |y | = s and let z = xy = yx, so zs = xs and (x} < (z). Then z e AH (G), but this is incompatible with property (*). □

Lemma 2.2. Let H be a proper subgroup of a finite group G, let N be a normal subgroup ofG such that G = NH, and let Kbea proper subgroup ofN containing H n N. Then Ak (N) C A h (G).

Proof. Let x e AK (N) and assume that x <£ AH (G). Then xg e H for some g e G. Since g e G = NH, we may write g = nh for some n e N and h e H. Then xg = (xn )h e H which implies that xn e Hh = H. Since both x and n are in N, we deduce that xn e H n N < K, contradicting the fact that x e Ak (N). □

Remark 2.3. Recall that the prime graph (or Gruenberg-Kegel graph) of a finite group G is the graph T(G) with vertex set n(G) and the property that two distinct vertices p and q are adjacent if and only if G contains an element of order pq. Now, a transitive permutation group G with point stabilizer H has property (*) only if one of the following holds:

(a) r is an isolated vertex in r(G);

(b) n(G) = n(H).

The finite simple groups with a disconnected prime graph are recorded in [39, Tables 1-3], and a similar analysis for almost simple groups is given in [41,42]. In particular, one could use these results to study the almost simple permutation groups for which (a) holds. Similarly, if G is almost simple and (b) holds, then the possibilities for G and H can be read off from [40, Corollary 5]. However, this is not the approach that we will pursue in this paper.

The next result is a special case of [31, Lemma 3.3].

Lemma 2.4. Let G be a finite permutation group and let N be a transitive normal subgroup ofG such that G/N = (Ng) is cyclic. Then Ng n A(G) is empty if and only if every element of Ng has a unique fixed point.

We will also need several number-theoretic lemmas. Given a positive integer n we write n2 for the largest power of 2 dividing n. In addition, recall that (a, b) denotes the greatest common divisor of the positive integers a and b. The following result is well known.

Lemma 2.5. Let q > 2 be an integer. For all integers n, m > 1 we have

(qn - 1, qm - 1) = q (n,m) - 1

q(n,tn) + 1 if 2m2 < n2

(qn - 1, qm + 1) = (qn + 1, qm + 1) =

(2, q - 1) otherwise

q (n'm) + 1 if m2 = n2 (2, q - 1) otherwise

Let q = pf be a prime power, let e > 2 be an integer and let r be a prime dividing qe - 1. We say that r is a primitive prime divisor (ppd for short) of qe - 1 if r does not divide ql - 1 for all 1 < i < e. A classical theorem of Zsigmondy [53] states that if e > 3 then qe - 1 has a primitive prime divisor unless (q, e) = (2, 6).

Primitive prime divisors also exist when e = 2, provided q is not a Mersenne prime. Note that if r is a ppd of qe — 1 then r = 1 (mod e). Also note that if n is a positive integer, then r divides qn — 1 if and only if e divides n. If a pdd of qe — 1 exists, then we will write le(q) to denote the largest pdd of qe — 1. Note that te(q) > e.

Lemma 2.6. Let r, s be primes, and let m, n be positive integers. If rm + 1 = sn, then one of the following holds:

(i) (r, s, m, n) = (2, 3, 3, 2);

(ii) (r, n) = (2, 1),m is a 2-power and s = 2m + 1 is a Fermat prime;

(iii) (s, m) = (2, 1), n is a prime and r = 2n — 1 is a Mersenne prime.

Proof. This is a straightforward application of Zsigmondy's theorem [53]. For completeness, we will give the details.

First assume that m = 1, so r = sn — 1 is a prime. If s is odd, then r is even, so r = 2 and sn = 3, which implies that n = 1 and s = 3. This case appears in (ii). Now assume s = 2, so r = 2n — 1 is prime. It follows that n must also be a prime and thus r is a Mersenne prime. This is (iii).

For the remainder, we may assume that m > 2.Notice that r 2m — 1 = sn (rm — 1). If (m, r) = (3, 2), then sn = 23 + 1 = 32 and thus (s, n) = (3, 2) as in (i). Now assume that (m, r) = (3, 2). By Zsigmondy' s theorem [53], the ppd t2m(r) exists and divides r2m — 1 = sn (rm — 1), but not rm — 1, hence s = t2m(r) > 2m > 4. Therefore s > 5 is an odd prime and rm = sn — 1 is even, so r = 2. We now consider three cases.

If n = 1, then s = rm + 1 = 2m + 1 is an odd prime, which implies that m is a 2-power as in case (ii). Next assume that n = 2. Here2m = s2 — 1 = (s — 1)(s +1) and thus s — 1 = 2a and s + 1 = 2b for some positive integers a and b. Then 2b — 2a = (s + 1) — (s — 1) = 2 and thus 2b—1 = 2a—1 + 1, which implies that (a, b) = (1, 2), so s = 3 and thus m = 3. Therefore, (r, s, m, n) = (2, 3, 3, 2) as in case (i). Finally, let us assume that n > 3. Now 2m = sn — 1 and Zsigmondy's theorem implies that the ppd tn (s) > n > 3 exists and divides 2m, which is absurd. □

Lemma 2.7. Let q be a prime power and let (a, e), (b, 8) e N x {±1}, where b > a > 2 and (a, e) = (2, —1). Let N = (qa + e)(qb + 8). Then one of the following holds:

(i) N has two distinct prime divisors that do not divide q2 — 1 ;

(ii) (a, e) = (2, 1), (b, 8) = (4, —1) andq2 + 1 = (2, q — 1)re for some prime r and positive integer e;

(iii) q = 3, (a, e) = (2, 1) and (b, 8) = (3, 1);

(iv) q = 2, (a, e) = (3, 1) and 2b + 8 is divisible by at most two distinct primes, one ofwhich is 3;

(v) q = 2, a = 3 and (b, 8) = (6, —1).

Proof. There are four cases to consider, according to the possibilities for the pair

(e,8).

First assume that (e, 8) = (1, 1). Suppose that neither (a, q) nor (b, q) is equal to (3, 2). Then the primitive prime divisors t2a(q) and t2b(q) exist, and they both

Table 2. The integers N in Lemma 2.8

N (e, q)

q6— l)/(7, q — e) none

q6 — 1)/(q — e)(6, q — e) (—, 2)

q5 — e)/(6, q — e) (+, 2), (+, 3), (+, 7), (—, 2), (—, 5)

q4 — 1)/(5, q — e) none

q4 — 1)/(q — e)(4, q — e) (—, 2), (—, 3)

q3 — e)/(4, q — e) (+, 2), (+, 3), (+, 5), (—, 2), (—, 3)

q3 — 1)(q + 1)/(5, q — e) none

divide N. Moreover, these primes are distinct since 2a < 2b, and neither of them divides q2 — 1 since 2b > 2a > 4. If (a, q) = (3, 2) then b > 4, N = 32(2b + 1) and either (i) or (iv) holds. If (b, q) = (3, 2), then a = 2, N = 32 ■ 5 and (iii) holds.

Next suppose that (e, 8) = (—1, —1), so a > 3. If neither (a, q) nor (b, q) is equal to (6, 2), then N is divisible by the distinct primes la(q) and lb(q), neither of which divide q2 — 1. If (a, q) = (6, 2), then N = 32 ■ 7(2b — 1) is divisible by 7 and lb (2) > b > 7. Finally, suppose that (b, q) = (6, 2), so N = 32 ■ 7(2a — 1) and 3 < a < 5. It is easy to check that (i) holds if a = 4 or 5, and that (v) holds if a = 3.

Now assume that (e, 8) = (1, —1). If (a, q) = (3, 2) then (i) or (iv) holds, so we may assume that (a, q) = (3, 2). If (b, q) = (6, 2) then N = 32 ■ 7(2a + 1), a e {2, 4, 5} and (i) holds. In each of the remaining cases, the primitive prime divisors l2a (q) and lb (q) exist, and they divide N, but not q2 — 1. Clearly, if b = 2a then these two primes are distinct and (i) holds, so let us assume that b = 2a, so N = (qa + 1)2(qa — 1). If (a, q) = (6, 2) then (i) holds. If (a, q) = (6, 2) and a > 3 then we can take the primitive prime divisors la (q) and l2a (q), so once again (i) holds. Finally, if a = 2 and b = 4 then N = (q2 — 1)(q2 + 1)2 and either (i) or (ii) holds.

Finally, let us assume that (e, 8) = (—1, 1). Here we may assume that a > 3. If (a, q) = (6, 2) then take la(q) and l2b(q), otherwise N = 32 ■ 7(2b + 1) is divisible by 7 and l2b (2). In both cases, (i) holds. □

Lemma 2.8. Let q be a prime power and let N be one of the integers in Table 2, where e = ±1. Then N is a prime power if and only if (e, q) is one of the cases recorded in the second column of the table.

Proof. This is entirely straightforward. For example, suppose that N = (q5 — 1)/(6, q — 1). Let d = (6, q — 1) and suppose that N = re for some prime number r and positive integer e. Then r = l5(q) and

(q — 1)(q4 + q3 + q2 + q + 1) = dre.

Since r does not divide q — 1, we must have q — 1 = d and thus q — 1 e{1, 2, 3, 6}. If q = 4 then N = 341 = 11 ■ 31 is not a prime power, but one checks that N is a prime power if q e {2, 3, 7}. The other cases are very similar. □

We will also need the following result, which follows from a theorem of Nagell [44].

Lemma 2.9. Let q = pf be a prime power and let r be a prime.

(i) If e is a positive integer such that q2 + q + 1 = re, then q ^ 1 (mod 3) and e = 1.

(ii) If e is a positive integer such that q2 + q + 1 = 3re, then q = 1 (mod 3) and e e{1, 2}.

(iii) Ifq2 + q +1 = (3, q - 1)re for some positive integer e, then either (q, r, e) = (4, 7, 1), or f = 3a for some integer a > 0.

Proof. Parts (i) and (ii) follow from [44]. For (iii), let d = (3, q - 1) and write f = 3am with (3, m) = 1 and a > 0. We may assume that q = 4. Seeking a contradiction, suppose that m > 1. Notice that

p3a+1m _ i

re = ^

d(p3"m — 1)

Since q = 4, the ppd t3 f (p) exists and divides q2 + q + 1, so r = t3 f (p). Let s = t3a+i (p). Since f = 3am is indivisible by 3a+\ it follows that (s, q — 1) = 1, so s does not divide d(q — 1) and thus s divides re, so r = s. But m > 1, so 3 f > 3a+l and thus r = s. This is a contradiction and the result follows. □

Remark 2.10. By a theorem of van der Waall [50], the Diophantine equation x2 + x + 1 = 3y2 has infinitely many integer solutions; the smallest nontrivial solution is (x, y) = (313, 181). Here x and y are both primes, and another solution in the primes is (x, y) = (2288805793, 1321442641).

3. A reduction theorem

The following theorem reduces the study of primitive permutation groups with property (*) to almost simple and affine groups.

Theorem 3.1. Let G < Sym(^) be a primitive permutation group with point stabilizer H. If (*) holds, then either

(i) G is almost simple; or

(ii) G = HN is an affine group with socle N = (Zr )k for some integer k > 1.

Moreover, if (ii) holds and | H | is indivisible by r, then G is a Frobenius group with kernel N and complement H.

Proof. Let N be a minimal normal subgroup of G, so N = x S2 x ■■ ■ x Sk, where Si = S for some simple group S and integer k > 1. Then G = HN and N is transitive on Let us assume that (*) holds.

First assume that H n N = 1, so N is regular and every nontrivial element in N is a derangement. If N is abelian, then we are in case (ii). Moreover, if | H | is indivisible by r, then N is a Sylow r-subgroup of G and thus A(G) c N. In this situation, [12, Lemma 4.1] implies that G is a Frobenius group with kernel N and complement H. Now, if N is nonabelian then S is a nonabelian simple group and

thus | S| is divisible by at least three distinct primes, whence S (and thus N) contains derangements of distinct prime orders, which is incompatible with property (*).

For the remainder, we may assume that H n N is nontrivial. It follows that N = Sk, where S is a nonabelian simple group and k > 1. If k = 1, then G is almost simple and (i) holds. Therefore, we may assume that k > 2.

Let T < N be a maximal subgroup of N containing H n N .By Lemma 2.2, we have AT(N) c AH(G). Since k > 2, there exist integers i and j such that 1 ^ i < j ^ k and L := Si x Sj ^ T. By relabelling the Sl, if necessary, we may assume that L = S1 x S2. Now L ^ N, so N = TL and thus

where K is a maximal subgroup of L containing L n T. Therefore, every derangement of L = Si x S2 on the right cosets L /K is an r-element.

By [49, Lemma 1.3], there are essentially two possibilities for K; either K is a diagonal subgroup of the form {(s, 0(s)) : s e S1} for some isomorphism 0 : S1 ^ S2, or K is a standard maximal subgroup, i.e., K = S1 x K2 or K1 x S2, where Ki < Si is maximal. In the diagonal case, every element in L of the form (s, 1) with 1 = s e S1 is a derangement on L/K. Clearly, this situation cannot arise. Now assume K is a standard maximal subgroup. Without loss of generality, we may assume that K = K1 x S2, where K1 is maximal in S1. Let s e S1 be a derangement on S1/K1 of prime power order, say pe for some prime p and integer e > 1 (such an element exists by the main theorem of [18]). Since |^(S)| > 3, choose t e S2 of prime order different from p. Then (s, t) e L is a derangement on L/K of non-prime power order, so once again we have reached a contradiction. □

This completes the proof of Theorem 1.

4. Almost simple groups

In this section we prove Theorem 2. We fix the following notation. Let r be a prime and let G < Sym(^) be an almost simple primitive permutation group with socle Go and point stabilizer H. Set H0 = H n Go and let M be a maximal subgroup of G0 containing H0. As before, let A(G) be the set of derangements in G, and let E(G) be the set of orders of elements in A(G). By Lemma 2.2, we have

Recall that if X is a finite set, then n(X) denotes the set of prime divisors of | X |.

Let us assume that (*) holds, so every derangement in G is an r-element, for some fixed prime r. Clearly, every derangement of G0 on ^ is also an r-element. Now, if s e n(G0)\n(M) then every nontrivial s-element in G0 is a derangement, so n(G0) = n(M) or n(M) U {r}. In particular, if we set n0 := n(G0) \n(M), then |n01 < 1.

Ak(L) c At(N) c Ah(G) = A(G),

Am(Go) C Aho (Go) C Ah (G) = A(G).

4.1. Sporadic groups

Proposition 4.1. Theorem 2 holds ifG0 is a sporadic group or the Tits group.

Proof. First assume that Go is not the Monster. The maximal subgroups of G0 are available in GAP [25], and it is easy to identify the cases (G0, M) with |n0| < 1. For the reader's convenience, the cases that arise are listed in Table 3. We now consider each of these cases in turn. With the aid of GAP [25], we can compute the permutation character x = 1M°, noting that

Am(Go) = {x e Go : x(x) = 0}.

In this way, we deduce that property (*) holds if and only if (G0, M) = (Mn, L2(11)).Here n(M) = n(G0), G = Mn, H = L2(11) and E(G) = {4, 8}. This case is recorded in Table 1.

Now assume G = M is the Monster. As noted in [6,45], there are 44 conju-gacy classes of known maximal subgroups of M (these subgroups are conveniently listed in [6, Table 1], together with L2(41)). Moreover, it is known that any additional maximal subgroup of M is almost simple with socle L2(13), U3(4), U3(8) or 2B2(8). It is routine to check that |n0| > 2 in each of these cases. □

4.2. Alternating groups

Proposition 4.2. Theorem 2 holds if G0 = An is an alternating group.

Proof. If n < 12 then the result can be checked directly using GAP [25]; the only cases (G, H) with property (*) are the following:

(A6, 32:4), (A5, D10), (A5, A4), (A5, S3),

which are recorded in Table 1 as

(L2(9), P1), (L2(4), D10), (L2(4), P1), (L2(4), D6)

respectively (see Remark 3). For the remainder, we may assume that n > 12. Seeking a contradiction, let us assume that there is a fixed prime r such that every derangement in G is an r-element.

Let s be a prime such that n/2 < s < n — 2 and let x e G0 be an s-cycle (such a prime exists by Bertrand's postulate). Since CG (x) is not an r-group, Lemma 2.1 (ii) implies that x has fixed points and thus H contains s-cycles. By applying a well known theorem of Jordan (see [52, Theorem 13.9]), we deduce that H is either intransitive or imprimitive, and we can rule out the latter possibility since s divides | H |. Therefore, H is the stabilizer of a k-set for some k with 1 < k < n/2.

Suppose n is even and let xi e G0 be an element with cycles of length i and n — i for i e {3, 5, 7}. Then at least two of the Xi are derangements, so we have reached a contradiction. Now assume n is odd. An n-cycle does not fix a k-set, so n must be an r-power. Therefore, any element with cycles of length (n — 1)/2, (n — 1)/2 and 1 must fix a k-set (since its order is not an r -power), so k = 1 or (n — 1)/2. It follows that any element with cycles of length 2, 3 and n — 5 is a derangement, and this final contradiction completes the proof of the proposition. □

Table 3. Maximal subgroups of sporadic simple groups, |no | < 1

Go M n0

M11 A6.23, S5 11

L2 (11) —

M12 M11, L2(11) —

A6.22, 2 x S5 11

M22 A7, L3(4) 11

L2 (11) 7

M23 M22 23

M24 M23 —

M22.2 23

J2 L3(2).2, U3(3) 5

3.A6.22, 21+4:A5, A4 x A5, A5 x Di0, 52:Di2, A5 7

J3 L2(16).2 19

L2 (19) 17

Co1 3.Suz.2 23

Co2, Co3, 211:M24 13

Co2 M23 —

McL, HS.2, U6(2).2, 210:M22.2 23

Co3 M23 —

McL.2, HS 23

F122 2.U6(2), 210:M22 13

«7(3) 11

Fi' F124 Fi23 29

HS M22 —

U3(5).2, L3(4).2x, S8 11

McL M22 —

U4(3), U3(5), L3(4).22, 2.A8, 24:A7 11

Suz G2 (4) 11

He Sp4(4).2 7

22.L3(4).S3, 3.S7 17

HN 2.HS.2, A12 19

O'N J1 31

Ru (22 x 2B2(8)):3 29

2F4(2)' L2(29) 13

L2(25) —

L3(3).2 5

A6.22, 52:4A4 13

4.3. Exceptional groups

Now let us assume that G0 is a simple exceptional group of Lie type over Fq, where q = pf and p is a prime. For x e G0, let M(x) be the set of maximal subgroups of G0 containing x. We will write ^ for the ith cyclotomic polynomial evaluated at q, so qn — 1 = nd\n . Recall that if e > 2 and qe — 1 has a primitive prime divisor, then we use the notation te (q) to denote the largest such divisor of qe — 1.

Proposition 4.3. Theorem 2 holds if G0 is a simple exceptional group of Lie type.

Proof. Recall the notational set-up introduced at the beginning of Sect. 4: H is a point stabilizer in G, and H0 = H n G0. In view of (3), in order to show that (*) does not hold we may assume that G = G0. Seeking a contradiction, suppose that every derangement in G is an r-element, for some fixed prime r. We will consider each possibility for G in turn. Case 1 G = 2B2(q), with q = 22m+1 and m > 1.

Let $4 = q + V2q + 1 and $4' = q - V2q + 1 (note that $4$4' = q2 + 1). By inspecting [2, Table II], [29, Table 6] and [30, Table 1], we see that G has two cyclic maximal tori T = X), i = 1, 2, of order $4 and $4', respectively, such that |Ng(Ti)/Ti | =4, (|X11, |X2|) = 1 and M(xt) = {Ng(T)}. Since no maximal subgroup of G can contain conjugates of both x1 and x2, it follows that xG n H is empty for some i = 1, 2. Therefore, xi e A(G) and thus x | is a power of r. Let j = 3 - i. Then ^j | is indivisible by r, so H contains a conjugate of xj and thus H = Ng(Tj) is the only possibility (up to conjugacy). Now G has a cyclic maximal torus of order q - 1, so let x e G be an element of order q - 1 > 7. Since |H| is indivisible by q - 1, it follows that x e A(G). But r does not divide q - 1, so we have reached a contradiction. Case 2 G = 2G2(q), with q = 32m+1 and m > 1.

This is very similar to the previous case. Here we take two cyclic maximal tori Ti = {xi), i = 1, 2, of order $6 = q + J3q + 1 and $6 = q - J3q + 1, respectively, such that |Ng(Ti)/Ti | = 6, (|x11, |x2 D = 1 and M(xi) = {Ng(Ti)}. Note that $6$6' = q2 - q + 1. By arguing as in Case 1, we deduce that x | is a power of r and H = NG(Tj) for some distinct i, j. Let x e G be an element of order 9 (see part (2) in the main theorem of [51], for example). Since |H| is indivisible by 9, it follows that x is a derangement, but this is a contradiction since r = 3.

Case 3 G = 2F4(q), with q = 22m+1 and m > 1.

Again, we proceed as in Case 1. Here G has two cyclic maximal tori Ti = {xi), i = 1 , 2, where

and |Ng(Ti)/Ti \ = 12, (\xi\, \X2\) = 1 and M(xi) = (NG(Ti)}. Note that ^12^12 = q4 - q2 + 1. As in Case 1, we see that \xi \ is a power of r and H = Ng (Tj ) for some distinct i, j .Let x e G be an element of order l4(q ). Since \H\ is indivisible by i4(q), it follows that x e A(G), but this is a contradiction since r = l4(q ). Case 4 G = Es (q).

Again, we can proceed as in the previous cases, working with cyclic maximal tori Ti, T2 and an element x e G of order l24(q), where

\Ti | = $15 = q8 - q7 + q5 - q4 + q3 - q + 1 \T2\ = $30 = q8 + q7 - q5 - q4 - q3 + q + 1

and |Ng (Ti)/Ti I =30, i = 1, 2. We omit the details (note that t24(q) e n(G) \

n(NG (Ti))).

Case 5 G = 3D4(q).

As indicated in [29, Table 6], G has a maximal torus T = (x} of order $12 = q4 — q2 + 1 such that |Ng(T)/T| = 4 and M(x) = {Ng(T)}.

Suppose that x £ A(G). Then xG n H is non-empty, and without loss of generality we may assume that x e H and thus H = NG (T). If q = 2 then |H| = 52 and |n(G)\n(H)| = 2, so we must have q > 2. Let y e G (i = 1, 2) be elements of order t := tm (q) > 5, where m1 = 3 and m2 = 6. Since | H| is indivisible by t1 and t2, it follows that y1, y2 e A(G). But this is a contradiction since t1; t2 are distinct primes.

Now assume that x e A(G), so |x | = $12 is a power of r .If q = 2 then r = 13 and H must contain elements of order 7, 8, 9, 14, 18, 21 and 28, but no maximal subgroup of G has this property (see [15], for example). Therefore, q > 2. Following [30, p. 698], let y e G be an element of order $3 such that |CG(y)| divides and

M(y) = {G2(q), PGL3(q), ($6 ◦ SL3(q)).2d, $2.SL2(3)},

where d = (3, $3).Now($12, $3) = 1,so y e A(G) andthus we may assume that H e M(y). Let z e G be an element of order $1$2$6 = (q2 — 1)(q2 — q + 1). Then |H| is indivisible by |z|, so z e A(G). But this is a contradiction since ($12, $1$2$6) = 1. Case 6 G = 2E6 (q).

Let d = (3, q + 1). As indicated in [29, Table 6] and [30, Table 1], G has two cyclic maximal tori Ti = (xi}, i = 1, 2, of order $18 /d and $6$12/d, respectively. Then (|x1|, |x2|) = 1 and

No maximal subgroup of G contains both x1 and x2 (see [40, Table 10.5]), so xi e A(G) for some i, and thus |x; | is a power of r.

First assume that q = 2, so |x1| = 19, |x2| = 13 and thus r e {13, 19}. If r = 13, then H contains a conjugate of x1,so H = SU3(8).3 is the only option, but this is not possible since |n(G)\n(H)| = 4. Similarly, if r = 19 then H e M(x2) must contain elements of order 11, 13 and 17, but it is easy to check that this is not the case.

Now assume that q > 2. Let x e G be an element of order 110 (q). Both |SU3 (q 3).3| and |$6.3D4(q ).3/d | are indivisible by t10(q ),sox e A(G). However, this is not possible since t10(q) and |x; | are coprime. Case 7 G = G2(q), q > 3.

We can use GAP [25] to rule out the cases q < 5, so we may assume that q > 7.

First assume that q = 7. By inspecting [29, Table 6] and [30, Table 1], we see that G has two cyclic maximal tori Ti = (xi}, i = 1, 2, of order $6 = 43 and $3 = 57, respectively, with M(x1) = {SU3 (7).2} and M(x2) = {SL3(7).2}.From [40, Table 10.5], it follows that xi e A(G) for some i, so H contains a conjugate

M(x1) = {SU3(q3).3}, M(x2)

{$6.3D4(q).3/d} if q > 2

{$6.3D4(2), F4(2), Fi22} if q = 2.

of xj, where j = 3 - i. Therefore, H = SL3(7).2 for some e = ±. As noted in [37, Table A.7], G contains elements of order 72 + 7 = 56 and 72 + 7 + 1 = 57. Now SU3(7).2 contains no element of order 57, and SL3(7).2 has no element of order 56. Therefore, G always contains a derangement of non-prime power order, which is a contradiction.

For the remainder, we may assume that q > 7. We use the set-up in [17, Section 5.7]. Choose a 4-tuple (k1, k2, k3, k6) such that (k1, k2) = 1, ki divides $i for i e {1, 2}, k3 = $3/(3, $3) and k6 = $6/(3, $6). Note that the numbers k1, k2, k3 and k6 are pairwise coprime. Let y1 e G be an element of order k6, and fix a regular semisimple element y2 e G of order k1. Similarly, fix zi e G, i = 1, 2, where |z11 = k3 and z2 is a regular semisimple element of order k2.

From [40, Table 10.5], it follows that either y1 or z1 is a derangement. Suppose that y1 e A(G). Then H contains a conjugate of z1, so [17, Lemma 5.27] implies that H = SL3 (q).2 is the only possibility. If H also contains a conjugate of z2, then H = G by [17, Corollary 5.28], a contradiction. Therefore z2 e A(G), but once again we reach a contradiction since (k2, k6) = 1. An entirely similar argument applies if z1 e A(G). Case 8 G e {E6(q), E7(q)}.

First assume that G = E7(q). Let d = (2, q - 1). As in [17, Section 5.2], let y1, y2 e G be elements of order $18 and $2$14/d = (q7 + 1)/d, respectively, and let z1, z2 e G be elements of order $9 and $1$7/d = (q7 - 1)/d, respectively. From [17, Corollary 5.6], we deduce that yi, zj e A(G) for some i, j e {1, 2}. However, it is easy to check that (| yi |, ^j |) = 1 for all i, j, so this is a contradiction.

The case G = E6(q) is entirely similar, using [17, Corollary 5.11] and elements yi, zi e G with |y11 = $9/d, |y2| = $4, |z11 = $3$12 and |z21 = $5 (where d = (3, q - 1)). Case 9 G = F4(q).

For q > 2, we can proceed as in Case 8, using the information in [17, Section 5.5]. The reader can check the details.

Now assume that q = 2. By inspecting [29, Table 6] and [30, Table 1], we see that G has two cyclic maximal tori Ti = {xi), i = 1, 2, of order $12 = 13 and $8 = 17, respectively, such that M(x1) = {3D4(2).3, 2F4(2), L4(3).22} and M(x2) = {Sp8 (2)}. Therefore, r e {13, 17}.Ifr = 13, then H contains a conjugate of x2, so H = Sp8(2). However, [15] indicates that G has an element of order 28, but Sp8(2) does not, so this case is ruled out. Therefore, r = 17 and H contains a conjugate of x1, so H e M(x1). However, in each case one can check that H does not contain an element of order 30, but G does. This final contradiction eliminates the case G = F4(q).

This completes the proof of Proposition 4.3. □

4.4. Classical groups

In order to complete the proof of Theorem 2, we may assume that G0 is a classical group over Fq .Due to the existence of certain exceptional isomorphisms involving low-dimensional classical groups (see [38, Proposition 2.9.1], for example), and in

Table 4. Finite simple classical groups

Go Conditions

Ln (q) n > 2, (n, q) = (2, 2), (2, 3), (2, 4), (2, 5), (2, 9), (3, 2), (4, 2)

Un (q ) n > 3, (n, q) = (3, 2)

PSpn (q ) n > 4 even, (n, q) = (4, 2), (4, 3)

P«n (q ) n > 7

view of our earlier work in Sects. 4.1, 4.2 and 4.3, we may assume that G0 is one of the groups listed in Table 4.

We will focus initially on the low-dimensional classical groups with socle L2 (q) and L3 (q), which require special attention. As before, if a primitive prime divisor of qe — 1 exists, then te (q) denotes the largest such prime divisor (as noted in Sect. 2, if e > 2, then te(q) exists unless (q, e) = (2, 6), or e = 2 and q is a Mersenne prime).

Lemma 4.4. Theorem 2 holds ifG = L2 (q) and q is even.

Proof. Write q = 2f, where f > 3 (since L2(4) = A5, we may assume that f > 3). The maximal subgroups of G were originally classified by Dickson [16] (also see [5, Tables 8.1 and 8.2]); the possibilities for H are as follows:

(a) H = (Z2)f :Zq—1 = P1 is a maximal parabolic subgroup of G;

(b) H = D2(q±1);

(c) H = L2(q0) with q = q0, where e is a prime and q0 = 2.

The case f = 3 can be handled using GAP [25], and we find that (*) holds if and only if (H, r, E(G)) is one of the following (recall that E(G) denotes the set of orders of derangements in G):

(P1, 3, {3, 9}), (D18, 7, {7}), (D14, 3, {3, 9}).

For the remainder, we may assume that f > 4.

Note that a Sylow 2-subgroup of G is self-centralizing and elementary abelian. In particular, if x e G then either |x | = 2, or |x | divides q ± 1. Also note that G contains elements of order q ± 1, and it has a unique class of involutions. Case 1 H = P1.

We claim that (*) holds if and only if r = q + 1 is a Fermat prime. To see this, first observe that |G : H | = q + 1 and | H | = q (q — 1) are relatively prime, so any element x e G of order q + 1 is a derangement. Therefore, if (*) holds then q + 1 = re for some e > 1, and thus Lemma 2.6 implies that f is a 2-power and e = 1 (so r = q + 1 is a Fermat prime).

For the converse, suppose that q + 1 is a Fermat prime. We need to show that every derangement in G has order r = q + 1. Let y e A(G), so |y| divides 2 or q ± 1. But q + 1 = r is a prime, so either |y| e {2, r} or |y | divides q — 1. Every involution has fixed points since G has a unique class of involutions, so |y| > 2. If | y| divides q — 1, then y belongs to a maximal torus that is G-conjugate to the subgroup Zq—1 < H. Again, this implies that y has fixed points. Therefore, | y| = r is the only possibility and the result follows.

Case 2 H = D2(q±1).

The case H = D2(q-1) is identical to the previous one, and the same conclusion holds. A very similar argument also applies if H = D2(q+1). Here any element of order q - 1 is a derangement and by applying Lemma 2.6 we deduce that (*) holds if and only if r = q - 1 is a Mersenne prime. Case 3 H = L2(q0), where q = q0, e prime, q0 = 2.

Finally, observe that subfield subgroups are easily eliminated since elements of order q ± 1 are derangements. □

Lemma 4.5. Theorem 2 holds ifG0 = L2 (q) and q is even.

Proof. As before, write q = 2f, where f > 3. In view of Lemma 4.4, we may assume that

G = G0.{$) < TL2(q) = Aut(G0),

where $ is a nontrivial field automorphism of G0, so the order of $ divides f. The case f = 3 can be handled directly, using [25] for example. Here G = TL2(8) and we find that (*) holds if and only if (H, r, E(G)) is one of the following:

(Ng(P1), 3, {3, 9}), (Ng(D18), 7, {7}), (Ng(Dm), 3, {3, 9}).

For the remainder, we may assume that f > 4.

Since G0 ^ H, we have G = G0 H. Set H0 = H n G0 and note that H0 is a maximal subgroup of G0 (see [5, Table 8.1]). As in (3), we have AH0(G0) c AH (G), whence Lemma 4.4 implies that (*) holds only if one of the following holds:

(a) H0 = P1, r = q + 1 is a Fermat prime;

(b) H0 = D2(q+1), r = q - 1 is a Mersenne prime;

(c) H0 = D2(q-1), r = q + 1 is a Fermat prime.

We consider each of these cases in turn.

Case 1 H0 = D2(q+1), r = q - 1 is a Mersenne prime.

Here f > 5 is aprime, so G = TL2(q) = G0.{$) and H = H0.{$) is the only possibility, where $ has order f. Note that

Cg ($) = L2(2) x{$) = S3 x Zf,

so if x e G then either ^ | e {2, r, f, 2 f, 3 f}, or ^ | divides q + 1. We claim that E(G) = {r}. Note that {$) is a Sylow f -subgroup of G.

Let y e G be a nontrivial element. If |y| e {2, f}, or if |y| divides q + 1, then y is conjugate to an element of H and thus y has fixed points. Next suppose that |y| = kf and k e {2, 3}. Then |yk| = f and thus yk is G-conjugate to $i for some 1 < i < f. Without loss of generality, we may assume that yk = $, so y e Cg($). Since |H| = 2(q + 1) f, H has a Sylow 2-group R = {u)= Z2 and a normal 2-complement V {$) of order (q + 1) f, where V = Zq+1. Since $ normalizes H0 = VR, we deduce that $ centralizes R. Now q + 1 is divisible by 3, so V has a unique subgroup of order 3, say {x). Then the involution u inverts x, and $ centralizes x since |$| = f > 5 is odd. Thus S3 = {u, x) < CG0($),

which implies that Cg (0) = (u, x) x (0) < H. Therefore, y e H. We conclude that every derangement in G has order r, as required.

In the two remaining cases, r = q + 1 is a Fermat prime and f = 2m for some integer m > 2. In both cases, we claim that (*) does not hold. In order to see this, we may assume that the index of Go in G is a prime number, which in this case implies that |G : G01= 2, so G = G0.(0) and 0 is an involutory field automorphism of G0. Indeed, if G0 ^ G1 ^ G then G = HG1 and Lemma 2.2 implies that AL(G1) c A(G) for any subgroup L of G1 containing G1 n H. Case 2 H0 = D2(q-1), r = q + 1 is a Fermat prime.

By the above comments, we may assume that G = G0.(0) and H = D2(q-1). (0), where 0 has order 2. Note that Cg (0) = L2(2f/2) x (0). Since Cg(0) does not contain a Sylow 2-subgroup of G, we deduce that the Sylow 2-subgroups of G are nonabelian. Therefore G contains an element z of order 4. However, the Sylow 2-subgroups of H are isomorphic to C2 x C2, so z e A(G). We conclude that G contains derangements of order r and 4, so (*) does not hold. Case 3 H0 = P1, r = q + 1 is a Fermat prime.

Finally, let us assume that H = NG(P1) = P1.(0} = H0.(0), where |0| = 2. As above, we have CG (0) = L2(2f/2) x (0), so CG (0) contains an element of order 2(q0 + 1), where q0 = 2f/2. We claim that H does not contain such an element. Seeking a contradiction, suppose x e H has order 2(q0 +1). Since H = H0 U H00 and H0 = Pi has no element of order 2(q0 + 1), we deduce that x e H00 and we may write x = u0 with u e H0. In terms of matrices (and a suitable basis for the natural L2(q)-module), we have

( X a \ u = ( 0 X-1)

where X, a e Fq and X = 0. Then x2 = (u0)(u0) = uu0 has order q0 + 1. We may assume that 0 is the standard field automorphism of order 2 with respect to this basis, so

2 _ 0 _ (X a \ /Xq° aq0 \ _ (X1+q° b \

x =uu ^0 X-V v 0 x-q0) ^ 0 x-1-q0)

with b = Xq0 + aX-q0. Since x2 has order q0 + 1 we deduce that x2(q0+1) = 1, which implies that Xq0+1 = 1 since Fq has characteristic 2. Therefore

2 /1 b' x 2 =

has order q0 +1, which is absurd. This justifies the claim, and we deduce that A(G) contains elements of order 2(q0 + 1). In particular, (*) does not hold. □

Lemma 4.6. Theorem 2 holds ifG0 = L2 (q) and q is odd.

Proof. Write q = pf, where p is an odd prime. In view of the isomorphisms L2(5) = A5 and L2(9) = A6, we may assume that q > 7 and q = 9. The case

q = 7 can be checked directly using GAP, and we find that (*) holds if and only if (G, H, r, E(G)) is one of the following:

(L2(7), P1, 2, {2, 4}), (L2(7), S4, 7, {7}), (PGL2(7), Ng(P1), 2, {2, 4, 8}).

For the remainder, we may assume that q > 11. Case 1 G = G0.

First assume that G = L2(q). The maximal subgroups of G are well known (see [5, Tables 8.1 and 8.2]); the possibilities for H are as follows:

(a) H = (Zp)f :Z(q-1)/2 = P1 is a maximal parabolic subgroup of G;

(b) H = Dq-e, where q > 13 if e = 1;

(c) H = L2(q0), where q = q0 for some odd prime e;

(d) H = PGL2(q0), where q = q2;

(e) H = A5, where q = ±1 (mod 10) and either q = p, or q = p2 and p = ±3 (mod 10);

(f) H = A4, where q = p = ±3 (mod 8) and q = ±1 (mod 10);

(g) H = S4, where q = p = ±1 (mod 8).

Note that G contains elements of order (q ± 1)/2, and a unique conjugacy class of involutions.

If H is a subfield subgroup (as in (c) or (d) above), then it is clear that any element of order (q ± 1)/2 is a derangement, so property (*) does not hold in this situation. Similarly, it is straightforward to handle the cases H e {A5, A4, S4}. For example, suppose H = A5, so q = ±1 (mod 10) and either q = p, or q = p2 and p = ±3 (mod 10). Note that every nontrivial element of H has order 2, 3 or 5. If q > 19 then any element of order (q ± 1)/2 is a derangement; if q = 11, then elements of order 6 are derangements. The cases H = A4 and S4 are just as easy.

If H = Dq-1 then any element in G of order p or (q + 1 )/2 is a derangement, and the dihedral groups of order q + 1 can be eliminated in a similar fashion.

Finally, let us assume that H = P1 = (Zp)f :Z(q-1)/2, so |H| = q(q - 1)/2 and G : H ^ q + 1. We claim that (*) holds if and only if q = 2re - 1 for some positive integer e.

First observe that any element of order (q + 1)/2 is a derangement, so if (*) holds then q = 2re - 1for some e e N. For the converse, suppose that q = 2re -1. We claim that

E(G) = {ri : 1 < i < e}.

Since | H| is indivisible by r, the inclusion {ri : 1 < i < e}c E(G) is clear. To see that equality holds, let y e G be a nontrivial element, and suppose that | y| is divisible by a prime s = r. Since a Sylow p-subgroup of G is self-centralizing, it follows that either | y| = p,or |y| =2 and r is odd, or | y | is a divisor of (q -1)/2. In the first two cases, it is clear that y has fixed points, so let us assume that | y| divides (q - 1)/2. Then y is conjugate to an element of the maximal torus Z(q-1)/2 < H, so once again y has fixed points. This justifies the claim. Case 2 G = G0.

To complete the proof of the lemma, we may assume that G = G0, q > 11 and H0 = H n G0 = P1, in which case (*) holds only if q = 2re - 1 for some

positive integer e (note that H n G0 is a maximal subgroup of G0). There are several possibilities for G.

First assume that G = PGL2(q), so H = (Zp)f :Zq—1. We claim that (*) holds if and only if r = 2 and q = 2e+1 — 1 is a Mersenne prime. As above, any element of order (q + 1)/2 is a derangement. Now G also contains elements of order q + 1, and they are also derangements. Therefore, if (*) holds then r = 2 is the only possibility, so pf + 1 = 2e+1 and Lemma 2.6 implies that q = p = 2e+1 — 1 is a Mersenne prime.

For the converse, suppose that q = p = 2e+1 — 1 is a Mersenne prime. We claim that

E(G) = {2i : 1 < i < e + 1}.

As above, any involution in G0 is a derangement, and so is any element in G of order 2l with 1 < i < e + 1 since |H|2 = 2, hence {2' : 1 < i < e + 1}c E(G). To see that equality holds, suppose that y e G has order divisible by an odd prime. Then either | y| = p, or y is conjugate to an element of the maximal torus Zq—1 < H; in both cases, y has fixed points. The result follows.

To complete the proof of the lemma, we may assume that G = G0.(0) or G0.(50), where 0 is a nontrivial field automorphism of G0 of order e (so e divides f) and 5 = diag^, &>2) e PGL2(q) (modulo scalars) is a diagonal automorphism of G0. Recall that (q +1)/2 = re for some prime r and positive integer e. Our goal is to show that (*) does not hold.

First observe that r is odd. Indeed, if r = 2 then pf + 1 = 2e+1 and thus Lemma 2.6 implies that f = 1, which is false. Next we claim that f is a 2-power. To see this, first assume that f is odd and p = 2l — 1 is a Mersenne prime. Then re = (pf + 1)/2 is divisible by (p + 1)/2 = 2l—1, but r is odd so this is not possible. For the general case, suppose that f = 2am where a > 0 and m > 1 is odd (and we may assume that a > 0 if p is a Mersenne prime). We now proceed as in the proof of Lemma 2.9(iii). We have

e q2 — 1 p2°+1m — 1

2(q — 1) 2(p2am — 1)

and thus r = t2f (p). Set s = l2a+i (p) (note that s exists since a > 0 if p is a Mersenne prime). Now f = 2am is indivisible by 2a+1, so (s, q — 1) = 1 and thus s does not divide 2(q — 1). Therefore, r = s is the only possibility, but this is a contradiction since 2 f = 2a+1 m > 2a+1. This justifies the claim.

Therefore, in order to show that (*) does not hold, we may assume that \G : Go \ = 2. Write G = Go U Go y .

If we identify ^ with the set of 1-dimensional subspaces of the natural L2(q)-module, then 0 and 8$ fix the 1-spaces ((1, 0)> and ((0, 1)>. Therefore, Lemma 2.4 implies that the coset G0y contains derangements. But every element in this coset has even order, which is incompatible with property (*). □

To summarize, we have now established the following result. (Note that the case appearing in the final row of Table 5 is recorded as (G, H) = (L3(2), Pi) in Table 1).

Table 5. The cases (G, H, r) in Proposition 4.7

r S (G)

Conditions

rL2(i) Ng (D2(?+1))

rL2(8) Ng (Pi), Ng (Di4)

PGL2(?) Ng (Pi)

L2(q) Pi

2 2', 1 < i < e + 1 r r', 1 < i < e

q = 2e+1 — 1 Mersenne prime q = 2re — 1 r = q + 1 Fermât prime r = q — 1 Mersenne prime

r = q — 1 Mersenne prime

P1, D2(q—1)

D2(q+1)

L2(9) P1 L2(8) P1, D14 L2(7) S4

3 3,9 7 7

Proposition 4.7. Let G be a finite almost simple primitive permutation group with point stabilizer H and socle L2(q), where q > 4 and q = 5. Then (*) holds if and only if (G, H, r) is one of the cases in Table 5.

Lemma 4.8. Theorem 2 holds ifG0 = L3 (q).

Proof. Set d = (3, q-1) andnotethat G0 contains elements of order (q2+q+ \)/d and (q2 - 1)/d. We may assume that q > 3 since L3(2) = L2(7). If 3 < q < 7, then we can use GAP to verify the desired result; we find that (*) holds if and only if G = L3(q), H e {Pi, P2} and r = (q2 + q + 1)/d, in which caseE(G) = {r} (note that (q2 + q + 1)/d is a prime number for all q e {3, 4, 5, 7}). For the remainder, we will assume that q > 8. In particular, note that (q2 - 1)/d is not a prime power (indeed, it is easy to check that (q2 - 1)/d is a prime power if and only if q = 3 or 7).

Case 1 G = G 0.

First assume that G = L3(q). The possibilities for H are given in [5, Tables 8.3 and 8.4]. We can immediately eliminate any subgroup H that does not contain an element of order (q2 - 1)/d, so this implies that H is either a maximal parabolic subgroup, or H = SO3 (q) (with q odd).

Suppose that H is a maximal parabolic subgroup. Without loss of generality, we may assume that H = P1 (the actions of G on 1-spaces and 2-spaces are permutation isomorphic), so | H | = q3(q - 1)(q2 - 1)/d. We claim that G has property (*) if and only if one of the following holds:

To see this, first notice that any element x e G of order (q2 + q + 1)/d is a derangement. Therefore, if (*) holds then (q2 + q + 1)/d = re for some positive integer e, and by applying Lemma 2.9 we deduce that (a) or (b) holds. Conversely, suppose that (a) or (b) holds. We claim that

(a) d = 1 and q2 + q + 1 = r ; or

(b) d = 3 and q2 + q + 1 e{3r, 3r2}.

E (G )

{r, r2} if d = 3 and q2 + q + 1 = 3r2 {r} otherwise.

To see this, we use the fact that the action of G on 1-spaces is doubly transitive, so the corresponding permutation character has the form 1H = 1 + x for some irreducible character x e Irr(G) of degree q(q + 1). By inspecting the character table of G (see [20, Table 2], for example), we see that x(x) = — 1 if and only if x has order r (or r2 if d = 3 and q2 + q + 1 = 3r2). This justifies the claim.

Now assume that H = SO3 (q), so q is odd. Here elements of order (q2 + q + 1)/d are derangements, and so is any unipotent element with Jordan form [ J2, J1] (where Ji denotes a standard unipotent Jordan block of size i). Therefore, (*) does not hold in this situation. Case 2 G = G0.

To complete the proof of the lemma, we may assume that G = G0 and q > 8. Let M be a maximal subgroup of G0 containing H0 := Hn G0. From the analysis in Case 1, we may assume that M = P1, in which case H0 is either equal to P1, or it is a non-maximal subgroup of typeP1i2 (aBorel subgroup of G0) orGL2(q) x GL1 (q). We can quickly eliminate the latter two possibilities. For instance, if H0 is a Borel subgroup then AH0 (G0) contains all elements of order (q2 — 1)/d, so (*) does not hold (see (3)). Similarly, if H0 is of type GL2(q) x GL1 (q) then AH0 (G 0) contains elements of order (q2+q+1 )/d, and also unipotent elements with Jordan form [ J3].

Therefore, we may assume that H0 = P1, with q > 8. To show that (*) does not hold, we may as well assume that we are in one of the two cases (a) and (b) in (4) above (otherwise the conclusion is clear). Note that the condition H0 = P1 implies that G < TL3 (q) (that is, G does not contain a graph or graph-field automorphism). Also note that we may identify ^ with the set of 1-dimensional subspaces of the natural L3 (q)-module. Note that r > 3.

First assume that G = PGL3 (q), so d = 3 since we are assuming that G = G0. Here G has a cyclic maximal torus (x) of order q2 + q + 1. Then x is a derangement and thus (*) does not hold since q2 + q + 1 is not a prime power (note that (q2 + q + 1)3 = 3).

For the remainder, we may assume that q = pf and f > 2 (also recall that q > 8). In view of (4), Lemma 2.9(iii) implies that f is a 3-power. To deduce that (*) does not hold, we may assume that | G : G01 is a prime number. Since G < TL3 (q) and f is a 3-power, we may assume that |G : G01 = 3 and thus G = G0.(0) or G0. (50), where 0 is a field automorphism of order 3 and 5 is an appropriate diagonal automorphism diag(«1, rn2, rn3) e PGL3(q) (modulo scalars). In both cases, the result follows by applying Lemma 2.4. For example, 50 has more than one fixed point on so Lemma 2.4 implies that the coset G050 contains derangements, none of which has r -power order. In view of this final contradiction, we conclude that (*) does not hold if G = G0. □

Lemma 4.9. Theorem 2 holds ifG0 = U3 (q).

Proof. Set d = (3, q + 1) and observe that G0 contains elements of order (q2 — q + 1)/d and (q2 — 1)/d. Note that (q2 — 1)/d is a prime power if and only if q e {3, 5}. In order to show that (*) does not hold, we may assume that G = G0.

The cases q e {3, 4, 5} can be handled directly, using GAP, so for the remainder we will assume that q > 7. Let V be the natural G0-module, and let P1 (respectively

N1) be the G0-stabilizer of a 1-dimensional totally isotropic (respectively, non-degenerate) subspace of V. Note that N1 is a subgroup of type GU1(q) x GU2(q).

We can immediately rule out any subgroup H that does not contain elements of order (q2 - 1)/d, which means that we may assume H is of type P1, N1 or O3 (q) (q odd). In all three cases, elements of order (q2 - q + 1)/d are derangements. In addition, if H = N1 (respectively, SO3 (q)) then unipotent elements with Jordan form [ J3] (respectively, [ J2, J1]) are derangements. Finally, suppose that H = P1. Let m e Fq2 be an element of order q + 1 and set x = diag(1, m, m-1) e G (modulo scalars) with respect to an orthonormal basis for V. Then x does not fix a totally isotropic 1-space, whence x is a derangement of order q + 1. □

Having handled the low-dimensional groups, we are now in a position to complete the proof of Theorem 2 for linear and unitary groups.

Lemma 4.10. Theorem 2 holds ifG0 = Lfn(q).

Proof. We may assume that n > 4. Set d = (n, q - e) and e = (q - e)d. Let V be the natural G0-module. Let Pj be the G0-stabilizer of a totally isotropic i-dimensional subspace of V (so Pi is a maximal parabolic subgroup of G0, and we can take any i-space if e = +). Similarly, if e = - then let Nj denote the G0-stabilizer of an i-dimensional non-degenerate subspace of V (so Nj is of type GUj (q) x GUn-i (q)). In order to show that (*) does not hold, we may assume that G = G0. There are several cases to consider. Case 1 n = 2m and m > 4 - e is odd.

First assume that m > 5. As in the proof of [12, Proposition 3.11], let x e G be an element of order (qm+2 - e)(qm-2 - e)/e. Then lx| is not a prime power (see Lemma 2.7), and [28, Table II] indicates that x is a derangement unless one of the following holds:

(a) e = + and H = Pm-2 (or Pm+2);

(b) e = - and H = Nm-2.

In (a), any element of order ln(q) or ln-1(q) is a derangement, and elements of order ln(q) and l2(n-1)(q) are derangements in case (b).

Now assume m = 3, so (e, n) = (+, 6). Let x e G be an element of order (q6 - 1)/e, which is not a prime power by Lemma 2.8. Here x is a Singer element, and the main theorem of [3] implies that x is a derangement, unless H is a field extension subgroup, so we have reduced to the case where H is of type GL3 (q2) or GL2(q3). In this situation, elements of order l5 (q) are derangements, and so are unipotent elements with Jordan form [ J2, Jx4]. Case 2 n = 2m and m > 3 - e is even.

First assume that m > 4. Let x e G be an element of order (qm+1 - e)(qm-1 -e)/e. Then Lemma 2.7 implies that lx| is not a prime power, and from [28, Table II] we deduce that x is a derangement unless one of the following holds:

(a)' e = + and H = Pm-1 (or Pm+0;

(b); e = - and H = Nm-1.

To deal with these cases, we can repeat the argument in Case 1.

Now assume m = 2, so (e, n) = (+, 4). By applying the main theorem of [3], we deduce that elements of order (q4 — 1)/e are derangements unless H is a field extension subgroup of type GL2(q2). Moreover, since (q4 — 1)/e is not a prime power (see Lemma 2.8), we can assume that H is of type GL2(q2). Here elements of order t3 (q) and unipotent elements with Jordan form [ J2, J2] are derangements. Case 3 e = +, n = 2m + 1 and m > 2.

If G = L11(2), then any element of order 211 — 1 = 23 ■ 89 is a derangement, unless H is a field extension subgroup of type GL1(211), in which case elements of order 210 — 1 are derangements. For the remainder, we may assume that (n, q) = (11, 2).

Let x e G be an element of order (qm+1 — 1)(qm — 1)/e. By Lemmas 2.7 and 2.8, \x \ is not a prime power, so we may assume that H = Pm (see [28, Table II]). If m > 3, then elements of order tn(q) or tn—2(q) are derangements. Similarly, if m = 2 then we can take elements of order t5(q) or t4(q). Case 4 e = —, n = 2m + 1 and m > 4.

Fix x e G, where

(qm+1 + 1)(qm — 1)/e m even |X 1 (qm+2 + 1)(qm—1 — 1)/e m odd.

By Lemma 2.7, \x \ is not a prime power, and [28, Table II] indicates that x has fixed points only if H stabilizes a subspace U of V with dim U > 2. Therefore, we may assume that H has this property, in which case any element of order t2n (q) or tn—1(q) is a derangement. Case 5 e = — and n e {4, 5, 6, 7}.

First assume that n = 7. Let x e G be an element of order (q6 — 1)/d. Since \x \ is not a prime power, by inspecting the list of maximal subgroups of G (see [5, Tables 8.37 and 8.38]) it follows that we can assume that H e {P3, N1, SO7 (q)}. In all three cases, any element of order t14(q) is a derangement. Similarly, elements of order t10(q) are derangements, unless H = N1, in which case any unipotent element with Jordan form [ J7] is a derangement. The case n = 5 is entirely similar.

Next assume that n = 6. For now, let us assume that q e {2, 5}. Let x e G be an element of order (q5 + 1)/d. Then \x \ is not a prime power (see Lemma 2.8) and H = N1 is the only maximal subgroup of G containing such an element (see [28, p. 767]). Now, if H = N1 then any element of order (q6 — 1)/e is a derangement of non-prime power order.

Suppose that n = 6 and q e {2, 5}. The case q = 2 can be checked directly, using GAP for example, so let us assume that q = 5. Let x e G be an element of order (56 — 1)/e = 434. By inspecting the list of maximal subgroups of G (see [5, Tables 8.26 and 8.27]), we deduce that x is a derangement unless H is of type P3, GL3(52) or GU2(53), so we may assume that H is one of these subgroups, in which case any element of order t10(5) = 521 is a derangement. Suppose that H = P3. Fix an orthonormal basis for V and let y = diag(1, 1, w, w—1, w2, w—2) (modulo scalars), where w e F25 is an element of order 6. Then y is a derangement. Similarly, if H is of type GL3(52) or GU2(53), then any unipotent element with Jordan form [ J2, J^] is a derangement. This eliminates the case G = U6(5).

A very similar argument applies if n = 4. Here the cases q e {2, 3} can be checked directly, so let us assume that q > 4. Let x e G be an element of order (q3 + i)/d. Then \x \ is not a prime power (see Lemma 2.8) and again we reduce to the case H = Ni (see [28, p. 767]). We can now take any element of order (q4 - i)/e, which will be a derangement of non-prime power order. □

Next, we turn our attention to symplectic groups. Let G0 = PSpn (q) be a symplectic group with natural module V. As before, we will write Pi (respectively, Nj) for the Go-stabilizer of an i -dimensional totally isotropic (respectively, non-degenerate) subspace of V. We will also use n = m ± (n - m) to denote an orthogonal decomposition of V of the form V = Vi ± V2, where Vi is a non-degenerate m-space. Further, we will say that a semisimple element x e G0 is of type m ± (n - m) if it fixes such an orthogonal decomposition of V, acting irreducibly on Vi and V2. Similar notation is used in [7,12,28].

To begin with, we will assume that n > 6; the special case G0 = PSp4(q) will be handled separately in Lemma 4.i2.

Lemma 4.11. Theorem 2 holds ifG0 = PSpn (q) and n > 6.

Proof. Set d = (2, q - 1) and write n = 2m with m > 3. As before, we may assume that G = G0. Case 1 m odd.

The case (n, q) = (6, 2) can be handled directly (using GAP, for example), so let us assume that (n, q) = (6, 2). Let x e G be an element of order (qm + i)/d. If q is even, then Lemma 2.6 implies that \x \ is not a prime power, and it is easy to see that the same conclusion also holds if q is odd. By the main theorem of [3], x is a derangement unless one of the following holds:

(a) H is a field extension subgroup of type Spn/k (qk) for some prime divisor k of n;

(b) q is even and H = O~(q).

In (a), elements of order in-2(q) are derangements, and so are unipotent elements with Jordan form [J2, J^-2]. Similarly, if (b) holds then elements of order lm (q) are derangements, and so are semisimple elements of type (n - 2) ± 2 and order (qm-i + i)(q + 1). Case 2 m > 6 even.

First assume that q is odd. Let x e G be a semisimple element of type (n - 4) ± 4, so

qm-2 + i if m = 0 (mod 4)

|x 1 = [ (qm-2 + i)(q2 + i)/2 if m = 2 (mod 4).

Clearly, if m = 2 (mod 4) then \x\ is divisible by i4(q) and in-4(q), so \x\ is not a prime power. The same conclusion also holds if m = 0 (mod 4) (see Lemma 2.6). By [7, Proposition 5.i0], we may assume that H is of type N4 or Spn/2(q2). In both cases, elements of order in-2(q) are derangements. In addition, unipotent elements with Jordan form [ Jn] (respectively, [ J2, J^-2]) are derangements if H is of type N4 (respectively, Spn/2(q2)).

Now assume that q is even. Let x e G be a semisimple element of type (n—2) ± 2 and order qm—1 + 1. Now Lemma 2.6 implies that |x | is not a prime power, and by applying the main theorem of [32] we deduce that x is a derangement unless H e {N2, O+(q)}. In both cases, elements of order ln (q) are derangements. Also, unipotent elements with Jordan form [Jn] are derangements if H = N2. Now, if H = O+(q) then let y e G be a block-diagonal element of the form y = [y1, y2] (with respect to an orthogonal decomposition n = (n — 2) ± 2), where y1 e Spn—2 (q) has order lm—1(q) and y2 e Sp2 (q) has order q + 1. Then y is a derangement and the result follows. Case 3 m = 4.

The case q = 2 can be checked directly, so we may assume that q > 3. If q is even, then we can repeat the relevant argument in Case 2. Now assume q is odd. Let x e G be a semisimple element of type 6 ± 2 and order q3 +1. By Lemma 2.6, |x | is not a prime power. The maximal subgroups of G are listed in [5, Tables 8.48 and 8.49], and we deduce that x is a derangement unless H is of type N2, GU4(q) or L2 (q3) (in the terminology of [5,38], the latter possibility is an almost simple irreducibly embedded subgroup in the collection S). In each of these exceptional cases, any element of order (q4 + 1)/2 is a derangement. In addition, if H = N2 then unipotent elements with Jordan form [ J8] are derangements. Similarly, if H is of type GU4(q) or L2(q3) then elements with Jordan form [J2, Jf] are derangements. □

Lemma 4.12. Theorem 2 holds ifG0 = PSpn (q).

Proof. We may assume that n = 4. The result can be checked directly if q < 7, so let us assume that q > 8.

First assume that q is odd. In terms of an orthogonal decomposition 4 = 2 ± 2, let x = [x1, x2] e G (modulo scalars) be an element of order p(q + 1), where x1 e Sp2(q) is a unipotent element of order p, and x2 e Sp2(q) is irreducible of order q + 1. By inspecting the list of maximal subgroups of G (see [5, Tables 8.12 and 8.13]), we deduce that x is a derangement unless H is of type P1 or Sp2(q) i S2. In both of these cases, any element of order l4(q) is a derangement. Similarly, unipotent elements with Jordan form [ J4] are derangements if H is of type Sp2(q )iS2. Finally, suppose that H = P1.NowSp2(q) hasprecisely y(q+1)/2 > 2 distinct classes of elements of order q + 1 (where y is the Euler totient function); if y1, y2 e Sp2 (q) represent distinct classes, then y = [y1, y2] e G (modulo scalars) is a derangement since it does not fix a totally isotropic 1-space.

Now assume q is even. As above, let x e G be an element of order 2(q + 1). The maximal subgroups of G are listed in [5, Table 8.14], and we see that x is a derangement unless H is of type P1, Sp2 (q) i S2 = O+(q) or O—(q). For H = P1, we can repeat the argument in the q odd case, so let us assume that H = O4(q). If e = + then any element of order l4(q) is a derangement, and we can also find derangements of order 4 (with Jordan form [J4]), since there are two conjugacy classes of such elements in G, but only one in H. Finally, if e = — then we can find derangements of order 2 (with Jordan form [ J|]; these are a2-type involutions in the sense of Aschbacher and Seitz [1]), and also derangements of order q + 1 of the form [y1, y2] as above. □

To complete the proof of Theorem 2, we may assume that G0 = PQfn (q) is an orthogonal group, where n > 7. The low-dimensional groups with n e {7, 8} require special attention. We extend our earlier notation for orthogonal decompositions by writing m± to denote a non-degenerate m-space of type ± (when m is even). Similarly, we write N± for the G0-stabilizer of such a subspace of the natural G0-module V. If q is even, we will also adopt the standard Aschbacher-Seitz notation for involutions (see [1]).

Lemma 4.13. Theorem 2 holds ifG0 = ^7(q).

Proof. We may assume that G = G0. The case q = 3 can be checked directly, so we may assume that q > 5 (recall that q is odd). Let x e G be an element of order (q3 +1 )/2, which is not a prime power. By [7, Proposition 5.20], x is a derangement unless H = N-, in which case any element of order l3(q) is a derangement, and so are unipotent elements with Jordan form [J7]. □

Lemma 4.14. Theorem 2 holds ifG 0 = P^+(q).

Proof. As usual, we may assume that G = G0. Let V be the natural module for G0. The case q = 2 can be checked directly, using GAP. Next suppose that q = 3. Let x e G be an element of order 20, fixing a decomposition of V of the form 8 = 4- ± 4-. As indicated in [7, Table 3], x is a derangement unless the type of H is one of the following:

P4, 07(3), O-(3) i S2, GU4(3), Sp4(3) ® Sp2(3)

where 07 (3) is irreducible and P4 is the stabilizer in G of a maximal totally singular subspace of V.

By considering elements of order 14, we can immediately eliminate the cases P4, 04- (3) i S2 and Sp4(3) ® Sp2(3). Similarly, G contains derangements of order 15 if H is of type GU4(3). Finally, suppose that H is an irreducible subgroup of type 07(3). To see that (*) does not hold, we may replace H by a conjugate HT, where t e Aut (G) is an appropriate triality graph automorphism such that HT is the stabilizer in G of a non-degenerate 1-space. For this reducible subgroup, elements with Jordan form [ J4] are derangements, and so are elements y e G of order 5 of the form y = yZ, where Z = Z(^+(3)) and CV(y) is trivial (the eigenvalues of y (in F34) are the nontrivial fifth roots of unity, each occurring with multiplicity 2).

For the remainder, we may assume that q > 4. Let x e G be an element of order (q3 + 1)/(2, q - 1), fixing an orthogonal decomposition 8 = 6- ± 2-. Then \x | is not a prime power, and x is a derangement unless H is of type N- or GU4 (q) (see [28, p. 767]). In both of these cases, elements of order l3 (q) are derangements. Similarly, if q is odd then unipotent elements with Jordan form [J7, J\] are also derangements. Finally, if q is even and H is of type N- (respectively, GU4 (q)) then unipotent elements with Jordan form [ J|] (respectively, [ J%, Jx4]; ^-involutions in the terminology of [1]) are derangements. The result follows. □

Lemma 4.15. Theorem 2 holds ifG 0 = P^-(q).

Proof. Again, we may assume that G = G0. If q < 3 then we can use GAP to verify the result, so let us assume that q > 4. The maximal subgroups of G are listed in [5, Tables 8.52 and 8.53]. By considering elements of order l8(q) and l6(q), we can eliminate subfield subgroups, together with the reducible subgroups of type P2, P3 ,N—, N3 and N+. Similarly, elements of order l8 (q) and l4 (q) are derangements if H is a non-geometric subgroup of type L3 (q). Therefore, to complete the proof, we may assume that H is either a field extension subgroup of type O—(q2), or a reducible subgroup of type P1, N+, O7(q) (q odd) or Sp6(q) (q even).

If H is of type O—(q2), then elements of order l6(q) are derangements, as well as unipotent elements with Jordan form [ J7, J1] if q is odd, and unipotent elements with Jordan form [ Jf] (a2-type involutions) if q is even. Similarly, if H = P1 or N+ then elements of order l8 (q) and l3 (q) are derangements (note that an element of order l3 (q) fixes a 2—-space, but not a 2+-space). Finally, suppose H is of type O7(q) (q odd) or Sp6 (q) (q even). In both cases, elements of order i8(q) are derangements. In addition, there are derangements with Jordan form [J5, J3] (q odd) and [ J|] (q even). □

Lemma 4.16. Theorem 2 holds ifG0 = P^n (q).

Proof. We may assume that G = G0 and n > 9. We have three cases to consider. Case 1 G0 = (q) and n > 10.

Write n = 2m and first assume that m is odd. Let x e G be an element of order (q(m—1)/2 + 1)(q(m+1)/2 + 1)/(4, q — 1), fixing an orthogonal decomposition of the form (m + 1)— ± (m — 1) —. Then Lemma 2.7 implies that |x| is not a prime power, so by [7, Proposition 5.13] we may assume that H = N——1. In this situation, elements of order ln—2(q) are derangements, and so are unipotent elements with Jordan form [Jn—1; J1] (q odd) or [Jn—2, J2] (q even).

A similar argument applies if m is even. Here we take an element x e G of order (q(m—2)/2 + 1)(q(m+2)/2 + 1/4, q — 1), fixing a decomposition (m + 2)— ± (m — 2)—. Then |x | is not a prime power, and [7, Proposition 5.14] implies that x is a derangement unless H is of type Nm——2 or On+/2(q2). In the former case, we complete the argument as above, so let us assume that H is of type O+/2(q2). Any element of order ln—2(q) is a derangement, and so are unipotent elements with Jordan form [ Jn—1; J1] if q is odd. Finally, if q is even then a2-type involutions are derangements.

Case 2 G0 = (q) and n > 10.

Again, write n = 2m. First assume that m > 11. Let x e G be an element of order

lcm(qm—5 + 1, q3 + 1, q2 + 1)/(2, q — 1)

fixing a decomposition (n — 10)— ± 6— ± 4—. Then |x| is not a prime power, and [7, Proposition 5.16] implies that x is a derangement unless H is of type N—, N— or N+0. In each of these cases, it is clear that elements of order in (q) and ln—2(q) are derangements.

Next suppose that m e {5, 6, 7, 9, 10}. Let x e G be an irreducible element of order (qm + 1)/(2, q — 1). We claim that |x| is not a prime power (here we

require m = 8). If q is even, this follows immediately from Lemma 2.6, so let us assume that q is odd. Suppose m = 5 and q5 + 1 = 2re for some prime r and positive integer e. Then (q + 1)(q4 - q3 + q2 - q + 1) = 2re and r = l10(q). Therefore, q + 1 = 2 is the only possibility, which is absurd. Similarly, if m = 6 andq6 +1 = (q2 + 1)(q4 -q2 + 1) = 2re,then r = l12(q) andq2 + 1 = 2, which is not possible. The other cases are entirely similar. Now, by the main theorem of [3], x is a derangement unless H is a field extension subgroup of type O-/k (qk) (k a prime divisor of n, n/k > 4 even) or GUn/2(q) (n/2 odd). In both cases, elements of order ln-2(q) are derangements. In addition, there are unipotent derangements; take [ Jn-i, J1] if q is odd, an a2-involution if q is even and H is of type 0-/k (qk), and a c2-involution if q is even and H is of type GUn/2(q).

Finally, let us assume that m = 8. As in [28, Table II], let x e G be an element of order lcm(q5 + 1, q2 + 1, q + 1)/(2, q - 1), fixing an orthogonal decomposition of the form 10- ± 4- ± 2-. Note that \x \ is divisible by li0 (q) and l4(q), so it is not a prime power. As indicated in [28, Table II], x is a derangement unless H is of type N-, N- or N+. In each of these cases, elements of order 116 (q) and 114 (q) are derangements.

Case 3 G0 = Un (q) and n > 9 is odd.

Write n = 2m + 1 and note that q is odd. First assume m is odd. Let x e G be an element of order

lcm(q(m+1)/2 + 1, q(m-1)/2 + 1)/2 = (q(m+1)/2 + 1)(q(m-1)/2 + 1)/4,

fixing an orthogonal decomposition (m + 1)- ± (m - 1)- ± 1. Note that \x\ is divisible by lm+1(q) and lm-1(q), so \x \ is not a prime power. Let H be a maximal subgroup of G containing x. By carefully applying the main theorem of [32], we deduce that H e {Nm+1, Nm_1, }. For example, the order of x rules out subfield subgroups and imprimitive subgroups of type 01(q) i Sn (see [7, Remark 5.1(i)]), and the dimensions of the irreducible constituents of x are incompatible with field extension subgroups of type 0n/k (qk). Now, if H is one of these reducible subgroups, then elements of order ln-1(q) and ln-3(q) are derangements. The result follows.

A similar argument applies if m is even. Here we take x e G to be an element of order

lcm(q(m+2)/2 + 1, q(m-2)/2 + 1)/2 = (q(m+2)/2 + 1)(q(m-2)/2 + 1)/4,

fixing a decomposition (m + 2)- ± (m - 2)- ± 1. We claim that \x\ is not a prime power. This is clear if m > 6, or if m = 4 and q is not a Mersenne prime, since \x\ is divisible by lm±2(q). Suppose that m = 4 and q is a Mersenne prime. If q = 3 then \x \ =28 and the claim holds, and if q > 3 then \x \ is divisible by 2 and l6(q). This justifies the claim. Using [32] one can check that the only maximal subgroups of G containing x are of type Nm+2, N--2 or N+„, so we may assume that H is one of these subgroups. Here we observe that elements of order ln-1(q) are derangements, and so are unipotent elements with Jordan form [ Jn]. □

This completes the proof of Theorem 2.

5. Affine groups

Let G be a finite primitive permutation group. By Theorem 1, if (*) holds then G is either almost simple or affine. In the previous section, we determined all the almost simple examples, and we now turn our attention to the affine groups with property (*). Our main aim is to prove Theorem 4.

Let G = HV < AGL(V) be a finite affine primitive permutation group with point stabilizer H = G0 and socle V = (Zp )k. As an abstract group, G is a semidirect product of V by H. Therefore, we will begin our analysis by studying the structure of a general semidirect product G = H k N with property (*), so G is a finite group, H is a proper subgroup and N is a normal subgroup of G such that G = HN and H n N = 1.

We will need some additional notation. If K is a subgroup of G and g e G, then we set

[K, g] = {[k, g]=k—1 g—1kg : k e K}. We also write K * for the set of all nontrivial elements of K. Lemma 5.1. Let G = H k N. The following hold:

(i) CG(x) = CH(x)CN(x)forallx e H.

(ii) IfK < H, then K n Kn = CK (n) for all n e N*.

(iii) IfN isabelian, then AH(G) = {tv : t e H,v e N\[N, t]}.

(iv) If property (*) holds, then N is an r-group.

Proof. First consider part (i). The result is clear if x = 1, so assume that x e H*. The inclusion CH (x)CN(x) c CG(x) is clear. Conversely, suppose that g = hn e CG (x) where h e H, n e N. Then hnx = xhn. Multiplying both sides by (xh)—1 = h—1 x—1, we obtain

hnxh—1 x—1 = (xh)n(xh)—1

which implies that

(hnh—1)(hxh—1 x—1) = (xh)n(xh)—1.

Since n e N ^ G and h, x e H, we deduce that

hxh—1 x—1 = (hn—1h—1)(xh)n(xh)—1 e H n N = 1

so h e CH (x). Since hnx = xhn = hxn, we deduce that nx = xn and thus n e CN(x). Therefore, g = hn e CH(x)CN(x) and part (i) follows.

For part (ii), let K < H and let n e N*. Assume that y e K n Kn. Then y = kn e K for some k e K, so

k—1 y = k—1n—1kn = (k—1n—1k)n e K n N = 1,

which implies that kn = nk and y = k, or equivalently y e CK (n). Therefore, K n Kn < CK (n). Conversely, if y e CK(n) then y e K and y = n—1 yn e Kn, so y e K n Kn and thus Ck (n) < K n Kn. The result follows.

Now consider part (iii). Assume that N is abelian. Set

r := {tv : t e H,v e N\[N, t]}.

First we claim that r c A H (G). Let g e r, say g = hn with h e H and n e N\[N, h]. Seeking a contradiction, suppose that g e AH(G). Then g e H1 for some t e G. Since t e G = HN, we may write t = h1m1 with h1 e H and m1 e N. It follows that g e Ht = Hm1, so m 1 gm—1 e H. Let m := m—1 e N. Then m—1 gm = m—1hnm = hmn e H (note that nm = mn since N is abelian) and thus h—1hmn = [h, m]n e H. We also have [h, m]n = (h—1m—1h)mn e N, so [h, m]n e H n N = 1 and we deduce that n = [m, h] e [N, h], contradicting our choice of n. We have now shown that r c AH (G). Conversely, suppose that g = hn e AH (G) with h e H, n e N .We claim that n e N \[N, h]. Seeking a contradiction, suppose that n e [N, h], say n = [m, h] for some m e N. Then m—1(hn)m = h, or equivalently gm e H, which is a contradiction.

Finally, let us turn to part (iv). If x e N * then xG c N ,so xG n H c N n H = 1 and thus xG n H = 0 since x = 1. Therefore N * c A H (G). In particular, if every element of AH (G) is an r-element (for some fixed prime r), then every element of N is also an r-element and thus N is an r-group. □

Lemma 5.2. Let G = H k N, where N is an r-group for some prime r. Then the following are equivalent:

(i) Property (*) holds.

(ii) CH (n) = H n Hn is an r-group for all n e N *.

(iii) CN (x) = 1 for every nontrivial r'-element x e H. In other words, every nontrivial r '-element of H induces a fixed-point-free automorphism of N via conjugation.

Proof. First we will show that (i) implies (ii). Suppose that (*) holds. Let n e N*. We claim that CH (n) is an r-group. Notice that Ch (n) = H n Hn by Lemma 5.1 (ii). Seeking a contradiction, suppose that |CH (n)| is divisible by a prime s = r. Choose y e CH(n) with |y| = s and let g := ny = yn e G. We claim that g e AH(G), which would be a contradiction since |g| = |n|s is not a power of r. Assume that g e AH(G), so g e Ht for some t e G. Since G = HN, we may assume that t e N. Then

gs = (ny)s = nsys = ns e H

and ns e N ^ G ,sot (ns )t—1 e H n N = 1 and thus ns = 1, which is not possible since n is a nontrivial r-element. Therefore, g = ny e AH(G) as required.

Next we will show that (ii) implies (i). Suppose that CH (n) is an r-group for all n e N*. Let g e AH(G), say g = hn with h e H and n e N*. We claim that g is an r-element. Seeking a contradiction, suppose that m := |g| is divisible by a prime s = r. Set x := gm/s e G and let S be a Sylow s-subgroup of H. Then |x| = s and S is also a Sylow s-subgroup of G since |G : H| = |N| is coprime to s. By Sylow's theorem, xt e S < H for some t e G. Since gG c AH (G), replacing g by gt we may assume that x e H. Then g e CG (x) = CH (x)CN (x) by Lemma 5.1(i).

Suppose that CN (x) = 1, say 1 = n e CN (x). Then x e CH (n), but this is a contradiction since \x\ = s and we are assuming that CH(n) is an r-group. Therefore, CN (x) = 1 and thus CG (x) = CH (x). Hence g e CH (x) < H, which contradicts the fact that g e AH (G). This final contradiction shows that g is an r-element, so (*) holds.

Now let us show that (ii) implies (iii). Suppose that CH (n) is an r -group for all n e N*. Let x e H* be an r'-element. We claim that CN(x) = 1. Seeking a contradiction, suppose that 1 = n e CN(x). Then x e CH(n), so \CH(n)\ is divisible by \x\, which is not an r-power. This contradicts the assumption that CH (n) is an r-group.

To complete the proof, it remains to show that (iii) implies (ii). Suppose that CN(x) = 1 for every nontrivial r'-element x e H. Let n e N*. If CH(n) is not an r-group, then there exists an element x e CH (n) with \x \ = s, where s = r is a prime. Therefore, 1 = n e CN (x), which is not possible since CN (x) = 1. □

We are now in a position to prove Theorem 4.

Proof of Theorem 4. Let G = HV < AGL (V) be a finite affine primitive permutation group with point stabilizer H = G0 and socle V = (Zp )k, where p is a prime and k > 1. If property (*) holds, then r = p and Lemma 5.2 implies that no nontrivial r'-element of H has fixed points on V \{0}. Therefore, the pair (H, V) is r'-semiregular in the sense of [19]. Conversely, if r = p and (H, V) is r'-semiregular, then CV (x) = 0 for every nontrivial r'-element x e H, so Lemma 5.2 implies that G has property (*). □

Remark 5.3. Note that the equivalence of (i) and (ii) in Lemma 5.2 implies that an affine group G = HV < AGL(V) has property (*) if and only if every two-point stabilizer in G is an r-group.

If G = HV < AGL(V) is an affine group (with V = (Zr)k) and r £ n(H), then G is a Frobenius group and property (*) clearly holds. Therefore, we may focus on the case where r e n(H). As noted in the Introduction, detailed information on r'-semiregular pairs (H, V) was initially obtained by Guralnick and Wiegand in [33, Section 4], where this notion arises naturally in their study of the multiplicative structure of field extensions. Similar results were established in later work by Fleischmann et al. [19]. In both papers, the main aim is to determine the structure of H. For solvable affine groups, we have the following result (in the statement, Ori (Y) denotes the largest normal r'-subgroup of Y):

Proposition 5.4. Let G = HV < AGL(V) be a finite affine primitive permutation group with point stabilizer H = G0 and socle V = (Zr )k. Assume that H is solvable andr e n(H). Then G has property (*) only if H = X x Y or (X x Y):2, where X e {1, SL2(3)}, Y = O r 1 (Y)R and R is a Sylow r-subgroup ofY.

Proof. This follows from [19, Theorem 2.1]. □

The main result for a perfect group H is Proposition 5.5 below (see [19, Theorem 4.1]; also see [33, Theorem 4.2]). In part (iv), S = {5, 13, 37, 73,...} is the set of all primes s satisfying the following conditions:

(a) s = 2a3b + 1, where a > 2 and b > 0;

(b) (s + 1)/2 is a prime.

It is not known whether or not S is finite.

Proposition 5.5. Let G = HV < AGL(V) be a finite affine primitive permutation group with point stabilizer H = G 0 and socle V = (Zr )k. Assume that H is perfect and r e n(H). Then G has property (*) only if one of the following holds:

(i) H = SL2 (ra), where a > 1 andra > 3;

(ii) H = 2B2(22a+1), r = 2 and a > 1;

(iii) H = 2B2(22a+1) x SL2(22b+1), r = 2 and a, b > 1 such that (2a + 1, 2b + 1) = 1;

(iv) H = SL2(s), r = 3 ands e S U{7, 17}.

For instance, H = SL2(7) has a 12-dimensional faithful, irreducible module V over F3, and the corresponding affine group G = HV has property (*) (with E(G) = {3, 9}). In the general case, we refer the reader to [19, Theorem 6.1] for a detailed description of the structure of H.

Finally, let us suppose that G = HV < AGL(V) is a finite affine primitive permutation group with property (*). Set

E(G) = Eh(G) = {|x| : x e Ah(G)}.

Can we determine when E(G) = {r}? In order to address this question, let P be a Sylow r-subgroup of G. Then V < P since V is a normal r-subgroup of G, and we have P = (H n P)V = KV with K := H n P. Note that P = KV is a semidirect product.

Proposition 5.6. Let G = HV < AGL(V) be a finite affine primitive permutation group with point stabilizer H = G 0 and socle V = (Zr )k. Assume that property (*) holds. Let P be a Sylow r-subgroup of G and set K = H n P. Then the following hold:

(i) P = KV is a transitive permutation group on P/K.

(ii) A(G) = UgSG Ak(P)g andE(G) = Ek(P).

Proof. As above, P = KV is a semidirect product. For part (i), it suffices to show that the core L of K in P is trivial. We have L < K < H and L ^ P, so [L, V] < L n V < K n V = 1 and thus L < CK(V) < CH(V) = 1 (here we are using the fact that V is a faithful irreducible H-module). This proves (i).

Now consider part (ii). Clearly, it suffices to show that the first equality holds. By applying Lemma 5.1(iii) we have

AK(P) = {tv : t e K,v e V\[V, t]}.

Since K < H, a further application of Lemma 5.1 (iii) (this time for G = HV) shows that AK(P) c A(G). As A(G) is a normal subset of G, it follows that

U Ak(P)g c A(G).

Since property (*) holds, every g e A(G) is an r-element, so some G-conjugate of g is in P. Without loss of generality, we may assume that g e P = KV. By Lemma 5.1 (iii) we have g = hn, with h e H and n e V \[V, h]. Moreover, since V < P and g e P, we have h = gn-1 e H n P = K. Therefore, by applying Lemma 5.1(iii) once again, we conclude that g = hn e AK (P), so A(G) = UgeG AK (P)g and the proof is complete. □

Now, if we assume that G = HV has property (*), then part (ii) of Proposition 5.6 implies that E(G) = {r} if and only if EK (P) = {r}. Clearly, if P has exponent r, then EK (P) = {r}. Conversely, if EK (P) = {r} with r = 2 or 3, then a theorem of Mann and Praeger [43, Proposition 2] implies that P has exponent r. In fact, for this specific transitive group P we can show that the same conclusion holds for any prime r (we thank an anonymous referee for pointing this out).

Theorem 5.7. Let G = HV < AGL(V) be a finite affine primitive permutation group with point stabilizer H = G0 and socle V = (Zp )k, where p is a prime and k > 1. Then every derangement in G has order r, for some fixed prime r, if and only if r = p and the following two conditions hold:

(i) Every two-point stabilizer in G is an r-group;

(ii) A Sylow r-subgroup of G has exponent r.

Proof. Let P be a Sylow r-subgroup of G. First assume that r = p and (i) and (ii) hold. By (i), the pair (H, V) is r'-semiregular so Theorem 4 implies that property (*) holds. Therefore, E(G) = Ek (P) by Proposition 5.6(ii) (with K = H n P) and thus condition (ii) implies that E (G) = {r} as required.

Conversely, let us assume that E(G) = {r },so r = p and property (*) holds. By Theorem 4, every two-point stabilizer in G is an r-group and so it remains to show that P has exponent r. Seeking a contradiction, suppose that exp(P) > r2. Note that r divides \H\. Let Q be a Sylow r-subgroup of H. Let x e P be an element of order r2 and observe that x belongs to a conjugate of H (since E(G) = {r}), so exp(Q) > r2. We may assume x e H and we choose an element v e V\[V, x]. Then xv e G is a derangement by Lemma 5.1(iii), but \xv\ > r2 so we have reached a contradiction. □

Acknowledgements. This work was done while the second author held a position at the CRC 701 within the project C13 'The geometry and combinatorics of groups', and he thanks B. Baumeister and G. Stroth for their assistance. Part of the paper was written during the second author's visit to the School of Mathematics at the University of Bristol and he thanks the University of Bristol for its hospitality. Burness thanks R. Guralnick for helpful comments. Both authors thank an anonymous referee for suggesting several improvements to the paper, including a simplified proof of Proposition 4.2 and a proof of Theorem 5.7.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), 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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