Result.Math. 52 (2008), 75–89 c 2008 Birkh¨  auser Verlag Basel/Switzerland 1422-6383/010075-15, published online July 21, 2008 DOI 10.1007/s00025-008-0295-3

Results in Mathematics

The Doubly Transitive t-Parallelisms Norman L. Johnson and Alessandro Montinaro Abstract. Johnson showed that the only doubly transitive parallelisms of P G(3, q) are exactly the two regular parallelisms in P G(3, 2). This article completely generalizes this result to doubly transitive t-parallelisms of arbitrary finite projective spaces. Mathematics Subject Classification (2000). 51E23, 51A40. Keywords. Parallelisms, t-parallelism, doubly transitive group.

1. Introduction A ‘parallelism’ of a projective space is an equivalence relation on the set of lines that satisfies the Euclidean parallelism postulate. Hence, the equivalence classes are sets of lines that cover the line set. To be clear, we also use the term ‘line parallelism or 1-parallelism’ for this concept. Although there are interesting line parallelisms of infinite projective spaces, such as line parallelisms over the real quaternions, in this article, we shall be concerned with finite parallelisms only. For a line parallelism in P G(3, q), the associated equivalence classes are called ‘spreads’ and each spread corresponds to an affine translation plane of order q 2 and kernel containing GF (q). A ‘regular spread’ is a spread with the property that given three mutually distinct lines of the spread, the unique regulus of P G(3, q) containing these three lines is also contained in the spread. Regular spreads in P G(3, q) correspond to Desarguesian affine planes of order q 2 . Given any parallelism in P G(3, q 2 ), one of whose spreads is regular, Beutelspacher [1] has shown that there is an associated line parallelism in P G(23 −1, q). This construction may be repeated so that there are line parallelisms in any P G(2z − 1, q). Given a regular parallelism in P G(3, q), (i.e., all spreads are regular), a construction of Walker [7] and Lunardon [6] shows that there is an associated affine The authors wish to thank the University of Iowa for support on this research.

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translation plane of order q 4 with spread in P G(7, q) that admits exactly 1 + q + q 2 derivable nets containing a regulus net of degree 1 + q. Note that the number of spreads of a parallelism in P G(3, q) is also 1 + q + q 2 . A duality of a parallelism in P G(3, q) produces another parallelism ‘the dual parallelism’ and when the original parallelism is regular, there are two associated translation planes of order q 4 with spreads in P G(7, q). There are exactly two regular parallelisms in P G(3, 2), dual to each other. These parallelisms produce and are equivalent to the translation planes of order 16 of Johnson–Walker and Lorimer–Rahilly. The associated translation planes admit a collineation group isomorphic to P SL(2, 7) that actually acts doubly transitively on the set of 1 + 2 + 22 derivable nets. This group action induces a collineation group acting on the regular parallelism and acting doubly transitive on the spreads of the parallelism. Pointing to the unique quality of both these two affine planes of order 16 and the two regular parallelisms of P G(3, 2), the first author proved the following theorem. Theorem 1.1 (Johnson [3]). If P is a parallelism of P G(3, q) that admits a collineation group acting doubly transitively on the spreads of P then q = 2 and P is one of the two regular parallelisms in P G(3, 2). In this article, we generalize these ideas by considering groups on t-parallelisms on vector spaces and projective spaces. Definition 1.1. A ‘t-spread’ of a finite vector space V over GF (q) is a partition of the non-zero vectors by t + 1-dimensional GF (q)-subspaces. Any t-spread of V induces a corresponding partition of the associated projective space, as the lattice of vector subspaces, by projective t-spaces. We use the term ‘t-spread’ in both the vector space and projective space settings. Remark 1.1. It is possible to have t-spreads when V has dimension (t + 1)k over GF (q). In general, t-spreads produce ‘translation Sperner spaces’, and if k = 2, produce translation planes, by defining ‘lines’ to be the translates of the spread subspaces. Definition 1.2. A ‘t-parallelism’ of a vector space of dimension (t+1)k over GF (q) is a partition of the (t+1)-vector subspaces by t-spreads. Equivalently, a t-parallelism of P G((t + 1)k − 1, q) is a partition of the t-dimensional projective subspaces by t-spreads. Actually, there are no known examples of t-parallelisms for t > 1, although there are many varieties of t-spreads and partial t-parallelisms (sets of t-spreads mutually disjoint on t-subspaces). We establish the following result, lending even more exceptional quality to the two parallelisms of P G(3, 2) and the associated Johnson–Walker and Lorimer–Rahilly planes of order 16. Theorem 1.2. Let P be a t-parallelism of P G(n, q) that admits a collineation group acting doubly transitive on the t-spreads of P. Then t = 1, n = 3, q = 2, and P is one of the two regular parallelisms in P G(3, 2).

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2. Background We begin by working out of the combinatorics of t-parallelisms. Note that a tspread in P G(n, q) occurs only if t+1 | n+1 (see [2]). Hence n+1 = (s+1)(t+1) for some positive integer s. Furthermore, if W is a t-spread of P G((s + 1)(t + 1) − 1, q), then q (s+1)(t+1) − 1 . |W| = q t+1 − 1 If P is a t-parallelism of P G(n, q), it is easily seen that |P| =

t  q (s+1)(t+1) − q j q t+1 − q j j=1

(2.1)

(see [2]). Using Theorem 1.1, we may henceforth consider either 2-parallelisms in P G(3, q) or t-parallelisms in P G(n, q) with t ≥ 1 and n > 3. Our main tool in the proof of our main result is the list of the possible finite doubly transitive groups, which depends on the classification theorem of finite simple groups for its completeness. Theorem 2.1. Let H be a finite group with a 2-transitive permutation representation of degree v and let S = soc(H) be the socle of H. Then one of the following occurs: 1. S is (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) 2. S is

non-Abelian simple, and S ≤ H ≤ AutS where S and v are as follows: Av with v ≥ 5; P SL(d, l), d ≥ 2, v = (ld − 1)/(l − 1) and (d, l) = (2, 2), (2, 3); P SU (3, l), v = l3 + 1, l > 2; Sz(l), v = l2 + 1, l = 22e+1 > 2; 2 G2 (l) , v = l3 + 1, l = 32e+1 ; Sp(2n, 2), n ≥ 3, v = 22n−1 ± 2n−1 ; P SL(2, 11), v = 11; Mathieu groups Mv , v = 11, 12, 22, 23, 24; M11 , v = 12; A7 , v = 15; HS (Higman–Sims group), v = 176; Co3 (Conway’s smallest group), v = 276. an elementary Abelian group of order v = pd , where p is a prime.

See for example [4]. Definition 2.1. A finite 2-transitive group is said either ‘almost simple’ or of ‘affine type’ according to whether its socle is a non-Abelian simple group or an elementary Abelian p-group, for some prime p, respectively. We also require the theorem of Zsigmondy.

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Theorem 2.2 (Zsigmondy, see, e.g., [5, 5.2.14, p. 182]). Let q and n be integers with q ≥ 2 and n ≥ 3. Then provided (q, n) = (2, 6), there is a prime u such that u | q n − 1 but u does not divide q i − 1 for i < n. Notation 2.1. Any such u above shall be denoted by qn . The following results follow immediately from the previous theorem (see, e.g., Kleidman–Liebeck [5, 5.2.15, p. 182]). Theorem 2.3. Assume that q ≥ 2, n ≥ 3 and (q, n) = (2, 6). (i) If qn | q m − 1 then n | m. (ii) qn ≡ 1 mod n.

3. The almost simple case In this section, we assume that G is an almost simple group. Let G be a subgroup of P ΓL((s + 1)(t + 1), q) acting doubly transitively on a t-parallelism P of P G((s + 1)(t + 1) − 1, q). Hence |P| = v, where v is the degree of a doubly transitive permutation representation. Therefore, we have to solve the following Diophantine equation: t  q (s+1)(t+1) − q j = v, q t+1 − q j j=1

(3.1)

by (2.1). Furthermore, since it must be that   G ≤ P ΓL (s + 1)(t + 1), q , it follows that the minimum degree of G is at most (s + 1)(t + 1) ,

(3.2)

which provides a strong restriction on the previous Diophantine equation. Lemma 3.1. v ≥ 31 and v = 63, 176 or 276. t (s+1)(t+1)−j 2s+1 2s+1 Proof. Clearly j=1 q qt+1−j −1 −1 ≥ q q−1−1 . On the other hand, q q−1−1 ≥ 25 −1 as s > 1 by our assumptions. Hence v ≥ 31 by (3.1). Finally, it is a straightforward computation to show that t  q (s+1)(t+1)−j − 1 q t+1−j − 1 j=1

cannot be equal to either 63, 176, or 276. Lemma 3.2. v = ur + 1 where u is a prime and r ≥ 1.



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Proof. Assume that v = ur + 1 where u is a prime and r ≥ 1. Then t  q (s+1)(t+1) − q j = ur + 1 . t+1−j − q j q j=1

First we note that it is easy to check that t  q (s+1)(t+1)−j − 1 ≡ 1 mod q . q t+1−j − 1 j=1

Hence, if

t  q (s+1)(t+1)−j − 1 = ur + 1 = kq + 1 t+1−j − 1 q j=1

q divides ur say q = uz . Then ur is a sum of terms of the form βuc , where 1 ≤ β < u and is larger of equal to any of these terms. By dividing by the smallest uc , it follows we have ur−c is a sum of say β < u and a set of elements each of which is divisible by u. This is a contradiction, so we must have that ur is a sum of exactly one term. In other words, t  uz(s+1)(t+1)−j − 1 −1 uz(t+1−j) − 1 j=1 has exactly one term ur−c . This means that j = 1 = t = s, which we have assumed is not the case. This completes the proof of the lemma.  Lemma 3.3. The subgroup soc(G) cannot be isomorphic to Av , v ≥ 5. Proof. Assume that soc(G) ∼ = Av , v ≥ 5. Actually v ≥ 31 by Lemma 3.1. On the other hand soc(G) ≤ P ΓL((s + 1)(t + 1), q). By Kleidman–Liebeck [5, 5.3.7 (i), p. 186], for v ≥ 9 then the minimum degree of Av is v−2. Therefore, (s+1)(t+1) ≥ v − 2. Then t  q (s+1)(t+1)−j − 1 ≤ (s + 1)(t + 1) + 2 , q t+1−j − 1 j=1 by (3.1). This, in particular, yields q (s+1)(t+1)−t − 1 ≤ (s + 1)(t + 1) + 2 . q−1 This implies, in turn, q (s+1)(t+1)−(t+1) + 1 ≤ (s + 1)(t + 1) + 2 . That is, q s(t+1) ≤ (s + 1)(t + 1) + 1 and hence, q s(t+1) ≤ s(t + 1) + t + 2 < 2s(t + 1) . Set z = s(t + 1). Then, since the function Fq (z) = q z − 2z is positive on [4, ∞) , we have a contradiction. 

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Lemma 3.4. If soc(G) ∼ = P SL(d, ur ), (d, ur ) = (2, 2), (2, 3), then t = 1 and n = 3. Proof. Assume that soc(G) ∼ = P SL(d, ur ), (d, ur ) = (2, 2), (2, 3). Then t  urd − 1 q (s+1)(t+1)−j − 1 = r t+1−j q −1 u −1 j=1

(3.3)

by (3.1) and Theorem 2.1. Note that d ≥ 3 by Lemma 3.2. We first assume that u = p and that t ≥ 2. Assume also that, for each i = 1 or 2, there exists a primitive prime divisor bi of q (s+1)(t+1)−i − 1. Clearly bi does divide q t+1−j − 1 for each 1 ≤ j ≤ t, as (s + 1)(t + 1) − i > t. Furthermore, b1 = b2 . Nevertheless, b1 b2 | urd − 1. This forces    (s + 1)(t + 1) − 1 (s + 1)(t + 1) − 2 rd , by Theorem 2.3. On the other hand, rd | (s + 1)(t + 1) − ¯j for some 1 ≤ ¯j ≤ t again by Theorem 2.3, which is a contradiction. Now since (s + 1)(t + 1) − 2 ≥ 4, being s ≥ 1 and t ≥ 2, and since d ≥ 3, as we noted above, we have one of the following three equalities:   q, (s + 1)(t + 1) − 1 = (2, 6) ,   q, (s + 1)(t + 1) − 2 = (2, 6) , or (u, rd) = (2, 6) by Theorem 2.2. If



 q, (s + 1)(t + 1) − 1 = (2, 6) , then (s + 1)(t + 1) = 7 with s ≥ 1 and t ≥ 2, which is clearly impossible. If (u, rd) = (2, 6) , then either

urd − 1 < 31 ur − 1

or

urd − 1 = 63 . ur − 1

t  q (s+1)(t+1)−j − 1 < 31 q t+1−j − 1 j=1

or

t  q (s+1)(t+1)−j − 1 = 63 , q t+1−j − 1 j=1

Therefore,

respectively. However, both are contradictions by Lemma 3.1. Hence, (s + 1)(t + 1) − 2 = 6 . This implies s = 1 and t = 3, as s ≥ 1 and t ≥ 2. Then (3.3) becomes urd − 1 = 3(27 − 1)(25 − 1) . ur − 1 At this point, since the primitive divisors of (25 − 1) and (27 − 1) divide urd − 1, we have 35 | rd by Theorem 2.3. On the other hand, rd divides 3, 5, or 7 by another application of Theorem 2.3, since we have ruled out the case (u, rd) = (2, 6) and

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since d ≥ 3. This contradiction shows that t = 1. Then (3.3), bearing in mind that q = ph , h ≥ 1, becomes prd − 1 ph(2s+1) − 1 = r . h p −1 p −1 Note that s > 1 as t = 1. Hence, h(2s + 1) > 3 and h(2s + 1) = 6. Furthermore, rd ≥ 3, and (p, rd) = (2, 6) by Lemma 3.1. Thus, by Theorem 2.3, h(2s + 1) | rd

and rd | h(2s + 1) .

This implies that rd = h(2s + 1) and hence, r = h. Thus, soc(G) ∼ = P SL(2s + 1, q) . Moreover,

  soc(G) ≤ P SL 2(s + 1), q . Note that since t = 1, our dimension is 2(s + 1). Hence, the minimum degree over fields of characteristic p of (soc(G)) is at most 2(s + 1). Note that 1 2s + 1 < 2(s + 1) ≤ (2s + 1)(2s + 2) . 2 We now apply Kleidman–Liebeck [5, 5.4.11, Table 5.4.A, p. 199], which says that for d = 2s+1, in this situation, then the GF (q)-module is a specific module having dimension one of the following: d(d + 1) d(d − 1) , , or 2 2 This yields one of the following equalities:

20 (where d = 6) .

2(s + 1) = s(2s + 1) , 2(s + 1) = (s + 1)(2s + 1) , 2(s + 1) = 20

for

or

2s + 1 = 6 .

A contradiction in either case as s > 1. Therefore, we may assume that u = p. If t = 1, then q 2s+1 − 1 urd − 1 = r . q−1 u −1 By a rough estimate, we obtain q 2s + 1 ≤

q 2s+1 − 1 ≤ urd − 2 . q−1

Therefore, q 2s ≤ urd − 3 . Since u = p and since

(3.4)

  soc(G) ≤ P SL 2(s + 1), q , then the minimum degree of (soc(G)) is at most 2(s + 1), over a field of characteristic not p. Let d = 2s + 1, then by Kleidman–Liebeck [5, 5.3.9, Table 5.3.A, p. 188]g, if the degree d ≥ 3 and q not 2 or 4 the minimum degree of P SL(d, q) is

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at least q d−1 − 1 (over a field of characteristic not p). Then, since d = 2s + 1 ≥ 3, we have, 2(s + 1) ≥ urd−1 − 1 . So, 2s ≥ urd−1 − 3 and composing this inequality with (3.4), we obtain rd−1

−3

≤ urd − 3 ,

(3.5)

≤ u(urd−1 − 3) + 3u − 3 .

(3.6)

qu which is equivalent to rd−1

qu Set z = u

rd−1

−3

− 3. Then (3.6) becomes q z ≤ uz + 3u − 3 .

Now, as z ≥ u and q ≥ 2 2z ≤ z 2 + 3z − 3 .

(3.7)

Finally, consider the function F (z) = 2z − z 2 − 3z + 3 . As d ≥ 3, it must be ur ≥ 2, so we have z ≥ 6. It is a straightforward computation to show that F (z) > 0 on [6, ∞). Hence, we have a contradiction to t = 1, when u = p. Therefore, assume that t ≥ 2 and u = p. Now observe that t  q (s+1)(t+1)−t − 1 q (s+1)(t+1)−(t−1) − 1 q (s+1)(t+1)−j − 1 ≥ · . t+1−j q −1 q−1 q2 − 1 j=1

(3.8)

For j = t, we have the following rough estimate: q (s+1)(t+1)−t − 1 > q s(t+1) . q−1 For j = t − 1, we have q (s+1)(t+1)−(t−1) − 1 ≥ q t+1 , q2 − 1 which is equivalent to q (s+1)(t+1)−(t−1) − 1 ≥ q t+3 − q t+1 , and, in turn, is equivalent to q (s+1)(t+1)−(t−1) > q t+3 − q t+1 . That is, q st+s+2 ≥ q t+3 − q t+1 , as s ≥ 3 by Theorem 1.1. As a consequence, we have q (s+1)(t+1)−t − 1 q (s+1)(t+1)−(t−1) − 1 ≥ q (s+1)(t+1) . q−1 q2 − 1

(3.9)

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Combining (3.3) with (3.9), we obtain q (s+1)(t+1) ≤

urd − 1 ≤ urd − 2 . ur − 1

(3.10)

As t ≥ 2 then q (s+1)(t+1) ≥ q 2s + 1 . Then, by (3.10), we obtain q 2s ≤ urd − 3 . However, we have seen this inequality previously in (3.4), and, as this leads to a contradiction, this completes the proof of the lemma.  Lemma 3.5. The subgroup soc(G) cannot be isomorphic to Sp(2n, 2) with n ≥ 3. Proof. Deny the assertion. Then t  q (s+1)(t+1)−j − 1 = 2n−1 (2n ± 1) , t+1−j − 1 q j=1

(3.11)

since v = 2n−1 (2n ±1). Clearly q must be odd. If t = 1, then the first part of (3.11) 2s+1 becomes q q−1−1 which is odd since 2s+1 odd, while 2n−1 (2n ±1) is even as n ≥ 3. This contradiction shows that t ≥ 2. Now, since   soc(G) ≤ P ΓL (s + 1)(t + 1), q and since q is odd, then (s + 1)(t + 1) is larger than the minimum degree of Sp(2n, 2) over fields of characteristic not 2. By Kleidman–Liebeck [5, 5.3.9, Table 5.3.A, p. 188], if d ≥ 2 and q not 2 then the minimum degree (over fields not of characteristic p) of P Sp(2d, q) is larger than or equal to

qd − 1 . 2

Therefore, qd − 1 ≥ 2n−2 (2n−1 − 1) . 2 As 2n−1 − 1 ≥ n + 1, being n ≥ 3, so (s + 1)(t + 1) ≥

(s + 1)(t + 1) ≥ n2 . On the other hand, arguing as in the final part of Lemma 3.4, we have t  q (s+1)(t+1)−j − 1 ≥ q (s+1)(t+1) . t+1−j − 1 q j=1

Since (s + 1)(t + 1) ≥ (n + 1)2 , then 2

q (s+1)(t+1) ≥ 2(n+1) .

(3.12)

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Hence, by combining the previous inequality with (3.12), we obtain t  2 q (s+1)(t+1)−j − 1 ≥ 2(n+1) , t+1−j q −1 j=1

which is a contradiction by (3.11).



Theorem 3.1. Let P be a t-parallelism of P G(n, q) that admits a collineation group that is almost simple and acts doubly transitively on the t-spreads of P. Then t = 1, n = 3, q = 2, and P is one of the two regular parallelisms in P G(3, 2). Proof. Here we may use Theorem 2.1. Moreover, the cases (1g)–(1l) are ruled out by Lemma 3.1. The cases (1c)–(1e) cannot occur by Lemma 3.2. Finally the cases (1a), (1b) when tn > 3, and (1c) are ruled out by Lemma 3.3, 3.4 and 3.5, respectively. Thus, tn ≤ 3. That is, t = 1 and n = 3 and hence the assertion follows by Theorem 1.1. 

4. The affine case Lemma 4.1. If G is of affine type then t = 1. Proof. Suppose that G is of affine type. Then we have t  q (s+1)(t+1) − q j = |T | , q t+1 − q j j=1

where T = soc(G) and T is an elementary abelian u-group for some prime u. That is, t  q (s+1)(t+1) − q j = ur , t+1 − q j q j=1 where r ≥ 1. Then, by dividing by q j , we have t  q (s+1)(t+1)−j − 1 = ur . t+1−j − 1 q j=1

(4.1)

Assume that t ≥ 2. Assume also that, for each i = 1 or 2, there exists a primitive prime divisor bi of q (s+1)(t+1)−i − 1. Then bi does not divide q t+1−j − 1 for each 1 ≤ j ≤ t, as (s + 1)(t + 1) − i > t. Furthermore b1 = b2 . Hence, b1 b2 divides t  q (s+1)(t+1)−j − 1 . q t+1−j − 1 j=1

That is, b1 b2 divides ur with u prime. However, this is a contradiction as b1 = b2 . Thus, we must have one of the following equalities on pairs:     q, (s + 1)(t + 1) − 1 = (2, 6) or q, (s + 1)(t + 1) − 2 = (2, 6)

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by Theorem 2.2, since (s + 1)(t + 1) − 2 ≥ 4, being s ≥ 1 and t ≥ 2. If the first alternative occurs, then (s + 1)(t + 1) = 7 with s ≥ 1 and t ≥ 2, which is clearly impossible. Hence (s + 1)(t + 1) − 2 = 6 so that (s + 1)(t + 1) = 8. Thus s = 1 and t = 3, as s ≥ 1 and t ≥ 2. Then ur = 3(27 − 1)(25 − 1) by (4.1), which cannot occur. Thus, the Diophantine equation (4.1) has no solutions for t ≥ 2, so we are reduced to the case t = 1.  Lemma 4.2. If ur =

q 2s+1 −1 q−1

then r = 1 and 2s + 1, q are primes.

Proof. Note that (p2s+1 − 1)k/(q − 1) = (q 2s+1 − 1)/(q − 1) , for some integer k. Now ur = (p2s+1 − 1)k/(q − 1) and u cannot divide p2s+1 − 1, unless q = p, since u is a p-primitive divisor of q 2s+1 − 1. Hence, ur divides k, a contradiction. Hence, q is a prime. Since u is the only p-primitive divisor of p2s+1 − 1, it now follows immediately that 2s + 1 is a prime. Our group is a subgroup of ΓL(2s + 2, p) = GL(2s + 2, p) , of order p(s+1)(2s+1)

2s+2 

(pi − 1) .

i=1

But, (2s + 1, i) = 1 , for all i = 1, 2, . . . , 2s + 2 , i = 2s + 1 . Hence, ur is the order of a Sylow u-subgroup of GL(2s + 2, p), since (p2s+1 − 1)(pi − 1) = p(2s+1,i) − 1 = p − 1 . Consider the vector space GF (p2s+1 ) ⊕ GF (p) , of dimension 2s + 2 over GF (p). There is a cyclic subgroup of order ur of the multiplicative group of GF (p2s+1 ) that induces a subgroup of GL(2s + 2, p). Hence, the Sylow u-subgroups of GL(2s + 2, p) are cyclic. But, in this setting, we have an elementary Abelian group of order ur . Therefore, r = 1. This proves the theorem.  Since r = 1 by Lemma 4.2, the group G is a subgroup of AGL(1, u). As AGL(1, u) is sharply 2-transitive, we infer that G = AGL(1, u). Therefore, we obtain the following lemma: Lemma 4.3. G0 = GL(1, u), where u = components.

p2s+1 −1 p−1 ,

and fixes a spread W0 of

We derive a contradiction by showing that p = 2 and p = 2.

p2(s+1) −1 p2 −1

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Lemma 4.4. No non-identity element of G0 can fix a 2-dimensional subspace over GF (p) that is not a component of W0 . Proof. Let g ∈ G0 fix a 2-dimensional subspace L over GF (p) that is not a component of W0 . Then L belongs to a unique spread W  of the parallelism P. By uniqueness, g fixes W  which implies g is 1 as the group G is sharply 2-transitive.  Lemma 4.5. p = 2. Proof. Assume that p is odd. By Theorem 2.2, let τu be an element in G0 of order u, where u is a primitive prime divisor of p2s − 1. Then τu fixes the spread 2(s+1) W0 , of the parallelism P. If τu acts semiregularly on W0 , then u | p p2 −1−1 as |W0 | = p p2 −1−1 , as t = 1. Then u | p2(s+1) − 1 and hence 2s | 2(s + 1) by Theorem 2.3. So s = 1, which is a contradiction by Theorem 1.1. Therefore, τu fixes at least one component L of W0 pointwise. If τu fixes two components L and M then it fixes a 1-space from each and therefore fixes the generated 2-space, a contradiction by Lemma 4.4. Hence, τu fixes a unique component L. There are 2(s+1)

p2(s+1) − 1 p2 (p2s − 1) −1= 2 p −1 p2 − 1 remaining spreads, which is an odd number as

2 2s p (p − 1) 2 , p − 1 = (p2 − 1, 2s + 1) . p2 − 1 Therefore, the involution σ in G0 fixes a second component M . Since σ fixes L and M , it induces a collineation on each of the two Desarguesian spreads of orders p, obtained from the spreads of 1-dimensional GF (p)-subspaces on L and M , respectively. Furthermore, σ cannot be the kernel involution by sharp double transitivity of G. We have the following possibilities: (1) σ fixes L and M pointwise. (2) σ fixes L pointwise, but acts faithfully on M . (3) σ acts faithfully on both L and M . Under (3), letting i denote the kernel involution, we have (3)(a) σ = i on both L and M , (3)(b) σ = i on L but not on M or (3)(c) σ is not i on either L or M . We use Lemma 4.4 to eliminate each case. If σ acts faithfully on one of the fixed components then since it is an involution on a Desarguesian affine plane, it can be either i or an affine central collineation (i.e., on either L or M ). In either case, σ would fix at least two 1-spaces on that component. If σ fixes a component pointwise, then σ fixes all 1-spaces on that component. Thus, in all cases, σ must fix at least two 1-spaces on each fixed component L and M , which cannot occur. This completes the proof of the lemma.  Finally, we show that p cannot be 2. Lemma 4.6. p cannot be 2. Proof. Assume that p = 2. Then 22s+1 − 1 = u, where 2s + 1 and u are primes and the group G0 = GL(1, u), of order u − 1 = 2(22s − 1). Let W0 denote the fixed spread which has (22s+2 − 1)/3 components. There are exactly 22s+2 − 1

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1-dimensional GF (2)-subspaces and the group GL(2s + 2, 2) is certainly transitive on this set. Hence, the stabilizer in GL(2s + 2, 2) of a 1-space X has order |GL(2s + 2, 2)| /(22s+2 − 1) . Note that 2(22s − 1) divides this order, from which it follows that G0 fixes a 1dimensional subspace X. There is a unique component L of W0 containing X, which, of course, implies that G0 fixes X and L. Note that G0 induces a subgroup in GL(2, 2) acting on L that fixes a non-zero point x0 , for X = x0 . Since GL(2, 2) has order 6 and L has exactly 3 1-dimensional subspaces, G0 can only induce a subgroup of order dividing 2. Hence, G0 contains a subgroup of order 22s − 1 that fixes L pointwise. G0 acts as a collineation group on the remaining (22s+2 − 1)/3 − 1 = 22 (22s − 1)/3 components. We have a cyclic subgroup of order 2(22s − 1)/3 in G0 and we claim that we have two orbits of length 2(22s − 1)/3 under this group. Consider a component J not equal to L and let an element g of G0 fix J. Recall that Lemma 4.4 shows that no non-identity element of G0 can fix a 2-dimensional GF (2)-subspace not on L. Since J has exactly 3 1-dimensional GF (2)-subspaces, it follows that the stabilizer of a component has order 1 or 3 and therefore the orbit lengths of a component are either 2(22s − 1) or 2(22s − 1)/3 under G0 . Therefore, we must have the equation a2(22s − 1) + b2(22s − 1)/3 = 22 (22s − 1)/3 , which implies that a3 + b = 2 , so that a = 0 and b = 2, as claimed. In each of the two orbits, there is a unique element of order 3 that fixes all components of that orbit and hence fixes all components. Indeed, we noted that the element of order 3 is transitive on the 1-spaces of each component other than L. Let J be an arbitrary component and let τ3 denote the element of order 3. J defines a Desarguesian spread of order 2. GL(2, 2) has a unique normal subgroup of order 3 and if we consider the points of J as a field GF (4) then τ3  is GF (4)∗ . Thus, on J, we can form a field isomorphic to GF (4) by adjoining a zero element to τ3 . Clearly, we may diagonalize τ3 as Diag(a1 , a2 , . . . , a2s , I) , where the ai are 2 × 2 matrices over GF (2) of order 3. Since there is a unique 2 × 2 matrix group of order 3, we see that all but one of the components may be considered 1-dimensional GF (4)-subspaces. Hence, we have a deficiency one partial Desarguesian spread D0 of 1-spaces over GF (4) in W0 . If we form GF (22(s+1) ) as a (s + 1) vector space over GF (4), there is clearly a Desarguesian spread Z1 of 1dimensional GF (4)-subspaces that contains D0 , simply by taking the set of all 1-dimensional GF (4)-subspaces. Therefore, the remaining 1-dimensional GF (4)subspace must be covered exactly by L. That is, L is also a GF (4)-subspace. So, τ3 must fix each 1-space over GF (4). However, this means that τ3 is Diag(a, a, a, a, . . . , a) , a contradiction. This completes the proof of lemma.



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Result.Math.

Therefore, this section and the previous section complete the proof of our main theorem.

5. Final remarks Although there are many varieties of partial t-parallelisms of large cardinalities, there are no known examples of t-parallelisms for t > 1. It may be that constructions of t-parallelisms are possible using transitive groups but in this article, we have noted that doubly transitive t-parallelisms are indeed very scarce – there are only the two in P G(3, 2) for t = 1. On the other hand, there are known transitive 1parallelisms in P G(n, 2), for any positive odd integer n ≥ 3. It would be expected that there are transitive 1-parallelisms in P G(n, q), for n > 3. Furthermore, it might be true that there are transitive t-parallelisms in P G(k(t + 1) − 1, q) for t > 1, as their construction might be defined using a particular transitive action. So, we leave the following open problem for consideration. Determine for what values of n and t it is possible to construct transitive t-parallelisms in P G(n, q).

Acknowledgements The authors are indebted to the referee for a very copious reading and many helpful suggestions.

References [1] A. Beutelspacher, Parallelisms in finite projective spaces, Geom. Dedicata, (1974), 35–40. [2] P. Dembowski, Finite geometries, Springer-Verlag, Berlin, 1997. [3] N. L. Johnson, Two-transitive parallelisms, Des. Codes Cryptogr. 22 (2001), 179–189. [4] W. Kantor, Homogeneous designs and geometric lattices, J. Combin. Theory Ser. A 38 (1985), 66–74. [5] P. B. Kleidman, M. Liebeck, The subgroup structure of the finite classical groups, Cambridge University Press, Cambridge, 1990, Acad. Press, Boston, 1994. [6] G. Lunardon, On regular parallelisms in PG (3, q), Discrete Math. 51 (1984), no. 3, 229–235. [7] M. Walker, On translation planes and their collineation groups, Ph.D. thesis, Univ. London, 1973. Norman L. Johnson Mathematics Department University of Iowa Iowa City, IA 52242 USA e-mail: [email protected]

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Alessandro Montinaro Dipartimento di Matematica Universit` a di Lecce Via Arnesano I-73100 Lecce Italy e-mail: [email protected] Received: July 31, 2007. Revised: February 22, 2008.

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