1989
DOI: 10.1007/bf00184160
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PSL(2,q) as a collineation group of projective planes of small order

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Cited by 11 publications
(19 citation statements)
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“…If |H X | = 3 k +1 2 , then F ix(σ ) has order 3 2k by (1), since X H = 3 k (3 k − 1), |l ∩ R k | = 3 k + 1 and √ n ≤ 3 2k . This is a contradiction, by [16], Theorem 1.2. (i), since H fixes l and H ∼ = PSL(2, 3 k ), k odd, k > 1.…”
Section: Theoremmentioning
confidence: 69%
See 1 more Smart Citation
“…If |H X | = 3 k +1 2 , then F ix(σ ) has order 3 2k by (1), since X H = 3 k (3 k − 1), |l ∩ R k | = 3 k + 1 and √ n ≤ 3 2k . This is a contradiction, by [16], Theorem 1.2. (i), since H fixes l and H ∼ = PSL(2, 3 k ), k odd, k > 1.…”
Section: Theoremmentioning
confidence: 69%
“…by [16], relation (9). Finally, denote by λ(H X ) the number of the H -orbits on l ∩ F ix(σ ) having the stabilizer of a point isomorphic to a given H X .…”
Section: Theoremmentioning
confidence: 99%
“…Thus H = H (Q, l) H = (Q, l) is ruled out by the above argument. Then there exists a nontrivial element in Z which is a homology of Π by [47,Proposition 2.6]. A contradiction, since each involution in S must be a Baer collineation of Π .…”
Section: Lemma 14mentioning
confidence: 78%
“…Let ρ be a subgroup of G of order 3. Then ρ fixes exactly 2 points in each G-orbit of length 56 or 8 by [41,Relation (9)]. Thus ρ fixes exactly 12 points on E, as x 2 = 5 and x 3 = 1.…”
mentioning
confidence: 98%
“…Clearly Q ∈ E, since |E| = 288 and o(γ) = 7. On the other hand, by [41,Relation (9)], the group γ fixes 3, 0 or 1 points on P G according to whether P G has length 24, 56 or 8, respectively. This in conjunction with the fact that γ fixes exactly one point in E yields (x 1, x 2 , x 3 ) = (0, 5, 1) in the above Diophantine equation.…”
mentioning
confidence: 99%