2000
DOI: 10.1007/bf01237471
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The Lunelli-Sce hyperoval inPG(2, 16)

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Cited by 8 publications
(13 citation statements)
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“…If both G B and G C are isomorphic to M 11 , then, since |G : G P | = (s + 1)(11s + 1) > 12 = (t + 1) = |G : G C |, we must have G P < G C . The only proper subgroup of M 11 with a 2-transitive action on 12 elements is PSL(2, 11), so we conclude that G P ∼ = PSL (2,11), and so (s + 1)(11s + 1) = |P| = |G : G P | = 144, which is a contradiction to s ∈ N. If G C = PSL(2, 11), then, since PSL(2, 11) has no proper subgroups with a 2-transitive action on 12 elements, we conclude that G P = G C and reach a contradiction as in the previous case. Finally, if G C = PGL(2, 11), then G = M 12 .2 and, since s > 2, we again conclude that G P = G C and reach a contradiction as in the previous cases.…”
Section: Imprimitive On Both Points and Linesmentioning
confidence: 76%
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“…If both G B and G C are isomorphic to M 11 , then, since |G : G P | = (s + 1)(11s + 1) > 12 = (t + 1) = |G : G C |, we must have G P < G C . The only proper subgroup of M 11 with a 2-transitive action on 12 elements is PSL(2, 11), so we conclude that G P ∼ = PSL (2,11), and so (s + 1)(11s + 1) = |P| = |G : G P | = 144, which is a contradiction to s ∈ N. If G C = PSL(2, 11), then, since PSL(2, 11) has no proper subgroups with a 2-transitive action on 12 elements, we conclude that G P = G C and reach a contradiction as in the previous case. Finally, if G C = PGL(2, 11), then G = M 12 .2 and, since s > 2, we again conclude that G P = G C and reach a contradiction as in the previous cases.…”
Section: Imprimitive On Both Points and Linesmentioning
confidence: 76%
“…If G = soc(G) = M 11 , then t + 1 = 11, G B = M 10 , and G C = PSL (2,11). Since PSL (2,11) has no proper subgroups with a 2-transitive action on 11 elements, we conclude that G P = G C . However, this means that (s + 1)(10s + 1) = |P| = |G : G P | = 12, a contradiction.…”
Section: Imprimitive On Both Points and Linesmentioning
confidence: 87%
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“…Therefore, there are two related ovals (consequently two Niho bent functions) up to equivalence, obtained from functions g(u) = 1 and g ′ (u) = 1 + u 4 + ū4 +u 5 + ū5 +u 8 + ū8 . The automorphism group of the Lunelli-Sce hyperoval is transitive on the points of the hyperoval [10,35]. Therefore, it determines only one oval (consequently one Niho bent function) with the function g ′′ (u) = 1+u 5 + ū5 (the Lunelli-Sce hyperoval is the first non-trivial member of the Subiaco [10,22] and the Adelaide families [24]).…”
Section: Niho Bent Functions In Small Dimensionsmentioning
confidence: 99%