1988
DOI: 10.2307/2046834
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Large Automorphism Groups of Hyperelliptic Klein Surfaces

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Cited by 3 publications
(10 citation statements)
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“…x m = y k ρ δ , c 2 ρ µ , y n ρ ε (8) for some ε, α, β, γ , δ, µ ∈ {1, 0}. For a i = (m i /p i ) − 1, we have cxc −1 = x −1 ρ a 1 +a 2 , cyc −1 = y −1 ρ a 1 +a 3 , xyx −1 y −1 = ρ a 1 +a 2 +a 3 +a 4 and so µ = a 1 , α ≡ a 1 + a 2 (2), β ≡ a 1 + a 3 (2) and γ ≡ a 1 + a 2 + a 3 + a 4 (2).…”
Section: Proof Let G Be a Group Of Automorphisms Of Type (3 3 3) Omentioning
confidence: 99%
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“…x m = y k ρ δ , c 2 ρ µ , y n ρ ε (8) for some ε, α, β, γ , δ, µ ∈ {1, 0}. For a i = (m i /p i ) − 1, we have cxc −1 = x −1 ρ a 1 +a 2 , cyc −1 = y −1 ρ a 1 +a 3 , xyx −1 y −1 = ρ a 1 +a 2 +a 3 +a 4 and so µ = a 1 , α ≡ a 1 + a 2 (2), β ≡ a 1 + a 3 (2) and γ ≡ a 1 + a 2 + a 3 + a 4 (2).…”
Section: Proof Let G Be a Group Of Automorphisms Of Type (3 3 3) Omentioning
confidence: 99%
“…The p-hyperelliptic (p 1) surfaces at large have been studied in [6][7][8][9][10][12][13][14], where the most attention has been paid to a study of groups of automorphisms of bordered Klein surfaces with the exception of [5] were the pairs of symmetries of compact p-hyperelliptic Riemann surfaces were classified.…”
Section: Introductionmentioning
confidence: 99%
“…, x 3 gz since otherwise we get a contradiction with lemata. So we get the sequence C 6 = (2, 4, 0, 0) for which σ g = (1, 3, 4) (2,5,6). By Lemma 1 and Corollary 3, G is generated by x, g and admits a normal cyclic subgroup H = x 4 .…”
Section: Theoremmentioning
confidence: 93%
“…However x 2 g(gz) = gx 3 gx 2 (gz) and so there exists one more point z 4 ∈ F such that g(x 3 gz) = z 4 . Thus g(z 4 ) = x 2 gz and σ g = (1,5,2,8,4,7,3,6). So for C 8 = (4; 4, 0, 0),G has the presentation 8.5.…”
Section: Theoremmentioning
confidence: 99%
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