Generator-offset property: Difference between revisions

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===== Some implications of the above =====

Suppose Q = (α, β, γ) ≠ R = (α, β’, γ’) are the two k-steps in S that project to P. Then T = (α’, β’’, γ’’) projects to I. Here the values in each component differ by at most 1, and α ≠ α’. Since Σ1 is a mos pattern of αa + (n-α)(b~c) and α’a + (n-α’)(b~c), the cyclic words Λ1 = the pattern of α and α’ must be a mos. Similarly, Λ2 = the pattern of β and β’, and Λ3 = the pattern of γ and γ’ must form mosses. In addition, Λ1 = α...αα’.

Suppose Q = (α, β, γ) ≠ R = (α, β’, γ’) are the two k-steps in S that project to P. Then T = (α’, β’’, γ’’) projects to I. Here the values in each component differ by at most 1, and α ≠ α’. Since Σ1 is a mos pattern of αa + (n − α)(b~c) and α’a + (n − α’)(b~c), the cyclic words Λ1 = the pattern of α and α’ must be a mos. Similarly, Λ2 = the pattern of β and β’, and Λ3 = the pattern of γ and γ’ must form mosses. In addition, Λ1 = α...αα’.

Suppose Λ2 is the mos λβ μβ’. Then Λ3 is the mos (λ ± 1)γ (μ ∓ 1)γ’. Since neither Λ2 nor Λ3 are multimosses, and at least one of μ and (μ ∓ 1) are even, it is now immediate that n is odd.

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 ===== Some implications of the above =====
-Suppose Q = (α, β, γ) ≠ R = (α, β’, γ’) are the two k-steps in S that project to P. Then T = (α’, β’’, γ’’) projects to I. Here the values in each component differ by at most 1, and α ≠ α’. Since Σ<sub>1</sub> is a mos pattern of αa + (n-α)(b~c) and α’a + (n-α’)(b~c), the cyclic words Λ<sub>1</sub> = the pattern of α and α’ must be a mos. Similarly, Λ<sub>2</sub> = the pattern of β and β’, and Λ<sub>3</sub> = the pattern of γ and γ’ must form mosses. In addition, Λ<sub>1</sub> = α...αα’.
+Suppose Q = (α, β, γ) ≠ R = (α, β’, γ’) are the two k-steps in S that project to P. Then T = (α’, β’’, γ’’) projects to I. Here the values in each component differ by at most 1, and α ≠ α’. Since Σ<sub>1</sub> is a mos pattern of αa + (n &minus; α)(b~c) and α’a + (n &minus; α’)(b~c), the cyclic words Λ<sub>1</sub> = the pattern of α and α’ must be a mos. Similarly, Λ<sub>2</sub> = the pattern of β and β’, and Λ<sub>3</sub> = the pattern of γ and γ’ must form mosses. In addition, Λ<sub>1</sub> = α...αα’.
 Suppose Λ<sub>2</sub> is the mos λβ μβ’. Then Λ<sub>3</sub> is the mos (λ ± 1)γ (μ ∓ 1)γ’. Since neither Λ<sub>2</sub> nor Λ<sub>3</sub> are multimosses, and at least one of μ and (μ ∓ 1) are even, it is now immediate that n is odd.