Generator-offset property: Difference between revisions

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(Note: We say “assume” as shorthand for either “assume without loss of generality” or “assume after excluding other possibilities”.)

Assume that S2's generator is also a k-step, and that Σ2's I is located at index n. Then Σ1 and Σ2 are the same mos and even the same mode. Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying a and b, and all instances of s in S1 come from c. Then the ~~corresponding~~ steps of S2 must be either all ~~a’s~~ or all b~c’s~~. Thus~~ these steps are ~~all a’s or~~ all b’s. ~~This contradicts~~ the assumption that S has exactly three step sizes. This shows that the maximum variety of S must be at least 3, and that P has at least two preimages in S.

Assume that S2's generator is also a k-step, and that Σ2's I is located at index n. Then Σ1 and Σ2 are the same mos pattern (up to knowing the order of the step sizes) and even the same mode:

S: a

S1: L ... L s

S2: A ... A a

Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S1 come from a. Then the steps of S2 corresponding to the L of S1 must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S1 and S2 are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting the assumption that S has exactly three step sizes. This shows that the maximum variety of S must be at least 3, and that P has at least two preimages in S.

At most two k-steps of S can project to P in S1, for if P has three preimages (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S (written in the basis a, b, c), then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S3, contradicting the mos property of S3.

@@ Line 125: / Line 125: @@
 (Note: We say “assume” as shorthand for either “assume without loss of generality” or “assume after excluding other possibilities”.)
-Assume that S2's generator is also a k-step, and that Σ2's I is located at index n. Then Σ1 and Σ2 are the same mos and even the same mode. Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying a and b, and all instances of s in S1 come from c. Then the corresponding steps of S2 must be either all a’s or all b~c’s. Thus these steps are all a’s or all b’s. This contradicts the assumption that S has exactly three step sizes. This shows that the maximum variety of S must be at least 3, and that P has at least two preimages in S.
+Assume that S2's generator is also a k-step, and that Σ2's I is located at index n. Then Σ1 and Σ2 are the same mos pattern (up to knowing the order of the step sizes) and even the same mode:
+ S:             a
+ S1:    L ... L s
+ S2:    A ... A a
+Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S1 come from a. Then the steps of S2 corresponding to the L of S1 must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S1 and S2 are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting the assumption that S has exactly three step sizes. This shows that the maximum variety of S must be at least 3, and that P has at least two preimages in S.
 At most two k-steps of S can project to P in S1, for if P has three preimages (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S (written in the basis a, b, c), then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S3, contradicting the mos property of S3.