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Extra info for Algebraic Structures [Lecture notes]

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G. 0 + 1 + 3 + 4 + 1 + 2 + 5 + 4 + 7 + 1 + 8 + 0 = 36 = 6. C) Does this allow to detect errors? Otherwise it is of no use. Certainly we will not be able to detect all errors, thus we have to distinguish certain types of errors! Some statistics tell us that the following two types are the most common ones. Type I: “Single Digit Errors” – i. e. just one digit is wrong. These are roughly 80% of the occuring errors. Type II: “Neighbour Transpositions” – i. e. two neighbouring digits have been interchanged.

We thus get for g, h ∈ G (gU) · (hU) = (Ug) · (hU) = U · (gh) · U = (gh) · U · U = ghU. 20 In order to show the equivalence of several statement we can do a so called ring closure. It suffices to show that “a. ⇒ b. ⇒ c. ⇒ d. ”, since “a. ” and “b. g. that “a. e. the seemingly missing implications follow as well. 50 d. : Let g ∈ G and n ∈ U be given, then g · n · g−1 = g · n · g−1 · e ∈ gU · g−1U = g · g−1 · U = eG · U = U. 26 The subgroup U := {id, (1 2)} ⊂ S3 is not a normal subgroup the S3 since for σ = (2 3) ∈ S3 we have σ ◦ (1 2) ◦ σ−1 = (2 3) ◦ (1 2) ◦ (2 3) = (1 3) ∈ U.

K−1(a) , (16) where k = min{l > 0 | σl(a) = a} = |a|. 8 the equivalence classes of ∼ form a partition of {1, . . , n}. We thus can choose integers a11, . . , at1 ∈ {1, . . , n} such that t {1, . . , n} = · i=1 ai1. 35 Set ki = |ai1| and aij = σj−1(ai1) then due to (16) we get t {1, . . , n} = · (17) {ai1, ai2, . . , aiki }. i=1 It remains to show that σ = σ1 ◦ · · · ◦ σt where σi = (ai1 · · · aiki ) is a ki-cycle. For this let b ∈ {1, . . , n} so that b = aij = σj−1(ai1) for some 1 ≤ i ≤ t and some 1 ≤ j ≤ ki.

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