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The important point was that ai = ai′ i. e. that the map ⇒ ωi · ai = ωi · ai′ , µωi : Z/10Z → Z/10Z : a → ωi · a is injective, and hence bijective since Z/10Z is a finite set. In other words, µωi is a permutation of the set Z/10Z. This leads to the following generalisation and definition. 1 Let (G, ·) be a group, g0 ∈ G a fixed element, and let π1, . . , πn ∈ Sym(G) be permutations. a. We call C = CG(π1, . . , πn, g0) = (g1, . . , gn)t ∈ Gn π1(g1) · · · πn(gn) = g0 a check digit code (CDC) of length n on the alphabet G.

By symmetry we also get x ∼ z and then x ∼ y by transitivity. Let now u ∈ x be any element. Then u ∼ x and again by transitivity we have u ∼ y. Hence, u ∈ y and therefore x ⊆ y. Exchanging the roles of x and y the same argument shows that x = y. 9 Let M be a finite set, ∼ be an equivalence relation on M, and M1, . . , Ms be the pairwise different equivalence classes of ∼. Then: s |M| = |Mi|. 8. 10 Let M = {(an)n∈N | an ∈ Q} be the set of all sequences of rational numbers. Show that (an)n∈N ∼ (bn)n∈N :⇐⇒ lim (an − bn) = 0 n→∞ defines an equivalence relation on M.

N) of the numbers 1, . . , n, where each of the numbers 1, . . , n occurs exactly once among the numbers σ(1), . . , σ(n). We now want to count how many possibilities there are for such a permutation. e. the image of 1. For this there are n choices. Once we have fixed σ(1), there are only n−1 choices for the image σ(2) of 2. Then for σ(3) there are only n − 2 choices. Going on like this for σ(i) we have n − i + 1 choices, and finally for σ(n − 1) there are n − (n − 1) + 1 = 2 choices and for σ(n) will be fixed.

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