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By Thomas Markwig Keilen

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42 If n ∈ Z>0 is a positive integer then U ≤ Zn ∃ m ∈ {1, . . , n} with m divides n : U = mZ/nZ = mn . ⇐⇒ In particular each subgroup of Zn is cyclic. 41 it suffices to find the subgroups U of Z with nZ ⊆ U. 39 such a subgroup has the form U = mZ for an integer m ≥ 0. e. m must lie between 1 and n and m is a divisor of n. We next want to compute the order of an element m ∈ Zn for positive integers m and n. For this we introduce the following notation. 43 For two integers a, b ∈ Z let lcm(a, b) := min{z > 0 | a and b divides z}, 0, if a, b = 0, if a = 0 or b = 0.

57 b. is called a semidirect product of g and h . One can show that if g · h = h · g then G is isomorphic to Z10 while otherwise G is isomorphic to D10. 58 Find all group homomorphisms α : Z10 −→ Zn with n ∈ {6, 13}. E) Cyclic Groups We want to close this section with the classification of all cyclic groups. 59 Let G = g be a cyclic group. a. If |G| = ∞ then we have the group isomorphism ∼ = α : Z −→ G : z → gz. 59 b. If |G| = n < ∞ then we have the group isomorphism ∼ = α : Zn −→ G : z → gz. Proof: For the map ∼ = α : Z −→ G : z → gz and two integers x, y ∈ Z we have α(x + y) = gx+y = gx · gy = α(x) · α(y).

I. e. the numbers will have to be similar! How can we get along with that problem? Idea: Store some redundant information which is not needed to identify the article, but only to detect possible errors. In the case of the EAN-13 only 12 digits characterise the article. Digit no. 13 is a so called check digit. B) How is the check digit related to the (real) article number? Basic Idea: It should be possible to calculate the check digit from the remaining digits in an easy way, but such that (common) errors are possibly detected.

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