By Thomas Markwig Keilen

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**Extra resources for Algebraic Structures [Lecture notes]**

**Example text**

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). e. α is surjective. e. in this case α is also injective. 55 Ker(α) = {z ∈ Z | gz = e} = nZ. The Homomorphism Theorem therefore implies that the map α is a group isomorphism. The classification of cyclic groups can be applied to compute the order of gn from the order of g. 60 If (G, ·) is a group, g ∈ G with o(g) < ∞ and 0 = m ∈ Z. Then o gm = lcm m, o(g) . 59. 61 implies therefore that o gm = o α(m =o m .

G. (1) or (3), and define each 1-cycle (a) (with a ∈ {1, . . , n}) to be the identity. 6 The permutations σ= 1 2 3 4 4 1 3 2 ∈ S4 and π= 1 2 3 4 5 4 1 3 2 5 ∈ S5 34 are all 3-cycles which permute the numbers 1, 4, 2 cyclically. In the above notation they are thus described as and σ = (1 4 2) π = (1 4 2). This shows the disadvantage of the new notation. It does not give any information on the set of definition of the permutation. Different maps are represented by the same scheme. However, we are willing to pay this price for a representation which is both shorter and better organised.

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. First Idea: Repeat the whole number! This is a bit too much redundancy and increases the risk of falsely scanned numbers. Second Idea: Take the cross sum of the digits of the real product number as check “digit”. E. g. if the product number is 013412547180, then the check digit would be 0 + 1 + 3 + 4 + 1 + 2 + 5 + 4 + 7 + 1 + 8 + 0 = 36. This will usually be several digits long, and is still too much redundancy.

### Algebraic Structures [Lecture notes] by Thomas Markwig Keilen

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