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

By Thomas Markwig Keilen

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

**Example text**

Mathematicians are lazy people, or maybe one should better say that they are efficient. Therefore, they have thought of a way to represent a permutation in such a way that each of the numbers 1 to n has to be written at most once, not twice. For this we need the notion of a cycle, which permutes k of the numbers 1, . . , n cyclically. 4 a. Let {1, . . , n} = {a1, . . , ak} ∪ {b1, . . , bn−k}, k ≥ 2, and σ= a1 a2 . . ak−1 ak b1 . . bn−k a2 a3 . . ak a1 b1 . . bn−k ∈ Sn, then we call σ a k-cycle, and we say that it cyclically permutes the numbers a1, .

0 + nZ 1 + nZ 2 + nZ = = = nZ, {1 + nz | z ∈ Z}, {2 + nz | z ∈ Z}, n − 1 = (n − 1) + nZ = {n − 1 + nz | z ∈ Z}. The index |Z : nZ| of nZ in Z is thus n. Proof: We have to show that each integer m ∈ Z belongs to one of the above mentioned equivalence classes and that they are pairwise different. Let m ∈ Z an arbitrary integer then there exists by division with remainder integers q, r ∈ Z such that m = qn + r with 0 ≤ r ≤ n − 1. But this implies19 m − r = qn = nq ∈ nZ = U. Thus m is equivalent to r, and therefore m ∈ r, where r is one of the above n equivalence classes.

Suppose i = j then j would be equivalent to i and hence j − i ∈ nZ would be a multiple of n. By assumption, 18 Note here that the group operation is addition, so that a left coset is not denoted by “g · U” but by “g+U”. Maybe this would not be so deceiving if not at the same time the subgroup U = nZ itself looked like a multiplicative left coset — which it is not! This is one of the main reasons why we prefer the notation k instead of k + nZ. 19 Note again that the group operation in (Z, +) is the addition.