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By Thomas Markwig Keilen

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24 For each n ∈ Z the subgroup nZ of (Z, +) is a normal subgroup. 25 Let G be a group and U ≤ G be a subgroup. The following statements are equivalent:20 a. U ✂ G is a normal subgroup of G. b. gUg−1 = U for all g ∈ G. c. gU = Ug for all g ∈ G. d. (gU) · (hU) = ghU for all g, h ∈ G. Proof: a. : By the assumption we have g · U · g−1 ⊆ U for any g ∈ G. Let’s now fix an arbitrary g ∈ G and apply this inclusion to g−1. We then get g−1 · U · g−1 −1 ⊆ U, U = e · U · e = g · g−1 · U · g−1 −1 · g−1 ⊆ g · U · g−1 ⊆ U.

Suppose we allow in the definition of ∼ all elements v, w ∈ R2, not only the non-zero ones. Is ∼ then an equivalence relation on R2? If so, what is the equivalence class of (0, 0)? 13 (The integers) Let M = N × N and let m = (a, b) ∈ M and m ′ = (a ′ , b ′ ) ∈ M be two elements in M. We define m ∼ m ′ ←→ a + b ′ = a ′ + b. Show that ∼ is an equivalence relation and that the map is bijective. 14 Let M be a set and σ ∈ Sym(M) a bijective map. a. By a∼b ⇐⇒ ∃ m ∈ Z : b = σm(a) for a, b ∈ M we define an equivalence relation on the set M.

N} then we can describe σ by the following scheme 1 2 ... n σ(1) σ(2) . . σ(n) respectively a1 a2 ... an σ(a1) σ(a2) . . σ(an) , if a1, . . , an is any alignment of the numbers 1, . . , n. 2 The group Sn is for n ≥ 3 not abelian since for the permutations 1 2 3 2 1 3 1 2 3 2 3 1 , ∈ S3 we have 1 2 2 1 3 3 ◦ 1 2 2 3 3 1 = 1 1 2 3 3 2 = 1 3 2 2 3 1 = 1 2 2 3 3 1 ◦ 1 2 2 1 3 3 Note that in the scheme it does not depend in which order the numbers 1 to n occur in the first row. g. 1 2 3 2 1 3 = 2 1 3 1 2 3 .

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