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By Thomas Markwig Keilen

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Extra info for Algebraic Structures [Lecture notes]

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51 that Ker(α) ≤ G a subgroup of G. Let u ∈ Ker(α) and g ∈ G then α gug−1 = α(g) · α(u) · α g−1 u∈Ker(α) = α(g) · eH · α g−1 = α(g) · α g−1 = α gg−1 = α(eG) = eH. Thus gug−1 ∈ Ker(α) and Ker(α) ✂ G. 20) sgn : Sn → {−1, 1} then Ker(sgn) = An is a normal subgroup of Sn. In general the product of two subgroups is not a subgroup any more. However, if one of the two subgroups is a normal subgroup their product will always be a subgroup again. 29 Let G be a group, U ≤ G and N ✂ G. Then a. UN ≤ G. b. N ✂ UN.

E. σ1 ◦ · · · ◦ σt (b) = σi(b) = σ(b). This proves the statement of the theorem. 8 Note that for two disjoint cycles σ = (a1 . . ak), π = (b1 . . bl) ∈ Sn we have obviously σ ◦ π = π ◦ σ. Sine for c ∈ {a1, . . , ak} we have σ(c) ∈ {a1, . . , ak} and thus necessarily c, σ(c) ∈ {b1, . . , bl} so that (σ ◦ π)(c) = σ π(c) = σ(c) = π σ(c) = (π ◦ σ)(c). (18) In both cases c ∈ {b1, . . , bl} and c ∈ {a1, . . , ak} ∪ {b1, . . , bl} we show (18) analogously which proves the above claim. Moreover, it is obvious that the cycle decomposition of σ is unique up the order of the cycles, since the elements of the cycles are cyclically permuted by σ.

C. ⇒ d. ”, since “a. ” and “b. g. that “a. e. the seemingly missing implications follow as well. 50 d. : Let g ∈ G and n ∈ U be given, then g · n · g−1 = g · n · g−1 · e ∈ gU · g−1U = g · g−1 · U = eG · U = U. 26 The subgroup U := {id, (1 2)} ⊂ S3 is not a normal subgroup the S3 since for σ = (2 3) ∈ S3 we have σ ◦ (1 2) ◦ σ−1 = (2 3) ◦ (1 2) ◦ (2 3) = (1 3) ∈ U. A good source to find normal subgroups are group homomorphisms. 27 If α : G → H is a group homomorphism then Ker(α) ✂ G is a normal subgroup of G.

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