New PDF release: Algebraic Structures [Lecture notes]

By Thomas Markwig Keilen

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30 We consider the subgroups U = (1 2) and V = (2 3) of S3. Then U · V = {id, (1 2), (2 3), (1 2 3)} and by the Theorem of Lagrange this product cannot be a subgroup of S3. 29 is essential. 31 Proof Part b. and c. 29. 32 Let G be a finite group and U ≤ G be a subgroup of index |G : U| = 2. Show that U is a normal subgroup of G. 33 Let G be a group and N ≤ G be the unique subgroup of G of order |N| = n. Show that N ✂ G is a normal subgroup. 34 Find all normal subgroups of the group D8 = (1 2 3 4), (2 4) .

N, where each of the numbers 1, . . , n occurs exactly once among the numbers σ(1), . . , σ(n). We now want to count how many possibilities there are for such a permutation. e. the image of 1. For this there are n choices. Once we have fixed σ(1), there are only n−1 choices for the image σ(2) of 2. Then for σ(3) there are only n − 2 choices. Going on like this for σ(i) we have n − i + 1 choices, and finally for σ(n − 1) there are n − (n − 1) + 1 = 2 choices and for σ(n) will be fixed. Altogether we get n · (n − 1) · (n − 2) · .

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 . In order to keep things well-arranged we advise, however, to write the numbers in the first row always in an ascending order.

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Algebraic Structures [Lecture notes] by Thomas Markwig Keilen

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