Conjugacy Classes in A_5 (Algebra 1: Lecture 12 Video 4) скачать в хорошем качестве

Conjugacy Classes in A_5 (Algebra 1: Lecture 12 Video 4)

Lecture 12: In this lecture we gave two proofs that A_5 is simple. We first gave one that involved the fact that a normal subgroup is a union of conjugacy classes. We determined the sizes of the 5 conjugacy classes of A_5 and noted that there was no way to take a union of these conjugacy classes (including the class of the identity) to get a set of size equal to a proper divisor of |A_5| = 60. In order to determine the conjugacy classes in A_5, we first talked about conjugacy of elements in S_n. We showed that two permutations are conjugate if and only if they have the same cycle type. The corresponding question about conjugacy in A_n is more complicated, but we did show that for n \ge 5, all 3-cycles in A_n are conjugate. We saw that the set of 24 5-cycles in A_n split into two distinct conjugacy classes of size 12. To prove this, we first counted the number of m-cycles in S_n and then computed the centralizer of an m-cycle in S_n and applied the orbit-stabilizer theorem. At the end of the lecture we gave a second proof that A_5 is simple where we argued by contradiction. We supposed that there was a proper nontrivial normal subgroup N of A_5. We showed that if N contains even one 3-cycle, then we got a contradiction. We then showed that starting from any non-identity element of N, we could produce a 3-cycle in N. We will begin the next lecture by generalizing this argument, arguing by induction that A_n is simple for all n \ge 5. Reading: Dummit and Foote discuss conjugacy in S_n near the end of Section 4.3, including the count for the number of m-cycles in S_n and the computation of the stabilizer of an m-cycle in S_n. The statement that A_5 is simple is Theorem 12 on page 128. We first gave a version of the proof given there. This material is also presented very nicely in Conrad's notes, 'The Simplicity of A_n'. We proved Lemma 2.2 given there. The second proof that A_5 is simple that we gave is given at the bottom of page 3 of these notes. We will start the next lecture by giving the first proof from these notes that A_n is simple for all n \ge 5. (See page 4 starting with Lemma 3.2.) This is a version of the proof given by Dummit and Foote in Section 4.6. At this point, you should read Dummit and Foote Sections 4.1-4.3 and 4.6. You should also read Conrad's notes on Group Actions and on The Simplicity of A_n.