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If |a| = n, then the cyclic subgroup generated by a^k is the same as the cyclic subgroup generated by a^(gcd(n,k)). "Contemporary Abstract Algebra", by Joe Gallian: https://amzn.to/2ZqLc1J. Check out my blog at: https://infinityisreallybig.com/ Abstract Algebra Playlist: • Abstract Algebra Course, Lecture 1: Introd... (0:00) Correction of the statement related to the lemma from the end of lecture 9A. (0:44) Make use of the lemma to prove that, if |a| = n, then the cyclic subgroup generated by a^k is the same as the cyclic subgroup generated by a^(gcd(n,k)). (6:55) Corollaries (the order of an element divides the order of the group and a^k is a generator of the cyclic subgroup generated by a if gcd(n,k) = 1. (8:25) Briefly discuss Fundamental Theorem of Cyclic Groups. (10:16) Make a subgroup lattice for a cyclic group of order 24. (16:05) Quick overview of group theory facts related to the Euler phi function. (18:57) Define permutations and permutation groups ("groups of permutations") on a set. (22:11) The symmetric group Sn on n objects is typically considered to be the collection of all permutations on the set A = {1,2,3,...,n}. (24:18) Elements of this group are often represented either as "arrays" or "cycles". Show just the array notation in this lecture. (26:19) The order of Sn is n! ("n factorial"). AMAZON ASSOCIATE As an Amazon Associate I earn from qualifying purchases.