HOMEWORK ASSIGNMENT 1 DUE: TUESDAY, JANUARY 15 The first week overlaps the group theory covered in Introduction to Abstract Algebra. It is quite likely you have seen some of these problems before. (1) An element g of a group G is called an involution if g2 = 1. (a) How many involutions are there in Sn? (b) Let G be a group such that every element is an involution. Prove that G is abelian. (2) (a) If ß is an n-cycle, prove that ßk is a product of gcd(n, k) disjoint cycles, each of length n/ gcd(n, k). (b) If p is a prime, then prove that every power of a p-cycle is either a p-cycle or 1. (c) A permutation a ? Sn is regular if either a = 1 or a has no fixed points and is the product of disjoint cycles of the same length. Prove that a is regular if and only if there exists an n-cycle ß and a positive integer m such that a = ßm. (Hint for the “only if” part: If a = (a1 . . . ak)(b1 . . . bk) · · · (z1 . . . zk), a product of m disjoint k-cycles and n = mk, then let ß = (a1b1 . . . z1a2b2 . . . z2 . . . akbk · · · zk).) (3) Show than an r-cycle is an even permutation if and only if r is odd. (4) Give an example of a group and elements in that group to show that the relation “x commutes with y” is not transitive. (5) (a) A permutation matrix P over a field F is an n×n matrix obtained from permuting the columns of the n×n identity matrix I (over F). In other words, if the columns of I are I = [ e1 e2 . . . en ], then P = [ ea1 ea2 . . . ean ] for some a ? Sn. Prove that the set of all n × n permutation matrices is a subgroup of GL(n,K) which is isomorphic to Sn. (b) Prove that every finite group is isomorphic to a group of matrices. (6) Let G be a group and fix a ? G. Define ?a : G ? G by ?a(x) = axa-1. (This mapping is called conjugation by a.) (a) Prove that ?a is an automorphism. (b) Define G : G ? Aut(G) by G(a) = ?a for all a ? G. Prove that G is a homomorphism. (7) Prove that a group G is abelian if and only if the map x ?? x-1 is an automorphism. (Since the mapping is obviously a bijection, it is only the property of being a homomorphism which is important here.)

1(b). We have given for any given a; b 2 G, we have a2 = e and b2 = e. This means a??1 = a and b??1 = b. We have (ab)(ba)??1 = (ab)(b??1a??1) = (a:b)(b:a) = a:(b2):a = a:a = e This gives (ab):(ba)??1 = e That is ab = ba Hence G is abelian. 3...

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1(b). We have given for any given a; b 2 G, we have a2 = e and b2 = e. This means a??1 = a and b??1 = b. We have (ab)(ba)??1 = (ab)(b??1a??1) = (a:b)(b:a) = a:(b2):a = a:a = e This gives (ab):(ba)??1 = e That is ab = ba Hence G is abelian. 3...