# Absract-Algebra-II - MAL 705 DISCRETE MATHEMATICAL...

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Chapter 3 / Exercise 21
Elementary Linear Algebra
Larson
Expert Verified
MAL 705: DISCRETE MATHEMATICAL STRUCTURES ABSTRACT ALGEBRA CLASS NOTE-II Rupam Barman DEPARTMENT OF MATHEMATICS INDIAN INSTITUTE OF TECHNOLOGY DELHI Octover, 2013
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Chapter 3 / Exercise 21
Elementary Linear Algebra
Larson
Expert Verified
Contents 1 Cyclic Groups and Lagrange’s Theorem 1 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Cyclic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2.1 Example of Cyclic Groups . . . . . . . . . . . . . . . . . 3 1.2.2 Generators of Cyclic Groups . . . . . . . . . . . . . . . . 4 1.2.3 Fundamental Theorem of Cyclic Groups . . . . . . . . . 6 1.3 Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.3.1 Definition and Examples . . . . . . . . . . . . . . . . . . 9 1.3.2 Properties of cosets . . . . . . . . . . . . . . . . . . . . . 10 1.3.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.4 Lagrange’s Theorem and Consequences . . . . . . . . . . . . . . 13 1.4.1 Consequences of Lagrange’s Theorem . . . . . . . . . . . 14 1.4.2 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . 17 1.5 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 i
Unit 1 Cyclic Groupsand Lagrange’sTheorem 1.1 Introduction Cyclic groups are the simplest groups to study. The beauty of a cyclic group is that information about the group can be obtained from a single element of the group. These elements are known as the generators. In the first section, we will study cyclic groups. In the second section, we prove the Lagrange’s theorem which is the single most important theorem in finite group theory. To prove this theorem, we introduce a new and powerful tool for analyzing a group- the notion of cosets. 1.2 Cyclic Groups In Unit-I, we discussed subgroup generated by an element of a group. Let X be a subset of a group G and let H be the intersection of all subgroups K of G such that X K . By Theorem ?? of Unit-I, H is a subgroup of G . Also, X H . Suppose that L H and X L . Then L is a subgroup of G which contains the elements of X . By definition of H , we obtain H L . Hence, L = H . This proves that the subgroup H is the smallest subgroup of G such that X H . This subgroup H is called the subgroup generated by X and 1
1.2 Cyclic Groups Unit 1 is denoted by ( X ) . Thus, ( X ) = intersectiondisplay X K,K G K. If X = , then we define ( X ) = { e } . The following result gives the structure of the elements of ( X ) . Theorem 1.2.1. Let X be a nonempty subset of a group G . Denote by X 1 = { x 1 : x X } . Then the subgroup ( X ) generated by X is the set of all finite products of elements of X X 1 . Proof. Let H denote the set of all finite products of elements of X X 1 . Clearly, X H . If x,y H , then x = u 1 u 2 · · · u r and y = v 1 v 2 · · · v s , where u i ,v j X X 1 for all i = 1 , 2 , · · · ,r and j = 1 , 2 , · · · ,s . Now, xy 1 = u 1 u 2 · · · u r v 1 s v 1 s 1 · · · v 1 1 which is again a finite product of elements of X X 1 , and hence, xy 1 H . Thus, H is a subgroup of G . To complete the proof, we need to show that H is the smallest subgroup such that X H . Let K be a subgroup of G such that X K . Since K is a subgroup, so X 1 K . Hence, X X 1 K . Again, since K is a subgroup, it contains all finite products of elements of X X 1 . Thus, H K . This proves that H = ( X ) . Definition 1.2.1. A subgroup H of G is said to be finitely generated if there exists a finite subset X = { x 1 ,x 2 , · · · ,x n } of G such that H = ( X ) . In this case we write H = ( x 1 ,x 2 , · · · ,x n ) . In particular, a group G is said to be finitely generated if G = ( x 1 ,x 2 , · · · ,x n ) for some x 1 ,x 2 , · · · ,x n G . Theorem 1.2.2. Let a be an element of a group G . If order of the element is infinite, then ( a ) is an infinite subgroup of G . If order of a is finite, say, o ( a ) = m , then ( a )