College Algebra Exam Review 83

College Algebra Exam Review 83 - H is already contained in...

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2.2. SUBGROUPS AND CYCLIC GROUPS 93 2.1.17. Let M be a set with an associative operation and let N be a sub- set which is closed under the operation. Show that N is also closed for the product of an arbitrary number of elements; i.e., if n ± 1 and a 1 ;a 2 ;:::;a n 2 N , then a 1 a 2 ²²² a n 2 N . 2.2. Subgroups and Cyclic Groups Definition 2.2.1. A nonempty subset H of a group G is called a subgroup if H is itself a group with the group operation inherited from G . We write H ³ G to indicate that H is a subgroup of G . For a nonempty subset H of G to be a subgroup of G , it is necessary that 1. For all elements h 1 and h 2 of H , the product h 1 h 2 is also an element of H . 2. For all h 2 H , the inverse h ± 1 is an element of H . These conditions also suffice for H to be a subgroup. Associativity of the product is inherited from G , so it need not be checked. Also, if conditions (1) and (2) are satisfied, then e is automatically in H ; indeed, H is nonempty, so contains some element h ; according to (2), h ± 1 2 H as well, and then according to (1), e D hh ± 1 2 H . These observations are a great labor–saving device. Very often when we need to check that some set H with an operation is a group,
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Unformatted text preview: H is already contained in some known group, so we need only check points (1) and (2). We say that a subset H of a group G is closed under multiplication if condition (1) is satised. We say that H is closed under inverses if condition (2) is satised. Example 2.2.2. An n-by-n matrix A is said to be orthogonal if A t A D E . Show that the set O .n; R / of n-by-n realvalued orthogonal matrices is a group. Proof. If A 2 O .n; R / , then A has a left inverse A t , so A is invertible with inverse A t . Thus O .n; R / GL .n; R / . Therefore, it sufces to check that the product of orthogonal matrices is orthogonal and that the inverse of an orthogonal matrix is orthogonal. But if A and B are orthogonal, then .AB/ t D B t A t D B 1 A 1 D .AB/ 1 ; hence AB is orthogonal. If A 2 O .n; R / , then .A 1 / t D .A t / t D A D .A 1 / 1 , so A 1 2 O .n; R / . n Here are some additional examples of subgroups:...
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