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Unformatted text preview: Homework 3 Solutions 1. (a) Not a group. Neither associative nor does it have an identity. Since a e = a we would need e = 0, but 0 a 6 = a (for a 6 = 0). Since no identity it has no inverses. (b) Is a group. It is nonabelian and the identity is: I = 1 0 0 1 (c) Not a group. Not every element has an inverse. For example: A = 2 0 0 2 ∈ GL (2 , Z ) but A 1 = 1 / 2 1 / 2 6∈ GL (2 , Z ) (d) Is a group. The identity is 1 (or, rather, the conjugacy class of 1). It is abelian. (e) Not a group. Not every element has an inverse. For example 2 has no inverse in Z 6 . (f) Is a group. The identity is 25. It is abelian. 2. Abelian group (symmetric over main diagonal) * e a b c d e e a b c d a a b c d e b b c d e a c c d e a b d d e a b c 3. If ∀ a ∈ G a 2 = e then we know that a 1 = a , b 1 = b and ( ab ) 1 = ab (since a,b,ab ∈ G ). So ab = ( ab ) 1 = b 1 a 1 = ba and thus G is abelian....
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This note was uploaded on 04/19/2011 for the course MATH 21373 taught by Professor Johnson during the Spring '11 term at Carnegie Mellon.
 Spring '11
 Johnson
 Algebra

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