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Unformatted text preview: Homework 2 Solutions Math 332, Spring 2010 Problem 1. Proposition. Every finite group with an even number of elements has at least one element of order two. Proof. Let G be a finite group, and suppose that G does not have any elements of order two. We shall prove that the order of G is odd. Since G does not have any elements of order two, we know that a 6 = a 1 for every non identity element a ∈ G . If we pair each element of G with its inverse, we get a partition of G of the following form: { e } , a 1 ,a 1 1 , a 2 ,a 1 2 , ... a n ,a 1 n . From this partition, we see that G must have 2 n + 1 elements, where n is the number of inverse pairs. In particular, the order of G must be odd. Problem 2. Proposition. Let G be a group, and let a be a fixed element of G . Define a new binary operation * on G by x * y = xay for all x,y ∈ G . Then G forms a group under the operation * . Proof. We must prove that * is associative and possesses an identity element, and that each element of G...
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This document was uploaded on 11/03/2010.
 Spring '09
 Math, Algebra

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