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Unformatted text preview: Homework 3 Solutions Math 332, Spring 2010 Problem 1. (a) Proposition. If n is odd, then all the reflections in D n lie in the same conjugacy class. Proof. Every reflection in D n has the form r k s for some power k . Conjugating such a reflection by r gives r ( r k s ) r 1 = rr k rs = r k +2 s. Therefore, if we start with s and repeatedly conjugate by r , we get the following sequence of conjugate elements: s, r 2 s, r 4 s, r 6 s, ... Since the order of r is odd, the powers of r 2 eventually cycle through all of the powers of r , and therefore all of the reflections in D n appear in this sequence. (b) Proposition. If n is even, then there are exactly two conjugacy classes of reflections in D n Proof. As in part (a), each reflection r k s is conjugate to r k +2 s . If we repeatedly conjugate s and rs by r , we find two sequences of conjugate elements: s, r 2 s, r 4 s, r 6 s, ... and rs, r 3 s, r 5 s, r 7 s, .......
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This document was uploaded on 11/03/2010.
 Spring '09
 Math, Algebra

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