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Unformatted text preview: the universal property for free groups, we may dene an epimorphism : F ( X ) Z r 2 by ( x i ) = ( ,..., , 1 , ,..., ) , where the 1 is in the i th position. This yields an epimorphism = : F ( Y ) Z r 2 . Therefore Z r 2 is generated by the s elements ( y i ) . Since Z r 2 is abelian and ( y i ) has order 1 or 2, we deduce that every element of Z r 2 can be written in the form ( ( y 1 )) 1 ... ( ( y s )) s where i = 0 or 1. We conclude that  Z r 2  2 s , which contradicts the fact that  Z r 2  = 2 r , and the proof is complete. Exercise 6.3.2 on page 220 Prove that if  S  > 1, then F ( S ) is nonabelian. Let x , y S be distinct. Then xy 6 = xy , hence F ( S ) is nonabelian....
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This note was uploaded on 01/02/2012 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.
 Fall '07
 PALinnell
 Math, Algebra

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