Unformatted text preview: the universal property for free groups, we may deﬁne an epimorphism α : F ( X ) → Z r 2 by α ( x i ) = ( ,..., , 1 , ,..., ) , where the 1 is in the i th position. This yields an epimorphism β = αθ : F ( Y ) → Z r 2 . Therefore Z r 2 is generated by the s elements β ( y i ) . Since Z r 2 is abelian and β ( y i ) has order 1 or 2, we deduce that every element of Z r 2 can be written in the form ( β ( y 1 )) ε 1 ... ( β ( y s )) ε s where ε 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....
View
Full Document
 Fall '07
 PALinnell
 Math, Algebra, Vector Space, Linear map, Morphism, Homomorphism, homomorphisms, free groups

Click to edit the document details