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Math 5125 Friday, September 16 September 16, Ungraded Homework Exercise 6.3.1 on page 220 (first half) Let F 1 and F 2 be free groups of finite rank. Prove that F 1 = F 2 if and only if they have the same rank. Suppose F 1 and F 2 are free groups of the same rank, say F 1 = F ( X ) , F 2 = F ( Y ) , and we have set maps α : X Y and β : Y X such that αβ and βα are the identity on Y and X respectively. Then there exist homomorphisms θ : F ( X ) F ( Y ) such that θ ( x ) = α ( x ) and φ : F ( Y ) F ( X ) such that φ ( y ) = β ( y ) for all x X and y Y . Note that θφ : F ( Y ) F ( Y ) is a homomorphism such that θφ ( y ) = αβ ( y ) = y for all y Y . By uniqueness of homomorphisms, we see that θφ is the identity map on F ( Y ) . Similarly φθ is the identity map on F ( X ) . This proves that F ( X ) = F ( Y ) . We haven’t used the hypothesis of finite rank in this part. Conversely suppose F 1 = F 2 , say F 1 = F ( X ) and F 2 = F ( Y ) where | X | = r and | Y | = s , and that F 1 and F 2 have different rank. Without loss of generality r > s . Also we have an isomorphism θ : F ( Y ) F ( X ) . Write X = { x 1 ,..., x r } and Y = { y 1 ,..., y s } . By using the universal property for free groups, we may define an epimorphism
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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....
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