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# sep30 - Math 5125 Friday September 30 September 30 Ungraded...

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Math 5125 Friday, September 30 September 30, Ungraded Homework Exercise 6.3.5 on page 220 Establish a finite presentation for A 4 using 2 generators. Let F denote the free group on { x , y } , let a = ( 1 2 )( 3 4 ) , b = ( 1 2 3 ) , and define θ : F A 4 by θ x = a , θ y = b . Let K = ker θ . It is easy to check that a , b = A 4 . Indeed by Lagrange’s theorem 6 | a , b | 12, so if a , b = A 4 , then | a , b | = 6. But a group of order 6 has a unique subgroup of order 3, so has exactly 2 elements of order 3. Since b , b 2 , aba are three distinct elements of order 3, we have a contradiction. Thus a , b = A 4 as asserted and we conclude that θ is onto. Therefore F / K = A 4 by the fundamental homomorphism theorem. Next note that θ x 2 = θ y 3 = θ ( xy ) 3 = 1. In view of this, we will try x 2 , y 3 , ( xy ) 3 for our set of relators. Let N denote the normal closure of { x 2 , y 3 , ( xy ) 3 } in F . Then x , y | x 2 = 1 , y 3 = 1 , ( xy ) 3 = 1 is by definition isomorphic to F / N . We want to show that F / N = A 4 , that is F / N = F / K . Since x 2 , y 3 , ( xy ) 3 K , we certainly have N K . We want to show equality,
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