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Unformatted text preview: 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 : F A 4 by x = a , y = b . Let K = ker . It is easy to check that h a , b i = A 4 . Indeed by Lagranges theorem 6 h a , b i 12, so if h a , b i 6 = A 4 , then h a , b i = 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 h a , b i = A 4 as asserted and we conclude that is onto. Therefore F / K = A 4 by the fundamental homomorphism theorem. Next note that x 2 = y 3 = ( 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 h x , y  x 2 = 1 , y 3 = 1 , ( xy ) 3 = 1 i is by definition isomorphic to...
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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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