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Unformatted text preview: MAT 322: Algebra with Galois Theory Professor Claus Sorensen Problem Set 2 Rik Sengupta October 2, 2009 1. Show that S n is generated by the tranpositions s i = ( i,i +1) , and that they satisfy the braid group relations. That is, s i s j = s j s i when  i j  > 1 , and s i s j s i = s j s i s j when  i j  = 1 . Solution. We will use induction on n . If n = 2 , the theorem is trivially true because the the group S 2 only consists of the identity and a single transposition. Suppose, then, that we know permutations of n numbers are generated by transpositions of successive numbers, i.e. by s i = ( i,i + 1). Let φ be a permutation of { 1 , 2 ,...,n + 1 } . If φ ( n + 1) = n + 1 , then the restriction of φ to { 1 , 2 ,...,n } is a permutation of n numbers, and so, by our hypothesis, it can be expressed as a product of transpositions. Suppose that, in addition, φ ( n + 1) = m with m 6 = n + 1 . Consider the following product of transpositions: ( n,n + 1)( n 1 ,n ) ··· ( m + 1 ,m + 2)( m,m + 1) It is easy to see that acting upon m with this product of transpositions produces n + 1 . Therefore, acting upon n + 1 with the permutation ( n,n + 1)( n 1 ,n ) ··· ( m + 1 ,m + 2)( m,m + 1) φ produces n +1 . Hence, the restriction of this permutation to { 1 , 2 ,...,n } is a permutation of n numbers, so, by hypothesis, it can be expressed as a product of transpositions. Since a transposition is its own inverse, it follows that φ may also be expressed as a product of transpositions. It is obvious that the transpositions follow the braid group relations. Sup pose  i j  > 1. This simply means that the ith place and the jth place have at least another place in between them. So, in that case, s i s j = ( i,i + 1)( j,j + 1) = ( j,j + 1)( i,i + i ) = s j s i , because s i and s j are independent of each other. So the order of operations is irrelevant in this case. Now suppose  i j  = 1. Without loss of generality suppose 1 j = i + 1. Then, s i and s j or any combination of them are going to affect only the elements in the three adjacent places i , i +1 = j and i +2 = j +1. So in order to check the final braid group relation, we can safely ignore the other elements and confine our attention only to { a,b,c } , where a , b and c occupy respectively the ith, jth and ( j +1)th places in an ordered sequence, and s i and s j correspond to swapping the first two and the last two elements respectively. Then, s i s j s i ≡ { c,b,a } ≡ s j s i s j . Thus the braid group relations are satisfied. 2. Is D 4 isomorphic to Q 8 ? Solution. No. We know, D 4 is the group of symmetries of a square. It is easy to see a square admits four different reflections, and since each of these reflections has order 2, it follows that D 4 has at least four elements that are of order 2. Specifically, there are five elements in D 4 – s,sr,sr 2 ,sr 3 ,r 2 which have order 2....
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 '09
 CLAUSSORENSEN
 Algebra, Normal subgroup, HK, Abelian group, Subgroup, Cyclic group

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