HW2 - Lenny T. Evans MATH 534 Homework 2 February 4, 2009...

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Unformatted text preview: Lenny T. Evans MATH 534 Homework 2 February 4, 2009 7.5.1 Each permutation of n elements in { 1 ,...,n } = J n , or every element in the symmetric group S n , is the product of transpositions. Proof: We will prove this by induction on n. Base case ( n = 2): There are two elements in J 2 to exchange, so the only possible permutations are the identity permutation that keeps the two elements the same, and the permutation that switches the two elements. The latter exchanges two elements and keeps the rest of J 2 the same (there is no rest of J 2 ) , so it is a transposition by the definition given. If this transformation is applied twice, it switches the two elements twice, effectively keeping the two elements the same, giving the other permutation. Thus, every permutation of J 2 is the product of transpositions. Now, assume that each permutation of a symmetric group S n- 1 is the product of transpositions. Let S n be a permutation of J n , and for 1 k n, k N , let ( n ) = k. Further, let S n be a transposition of J n such that ( n ) = ( k ) = n. Then, leaves n fixed, and can be seen as a permutation of J n- 1 . Then by assumption, there exists a set of transpositions 1 ... q such that = q Y i =1 i . The inverse of a transposition must be a transposition as it restores the two elements that it exchanged, and leaves everything else fixed. Then, = - 1 q Y i =1 i . Showing that each S n can be written as the product of transpositions, and thus, by induction, each S n can be written as the product of transpositions for any n N . 7.5.2 Every product of two transpositions of J n , in S n , for n 3 can be written as the product of 3-cycles. Further, the alternating group A n is precisely the products of these 3-cycles. Proof: Two transpositions can either have no elements in common, 1 element in common, or 2 elements in common. (If they have all elements in common, it is multiplying two identity permutations which are not transpositions). We see that if there are no elements in common ( a 6 = b 6 = c 6 = d ) , ( a b )( c d ) = ( a b c )( c d b ) If there is one element in common ( a = c,b 6 = d ), ( a b )( d a ) = ( a d b ) (or to make this a product of 3-cycles, ( a b )( d a ) = ( a d b )( a d b )( b d a )) And if they have two elements in common ( a = c,b = d ), ( a b )( a b ) = ( a m b )( b m a ) (where m is some other element in J n , which must exist since n 3)....
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This note was uploaded on 03/13/2011 for the course MATH 534 taught by Professor Kumar during the Spring '11 term at UNC.

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HW2 - Lenny T. Evans MATH 534 Homework 2 February 4, 2009...

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