aug29 - S 4 which commute with(1 2(3 4 Let n ∈ Z and let...

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Math 5125 Monday, August 29 August 29, Ungraded Homework Exercise 4.2.9 on page 130 Show that | Cen S n (( 1 2 )( 3 4 )) | = 8 · ( n - 4 ) ! for all n 4. Determine the elements in this centralizer explicitly. | K (( 1 2 )( 3 4 )) | = n ( n - 1 )( n - 2 )( n - 3 ) / 2 3 (because we can rotate each of the 2-cycles, and then we can interchange the two 2-cycles). Using the formula | K ( g ) || Cen S n ( g ) | = | S n | = n !, we deduce that | Cen S n (( 1 2 )( 3 4 )) | = 8 · ( n - 4 ) !. The elements of the centralizer are of the form ab , where b denotes the permutations which involve only 5 , 6 , . . . , n (a total of ( n - 4 ) ! permutations), and a = { (1), (1 2), (3 4), (1 2)(3 4), (1 3)(2 4), (1 4)(3 2), (1 3 2 4), (1 3 2 4) } . It is a little difﬁcult to motivate this last step; however this gives the correct number of elements, and we were trying to ﬁnd 8 elements in
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Unformatted text preview: S 4 which commute with (1 2)(3 4). Let n ∈ Z + and let g ∈ A n . Suppose 4-n and | g | ≤ 2. Considering A n as acting on { 1 , . . . , n } , prove that gi = i for some i where 1 ≤ i ≤ n . Since the order of g is 1 or 2, when we write it as a product of disjoint cycles, only 2-cycles can occur (and 1-cycles, which may be omitted). Suppose g has no ﬁxed points (i.e. no such i exists). Then g must be a product of n / 2 disjoint 2-cycles. Since n / 2 must be an integer and 4-n , we deduce that g is a product of an odd number of transpositions. This contradicts the fact that g ∈ A n ....
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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.

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