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Unformatted text preview: 21. Permutation groups II 21.1. Conjugacy classes. Let G be a group, and consider the following relation on G : given f,h G , we put f h there exists g G s.t. h = gfg 1 . Thus, in the terminolgy from Lecture 20, f h h is a conjugate of f . Definition. The relation is called the conjugacy relation. Lemma 21.0. The conjugacy relation is an equivalence relation on G . Proof. (i) Reflexivity: for every f G we have efe 1 = f , so f f (that is, f is a conjugate of itself). (ii) Symmetry: assume that f h , so that h = gfg 1 for some g G . Then f = g 1 hg = uhu 1 where u = g 1 G . Therefore, h f . (iii) Transitivity: assume that f h and h k , so that h = g 1 fg 1 1 and k = g 2 hg 1 2 for some g 1 ,g 2 G . Then k = g 2 g 1 fg 1 1 g 1 2 = gfg 1 where g = g 2 g 1 G , so f k . Having established that is symmetric, we can safely use the terminology f and h are conjugate instead of saying h is a conjugate of f . Definition. The equivalence classes with respect to the conjugacy relation are called the conjugacy classes of G . For each f G we denote its conjugacy class by K ( f ). Thus, K ( f ) = { h G : h = gfg 1 for some g G } . Note that by the general properties of equivalence classes, conjugacy classes form a partition of G , that is, distinct conjugacy classes are disjoint, and the union of all conjugacy classes of G is the entire group G . Warning: Conjugacy classes should not be confused with cosets. 21.2. Conjugacy classes in S n . Computation of conjugacy classes in a given group may be a complicated problem. However, conjugacy classes in permutation groups admit a very simple and explicit description. We start with an example showing how conjugation works in S n . 1 2 Example: Let n 3, let f = (1 , 2 , 3) S n , and let g be some element of S n . Let us compute the conjugate gfg 1 = g (1 , 2 , 3) g 1 . To do this we have to track the image of each i { 1 ,...,n } under the composed map. First we analyze where gfg 1 sends g (1), g (2) and g (3). We have g (1) g 1 1 f 2 g g (2) g (2) g 1 2 f 3 g g (3) g (3) g 1 3 f 1 g g (1) . Now take any i 6 = g (1) ,g (2) or g (3). Then (since g is bijective) we have g 1 ( i ) 6 = 1 , 2 or 3, and therefore f ( g 1 ( i )) = g 1 ( i ). Thus we get i g 1 g 1 ( i ) f g 1 ( i ) g i. So, gfg 1 maps g (1) to g (2), g (2) to g (3), g (3) to g (1), and fixes all other elements. Therefore, gfg 1 = g (1 , 2 , 3) g 1 = ( g (1) ,g (2) ,g (3)). It is not hard to see that similar formula is true in general: for any cycle ( i 1 ,...,i k ) and any g S n we have g ( i 1 ,...,i k ) g 1 = ( g ( i 1 ) ,...,g ( i k )) . (K1) In other words, if f is a cycle of length k , then gfg 1 is also a cycle of length k whose entries are obtained by applying g to the entries of f ....
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This note was uploaded on 01/18/2012 for the course INFORMATIK 2011 taught by Professor Phanthuongcang during the Winter '11 term at Cornell University (Engineering School).
 Winter '11
 PhanThuongCang

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