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Unit 11: Conjugate Elements




          Solution: When n is odd the center is trivial and elements of the form a  b are all conjugate.  Notes
                                                                      i
          Elements of the form a  are conjugate in pairs; a   a  since a   1. We can write the  class
                                                  m
                             i
                                                              2m
                                                      -m
          equation in the following form:
                                      |G| = 1 + ((n-1)/2) · 2 + n
          When n is even, the center has two elements. (The element a  is conjugate to itself since it is
                                                            n/2
          equal to a -n/2 . This shows that Z(G) = { 1, a  }.) Therefore, elements of the form a  b split into two
                                                                          i
                                           n/2
          conjugacy classes. In this case the class equation has the following form:
                                   |G| = 2 + ((n-2)/2) · 2 + 2 · (n/2)
                Example: Show that for all n  4, the centralizer of the element (1,2)(3,4) in S  has order
                                                                             n
          8· (n-4)!. Determine the elements in the centralizer of ((1,2)(3,4)).
          Solution: The conjugates of a = (1,2)(3,4) in S  are the permutations of the form (a,b) (c,d). The
                                               n
          number of ways to construct such a permutation is
                                     n(n-1)/2 · (n-2)(n-3)/2 · 1/2 ,
          and dividing this into n! gives the order 8 · (n-4)! of the centralizer.

          We first compute the centralizer of a in S . The elements (1, 2) and (3, 4) clearly commute with
                                           4
          (1, 2) (3, 4). Note that a is the square of b = (1, 3, 2, 4); it follows that the centralizer contains
          < b >, so b  = (1, 4, 2, 3) also belongs. Computing products of these elements shows that we must
                  3
          include (1, 3)(2, 4) and (1, 4)(2, 3), and this gives the required total of 8 elements.
          To find the centralizer of a  in S , any of the elements listed above can be  multiplied by any
                                    n
          permutation disjoint from (1, 2)(3, 4). This produces the required total |C(a)| = 8 · (n-4)!.

          Self Assessment

          1.   Let G be a group and let x be an elements of the G. Then L(x) is a ............... of G.
               (a)  Normal subgroup         (b)  Cyclic subgroup
               (c)  Subgroup                (d)  Permutation group

          2.   Any group of order p  is ...............
                                2
               (a)  permutation             (b)  abelian
               (c)  cyclic                  (d)  finite

          3.   If G is a ............... group and P is a prime divisor of the order of G, then G contains an
               element of order P.
               (a)  direct                  (b)  external

               (c)  internal                (d)  finite
          4.   Let P be a prime number. The center of any P-group is ...............
               (a)  trivial                 (b)  non-trivial
               (c)  finite                  (d)  infinite
          5.   A group of order p , with P is a prime number and n ............... is called a p-group.
                              n
               (a)  a = 1                   (b)  b > 1
               (c)  c < 1                   (d)  d  1





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