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13. Symmetric groups 13.1 13.2 13.3 Cycles, disjoint cycle decompositions Adjacent transpositions Worked examples 1. Cycles, disjoint cycle decompositions The symmetric group Sn is the group of bijections of {1, . . . , n} to itself, also called permutations of n things. A standard notation for the permutation that sends i −→ `i is   1 2 3 ... n `1 `2 `3 . . . `n Under composition of mappings, the permutations of {1, . . . , n} is a group. The fixed points of a permutation f are the element
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13. Symmetric groups 13.1 Cycles, disjoint cycle decompositions13.2 Adjacent transpositions13.3 Worked examples 1.  Cycles, disjoint cycle decompositions  The  symmetric group  S  n  is the group of bijections of   { 1 ,...,n }  to itself, also called  permutations  of   n things. A standard notation for the permutation that sends  i  −→   i  is  1 2 3  ... n 1   2   3  ...  n  Under composition of mappings, the permutations of   { 1 ,...,n }  is a  group .The  ﬁxed points  of a permutation  f   are the elements  i  ∈ { 1 , 2 ,...,n }  such that  f  ( i ) =  i .A  k -cycle  is a permutation of the form f  (  1 ) =   2  f  (  2 ) =   3  ... f  (  k − 1 ) =   k  and  f  (  k ) =   1 for distinct   1 ,..., k  among  { 1 ,...,n } , and  f  ( i ) =  i  for  i  not among the   j . There is standard notation forthis cycle:(  1   2   3  ...  k )Note that the same cycle can be written several ways, by cyclically permuting the   j : for example, it alsocan be written as(  2   3  ...  k   1 ) or (  3   4  ...  k   1   2 )Two cycles are  disjoint  when the respective sets of indices  properly moved   are disjoint. That is, cycles(  1   2   3  ...  k ) and (   1    2    3  ...   k  ) are disjoint when the sets  {  1 , 2 ,..., k }  and  {   1 ,  2 ,...,  k  }  aredisjoint. [1.0.1]  Theorem:  Every permutation is uniquely expressible as a product of disjoint cycles.191  192  Symmetric groups  Proof:  Given  g  ∈  S  n , the cyclic subgroup   g  ⊂  S  n  generated by  g  acts on the set  X   =  { 1 ,...,n }  anddecomposes  X   into disjoint  orbits  O x  =  { g i x  :  i  ∈    } for choices of orbit representatives  x  ∈  X  . For each orbit representative  x , let  N  x  be the order of   g  whenrestricted to the orbit   g  ·  x , and deﬁne a cycle C  x  = ( x gx g 2 x ... g N  x − 1 x )Since distinct orbits are disjoint, these cycles are disjoint. And, given  y  ∈  X  , choose an orbit representative x  such that  y  ∈  g  ·  x . Then  g  ·  y  =  C  x  ·  y . This proves that  g  is the product of the cycles  C  x  over orbitrepresentatives  x .  /// 2.  Transpositions  The  (adjacent) transpositions  in the symmetric group  S  n  are the permutations  s i  deﬁned by s i (  j ) =  i  + 1 (for  j  =  i ) i  (for  j  =  i  + 1)  j  (otherwise)That is,  s i  is a 2-cycle that interchanges  i  and  i  + 1 and does nothing else. [2.0.1]  Theorem:  The permutation group  S  n  on  n  things  { 1 , 2 ,...,n }  is generated by  adjacent transpositions   s i . Proof:  Induction on  n . Given a permutation  p  of   n  things, we show that there is a product  q   of adjacenttranspositions such that ( q   ◦  p )( n ) =  n . Then  q   ◦  p  can be viewed as a permutation in  S  n − 1 , and we doinduction on  n . We may suppose  p ( n ) =  i < n , or else we already have  p ( n ) =  n  and we can do the inductionon  n .Do induction on  i  to get to the situation that ( q   ◦  p )( n ) =  n  for some product  q   of adjacent transposition.Suppose we have a product  q   of adjacent transpositions such that ( q  ◦  p )( n ) =  i < n . For example, the emptyproduct  q   gives  q   ◦  p  =  p . Then ( s i  ◦  q   ◦  p )( n ) =  i  + 1. By induction on  i  we’re done.  /// The  length  of an element  g  ∈  S  n  with respect to the generators  s 1 ,...,s n − 1  is the smallest integer    suchthat g  =  s i 1  s i 2  ... s i  − 1  s i   Garrett: Abstract Algebra   193 3.  Worked examples  [13.1]  Classify the conjugacy classes in  S  n  (the  symmetric group  of bijections of   { 1 ,...,n }  to itself).Given  g  ∈  S  n , the cyclic subgroup   g   generated by  g  certainly acts on  X   =  { 1 ,...,n }  and thereforedecomposes  X   into  orbits  O x  =  { g i x  :  i  ∈    } for choices of orbit representatives  x i  ∈  X  . We claim that the (unordered!)  list of sizes   of the (disjoint!)orbits of   g  on  X   uniquely determines the conjugacy class of   g , and  vice versa  . (An unordered list that allowsthe same thing to appear more than once is a  multiset . It is not simply a  set  !)To verify this, ﬁrst suppose that  g  =  tht − 1 . Then   g   orbits and   h   orbits are related by  g  -orbit  O tx  ↔  h  -orbit  O x Indeed, g  ·  ( tx ) = ( tht − 1 )  ·  ( tx ) =  t ( h  ·  x )Thus, if   g  and  h  are conjugate, the unordered lists of sizes of their orbits must be the same.On the other hand, suppose that the unordered lists of sizes of the orbits of   g  and  h  are the same. Choosean ordering of orbits of the two such that the cardinalities match up: | O ( g ) x i  |  =  | O ( h ) y i  |  (for  i  = 1 ,...,m )where  O ( g ) x i  is the   g  -orbit containing  x i  and  O ( h ) y i  is the   g  -orbit containing  y i . Fix representatives asindicated for the orbits. Let  p  be a permutation such that, for each index  i ,  p  bijects  O ( g ) x i  to  O ( g ) x i  by  p ( g  x i ) =  h  y i The only slightly serious point is that this map is well-deﬁned, since there are many exponents    which maygive the same element. And, indeed, it is at this point that we use the fact that the two orbits have thesame cardinality: we have O ( g ) x i  ↔  g  /  g  x i  (by  g   g  x i  ↔  g  x i )where   g  x i  is the isotropy subgroup of   x i . Since   g   is cyclic,   g  x i  is necessarily   g N    where  N   is thenumber of elements in the orbit. The same is true for  h , with the same  N  . That is,  g  x i  depends exactly on   mod  N  , and  h  y i  likewise depends exactly on    mod  N  . Thus, the map  p  is well-deﬁned.Then claim that  g  and  h  are conjugate. Indeed, given  x  ∈  X  , take  O ( g ) x i  containing  x  =  g  x i  and  O ( h ) y i containing  px  =  h  y i . The fact that the exponents of   g  and  h  are the same is due to the deﬁnition of   p .Then  p ( gx ) =  p ( g  ·  g  x i ) =  h 1+  y i  =  h  ·  h  y i  =  h  ·  p ( g  x i ) =  h (  px )Thus, for all  x  ∈  X  (  p  ◦  g )( x ) = ( h  ◦  p )( x )Therefore,  p  ◦  g  =  h  ◦  p or  pgp − 1 =  h (Yes, there are usually many diﬀerent choices of   p  which accomplish this. And we could also have tried tosay all this using the more explicit cycle notation, but it’s not clear that this would have been a wise choice.)  194  Symmetric groups  [13.2]  The  projective linear group  PGL n ( k ) is the group  GL n ( k ) modulo its center  k , which is thecollection of scalar matrices. Prove that  PGL 2 (    3 ) is isomorphic to  S  4 , the group of permutations of 4things. ( Hint:  Let  PGL 2 (    3 ) act on  lines  in    23 , that is, on one-dimensional    3 -subspaces in    23 .)The group  PGL 2 (    3 ) acts by permutations on the set  X   of lines in    23 , because  GL 2 (    3 ) acts on non-zerovectors in    23 . The scalar matrices in  GL 2 (    3 ) certainly stabilize every line (since they act by scalars), soact trivially on the set  X  .On the other hand, any non-scalar matrix  a bc d  acts non-trivially on some line. Indeed, if   a bc d  ∗ 0  =  ∗ 0  then  c  = 0. Similarly, if   a bc d  0 ∗  =  0 ∗  then  b  = 0. And if    a  00  d  11  =  λ  ·  11  for some  λ  then  a  =  d , so the matrix is scalar.Thus, the map from  GL 2 (    3 ) to permutations Aut set ( X  ) of   X   has kernel consisting exactly of scalar matrices,so  factors through   (that is, is well deﬁned on) the quotient  PGL 2 (    3 ), and is  injective   on that quotient. (Since PGL 2 (    3 ) is the quotient of   GL 2 (    3 ) by the kernel of the homomorphism to Aut set ( X  ), the kernel of themapping induced on  PGL 2 (    3 ) is trivial.)Computing the order of   PGL 2 (    3 ) gives | PGL 2 (    3 ) |  =  | GL 2 (    3 ) | / | scalar matrices |  = (3 2 −  1)(3 2 −  3)3  −  1 = (3 + 1)(3 2 −  3) = 24(The order of   GL n (    q ) is computed, as usual, by viewing this group as automorphisms of     nq .)This number is the same as the order of   S  4 , and, thus, an injective homomorphism must be surjective, hence,an isomorphism.(One might want to verify that the center of   GL n (    q ) is exactly the scalar matrices, but that’s not strictlynecessary for this question.) [13.3]  An automorphism of a group  G  is  inner  if it is of the form  g  −→  xgx − 1 for ﬁxed  x  ∈  G . Otherwiseit is an  outer automorphism . Show that every automorphism of the permutation group  S  3  on 3 things is inner  . ( Hint:  Compare the action of   S  3  on the set of 2-cycles by conjugation.)Let  G  be the group of automorphisms, and  X   the set of 2-cycles. We note that an automorphism must sendorder-2 elements to order-2 elements, and that the 2-cycles are exactly the order-2 elements in  S  3 . Further,since the 2-cycles  generate   S  3 , if an automorphism is trivial on all 2-cycles it is the trivial automorphism.Thus,  G  injects   to Aut set ( X  ), which is permutations of 3 things (since there are three 2-cycles).On the other hand, the conjugation action of   S  3  on itself stabilizes  X  , and, thus, gives a group homomorphism f   :  S  3  −→  Aut set ( X  ). The kernel of this homomorphism is trivial: if a non-trivial permutation  p  conjugatesthe two-cycle  t  = (1 2) to itself, then(  ptp − 1 )(3) =  t (3) = 3so  tp − 1 (3) =  p − 1 (3). That is,  t  ﬁxes the image  p − 1 (3), which therefore is 3. A symmetrical argument showsthat  p − 1 ( i ) =  i  for all  i , so  p  is trivial. Thus,  S  3  injects to permutations of   X  .

Jul 23, 2017

#### CX20P

Jul 23, 2017
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