Documents

13.pdf

Description
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
Categories
Published
of 6
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Related Documents
Share
Transcript
  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  .

ASTER_FLAASH.pdf

Jul 23, 2017

CX20P

Jul 23, 2017
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks