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Symmetric group S3 with symbols

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the orders of all the elements for the symmetric group on 3 symbols S3.

    2. Relevant equations

    _______________________________________

    3. The attempt at a solution

    3 symbols : e,a,b

    I dont know how to do the S3 table using just these 3 letters
    I can do it using 6 letters (e,s,t,u,v,w) which is going to be like this:

    e s t u v w
    s e v w t u
    t u e s w v
    u t w v e s
    v w s e u t
    w v u t s e
     
  2. jcsd
  3. Feb 29, 2012 #2
    1. The problem statement, all variables and given/known data

    Determine the orders of all the elements for the symmetric group on 3 symbols S3.

    2. Relevant equations



    3. The attempt at a solution[/b

    3 symbols : e,a,b

    I dont know how to do the S3 table using just these 3 letters
    I can do it using 6 letters (e,s,t,u,v,w) which is going to be like this:

    e s t u v w
    s e v w t u
    t u e s w v
    u t w v e s
    v w s e u t
    w v u t s e
     
  4. Feb 29, 2012 #3

    Deveno

    User Avatar
    Science Advisor

    the "symbols" of S3 in this case aren't the elements, they are the elements of the domain of the functions that make up S3.

    ok, what S3 is, is the group of permutations of 3 "things". sometimes the things are called "letters", sometimes they are taken to be the elements of the set {1,2,3}. but what an element of S3 IS, is a certain kind of function: namely, a bijection on a 3-element set with itsef. all 3-element sets are pretty much alike (as sets), they just have different "element-names". so it doesn't really matter if you call the element (1 2) in S3 the function:

    1→2
    2→1
    3→3

    or:

    a→b
    b→a
    c→c

    what we actually mean by (1 2) is:

    "the function on a 3-element set that swaps the first and second elements".

    there are 6 possible bijections on the 3-element set A = {a,b,c} (you can think of a = 1, b= 2, and c = 3 if you want to, but the properties of 1,2 and 3 as integers aren't relevant here. it's better to think of them as:

    a = "element number 1"
    b = "element number 2"
    c = "element number 3").

    one of the 6 possible bijections is:

    a→a
    b→b
    c→c

    better known as: the identity function on A: f(x) = x, for all x in A.

    there are 3 bijections (permutations) of A that interchange 2 elements, and 2 bijections of A that send all 3 "to something else". see if you can puzzle out what these mappings might be. then give all 6 maps a name.

    multiplication on S3 is functional compostion. so if f is in S3, you want to answer the following question:

    "what is the smallest postive integer n, for which f°f°...°f(x) = fn(x) = x, for all x in A?"

    that integer is the order of f.

    for our example f =

    a→b
    b→a
    c→c,

    f2 =

    a→b→a
    b→a→b
    c→c→c

    so f2(x) = x, for all x in A, that is: f2 = idA, so the order of f is 2, in this case. 1 down, 5 to go.
     
  5. Feb 29, 2012 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I have merged the two threads. Please do not post the same question more than once.
     
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