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Presentations of groups

  1. Jun 25, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that [tex]S_3[/tex] has the presentation [tex]<x,y|x^3=y^2=(xy)^2=1>[/tex]


    2. Relevant equations
    [tex]x^{-1}=x^2,y^{-1}=y,xyxy=1[/tex]
    [tex]xyx=y^{-1}[/tex]


    3. The attempt at a solution

    Let H=<x>, has at most order 3.
    Then
    [tex] y^{-1}xy=yxy=x^{-1}=x\in < x >[/tex]
    [tex]x^{-1}xx=x\in < x > [/tex]

    so
    [tex]<x>\lhd G[/tex]

    Then let <y>=K
    and use
    If
    [tex] H,K\subseteq G ,H\lhd G[/tex]
    then
    [tex]G=<x><y> \subseteq G[/tex]
    [tex]G=<xy>=<x><y>[/tex]
    So
    [tex]|G|\leq 6[/tex]

    Or I can write out all possible elements of the group
    [tex]\{x,y,x^2,xy,x^2y,(xy)x^2\}[/tex]
    So the group presented has order of at most 6.(not sure if that's true)

    My trouble comes when I try to show that it IS 6.

    Do I list all of the cosets? How do I get equality so that I can show that this presentation is isomorphic to S3?

    CC
     
    Last edited: Jun 25, 2007
  2. jcsd
  3. Jun 25, 2007 #2

    NateTG

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    Well, if you're masochistic, you can write out the isomorphism and multiplication tables for [itex]S_3[/itex] and [itex]G[/itex], but that seems a bit excessive.

    An alternative might be to show that there are only two possible groups of order [itex]6[/itex] and one of them is commutative.

    Another option might be to show that [itex]G[/itex] and [itex]S_3[/itex] both have identical effective group actions on a set. Since you already know that [itex]|G|=6[/itex] showing that [itex]G<S_3[/itex] is sufficient, which, in turn can be showing by effective group action of [itex]G[/itex] on a set of order 3.

    P.S.
    Notation -- this is really not a big deal, but it bugs me.
    Has [itex]x[/itex] referring to two different values. Splitting it into
    [tex]y^{-1}xy=yxy=x^{-1}[/tex]
    and
    [tex]x^{-1} \in <x>[/tex]
    is probably clearer.
     
  4. Jun 25, 2007 #3

    Hurkyl

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    What do you know about kernels of homomorphisms?
     
  5. Jun 29, 2007 #4
    Ok here's what I came up with:
    Consider:
    [tex]a=(12) b=(123)[/tex]
    then
    [tex] a^2=(12)(12)=1;
    b^3=(123)(123)(123)=1;
    (ab)^2=(13)(13)=1[/tex]

    let [tex]N=<ab|a^2=b^2=(ab)^2=1>[/tex]


    so then let x------>(12) and y------>(123)

    F({xy})------->S3 is an onto homomorphism
    and
    G------->S3 is an onto homomorphism
    but this implies that [tex]|G| \geq 6[/tex]
    putting that together with what I showed above, that [tex]|G| \leq 6[/tex] this means that |G|=6 and the homomorphism from G to S3 is actually an isomorphism. (The kernel is contained in N)

    I don't know for sure if my reasoning here is correct, and I'm not sure if it's completely clear.
    Please give me some input.
    thanks,
    CC
     
    Last edited: Jun 29, 2007
  6. Jun 30, 2007 #5

    matt grime

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    You say you wrote out all the elements

    [tex]\{x,y,x^2,xy,x^2y,(xy)x^2\}[/tex]

    in G. I don't see the identity in there....

    The elements are:

    e

    then look at powers of x:

    x,x^2

    cos x has order 3

    powers of y

    y

    cos y has order 2. Now that leaves mixed powers:

    xy, x^2y

    and we know that yx=x^2y, hence we can always write a word as one of the 6 listed items, e,x,x^2,y,xy,x^2y, and that these are all distinct. Clearly there are only 2 groups of order 6 (prove this - you simply need to look at what expressions xy might be eqaul to), this is one of them it isn't abelian, so it is S_3.
     
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