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Homework Help: 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

    3. The attempt at a solution

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

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

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

    Or I can write out all possible elements of the group
    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?

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


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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.

    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]x^{-1} \in <x>[/tex]
    is probably clearer.
  4. Jun 25, 2007 #3


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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:
    [tex]a=(12) b=(123)[/tex]
    [tex] a^2=(12)(12)=1;

    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
    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.
    Last edited: Jun 29, 2007
  6. Jun 30, 2007 #5

    matt grime

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    You say you wrote out all the elements


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

    The elements are:


    then look at powers of x:


    cos x has order 3

    powers of 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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