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Math Help - Isomorphism question

  1. #1
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    Isomorphism question

    Let G=<x, y|x^4=y^4=e, xyxy^1=e>. Show that |G|≤16. Assuming |G|=16, show G/<y^2> is isomorphic to D4.
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  2. #2
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    Quote Originally Posted by aabsdr View Post
    Let G=<x, y|x^4=y^4=e, xyxy^1=e>. Show that |G|≤16. Assuming |G|=16, show G/<y^2> is isomorphic to D4.
    D_{2n} of order 2n have presentations

    1. <r, s |r^n=1, s^2=1, sr=r^{-1}s>,
    2. <x, y | x^2 = y^2 = {(xy)}^n=1>.

    2 is denoted as I_2(n), implying that D_{2n} can also be generated by two reflections instead of one rotation and one reflection. You can find this idea in the context of Coxeter groups.

    If you just play a little bit of your question, you will reduce your presentation to <a, b | a^2 = b^2 = {(ab)}^2=1>, which is D_4.

    I'll leave it to you to covert the presentation 1 into 2 and verify that they are indeed equivalent.
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  3. #3
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    Here is what I have:
    since xyxy^-1=e, we know that yxy^-1=x^-1=x^3, so we know that <x) is normal. So G= <x> union y<x> union y^2<x> union y^3<x> and |G|<= 16.
    Lets assume that |G|=16.
    So |G|/|<y^2>|=8, thus |G|/|<y^2>|>= |D4|.
    D4=<a,b|a^4=b^2=(ab)^2=e, or ab=ba^3>.
    Let w=x<y^2> , z=y<y^2> and q=e<y^2> (where q is are identity element in G/<y^2>.
    Need to show that G/<y^2>= <w,z|w^4=z^2=(wz)^2=e>.
    1. Since x and y are generates in G, then w and z are generates in G/<y^2>. I am not sure if this is right, could someone explain this to me? (I need to show that w and z generate G/<y^2>).
    2. w^4=e since (x<y^2>)^4=x^4<y^2>=e<y^2>=q.
    3. z^2=e since (y<y^2>)^2=y^2<y^2>=e<y^2>=q.
    4. (wz)^2=(wz)(wz)=e. Since wz=(z^-1)(w^-1), then wz=xy<y^2> and z^-1w^-1=y^3x^3<y^2>=yx^3<y^2>. Thus xy<y^2>=yx^3<y^2>.
    Therefore, G/<y^2> is isomorphic to D4.
    Last edited by aabsdr; May 23rd 2010 at 08:48 PM.
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  4. #4
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    how would I find the center for this group if the groups order was 16? I know that since xyxy^-1=e that xy=yx^3, so x=yx^3y^-1, so than x^2y=xyx^3=yx^6=yx^2. So x^2 and e is in the center. How would I show that nothing else is in the center besides <x^2>?
    Last edited by aabsdr; May 23rd 2010 at 10:32 PM.
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by aabsdr View Post
    how would I find the center for this group if the groups order was 16? I know that since xyxy^-1=e that xy=yx^3, so x=yx^3y^-1, so than x^2y=xyx^3=yx^6=yx^2. So x^2 and e is in the center. How would I show that nothing else is in the center besides <x^2>?
    Looking at the presentation you give, the third relation gives you something which looks like commutativity. Essentially, it allows you to write every element in the form x^iy^j for i, j \in \{0, 1, 2, 3\}. What you need to do is find the values of i,j such that x^iy^jx^ay^b = x^ay^ba^iy^j for all a, b. So just plug it in...

    Alternatively, noting that you are in D_{2n} you can do the same trick. And yes, your group will have a center of order 2 in this case as D_{2n} has a center of order 2 if and only if n is even. Otherwise, it has trivial center.
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  6. #6
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    Quote Originally Posted by aabsdr View Post
    Here is what I have:
    since xyxy^-1=e, we know that yxy^-1=x^-1=x^3, so we know that <x) is normal. So G= <x> union y<x> union y^2<x> union y^3<x> and |G|<= 16.
    Lets assume that |G|=16.
    So |G|/|<y^2>|=8, thus |G|/|<y^2>|>= |D4|.
    D4=<a,b|a^4=b^2=(ab)^2=e, or ab=ba^3>.
    Let w=x<y^2> , z=y<y^2> and q=e<y^2> (where q is are identity element in G/<y^2>.
    Need to show that G/<y^2>= <w,z|w^4=z^2=(wz)^2=e>.
    1. Since x and y are generates in G, then w and z are generates in G/<y^2>. I am not sure if this is right, could someone explain this to me? (I need to show that w and z generate G/<y^2>).
    2. w^4=e since (x<y^2>)^4=x^4<y^2>=e<y^2>=q.
    3. z^2=e since (y<y^2>)^2=y^2<y^2>=e<y^2>=q.
    4. (wz)^2=(wz)(wz)=e. Since wz=(z^-1)(w^-1), then wz=xy<y^2> and z^-1w^-1=y^3x^3<y^2>=yx^3<y^2>. Thus xy<y^2>=yx^3<y^2>.
    Therefore, G/<y^2> is isomorphic to D4.
    In G/<y^2>, simply plug y^2=1 into your presentation of G.

    Then, the original presentation reduces to <x, y|x^2 = y^2 = xyxy^{-1}=1>=<x, y|x^2 = y^2 = xyxy=1> , where y=y^{-1} since y^2=1. It is exactly the presentation of D_4 of order 4 (See my previous reply).

    To convert the presentation 1 into presentation 2 in my previous reply, substitue s for x and sr for y. Then you will see that both are equivalent presentations.
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by TheArtofSymmetry View Post
    In G/<y^2>, simply plug y^2=1 into your presentation of G.
    (It should perhaps be pointed out that relations in a presentation, homomorphic images and quotients are all equivalent. Thus, putting a relation into your presentation is the same as taking a quotient.

    This is because, formally, \langle X ; R=1 \rangle :\cong F(X)/<<R>> where F(X) is the free group over the alphabet X and <<R>> is the normal subgroup generated by the elements of R.)
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