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Tricky group problem!

  1. Apr 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider the group [itex]D_{4} = <x,y:x^2=1,y^4=1,yx=xy^3>[/itex] and the homomorphism [itex]\Phi : D_{4} \rightarrow Aut(D_{4})[/itex] defined by [itex]\Phi (g) = \phi _{g}[/itex], such that [itex]\phi _{g} = g^{-1}xg[/itex].

    (a) Determine [itex]K = ker(\Phi)[/itex]

    (b) Write down the cosets of [itex]K[/itex].

    (c) Let [itex]Inn(D_{4}) = \Phi (D_{4})[/itex]. Then, [itex]\Phi : D_{4} \rightarrow Inn(D_{4})[/itex] is surjective. Exhibit the correspondence in the Correspondence Theorem explicitly.

    2. Relevant equations

    • Definition of the Correspondence Theorem
    • Definition of the Kernel
    • Definition of the Cosets

    3. The attempt at a solution

    I first list the elements in [itex]D_{4}[/itex], which are:

    [itex]{e,x,y,y^2,y^3,xy,xy^2,xy^3}[/itex]

    The order of [itex]D_{4}[/itex] is 8 since there are 8 distinct elements. My goal to answer the first problem is to find the elements in [itex]D_{4}[/itex] such that if I substitute the element from [itex]D_{4}[/itex] into [itex]\phi _{g}[/itex], then I obtain the identity. I would assume that the kernel of Φ consists of the cyclic groups of x², y^4 and x²y^4 since they yield the identity, and so substituting these elements for the function would give the identity. I may be wrong.

    I didn't answer the second part since I need to get down the kernel of the whole function.

    The third part might be related to the first two parts, but I'm not sure if the solutions are actually different.
     
  2. jcsd
  3. Apr 21, 2013 #2

    Dick

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    x^2, y^4 and x^2y^4 are all equal to e. If ##\phi_{g}## is going to be the identity then g must be in the center of the group. It must commute with all other elements of the group. e certainly works. One other element works.
     
  4. Apr 21, 2013 #3
    So the hint to find the kernel is to determine all elements in [itex]D_{4}[/itex] such that the elements are the identity, and if you substitute them for [itex]\Phi _{g}[/itex], then the elements commute?

    So this means that the kernel doesn't consist of cyclic groups? So the kernel is this?

    [itex]{e, x^2, y^4, x^2y^4, y^2x^2}[/itex]
     
  5. Apr 21, 2013 #4

    Dick

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    No, you don't want to find all of the elements that are the identity. There is only ONE element that is the identity. That's e. There is no difference between x^2 and e. They are the same thing. What you want to do is find all of the elements that commute with all the other elements. Can you see why? e is certainly one. There is another one that isn't e. What is it?
     
    Last edited: Apr 21, 2013
  6. Apr 21, 2013 #5
    Let me see...

    e is the kernel because:

    [itex]\Phi (e) = \phi _{e} = e^{-1}xe = x[/itex]

    I believe another element x would be in the kernel since:

    [itex]\Phi (x) = \phi _{x} = x^{-1}xx = x^{-1}x^2 = x^{-1}e = x^{-1} = x [/itex]

    OR

    [itex]\Phi (g) = \phi _{g} = g^{-1}xg[/itex]
    [itex]\Phi (g)g^{-1} = g^{-1}x[/itex]
    [itex]g^{-1}xgg^{-1} = g^{-1}x[/itex]

    If g = e, then...

    [itex]e^{-1}x(ee^{-1}) = e^{-1}x[/itex]
    [itex](e^{-1}xe)e^{-1} = e^{-1}x[/itex]
    [itex]xe^{-1} = e^{-1}x[/itex]

    If g = x, then...

    [itex](x^{-1}xx)x^{-1} = x^{-1}x[/itex]
    [itex]xx^{-1} = x^{-1}x[/itex]

    Is this the correct reason?
     
    Last edited: Apr 21, 2013
  7. Apr 21, 2013 #6

    Dick

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    You are getting there. But if ##\phi_g## is the identity then ##\phi_g(a)=a## for ALL a in the group. That means ##g^{-1}ag=a## or multiplying both sides by g, that ##ag=ga##. Now g=e works, as you've shown. g=x doesn't work. Take a=y, then yx is not equal xy. Your choice of g has to be an element that commutes with both x and y.
     
  8. Apr 21, 2013 #7
    I believe I got another element. Is it y², which is the part of the center of [itex]D_{4}[/itex]? I believe the center of that group relates to the kernel.

    You said that if you select an element from the group and multiply any element by that element, then the product commutes. I am not sure if y² is the part of the kernel.
     
  9. Apr 21, 2013 #8

    Dick

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    I'm not quite sure what you are asking. Sure, ##y^2 a=a y^2## for any element in the group. That's easy enough to check. Doesn't that mean ##\phi_{y^2}## is the identity map?
     
  10. Apr 21, 2013 #9
    Yes, it is the identity map, so e and y² are in the kernel. These are the elements in the kernel.

    Then, to exhibit the cosets of K, do I need to select each of the 8 elements from [itex]D_{4}[/itex] and multiply that by the kernel?
     
  11. Apr 21, 2013 #10

    Dick

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    You could. But you are only going to get 4 different subsets. One of them will be {e,y^2}, you just have to figure out the other 3. You shouldn't have to do all 8 products.
     
  12. Apr 22, 2013 #11
    How would I set up the correspondence between subgroups in both groups [itex]D_{4}[/itex] and [itex]Inn(D_{4})[/itex] for the third part? It says:

    Let [itex]Inn(D_{4}) = \Phi (D_{4})[/itex]. Then, [itex]\Phi : D_{4} \rightarrow Inn(D_{4})[/itex] is surjective. Exhibit the correspondence in the Correspondence Theorem explicitly.

    So to do this, I need to take the kernel from the previous parts. The order of the kernel is 2, and the order of [itex]D_{4}[/itex] is 8. Then, the order of [itex]Inn(D_{4})[/itex] is 4 since [itex]\Phi[/itex] is surjective. I would say that there are 4 pairs of corresponding subgroups.

    I know that the kernel of [itex]D_{4}[/itex] corresponds with the trivial subgroup in [itex]Inn(D_{4})[/itex], and that the whole group [itex]D_{4}[/itex] corresponds with [itex]Inn(D_{4})[/itex]. However, I don't know other corresponding subgroups. Here is the question:

    Would I say that there are two proper subgroups of order 2 in [itex]Inn(D_{4})[/itex], corresponding to the subgroups of order 4 in [itex]D_{4}[/itex]?
     
  13. Apr 22, 2013 #12

    Dick

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    Something like that. I don't really know the Correspondence theorem that well. I looked it up and that sounds like what you are supposed to do. There should be a 1-1 correspondence between 4 element subgroups of D4 and 2 element subgroups of D4/kernel(phi). I don't think there are only two of them, though. Seems to me there's one more. Can you write down what they are?
     
  14. Apr 22, 2013 #13
    Why would there be one more subgroup? I thought there would be four pairs of corresponding subgroups total based on the Correspondence Theorem. I need to make the list of them as the third part is given. I got down two subgroups, which are:

    • [itex]D_{4} \leftrightarrow Inn(D_{4})[/itex]
    • [itex]Kernel \leftrightarrow e[/itex]

    I can't seem to find the working subgroup. Any more hint or some thing you may want to share to me?
     
  15. Apr 22, 2013 #14

    Dick

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    I still don't know exactly what you want to show. Those aren't subgroups. I only vaguely understand what you mean by <-->. D4 has 3 subgroups of order 4. D4/kernel which is isomorphic to Inn(D4) has 3 subgroups of order 2. You should be able to match them up with each other. Can you actually show me any subgroups of either one that you've figured out?
     
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