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Isomorphic: PLease Help

  1. Jul 11, 2006 #1
    The 6th dihedral group is as follows:

    D6={e, a, a^2, a^3, a^4, a^4, a^5, b, ab, a^2b, a^3b, a^4b, a^5b}
    where a^6=b^2=e abd ba^i for all i in Z. Now I need to show whether D6 is isomorphic to G:
    Here are G:

    G= Z2 X Z2
    G=Z4

    if they are isomorphic I need to list the elements if they are not I need to prove that no such subgroups exists.

    so here it goes:
    G= Z2 X Z2= {e, a, a^2, b, ab, a^2b}// is this the right list because if it is it's isomorphic to D6. I'm not sure about this but I know that G= Z2 X Z2 is isomorphic to Z8.
     
  2. jcsd
  3. Jul 11, 2006 #2

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    Are you supposed to look for subgroups of D6 which are isomorphic to Z2 X Z2 or Z4? Because clearly D6 itself isn't isomorphic to either, since it has order 12 while these each have order 4.

    I don't know what you mean here. Z2 X Z2 is isomorphic to neither D6 nor Z8, by order arguments alone (the groups have order 4, 12, and 8 respectively.) Also, the set you mention is not a subgroup of D6, as it is not closed under the group operation. Start of by trying to list all the subgroups of D6.
     
    Last edited: Jul 11, 2006
  4. Jul 11, 2006 #3
    Given the elements of D6 I'm supposed to state whether z2xz2 and/or Z4 is isomorphic to D6. If none, then I'm supposed to prove why they are not. Hope this is clearer. Isn't the list that I have right for D6?
     
  5. Jul 11, 2006 #4

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    As I said, the answer is clearly no, since the groups have different numbers of elements. An isomorphism is first and foremost a bijection, and so any isomorphic groups must have the same number of elements.
     
  6. Jul 11, 2006 #5
    Ok, so How do I start the prove that neither one is not isomorphic to D6.
     
  7. Jul 11, 2006 #6

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    I've already explained the answer. Isomorphic groups must have the same number of elements, and these groups do not. What part is confusing you?
     
  8. Jul 12, 2006 #7
    But I did find a list of elements for part a that are isomorphic to D6 here's the list:
    {e, a, a^3, b}
     
  9. Jul 12, 2006 #8

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    That is a proper subset of D6, so it cannot be isomorphic to it. Did you mean to say it is isomorphic to Z2XZ2 or Z4? The problem is, that isn't a group, because it isn't closed under the group operation (for example, a*a=a^2, but a^2 does not belong to the set). An example of subgroup of D6 is the set {e,a,a^2,a^3,a^4,a^5}, because the product of any elements of this set is another element of the set, and the inverse of any element in the set is in the set. Are you having trouble understanding what a subgroup is?

    I'm still unclear on what your question is. Are you trying to prove D6 is not isomorphic to either Z2XZ2 or Z4? If so, the order argument I've mentioned suffices, but I can expand on this if you still don't understand it. If not, please be clearer about what you're asking.
     
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