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The Kernel of Z (mod 24) X Z (mod 81)

  1. May 1, 2007 #1
    1. The problem statement, all variables and given/known data
    I want to find the kernel of PHI: Z-> Z (mod 24) X Z (mod 81)

    I am beginning to think that the kernel of this is actually just the set containing the identity element, or the trivial subgroup of Z mod 24. I am thinking this because none of the subgroups of Z mod 24 are equal to the identity of Z mod 81. Am I thinking about this correctly??
     
    Last edited: May 1, 2007
  2. jcsd
  3. May 1, 2007 #2

    Dick

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    PHI(0)=0X0. PHI(3)=0X0. Right? The group operation is the sum mod whatever, correct?
     
    Last edited: May 1, 2007
  4. May 1, 2007 #3
    Yes that is correct... therefore is the kernel of phi the set {0,3}?
     
  5. May 1, 2007 #4

    Dick

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    No. Because I goofed. I meant to say PHI(648)=0x0. Duh. Sorry, it's getting late. Where did I get 648? Could you check that, please? (Actually the real question is where did I get 3? Getting punchy?). And remember the kernel is a subgroup, I think we discussed this. What do subgroups of Z look like?
     
    Last edited: May 1, 2007
  6. May 1, 2007 #5
    Well the subgroup diagram for Z mod 24 is
    Z24
    / \
    2 3
    / \ /
    4 6
    / \ /
    8 12
    \ /
    24
     
  7. May 1, 2007 #6

    Dick

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    Too complicated. What's the identity of Z(mod24)XZ(mod81)? How could an element of Z map into it?
     
  8. May 1, 2007 #7

    StatusX

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    Note that phi(1) determines phi for all integers. There is a simple relation between the order of phi(1) and the kernel of phi. By the way, you don't have enough information to get one answer, but you can narrow down the possibilities.
     
  9. May 1, 2007 #8
    I dont know, I am totally confused right now... I would guess the identity element is 1.
     
  10. May 1, 2007 #9

    Dick

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    I was assuming the statement of the question was not only stating the domain and range but that literally phi(z)=(z mod 24)x(z mod 81). Do you think I'm wrong? I already gaffed one.
     
  11. May 1, 2007 #10

    Dick

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    If the group operation is sum, then the identity is not likely to be 1. Or 1x1.
     
  12. May 1, 2007 #11
    Ok.. the identity elements in the group Z mod 81 are 0, 81, 162, etc (81Z).
    Doesnt that mean that the ker(phi) must be {0} or the identity element of Zmod24??
     
  13. May 1, 2007 #12

    Dick

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    The elements of Z mod 81 are 0....80. 81 mod 81=0. Z mod 81 has only 81 elements. There is no element called 81. Same for Z mod 24. Maybe this is why you are finding this difficult. It's a quotient group, and you said you were vague about that. Maybe this will help. Z mod 81 is the quotient group Z/H where H is the subgroup of Z comprising all multiples of 81.
     
  14. May 1, 2007 #13

    StatusX

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    Maybe you're right. Usually the notation f:A->B just specifies the domain and range though. If you're right, then the notation the OP's using is so economical as to be confusing.
     
  15. May 1, 2007 #14

    Dick

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    The OP didn't state it was a homomorphism either. In which case you would have nothing to go on. rocky, can you clarify, if it's just given to be a homomorphism then as StatusX points out there are other possibilities. Like PHI(x)=0x0 for all x.
     
  16. May 1, 2007 #15
    Yes im sorry I didnt make the question more clear.... here is the complete question as given in the book...

    Let phi:Z->Zmod24 x Zmod81
    be the homomorphism defined by phi(m)=([m]mod24,[m]mod81) for m in Z.

    Find the kernel of phi.
     
  17. May 2, 2007 #16

    StatusX

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    Well my hint still stands: how does the order of phi(1) relate to the kernel of phi?

    Note that for any group G, and any g in G, there is a homomorphism psi from Z to G with psi(1)=g, from which you get that psi(n)=psi(1+1+...+1)=psi(1)*psi(1)*...*psi(1)=g*g*...*g=g^n (incidentally, this is because Z is what's known as a free group, which means there are no relations between its elements other than the bare minimum ones that make it a group). The image of Z under psi is <g>, the cyclic group generated by g, with size the order of g.
     
  18. May 2, 2007 #17

    Dick

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    It might also be a good exercise to write out PHI(n) for a number of values of n. Like PHI(5)=(5,5), PHI(6)=(6,6). Concentrate hard when you get to numbers like n=24,48,72,81 etc.
     
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