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Group homomorphism

  1. Aug 9, 2008 #1
    I'm looking for a formula isomorphism from the set of integers under multiplication mod 13 to the set of integers under addition mod 12. I know the other way around it's easily expressed as a power of class 2. But this way I have no idea if its expressible as a formula.
     
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  3. Aug 9, 2008 #2

    HallsofIvy

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    That should be straightforward. Have you written down the operation tables for the two sets?
     
  4. Aug 9, 2008 #3
    I have now, but actually now i'm even doubting the isomorphism at itself. The only thing i can think of is something with a logarithm, base 2 cause that converts a 2 into 1 and a product into a sum. But i cant verify it cause it doesnt invert the mapping of the power of class 2.
     
  5. Aug 9, 2008 #4

    HallsofIvy

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    Why "isomorphism"? Your original question was only about a "homomorphism".
     
  6. Aug 9, 2008 #5
    Okay i'm sorry, the title is wrong, i'm looking for a isomorphism. (In the end i'm looking for a homomorphism from the integers under multiplication mod13 to the complex numbers length 1.)
     
  7. Aug 9, 2008 #6

    HallsofIvy

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    Well, that looks straight forward. When I write down the two operation tables I see several things: first the identity of {Z13,*} is 1 while the identity of {Z12, +} is 0. Any isomorphism must map 1 into 0. I notice that 12*12= 1 mod 13 and that 6+ 6= 0 mod 12. That is, that 12 and 6 have the property that they are there own inverses in the respective groups. Since they are the only values that have that property, 12 must map into 6: f(12)= 6 for an isomorphism from {Z13,*} to {Z12, +}. I then notice that 5*5= 12 mod 13 and that 3+ 3= 6 mod 12. Again, that tells me that f(5)= 3. Unfortunately, there is no n such that n*n= 5 mod 13 but I do see that 12*5= 8 mod 13 and 6+ 3= 9 mod 12. We must have f(8)= 12. Continuing in that way you should be able to identify f(n) for all n in Z13.
     
  8. Aug 11, 2008 #7
    Hey thanks for the stated reasoning. Now, I'm asking if it is also correct to just take the inverse of all the couples I get from the relation k->2^k, but then I get a different mapping from yours: f={(1,0),(2,1),(3,4),(4,2),(5,9),(6,5),(7,11),(8,3),(9,8),(10,10),(11,7),(12,6)}
     
  9. Aug 11, 2008 #8
    okay i found out it is not okay, but can you explain me why not? i thought the decomposition theorem for group homomorphisms sais it should be correct?

    ps: i assume you meant f(8)=9
     
    Last edited: Aug 11, 2008
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