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I need a homomorphism equation from (Q\{0},*) to (Z,+)

  1. Feb 29, 2012 #1
    1. Let G=(ℚ-{0},⋅), and let H be the subgroup H={(a/b)|a and b are odd integers}. Use the Fundamental Theorem to show that G/H≅(ℤ,+).



    2.



    3. I'm sure I can get the whole thing except that I just can't find the homomorphism to start with...?
     
  2. jcsd
  3. Feb 29, 2012 #2

    Deveno

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    suppose we have a rational number q = a/b , where gcd(a,b) = 1, and b ≠ 0.

    write q = (±1)(2k1)(3k2)....(ptkt)

    (that is factor a and b into prime factors, where pt is the largest prime occuring in the factorization of a or b).

    can you think of how to define f:Q*→Z so that ker(f) = H (hint: powers of 2).
     
  4. Feb 29, 2012 #3

    Dick

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    Think about what the elements of G that aren't in H look like. In other words, can you describe the cosets of G/H?
     
  5. Feb 29, 2012 #4
    Ok, so q= (±1)(2^k1)(3^k2)....(pt^kt) / (±1)(2^m1)(3^m2)....(pt^mt)

    Then ψ(q)=k1+m1 would be the function you are referring to because a,b both odd would produce 0+0=0 and nothing else would do so, right?

    I believe that also does satisfy the definition of homomorphism as well, right?
     
  6. Feb 29, 2012 #5

    Dick

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    That's not quite it. If k1=m1=1 then your fraction is in H after you divide the 2 out. So ψ(q) should still be 0.
     
  7. Feb 29, 2012 #6
    Sorry, Dick, that explanation is probably perfect, it's just a little abstract for me. Probably appropriate for an "Abstract Algebra" class (haha) but I'm trying to get this as mechanically as possible. I will definitely think on that, though! I need to make sure I understand that approach before the test. :cool: Thanks for pointing it out.
     
  8. Feb 29, 2012 #7

    Deveno

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    how do we write 1/(pk) in exponent form?
     
  9. Feb 29, 2012 #8
    Dick,
    "That's not quite it. If k1=m1=1 then your fraction is in H after you divide the 2 out. So ψ(q) should still be 0."

    I'm assuming first of all, of course, that a/b is reduced. I think that's a reasonable assumption, right?
     
  10. Feb 29, 2012 #9

    Dick

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    I'm just saying your ψ wrong. Because 2*3/2*1=3/1 IS in your subgroup H. And a element in H should map to the identity of G/H. Which is zero. Change your ψ. Just a little bit.
     
  11. Feb 29, 2012 #10
    Oh... I see your point. the multiplicities would subtract top-bottom. Ok. So it's a bit simpler than I stated it. Thanks!
     
  12. Feb 29, 2012 #11

    Dick

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    You could assume that a/b is reduced, but you don't have to. The same ψ will work if it's not.
     
  13. Feb 29, 2012 #12
    But assuming that it's reduced takes away the issue you were talking about, with the (2*3)/(2*1).
     
  14. Feb 29, 2012 #13
    Or, I guess, redefining a/b as (±1)(2^{k₁})(3^{k₂})(5^{k₃})....(p_{t}^{k_{t}} without the division I put in there and allowing negative k's, also solves the issue you are referring to. Ido like that better.
     
  15. Feb 29, 2012 #14

    Dick

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    Yeah, I was just trying to get you to think of ψ(q)=m1-k1. Which also works for (2*3)/(2*1) giving 0.
     
  16. Feb 29, 2012 #15
    Well, no, not really (by the very example you used: (2*3)/(2*1)). a and b are both even, so a/b not in H, but by this function, they would still produce 0.
     
  17. Feb 29, 2012 #16

    Dick

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    3/1 IS in H. And it's EQUAL to (2*3)/(2*1). How can (2*3)/(2*1) NOT be in H? (2*3)/(2*1) and 3/1 are the same number.
     
  18. Feb 29, 2012 #17
    Ok, I guess what I'm saying is that I don't have to assume they are reduced if I do it the m1+k1 way.
     
  19. Feb 29, 2012 #18
    Sorry, we *have to* assume that it is reduced, otherwise, an element 6/2 could both be in H and simultaneously not. Got it. My bad. I'm straightened out now! :cool:
     
  20. Feb 29, 2012 #19

    Deveno

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    we have to "reduce" somewhere...we can either put the "reduced form" in the definition of Q, or put the "reduced form" in the definition of H, but we can't neglect it all-together or our mapping may not be well-defined.

    (the rationals are kinda funny that way: 1/2 isn't a rational number, it's a representative of an entire equivalence class of rational numbers: 1/2, 2/4, 3/6 and so on).
     
  21. Mar 8, 2012 #20
    Just wanted all of you to know that my professor's comment was:
    "It is rare that students come up with a solution of this problem."
    I think he was impressed.
    You guys are awesome. Thanks!

    My final submission was:
    First, we *must assume that (a/b) is always reduced. Otherwise, (6/2)=(3/1) is both in H and simultaneously not in H, which cannot be.
    First, we need to find an "onto" homomorphism ϕ:G→(ℤ,+) such that ker(ϕ)=H.
    Let any q∈G be expressed as its prime factorization q=(±1)(2^{k₁})(3^{k₂})(5^{k₃})....(p_{t}^{k_{t}}) where t,k₁,k₂,k₃,...,k_{t}∈ℤ.
    Then k₁ represents the multiplicity of the factor 2 in the prime factorization of q.
    If k₁=0, then we know that a and b are both odd.
    (Side note: If k₁>0, then a is even and b is odd. If k₁<0, then b is even and a is odd. There is no possibility where they are both even, since (a/b) is always reduced.)
    So, let ϕ(q)=k₁ where k₁ is defined as described above.
    ϕ(q)=k₁ is clearly onto (ℤ,+), since k₁ can be any integer, positive, negative, or zero.
    We know that ϕ(q) is a homomorphism, since: for any (c/d),(f/g)∈G,
    If (c/d)=(±1)(2^{k₁})(3^{k₂})(5^{k₃})....(p_{t}^{k_{t}}) and (f/g)=(±1)(2^{m₁})(3^{m₂})(5^{m₃})....(p_{t}^{m_{t}}), then (c/d)⋅(f/g)=(±1)(2^{k₁+m₁})(3^{k₂+m₂})(5^{k₃+m₃})....(p_{t}^{k_{t}+m_{t}}).
    So, ϕ((c/d)⋅(f/g))=k₁+m₁=ϕ((c/d))+ϕ((f/g)).
    Then, by Thm 13.2 (Fundamental Theorem on group homomorphisms), G/ker(ϕ)≅(ℤ,+). Therefore, G/H≅(ℤ,+).
     
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