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Homework Help: Homomorphisms from a free abelian group

  1. Mar 5, 2008 #1
    1. The problem statement, all variables and given/known data
    How many homomorphism are there of a free abelian group of rank 2 into a) Z_6 and b) S_3.


    2. Relevant equations



    3. The attempt at a solution
    Since the images of the generators completely determine a homomorphism, the upper bound for both is 36.
    Now a free abelian group of rank 2 is isomorphic to Z \times Z, which has basis {(1,0),(0,1)} and by the definition of a free abelian group, every nonzero element in the group can be uniquely expressed in the form a(1,0)+b(0,1) with a,b in Z. Therefore, given two arbitrary elements x and y in Z_6, we define phi((0,1)) = x and phi((1,0)) = y and for an arbitrary nonzero c = a(1,0)+b(0,1) in Z \times Z, we define [itex]\phi(c) = a\cdot x+b\cdot y[/itex]. And then define \phi(0) = 0.

    Now, why is that a homomorphism? Because if d and e are arbitrary nonzero elements in Z \times Z, such that d = a_1(1,0)+b_1(0,1) and e = a_2(1,0)+b_2(0,1), then we have
    [tex]\phi(d+e) = \phi((a_1+a_2)(1,0)+(b_1+b_2)(0,1)) = (a_1+a_2)x+(b_1+b_2)y = (a_1x+b_1y)+(a_2x+b_2y) = \phi(d)+\phi(e)[/tex]
    where in the second-to-last step I have used the fact that Z_6 is abelian. If d is 0 then we have [itex]\phi(d+e)=\phi(e)=\phi(e)+0=\phi(e)+\phi(0)[/itex].

    Therefore the answer to part a is 36. Please confirm that proof. I have absolutely no idea how to do part b since that group is not abelian.
     
  2. jcsd
  3. Mar 5, 2008 #2
    anyone?
     
  4. Mar 6, 2008 #3

    morphism

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    I quickly scanned your proof, and it looks alright. Although you can just use the universal property for free groups: if you send (1,0) and (0,1) to G, then this extends to a unique homomorphism of ZxZ into G. This is true for all free groups, not just abelian ones. So part (b) follows similarly.
     
  5. Mar 6, 2008 #4
    I am kind of confused about your last post. Please read the problem again. The answer to part b is 18 not 36 which means what you said cannot be true.
     
  6. Mar 8, 2008 #5

    morphism

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    Forget my last post. I'm not sure what was going through my head at the time.

    If f:ZxZ->S_3 is a homomorphism, then (ZxZ)/kerf =~ imf. What does this tell us?
     
    Last edited: Mar 8, 2008
  7. Mar 8, 2008 #6
    I am not sure what you are getting at. Oh wait--any quotient group of an abelian group must be abelian so the image of f must be abelian. Is that what you meant?
     
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