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Natural group homomorphism

  1. Jan 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Let G and G' be groups. Let H and H' be normal subgroups of G and G' respectively. Let
    f:G->G' be a group homomorphism.

    Show that if f(H) is a subset of H', then there exists a natural homomorphism f*:(G/H)->(G'/H'). (Hint: This fact is used constatntly in algebraic topology.)


    2. Relevant equations
    If X is a normal subgroup of a group Y, then the factor group Y/X can be formed where the group operation is coset multiplication.


    3. The attempt at a solution
    Using the assumptions that H is normal in G, I showed that f(H) is a normal subgroup of f(G).

    I formed the factor group f(G)/f(H) and defined a function f*:(G/H)->(f(G)/f(H)) by
    f*(aH) = (f(a))(f(H)) which turns out to be a homomorphism.

    This seems like a "natural" thing to do, so I think I'm on the right track. The problem is that my function f* is not mapping to the right place: I don't think that f(G)/f(H) is a subset of G'/H'.

    I also noticed that in proving the above, I did not make use of the hypotheses that H' is normal in G' or that f(H) is a subset of H'. Can these assumptions be used to show that
    f(H) = H'?

    Any hints would be greatly appreciated.
    Sam
     
  2. jcsd
  3. Jan 16, 2008 #2
    Admittedly, I haven't put anything down on paper here, but does a problem arise if you define your function as f*(aH) = f(a)H' ?
     
  4. Jan 16, 2008 #3

    morphism

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    Science Advisor
    Homework Helper

    Well, you are on the right track. Try drawing a diagram. We have 3 maps: f:G->G', and the two natural projection maps G->G/H and G'->G'/H'. Try to make the diagram commute.
     
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