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Math Help - Abelian groups and Homology groups

  1. #1
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    Abelian groups and Homology groups

    Given an abelian group G, and a natural number n, does there exist a space X such that H_n(X)=G?
    Where  H_n(X) is the n-dimensional homology group of X.
    I know that for every group G, there is a 2-dimensional cell complex X_G such that the fundamental group of X_G is isomorphic to G.
    Can we generalize this result to the homology groups, when G is abelian? If yes, how do you prove this
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  2. #2
    Senior Member Tinyboss's Avatar
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    The 1st homology group is the abelianization of the fundamental group, so as you said, you can find a space with arbitrary H1 (although I believe the group G needs to be finitely presented). The suspension SX of a space X has the property that H_{k+1}(SX)\cong H_k(X) for all k, which will finish it up for you. This is exercise 2.2.32 in Hatcher, and is proved with the Mayer-Vietoris sequence.
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  3. #3
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    So, since in this case, G is abelian, the same X_G would work for H_1. (i.e. H_1  (X_G) = G ).

    And for H_2, from the recursive formula with the suspension, SX_G should work, and for  H_3, S(SX_G) should work etc...., so for H_n , the space X we're looking for is S(S(S...(S X_G))) where S is applied n-1 times?

    What if n=0?
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  4. #4
    Senior Member Tinyboss's Avatar
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    Yes, that looks exactly right.

    You can't do it for H_0, since that group is just free abelian on k generators, where k is the number of path components of the space (or one fewer for reduced homology).
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