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Thread: Abelian groups and Homology groups

  1. #1
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    Abelian groups and Homology groups

    Given an abelian group $\displaystyle G$, and a natural number $\displaystyle n$, does there exist a space X such that $\displaystyle H_n(X)=G$?
    Where $\displaystyle H_n(X)$ is the n-dimensional homology group of X.
    I know that for every group $\displaystyle G$, there is a 2-dimensional cell complex $\displaystyle X_G$ such that the fundamental group of $\displaystyle X_G$ is isomorphic to $\displaystyle G$.
    Can we generalize this result to the homology groups, when $\displaystyle G$ 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 $\displaystyle H_{k+1}(SX)\cong H_k(X)$ 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 $\displaystyle X_G$ would work for $\displaystyle H_1$. (i.e. $\displaystyle H_1 (X_G) = G$ ).

    And for $\displaystyle H_2$, from the recursive formula with the suspension, $\displaystyle SX_G$ should work, and for$\displaystyle H_3$, S(SX_G) should work etc...., so for $\displaystyle H_n$ , the space $\displaystyle X$ we're looking for is $\displaystyle S(S(S...(S X_G)))$ 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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