Thread: Abelian groups and Homology groups

Given an abelian group , and a natural number , does there exist a space X such that ?
Where is the n-dimensional homology group of X.
I know that for every group , there is a 2-dimensional cell complex such that the fundamental group of is isomorphic to .
Can we generalize this result to the homology groups, when is abelian? If yes, how do you prove this

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 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.

So, since in this case, G is abelian, the same would work for . (i.e. ).

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

What if n=0?

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).