# Abelian groups and Homology groups

• Dec 4th 2010, 02:50 PM
math8
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
• Dec 5th 2010, 08:09 PM
Tinyboss
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.
• Dec 6th 2010, 08:43 AM
math8
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?
• Dec 6th 2010, 02:44 PM
Tinyboss
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).