# Abelian groups and Homology groups

• Dec 4th 2010, 01:50 PM
math8
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
• Dec 5th 2010, 07:09 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 \$\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.
• Dec 6th 2010, 07:43 AM
math8
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?
• Dec 6th 2010, 01:44 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).