# Finitely Generated Homology Groups

Let $$X$$ be a topological space and let assume that it's cohomology groups $$H^n(X; \mathbb{Z})$$ (therefore considered as $$\mathbb{Z}$$ modules) are finitely generated. That means by classification theorem for finitely genenerated abelian groups that for every $$n$$ we have $$H^n(X; \mathbb{Z}) \cong \mathbb{Z}^{r_n} \oplus \bigoplus_{p \text{ prime }} \oplus _{k \ge 1} (\mathbb{Z}/(p^k\mathbb{Z}))^{r_n ^{p^k}}$$

with appropiate $$r_n$$ and $$r_n ^{p^k} \in \mathbb{N}$$.

My question is if and how to see that in this case the homology groups $$H_n(X; \mathbb{Z})$$ are also finitely generated?

I tried to use the Universal Efficient Theorem (correctly U Cocoefficient Thm :) ) to get following exact sequence

$$0\to \operatorname {Ext} _{\mathbb{Z}}^{1}(\operatorname {H} _{i-1}(X;\mathbb{Z}),\mathbb{Z})\to H^{i}(X;\mathbb{Z})\,{\overset {h}{\to }}\,\operatorname {Hom} _{\mathbb{Z}}(H_{i}(X;\mathbb{Z}),\mathbb{Z})\to 0}$$

From this I can only conclude that $$\operatorname {Hom} _{\mathbb{Z}}(H_{i}(X;\mathbb{Z}),\mathbb{Z})$$ is also finitely generated, not more.

From here I can only conclude that the free part of $$H_{i}(X;\mathbb{Z})$$ finitely generated. But what about the torsion part $$T$$. Here I used the expression $$H_{i}(X;\mathbb{Z})= \mathbb{Z}^d \oplus T$$. By the way: Does there exits such splitting for non finitely generated abelian groups? If yes, what I know about $$T$$?

Or is there another way to show that $$H_n(X; \mathbb{Z})$$ are finitely generated?

• Also, keep in mind that an abelian group $A$ doesn't have a meaningful "free part and a torsion part". It has a torsion subgroup $T \subset A$, the set of elements with $na = 0$ for some $n > 0$. Then $A/T$ is torsion-free, but there is no need for it to be free. For instance, consider $A = \Bbb Q$. In that case, $\text{Hom}(A, \Bbb Z) = 0$, so you need Ext information to see that $H_i$ is finitely generated.
– user98602
Oct 13, 2018 at 19:41

If $$A$$ is not finitely generated, then either $$\text{Hom}(A,\Bbb Z)$$ or $$\text{Ext}(A,\Bbb Z)$$ is uncountable. Hence if $$H_n(X;\Bbb Z)$$ is not finitely generated, then either $$H^n(X;\Bbb Z)$$ or $$H^{n+1}(X;\Bbb Z)$$ is uncountable.