# Homology Groups are Abelian Groups?

I am talking about singular homology here. I can recall that a chain complex is sequence of groups $$C_n$$ along with maps $$f_n: C_n \to C_{n-1}$$ where $$f_{n+1}f_n=0$$ or in other words $$im(f_{n+1}) \subset ker(f_n)$$

Now given a Topological Space $$X$$. We define a singular-p-simplex to be a continuous map from $$\sigma_p \to X$$. Here $$\sigma_p$$ is $$p-simplex$$ which is defined to be the convex hull of $$\{e_1, e_2, \dots e_{p+1} \}\in \mathbb{R}^{p+1}$$.

Next we define $$C_n=S_n$$ to be the 'free' group generated by all $$n-$$ simplices. Since this is a free group then $$S_n$$ is not abelian. Further we have the map $$\delta:S_n\to S_{n-1}$$ to be the map $$\delta(f)=\sum_{i=0}^{i=n} d_i(f)$$ . This forms a chain complex.

Then we define the homology gr)/oup of $$X$$ to be $$\frac{ker{(\delta_n)}}{im(\delta_{n+1})}$$. However I don't see why $$H^n(X)$$ always turns out to be abelian group?

Further how can I find spaces with given homology groups?

• oh, so only $H^0$ is abelian. Can you give some example where the homology groups of a space isn't abelian? Commented Nov 12, 2022 at 18:52
• Sorry I was mistaken because I confused with another homology theory. The point here is that it is a free abelian group, not a free group. Thus it is abelian by construction. Commented Nov 12, 2022 at 18:57
• In the definition of singular homology, $C_n$ is not the free group generated by..., but the free abelian group generated by... As for your final question, you should post it separately. Commented Nov 12, 2022 at 18:57

Next we define $$C_n=S_n$$ to be the 'free' group generated by all $$n-$$ simplices. Since this is a free group then $$S_n$$ is not abelian.

No, this is incorrect. We define them as free abelian groups. Meaning the direct sum of copies of $$\mathbb{Z}$$, one for each simplex. And so it is abelian by definition.

The boundary maps don't work well over just free groups. In particular I don't think $$f_{n+1}f_n=0$$ holds. Moreover you wouldn't be able to form a quotient, because in non-abelian case $$ker f_n$$ doesn't have to be normal in $$im f_{n+1}$$.

Further how can I find spaces with given homology groups?

• can you explain me what is the difference between free group generated by $a,b$ and free abelian group generated by $a,b$. Is it that the free abelian group is quotiented by the commutator subgroup? That is does free abelian group mean $\frac{ a b }{ \langle aba^{-1} b^(-1} \rangle}$ ? Commented Nov 12, 2022 at 19:17
• @permutation_matrix the difference is precisely that in the first one $ab\neq ba$ while in the second one $ab=ba$. And in fact the free abelian group can be defined as abelianization (quotient by commutator) of free group. Typically though we just think about it as some $\bigoplus\mathbb{Z}$. Google and read about both on wiki. Commented Nov 12, 2022 at 19:19
You are confusing your 'free' groups. $$C_n$$ is the free abelian group, not the free group, generated by all $$n$$-simplices. A free abelian group on a set of generators $$I$$ is just the group $$\bigoplus_{i\in I} \mathbb{Z}i$$, its elements are simply formal integral linear combinations of the generators. As a consequence, homology groups are abelian as quotients of abelian groups.