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?


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.
