# Eilenberg-MacLane spaces with $G$ an abelian torsion group

let $G$ be an abelian torsion group. let $z\in H_i(K(G,1))$. throughout the coefficient group is $\mathbb Q$.

1) Why there exists a FINITE subcomplex $X$, such that $j:X\hookrightarrow K(G,1),\, j_*(x)=z$ for some $x\in H_i(X)$ ?

2) Why the image of $\pi_1(X)$ in $\pi_1(K(G,1))=G$ is finite?

MY guess:

1) $G=\oplus_p G_p$ where $G_p$ ranges over all $p$-subgroups of $G$ so by Kunneth $$H_i(K(G,1))=\oplus_{i_1+\cdots+i_l=i}H_{i_1}(K(G_{p_1},1) \otimes \cdots\otimes H_{i_l}(K(G_{p_l},1 )$$ and I don't know what does it mean for $z$ to be in this decomposition!!

2) if $X$ is a finite CW that implies that its $\pi_1$ is finitely generated and not finite !!! and what finite CW means here anyway: having a finite number of cells at each dimension or having no cells beyond a dimension $n$ so that $X$ equals its $n$-skeleton?

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In (2): it is not true that if $X$ is a finite CW-complex, then its $\pi_1$ is not finite. There are lots of simply connected finite CW complexes: balls, for example! –  Mariano Suárez-Alvarez May 15 '11 at 21:06
@Mariano Suárez-Alvarez : i clearly meant "$\pi_1$ not necessarely finite" –  studento May 15 '11 at 22:05
I only made that comment because it was not clear to me… –  Mariano Suárez-Alvarez May 16 '11 at 2:59

Pick the usual simplicial model for $K(G,1)$. If $x\in H_i(K(G,1))$, then $x$ is the class of a cycle which involves finitely many simplices, which in turn involve finitely many group elements. Let $H$ be the subgroup generated by all these. Then $x$ is in the image of $$H_i(K(H,1))\to H_i(K(G,1)).$$ Moreover, if $X$ is the $n$-skeleton of $K(H,1)$ with $n>i+1$, then $x$ is in the image of the composition $$H_i(X)\to H_i(K(H,1))\to H_i(K(G,1)).$$
Since $H$ is finitely generated and torsion, it is finite, so $X$ is finite.
1) $x\in H_i(K(G,1))$ is represented by a cycle $\alpha\in C_i(K(G,1)) – studento May 15 '11 at 22:28 @student: I don't understand what you mean with that comment—I did fix a bit what I think is the relevant phrase, though. – Mariano Suárez-Alvarez May 16 '11 at 3:01 @ Mariano Suárez-Alvarez: i'm sorry my comment was badly edited.I rewrite it here. 1)$x\in H_i(K(G,1))$is represented by a cycle$\alpha\in C_i(K(G,1))$where$C_i(K(G,1))$is the free ab group on the$i$-simplices$s_i$so you mean there are finitely many$\alpha_i\in\mathbb Z$such that$\alpha=\sum{\alpha_is_i}$but why and how does this imply that it involves finitely many elements of the group$G$? 2) if$G_1\subset G_2$is there a map$K(G_1,1)\rightarrow K(G_2,1) $that induces a surjection on$H_i$like you say here? 3) why$H$finite implies$X\$ finite? thanks alot in advance? –  studento May 16 '11 at 5:07