Higher homology group of Eilenberg-Maclane space is trivial

I’m trying to solve the following exercise from Algebraic Topology by Hatcher (self-study):

Show that $ H_{n+1}(K(G,n);\mathbb{Z}) = 0 $ if $ n > 1 $.

$ K(G,n) $ is the Eilenberg-Maclane space.

I’m following a hint suggesting starting with a Moore space $ M(G,n) $. The idea
I have is to kill homotopy groups at dimensions $ > n $ by attaching cells of dimension $ > n + 1 $. By Hurewicz theorem the resulting space is $ K(G, n) $. The problem I have is that attaching cells of dimension $ n + 2 $ affects $H_{n+1}$ so it’s not necessarily $ 0 $. I don’t know whether I need to do something extra to ensure it stays $ 0 $.

This exercise is later used to prove the surjectivity of the Hurewicz homomorphism at $ n + 1 $. Hence we cannot use this fact here.

Help is appreciated. Thanks

Solutions Collecting From Web of "Higher homology group of Eilenberg-Maclane space is trivial"

Attaching cells of dimension $(n+2)$ can indeed change $H_{n+1}(X)$ for some space $X$, but it can only make it smaller. That is, if $X^\prime$ is $X$ with the $(n+2)$ cells attached, then $H_{n+1}(X^\prime)$ is a quotient of $H_{n+1}(X)$.

This follows from looking at what happens in the long exact sequence of homology associated to the cofibre sequence $\bigsqcup S^{n+1} \rightarrow X \rightarrow X^\prime$, where the first map is the characteristic map of the cells being attached.

Also, in your case to suffices to observe that if a CW-complex $Y$ has no $k$-cells, then $H_{k}(Y) = 0$, since the cellular chain complex (which can be used to compute the singular homology) is generated in degree $k$ by the set of all $k$-cells of $Y$.