cohomology of Eilenberg-Maclane space

Question

In line 5, Page 394 of Allen Hatcher's book Algebraic Topology, it is claimed that $H^n(K(G,n);G)=Hom(H_n(K(G,n),\mathbb{Z});G)$ for any abelian group $G$. How to get it?

I have tried but cannot continue: by universal coefficient theorem, we need $Ext(H_{n-1}(K(G,n);\mathbb{Z}),G)=0$. This holds if $H_{n-1}(K(G,n);\mathbb{Z})$ is free.

Thanks a lot.

Answer 1

I think the group $H_{n-1}(K(G,n);\mathbb{Z})$ is actually zero. This follows from Hurewicz theorem.