If we have a simplical complex $K$, then we are able to define $C_i(K)$ as the free abelian group over $\mathbb Z_2$ with the basis of all $i$-dimensional simplices. By using the boundary map we are able to define a long sequence of $C_i(K)$ and as result we can compute the ith homology group $H_i(K,\mathbb Z_2)$.
Now let us go to the group homology for a given group $G$. $H_n(G,\mathbb Z_2)$ can be computed using a projective resolution of $\mathbb Z_2G$.
Now my question is How can I construct a simplicial complex $K$ associated to $G$ such that $H^i(K,\mathbb Z_2)$ is exactly the same as $H^i(G,\mathbb Z_2G)$?
I was thinking of the Cayley complex but unfortunately the maximum dimension of cayley complex is two and so it is not a good choice.