# rational cohomology of symmetric groups

Let $\Sigma_k$ be the $k$-th symmetric group and $B\Sigma_k$ be its classifying space. How to prove:

for any $n\geq 1$ and the $n$-skeleton $sk_n (B\Sigma_k)$, there exists a finite dimensional $CW$-complex $K$ such that

(i). $sk_n(B\Sigma_k)\subseteq K\subseteq B\Sigma_k$;

(ii). $H^*(K;\mathbb{Q})$ is trivial?

Could I just let $K=B\Sigma_k$?

• $B\Sigma_k$ is not finite dimensional. – Jim Conant Nov 18 '15 at 3:13
• @JimConant. Yes. How to prove the original question? – Quan Nov 18 '15 at 3:17

Take the $n$-skeleton. It has trivial rational homology except possibly in degree $n$. Now add enough $n+1$-cells from the $n+1$-skeleton to kill this top homology. You won't have created any $n+1$-dimensional homology since the boundary operator $\partial_{n+1}\colon C_{n+1}\to C_n$ is a rational isomorphism onto $\ker(\partial_n)$. This construction works for any finite group, not just the symmetric group.
If you want to prove that group cohomology $H^{\ast}(\Sigma_{n},\mathbb{Q})=0$ you use the fact that $\mathbb{Q}$ is a projective $\mathbb{Q}[\Sigma_{n}]$-module (Maschke’s theorem) therefore by definition $$H^{\ast}(\Sigma_{n},\mathbb{Q})=Ext^{\ast}_{\mathbb{Q}[\Sigma_{n}]}(\mathbb{Q},\mathbb{Q})=0$$ for $\ast >0$.
In general if $G$ is a finite group and $K$ a field such that its characteristic does not divide $|G|$, then $$H^{\ast}(G,K)=0$$ for $\ast >0$.