Let $G = Sym(a) \wr Sym(b)$ be a wreath product of symmetric groups - I'm particularly interested in the Weyl group of type $B$, $Sym(2) \wr Sym(n)$. Let $k$ be a field of characteristic $p$.


What is $H^\*(G,k)$? 


If $i \leq p-3$ and we're in the symmetric group case, then $H^i(Sym(n), k)=0$. 


If $i=1$ and $G$ is as above then $H^1(G, k)=0$, for all $a$ and $b$, unless we're in characteristic 2.


Is anything else possible to say? What I REALLY want is that the symmetric group result generalises so that:


If $i \leq p-3$, then $H^i(G, k)=0$ for $i \leq p-3$, for $G$ a wreath product as above.


Any ideas if this is true? It might be that it holds with fewer restrictions on $i$ and $G$ - the proof I know for the symmetric group case uses the Schur functor and tilting modules for $GL\_n$. However, the result proved is FAR more general (it concerns all Specht modules) - so maybe this $GL\_n$ approach isn't needed.



Cohomology $H^*(G,K)$ of wreath products
Set $H^\ast(-) := H^\ast(-,k)$ and $S\_a = Sym(a)$. Let $G$ be the wreath product that fits into the extension 
$$ 1 \to S\_a^b \to G \to S\_b \to 1.$$



> 
> **Claim:** $H^n(G) = 0$ for $1 \le n \le p-3.$
> 
> 
> 


*Proof:* The LHS spectral sequence corresponding to the extension is 
$$E\_2^{pq} = H^p(S\_b, (H^\ast(S\_a)^{\otimes b})^q)$$
where 
$$(H^\ast(S\_a)^{\otimes b})^q = \oplus\_{i\_1 + ... + i\_b = q} H^{i\_1}(S\_a) \otimes \cdots \otimes H^{i\_b}(S\_a).$$


Let $1 \le q \le p-3$. Then $i\_j \le p-3$ for $j=1,...,b$ and not all $i\_j$ can be zero. Hence $H^i(S\_a)=0$ for $1 \le i \le p-3$ implies $(H^\ast(S\_a)^{\otimes b})^q = 0$. 


Thus $E\_2^{\ast,q}=0$ for $1 \le q \le p-3$ and $E\_2^{p,0} = 0$ for $1 \le p \le p-3$. This shows $H^n(G) = 0$ for $1 \le n \le p-3$. 


**Remarks:** 1) Even more is true for wreath products: 
$$H^\ast(G) \cong H^\ast(S\_b,(H^\ast(S\_a)^{\otimes b})$$ 
as graded rings (cf. Nakaoka: Homology of the Infinite Symmetric Group, Ann. of. Math. 73(1961),229-257, Theorem 3.3). 


2) Since the extension splits, $H^\ast(S\_b)$ is a direct summand of $H^\ast(G)$. Hence the vanishing range for the cohomology of $H^\ast(G)$ stated above cannot be improved. 


