Group cohomology of $S_3$ in terms of its Sylow subgroups

Question

I am trying to understand $H^*(S_3, M)$ in terms of it's Sylow $p$ subgroups. From III.10.2 and III.10.3 in Brown we know that \begin{equation}H^n(G,M) = \bigoplus_p H^n(H,M)^G\end{equation} where $p$ ranges over primes dividing $|G|$ and $H^n(H,M)^G$ is the $G$-invariant submodule. We say $z \in H^n(H,M)$ is $G$-invariant if $\text{res}_{H \cap gHg^{-1}}^{H} z= \text{res}_{H\cap gHg^{-1}}^{gHg^{-1}}gz$ for all $g \in G$.

For $G = S_3$, the Sylow subgroups we have to consider are isomorphic to $C_2$ and $C_3$. The cohomology of the cyclic groups are well as $$H^i(C_n, M) = \begin{cases}M/nM \text{ for } i = 2,4,6,\ldots \\\text{Ann}_M(n) \text{ for } i = 1,3,5,\ldots\\ M \text{ for } i = 0\end{cases}$$ where $\text{Ann}_M(n)$ is the $n$-torsion submodule of $M$.

Then, from https://groupprops.subwiki.org/wiki/Group_cohomology_of_symmetric_group:S3 we know $$H^i(S_3,M) = \begin{cases}M \text{ for } i = 0\\ \text{Ann}_M(2) \text{ for } i \equiv 1 \mod 4\\ M/2M \text{ for } i \equiv 2 \mod 4\\\text{Ann}_M(6) \text{ for } i \equiv 3 \mod 4\\M/6M \text{ for } i > 0, i \equiv 0 \mod 4\end{cases}$$

My question is how the period 2 cohomology for the cyclic groups can turn into period 4 cohomology for $S_3$. More specifically, applying the formula from Brown at each $i$, we get $$H^1(S_3,M) = \text{Ann}_M(2)^G \oplus \text{Ann}_M(3)^G$$ So, it seems that $\text{Ann}_M(3)^G$ should be 0 here? Then $$H^2(S_3,M) = M/2M^G \oplus M/3M^G$$ which also seems to imply $M/3M^G = 0$ here. Then $$H^3(S_3,M) = \text{Ann}_M(2)^G \oplus \text{Ann}_M(3)^G$$ Assuming that it is true that $\text{Ann}_M(2) \oplus \text{Ann}_M(3) = \text{Ann}_M(6)$, then it seems like here, $\text{Ann}_M(3)^G$ should be $\text{Ann}_M(3)$. Finally $$H^4(S_3,M) = M/2M^G \oplus M/3M^G$$ Similarly, here it seems that $M/3M^G$ should be $M/3M$.

It comes down to fact that I don't understand how the $G$-invariant submodules of the same cohomology groups, except at a different indices, could be different. Is there with understanding the formula from Brown is wrong or is my understanding of computing $G$-invariants flawed?

Answer 1

Let's take $M$ to be $Z/3$ with trivial $S _3$ action for concreteness' sake. Then we have $$H^*(C_3,M)\cong Z/3[x_2]\otimes \Lambda (e_1).$$ Now, $S_3/C_3\cong F_3 ^*$ acts on this ring just by multiplication by constant on $x_2$. So as $a^2=1$ for all $a \in F_3 ^*$ by Fermat's small theorem, invariants are $ Z/3[x_2^2]\otimes \Lambda (e_1).$ Thus we pass from period 2 to period 4. In other words, the $S_3$ action differs on dim $4n$ and $4n+2$.