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?