primary decomposition for nonabelian cohomology of finite groups

Question

Let $G$ be a finite group, and let $M$ be a group on which $G$ acts (via a homomorphism $G\to \operatorname{Aut}(M)$).

If $M$ is abelian, hence a $\mathbb{Z}G$-module, there is a primary decomposition $$ H^k(G;M)=\bigoplus_p H^k(G;M)_{(p)} $$ for each $k>0$, where $p$ ranges over the primes dividing $|G|$ and $H^k(G;M)_{(p)}$ denotes the $p$-primary component of $H^k(G;M)$. It follows that if $H^1(G;M)$ is nonzero, then $H^1(P;M)$ is nonzero for some $p$-Sylow subgroup $P$ of $G$. See Section III.10 of K.S. Brown's "Cohomology of Groups".

If $M$ is nonabelian, then $H^1(G;M)$ is still defined, but is not a group in general, so that asking for a decomposition such as above would not make any sense. But one could still ask the following weaker question:

With $G$ and $M$ as above, suppose that $H^1(G;M)$ is non-trivial. Then must $H^1(P;M)$ be non-trivial for some $p$-Sylow subgroup $P$ of $G$?

Edit: Tyler Lawson's answer below is very informative, but unfortunately is not complete because the cohomology of a free product is not necessarily isomorphic to the direct product of the cohomologies. So the question still seeks an answer.

Answer 1

Yes, this is true.

Let $F$ be the free product of all the Sylow subgroups of $G$: $$ F = \mathop\ast_{P\text{ Sylow}} P $$ The inclusions $P \to G$ together give a group homomorphism $F \to G$. On cohomology, this induces the map $$ H^1(G;M) \to H^1(F;M) = \prod_{P} H^1(P;M) $$ and we are done if we can verify that this map is injective.

However, the map $F \to G$ is surjective: given any element $g \in G$ of order $n$, generating a copy of $\Bbb Z/n$, we can use the Chinese remainder theorem to write $\Bbb Z/n$ as a product $\prod \Bbb Z/p^r$ of primary groups. This allows us to conclude that $g = g^{e_1} g^{e_2} \dots g^{e_k}$ where each $g^{e_i}$ has order a prime power, hence is in a Sylow subgroup, and hence is in the image of $P$. Thus $g$ is in the image of the amalgamated product $F$.

This now reduces us to showing: If $F \to G$ is a surjective map of groups and $M$ is a nonabelian group with $G$-action, then the map $H^1(G;M) \to H^1(F;M)$ is injective. This is straightforward from the cocycle definition: if we have two cocycles $f, h: G \to M$ which become equal in $H^1(F;M)$, then by definition there is an element $m \in M$ such that $h(x) = m^{-1} \cdot f(x) \cdot {}^x m$ for all $x \in F$, but both sides of this identity only depend on the image of $x$ in $G$.

Answer 2

It turns out the answer is no. Here I'll sketch a counter-example in which $H^1(G;M)$ is non-trivial (in fact infinite), while $H^1(H;M)$ is trivial for all proper subgroups $H<G$.

Let $G=A_5$, the alternating group on $5$ letters. Then $G$ acts on the $2$-spine of the punctured Poincaré $3$-sphere without fixed points. Letting $L$ denote the barycentric subdivision of this $2$-spine, we obtain an acyclic $2$-dimensional flag complex with an admissible $G$-action for which $L^G=\emptyset$. This is a standard example in the theory of finite group actions, originally due to Floyd and Richardson. Since any proper subgroup $H<G$ is solvable, it follows from a theorem of Segev that $L^H$ is acyclic, and in particular non-empty. All of this can be found in this survey talk by Alejandro Adem.

Now let $\Gamma_L$ denote the right-angled Artin group associated to the $1$-skeleton $L^1$ of $L$, and let $M=M_L$ denote the kernel of the homomorphism $\Gamma_L\to \mathbb{Z}$ which sends every generator to $1$. This construction appears in work of Bestvina and Brady, who used it to exhibit groups with exotic finiteness properties.

The simplicial action of $G$ on $L$ induces an action of $G$ on the group $M$, so we can form the semi-direct product $M\rtimes G$. Now according to Theorem 3 of

Leary, Ian J.; Nucinkis, Brita E. A., Some groups of type $VF$, Invent. Math. 151, No. 1, 135-165 (2003). ZBL1032.20035,

since $L^G=\emptyset$ the group $M\rtimes G$ contains infinitely many conjugacy classes of subgroups which project isomorphically to $G$.

It follows that $H^1(G;M)$ is infinite. To see this, note that every such subgroup $\tilde{G}\leq G\rtimes M$ may be written as $\{(\varphi(g),g)\mid g\in G\}$ for some cocycle $\varphi:G\to M$. If cocycles $\varphi_1$ and $\varphi_2$ corresponding to $\tilde{G_1}$ and $\tilde{G_2}$ are cohomologous, then there exists $m\in M$ such that $m\varphi_1(g){}^g(m^{-1})=\varphi_2(g)$ for all $g\in G$, from which it follows that $(m,1)\tilde{G_1}(m,1)^{-1}=\tilde{G_2}$.

The same theorem of Leary and Nucinkis cited above also states that for any proper subgroup $H<G$, the fact that $L^H\neq\emptyset$ implies that all subgroups of $M\rtimes G$ which project isomorphically to a conjugate of $H$ in $G$ are themselves conjugate in $M\rtimes G$.

This implies that $H^1(H;M)$ is trivial. For let $\varphi:H\to M$ be a cocycle, and let $\tilde{H}=\{(\varphi(h),h)\mid h\in H\}$ be the corresponding subgroup. By assumption there exists $(m,g)\in M\rtimes G$ such that $(m,g)H(m,g)^{-1}=\tilde{H}$. Since $G$ is simple, $H$ is self-normalising and so $g\in H$, and one calculates that $$ (m,g)H(m,g)^{-1} = \{(m{}^{ghg^{-1}}(m^{-1})),ghg^{-1}\mid h\in H\} = \{(m{}^h(m^{-1}),h)\mid h\in H\}, $$ from which it follows that $\varphi=\varphi_m$ is principal.