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.



primary decomposition for nonabelian cohomology of finite groups
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](http://www.math.ubc.ca/~adem/bourbaki-talk.pdf) 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 $VF$**](http://dx.doi.org/10.1007/s00222-002-0254-7), Invent. Math. 151, No. 1, 135-165 (2003). [ZBL1032.20035](https://zbmath.org/?q=an:1032.20035),


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.


