Counterexample (§1, p. 2) — quaternion group

There exists a group homomorphism $a:S_3\to\operatorname{Aut}(Q_8)$, where $S_3$ is the permutation group of the three-element set $\{0,1,2\}$ and $Q_8$ is the quaternion group of order $8$, with the following simultaneous properties. The semidirect product $E=Q_8\rtimes_a S_3$, whose multiplication is $(n,s)(n',s')=(n\,a(s)(n'),ss')$, is isomorphic as an abstract group to the group of invertible $2\times2$ matrices over $\mathbb Z/3\mathbb Z$. Let $H^1_b(A,Q_8)$, for an action $b:A\to\operatorname{Aut}(Q_8)$, denote the quotient of all maps $f:A\to Q_8$ satisfying $f(xy)=f(x)b(x)(f(y))$ by the equivalence relation $f\sim g$ when there exists one $n\in Q_8$ with $g(x)=n^{-1}f(x)b(x)(n)$ for every $x\in A$; no continuity is imposed in this definition. Then $H^1_a(S_3,Q_8)$ has exactly two elements, while for every natural prime $p$ and every Sylow $p$-subgroup $P$ of $S_3$, any two elements of $H^1_{a|_P}(P,Q_8)$ are equal. In addition, writing $B=\{(n,1):n\in Q_8\}$ and $C=\{(1,s):s\in S_3\}$, there exists a subgroup $J'\le E$ such that every element of $E$ has a unique expression $bj'$ with $b\in B,j'\in J'$, for every natural prime $p$ there exist Sylow $p$-subgroups $P_p\le C$, $Q_p\le J'$ and $e_p\in E$ with $e_pP_pe_p^{-1}=Q_p$, and there is no $e\in E$ with $eCe^{-1}=J'$. Here a Sylow $p$-subgroup is a subgroup maximal among those whose every element is killed by some power of $p$. The quantifiers include primes other than $2$ and $3$, whose Sylow subgroups in $S_3$ are trivial. Every cohomology set just defined contains the class of the constant map $1$, so the assertion that any two restricted classes are equal is not an assertion about an empty set. The cardinality-two assertion concerns a finite set and therefore does not use the convention assigning natural-number cardinal $0$ to an infinite set. The isomorphism with the matrix group is asserted to exist, and no particular choice of it or of the action $a$ is specified.