Nontrivial finite group with trivial cohomology in prescribed degree For any non-trivial finite group $G$ there exists some $j > 0$ such that $H^{aj}(G) \neq 0$ for all $a = 1,2,3,\dots$, see e.g. this question: Non-vanishing of group cohomology in sufficiently high degree.
Furthermore, it is not known whether there exists a positive $N$ such that $H^i(G) \neq 0$ for $0 < i \leq N$ implies $G = 1$, see this question Nontrivial finite group with trivial group homologies?.
My question is: Given a positive integer $i$, does there always exist a non-trivial finite group $G$ with $H^i(G) = 0$? (All cohomology groups are meant to be with $\mathbb Z$ coefficients.)
 A: The qualifier "finite" is crucial, because you can always construct a general group with prescribed (co)homologies.
For $i$ odd, take any cyclic group $\mathbb{Z}_n$. In fact, its cohomology is trivial in all odd degrees and is isomorphic to the cyclic group in all even degrees. Or take $S_3$. It's a fun exercise to prove a more general result: $H^\text{odd}(G)=0$ if $G$ has periodic cohomology ($\mathbb{Z}_n$ has period 2 and $S_3$ has period 4). In particular, odd cohomology vanishes for any $p$-group $G$ which has a unique $\mathbb{Z}_p$ subgroup (it has period $2|G:\mathbb{Z}_p|$). But the exact opposite occurs in even degrees:
For $i$ even, $G$ cannot be a $p$-group. In fact, its cohomology is nontrivial in all even degrees (can explain why later). I currently cannot say more except for the following: The $p$-primary component of $H^{2k}(G)$ is isomorphic to the set of $G$-invariant elements of $H^{2k}(P)$, where $P$ is a Sylow $p$-subgroup. This makes our attempt more difficult. I will think more.
I briefly thought more: The binary icosahedral group $SL_2(\mathbb{F}_5)$ has trivial cohomology in degrees 2 mod 4. That leaves the case open for $i\in 4\mathbb{Z}$.
