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.)

  • 1
    $\begingroup$ A variant: given $i$, does there exist a finite group $G$ such that $H^j(G)=0$ for every $j$ with $0<j\le i$? $\endgroup$
    – YCor
    May 26, 2015 at 12:44
  • $\begingroup$ @YCor I'm pretty sure that's not known. Milgram has a paper calculating the cohomology of the Mathieu group $M_{23}$, which is zero in degrees $0<j\leq5$ and remarks that it's the first example of a finite group with vanishing cohomology in degrees $0<j\leq4$. As Chris Gerig says in his answer, there are small examples of finite groups with odd degree cohomology vanishing, and there are also examples (e.g. $SL(2,5)$) with cohomology only in degrees divisible by $4$. I'm far from an expert, but I don't think I know of any examples where $H^8(G)=0$. $\endgroup$ May 26, 2015 at 15:21
  • $\begingroup$ @ChrisGerig I think Milgram does show $H^5(M_{23})=0$. Note that it some places in the paper he talks about homology and in some about cohomology. $\endgroup$ May 27, 2015 at 8:08

1 Answer 1


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 $i\in 4\mathbb{Z}$.

  • $\begingroup$ Your formulae for the period doesn't seem to be correct: The quaternion group has order 8 but period 4. In general the quaternion group of order $2^n\,\,(n>2)$ has period $4$ (because it's a subgroup of $SU_2$ which is topologically a 3-sphere). $\endgroup$ May 26, 2015 at 9:14
  • $\begingroup$ True, what I wrote is not the minimal period. It does admit a periodic resolution of that period, which suffices as my intent. $\endgroup$ May 26, 2015 at 9:32

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