Let $Q\_8$ be the group of quaternions of order $8$. It is a non-abelian $2$-group such that $H^3(Q\_8,\mathbb{Z})=0$, where $\mathbb{Z}$ has the trivial action. For a proof, see the book "Homological Algebra" of Cartan and Eilenberg, Chapter XII, Section 7 (Examples), where the case of cyclic groups and generalized quaternions is also considered.


I am curious if more examples of this kind exist (for other primes). More precisely, let $p>2$ be an odd prime. Does there exists a finite (non-abelian) $p$-group $G$ such that $H^3(G,\mathbb{Z})=0$? I could not find anything even for groups of order $p^3$.



$p$-groups with trivial $H^3$
For $G$ a finite group, $H^3(G,\mathbb{Z})$ is isomorphic to the Schur multiplier, and you’ll find lots of examples using that as a search term (also, “Schur-trivial” is sometimes used to mean “having trivial Schur multiplier”).


For an example of order $p^3$, see *[The integral cohomology rings of groups of order $p^3$](https://doi.org/10.1090/S0002-9947-1968-0223430-6)* by Gene Lewis (Trans AMS, 132(2), p. 501-529, (1968)). The semidirect product $C\_{p^2}\rtimes C\_p$, with a generator of $C\_p$ acting on $C\_{p^2}$ by $x\mapsto x^{p+1}$ is an example for every $p$.


