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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 $p^3$.

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    $\begingroup$ I know you asked for odd primes, but do you happen to know what happens with the generalized quaternion groups, which are $2$-groups? $\endgroup$
    – Pedro
    Commented Aug 31, 2018 at 15:19
  • $\begingroup$ Thank you! Yes, I am aware of what happens for cyclic and generalized quaternion groups (I read the exposition of Cartan-Eilenberg, chapter XII). $\endgroup$
    – Bemu
    Commented Aug 31, 2018 at 15:24
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    $\begingroup$ Interesting question.. Classification of $p$ groups whose third cohomology is zero... Can tac.mta.ca/tac/volumes/23/8/23-08.pdf be of any use?? $\endgroup$
    – Praphulla Koushik
    Commented Aug 31, 2018 at 15:25
  • $\begingroup$ Can you give a definition of your $p$-group? $\endgroup$
    – wonderich
    Commented Sep 1, 2018 at 3:58
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    $\begingroup$ A group whose order is a power of $p$ $\endgroup$
    – Bemu
    Commented Sep 1, 2018 at 8:04

1 Answer 1

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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 $p^3$ 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 $p$.

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  • $\begingroup$ If you take $G$ with a normal cyclic subgroup of order $p^2$ and quotient cyclic of order $p$, I wuold say that Lewis' paper shows that $H^3(G,\mathbb{Z})$ has order $p^2$: see Remark 2.2 (2) on page 504. Am I wrong? $\endgroup$
    – Filippo Alberto Edoardo
    Commented Sep 1, 2018 at 14:51
  • $\begingroup$ @FilippoAlbertoEdoardo That remark only says it’s a subgroup of a group of order $p^2$, I think. The comments following Prop. 5.1 say it’s trivial (I think). $\endgroup$
    – Jeremy Rickard
    Commented Sep 1, 2018 at 16:53
  • $\begingroup$ Oh yes, you're right: I read $=$ instead of $\leq$ in Remark 2.2(2): thanks! $\endgroup$
    – Filippo Alberto Edoardo
    Commented Sep 1, 2018 at 17:49

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