most recent 30 from http://mathoverflow.net 2013-05-24T23:43:08Z http://mathoverflow.net/feeds/question/39881 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39881/classification-of-p-groups-of-order-pn-with-rank-n-1 Maurizio Monge 2010-09-24T18:00:09Z 2010-09-24T20:42:32Z

<p>Hello, i've been looking for a way to classify the non-trivial $p$-groups $G$ that live in an exact sequence of the form $ 0 \rightarrow \mathbb{Z}/p\mathbb{Z} \rightarrow G \rightarrow (\mathbb{Z}/p\mathbb{Z})^{n-1} \rightarrow 0 $. Was this question settled before? Or is there any explicit computation of $H^2((\mathbb{Z}/p\mathbb{Z})^{n-1}, \mathbb{Z}/p\mathbb{Z})$? Thanks!</p>

http://mathoverflow.net/questions/39881/classification-of-p-groups-of-order-pn-with-rank-n-1/39900#39900 Steve D 2010-09-24T20:42:32Z 2010-09-24T20:42:32Z

<p>Your group is such that $|G|=p^n$ and $|\Phi(G)|=p$. Since $(C_p)^{n-1}$ is completely reducible, there is a subgroup $H$ of $G$ such that $G=HZ(G)$ and $H\cap Z(G)=\Phi(G)$. Thus $H$ is an extra-special group (possibly trivial), and we are taking the central product with the abelian group $Z(G)$, which is either of the form $(C_p)^m$ or $(C_{p^2})\times(C_p)^m$. The first case is easy, since again, it is completely reducible, so we get a group of the form (extra-special) times (some copies of $C_p$). The second case also gives (some group) times (some copies of $C_p$). I believe the (some group) is uniquely determined by its order (that is the central product of either of the two non-abelian groups of order $p^3$ and $C_{p^2}$ are isomorphic), but I haven't checked any cases but $p=2$.</p> <p>Steve</p>