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!



Classification of $p$-groups of order $p^n$ with rank $n-1$
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 $p=2$.


Steve


