Hello, i've been looking for a way to classify the nontrivial $p$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})^{n1} \rightarrow 0 $. Was this question settled before? Or is there any explicit computation of $H^2((\mathbb{Z}/p\mathbb{Z})^{n1}, \mathbb{Z}/p\mathbb{Z})$? Thanks!

$\begingroup$ For $p=2$ this is a simple exercise, and for general $p$ I don't think it is much more difficult. Have you looked into Huppert? $\endgroup$– Franz LemmermeyerSep 24, 2010 at 18:40

1$\begingroup$ Maybe you are interested in this en.wikipedia.org/wiki/Extra_special_group $\endgroup$– Michele TriestinoSep 24, 2010 at 18:41

1$\begingroup$ Extraspecial groups don't tell the whole story: consider $C_2\times D_8$. $\endgroup$– Steve DSep 24, 2010 at 20:02

$\begingroup$ Thanks for the comments, i already know about extraspecial groups and maybe i had to state a few examples of families of groups that statisfy the above requirements. I don't have yet enough families to conjecture that they cover all cases, but for instance for G you can have all groups of the form $C_p^{nk} \times E_k$ where $E_k$ is the extraspecial of order $p^k$, for odd $k$. You also have $C_p^{n3} \times M$ where $M = C_{p^2} \rtimes C_p$, and the groups you can obtain making the amalgamated product of a certain number of copies of $M$, $E_3$ and $C_{p^2}$. $\endgroup$– Maurizio MongeSep 25, 2010 at 13:34

$\begingroup$ Sorry, for "extraspecial" i was actually meaning "extraspecial of exponent p", forgetting that those with exponent $p^2$ are also called extraspecial. $\endgroup$– Maurizio MongeSep 25, 2010 at 14:00
1 Answer
Your group is such that $G=p^n$ and $\Phi(G)=p$. Since $(C_p)^{n1}$ 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 extraspecial 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 (extraspecial) 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 nonabelian groups of order $p^3$ and $C_{p^2}$ are isomorphic), but I haven't checked any cases but $p=2$.
Steve

$\begingroup$ Seems to be all ok, thanks for the answer. It is indeed easy to verify that the amalgamated products of the two nonabelian $p$groups of order $p^3$ with $C_{p^2}$ give the same group $(C_p\times C_{p^2})\rtimes C_p$, the action of the last $C_p$ being described as adding $p$ times the first coordinate of the $C_p\times C_{p^2}$ to the second. $\endgroup$ Sep 25, 2010 at 14:42