MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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? – Franz Lemmermeyer Sep 24 '10 at 18:40
Maybe you are interested in this – Michele Triestino Sep 24 '10 at 18:41
Extra-special groups don't tell the whole story: consider $C_2\times D_8$. – Steve D Sep 24 '10 at 20:02
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 $C_p^{n-k} \times E_k$ where $E_k$ is the extraspecial of order $p^k$, for odd $k$. You also have $C_p^{n-3} \times M$ 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 $C_{p^2}$. – Maurizio Monge Sep 25 '10 at 13:34
Sorry, for "extraspecial" i was actually meaning "extraspecial of exponent p", forgetting that those with exponent $p^2$ are also called extraspecial. – Maurizio Monge Sep 25 '10 at 14:00
up vote 4 down vote accepted

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 $p=2$.


share|cite|improve this answer
Seems to be all ok, thanks for the answer. It is indeed easy to verify that the amalgamated products of the two non-abelian $p$-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. – Maurizio Monge Sep 25 '10 at 14:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.