Let $p$ be a prime number, $C_p$cyclic group of order $p$, and $G$ an elementary pgroup of order $p^n$. Let us denote by Cext$(G,C_p)$ the group of all central extensions of $C_p$ by $G$. Is the number of non isomorphic groups in Cext$(G,C_p)$ known as a function of $n$?

1$\begingroup$ Central extensions up to isomorphism would be classified by $H^2(C_p,G)\cong G$ whose cardinality is $p^n$. The nonisomorphic groups are probably classified by $H^2$ up to scalar operation of $C_p$. So I guess the answer is $n\mapsto p^{n1}$. $\endgroup$– Matthias WendtMay 17, 2014 at 20:12

$\begingroup$ Sorry, the arithmetic was wrong. The nontrivial extensions seem to be classified by $\mathbb{P}^{n1}(\mathbb{F}_p)$ whose cardinality is $\sum_{i=0}^{n1} p^i$. And then there is the trivial extension, one more. Sorry. $\endgroup$– Matthias WendtMay 17, 2014 at 20:23

1$\begingroup$ @MatthiasWendt When $n=2$ the answer is 4 regardless of $p$, while you claim $p+2$. (There are 5 isomorphism types of groups of order $p^3$ and all occur except the cyclic one.) $\endgroup$– YCorMay 17, 2014 at 20:56

1$\begingroup$ This question and it's answer might contain some useful information: mathoverflow.net/questions/75170/… $\endgroup$– Mark GrantMay 18, 2014 at 7:35

2$\begingroup$ You have not made it clear in your question whether you are looking for extensions with normal subgroup $C_p$ and quotient $C_p^n$, or the other way round. The expression "extension of $A$ by $B$" is used roughly equally often with both of those meanings. $\endgroup$– Derek HoltMay 18, 2014 at 9:47
2 Answers
I voted to close because I was unsure which way around the extension went but, as Yves said, the question is almost trivial if $C_p^n$ is the normal subgroup.
So, suppose that $N \unlhd G$ with $N=C_p$ and $G/N \cong C_p^n$.
Recall that a $p$group of this form is called extraspecial if $N=Z(G)$. It is a standard result that extraspecial groups have order $p^{2k+1}$ for some $k \ge 1$, and that for each $p$ and $k$ there are exactly two isomorphism classes of extraspecial groups. (They all arise as central products of extraspecial groups of order $p^3$. For $p$ odd, one of these groups has exponent $p$ and the other does not. For $p=2$, they are central products of $D_8$ and $Q_8$ and the isomorphism type depends on the parity of the number of $D_8$s or $Q_8$s.)
For each $k \ge 1$, we can define a group $S_{p,k}$, which has order $p^{2k+2}$, and is a central product of an extraspecial group of order $p^{2k+1}$ with $C_{p^2}$. It is not hard to show the two types of extraspecial groups give rise to isomorphic groups $S_{p,k}$.
So, in the problem, let $G= \langle z,x_1,x_2,\ldots,x_n \rangle$ with $z \in N$, and order the generators such that, for some $i$, $Z(G) = \langle z,x_{i+1},x_{i+2},\ldots,x_n \rangle$ .
If $i=0$, then $G$ is abelian and $G \cong C_p^{n+1}$ or $C_{p^2} \times C_p^{n1}$. So suppose that $i>0$.
Then $x_1,\ldots,x_i$ generate an extraspecial group, so $i=2k$ is even. Now there are two cases.
If $Z(G)$ is elementary abelian, then $G \cong E \times C_p^{n2k}$, where $E$ is extraspecial of order $p^{2k+1}$. For each $k$ with $0 < 2k \le n$, there are two isomorphism types of groups of this form, one for each of the two types of extraspecial group.
Otherwise $Z(G) \cong C_p^2 \times C_p^{n2k+1}$, and $G \cong S_{p,k} \times C_p^{n2k+1}$. For each $k$ with $0 \le 2k \le n1$, there is a single isomorphism class of groups of this form.

$\begingroup$ I answered this question some time ago on MO: mathoverflow.net/questions/39881/… $\endgroup$– Steve DMay 19, 2014 at 14:27

$\begingroup$ Ah right! I waited some time, hoping someone else would answer. I should have waited a little longer. $\endgroup$ May 19, 2014 at 14:39
I'd like to thank everyone who has responded to my query. The question was whether $I(n,p)$= # of non isomorphic groups in $\mathrm{Cext}(G,C_p)$ is known. My impression so far is that it is not. Regarding discussion, first $\mathrm{Cext}(G,A)$, with $A$ abelian stands for $\mathrm{Opext}(G,A,\text{triv})$, hence by the standard convention $A$ is a normal subgroup of extension. The number in question $I(n,p)=\lfloor\frac{3n+2}{2}\rfloor$ for odd $p$. The reason for exclusion of $p=2$ lies in the fact that $H^2(G,C_p)$ fits into the exact sequence $$0\to\widehat G\to H^2(G,C_p)\to\mathrm{Alt}(G)\to 0,$$ where $\widehat G$ is the dual group, splits up as $\mathrm{Aut}(G)$module iff $p$ is odd. I also show, as a consequence of a general theorem on abelian extensions of Hopf algebras, that the socalled weak isoclasses of some previous posts coincide with isoclasses of noncommutative extensions. Commutative extensions labeled by $\mathrm{Ext}(G,C_p)$ are also in bijection with the orbits, this is trivial. It follows that one has to classify the orbits of $\mathrm{Aut}(G)\times\mathrm{Aut}(C_p)$ in $H^2(G,C_p)$. Roughly speaking, for a pair $(f,\alpha)$ its invariant is $(\ker f,\text{rad}(\alpha)$. From this one gets the number. George Glauberman seems to have a proof that the formula for $I(n,p)$ holds for $p=2$. Details can be found at arXiv:1211.5621, look up the latest version (should appear shortly)

$\begingroup$ You say it is not known, but as Steve D pointed out in his comment to my answer, the question was answered completely in September 2010 in mathoverflow.net/questions/39881. The argument there is more or less the same as in my answer, and it covers all primes, not just odd primes. $\endgroup$ May 20, 2014 at 8:51

$\begingroup$ I agree the number in question should be easy to derive from Steve D's or your argument. My solution is along the lines of Matthias Wendt post(s) and earlier posts on this subject, and is complimentary to your. $\endgroup$ May 20, 2014 at 16:49