The number of non isomorphic groups in Cext(G,C_p)

Question

Let $p$ be a prime number, $C_p$-cyclic group of order $p$, and $G$ an elementary p-group 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 $n$?

Answer 1

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 $C_{p^2} \times C_p^{n-1}$. 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 $G \cong E \times C_p^{n-2k}$, 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 $Z(G) \cong C_p^2 \times C_p^{n-2k+1}$, and $G \cong S_{p,k} \times C_p^{n-2k+1}$. For each $k$ with $0 \le 2k \le n-1$, there is a single isomorphism class of groups of this form.

Answer 2

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 so-called 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)