Writing a paper on algebraic surfaces, I was led to consider the finite group $\mathsf{H}(A)$ whose presentation is the following.


I start with an anti-symmetric matrix $A=(a\_{ij})$ of order $2n$ over $\mathbb{Z}\_p$ (the finite group/field with $p$ elements), $p \geq 3$, and I consider the following system of generators and relations:


**Generators:**


$x\_1, \ldots, x\_{2n}, \; z$


**Relations:**
\begin{equation}
\begin{split}
x\_1^p &= \ldots=x\_{2n}^p=z^p=1 \\
[x\_1, \, z] &= \ldots = [x\_{2n}, \, z]=1\\
[x\_i, \, x\_j]& =z^{a\_{ij}}
\end{split}
\end{equation}
where, by slight abuse of notation, the exponent $a\_{ij}$ stands for any lifting in $\mathbb{Z}$ of $a\_{ij} \in \mathbb{Z\_p}$.


I can prove the following



> 
> **Proposition 1.** If $\det A \neq 0$, then the group $\mathsf{H}(A)$ presented as above is extra-special of order $p^{2n+1}$ and exponent $p$.
> 
> 
> 


*Proof*. Calling $\omega$ the symplectic form on the $\mathbb{Z}\_p$-vector space $V:=(\mathbb{Z}\_p)^{2n}$ whose matrix (with respect to the standard basis) is $A$, one checks that $\mathsf{H}(A)$ is isomorphic to the [Heisenberg group](https://en.wikipedia.org/wiki/Heisenberg_group) $\mathsf{H}(V, \, \omega)$. By standard linear algebra, up to a change of coordinates the symplectic form $\omega$ can be transformed into the standard symplectic form $\omega\_{\mathrm{st}}$, so our group is also isomorphic to $\mathsf{H}(V, \, \omega\_{\mathrm{st}})$, that has the desired properties, being the central product of $n$ copies of the $p$-group of order $p^3$ and exponent $p$. $\square$


Now, I was asked by the referee to give a *direct* proof of Proposition 1, namely a proof not involving symplectic forms. The motivation for this request is that the present proof is a bit far from the spirit of the paper, whose emphasis is mainly on group presentations.


Of course, I can give an equivalent proof involving nonsingular anti-symmetric matrices instead of symplectic forms, but I suspect that this is not what he/she/other pronoun was asking for. So let me ask the following two questions.



> 
> **Question 1.** Is there a proof of Proposition 1 essentially different from the one I gave?
> 
> 
> **Question 2.** Is there any precise reference for Proposition 1?
> 
> 
> 


I suspect that all of this is very well-known to the experts. Any hint to orientate myself in the extensive literature on $p$-groups would be greatly appreciated.



Direct proof (or reference) that a given $p$-group is extra-special
It is clear from standard properties on commutators that $H(A)$ is nilpotent of class at most $2$ with derived group and Frattini subgroup contained in $\langle z \rangle$
(just consider $H(A)/\langle z \rangle$, which is elementary Abelian of order $p^{2n})$.


The general property of commutators that you need is that we always have $[a,bc] = [a,b]^{c}[a,c]$. It is this, combined with the last relation(s), that ensures that the derived group of $H(A)$ is containedd in $\langle z \rangle$. It is necessary $A$ is antisymmetric, since we have $[x\_{j},x\_{i}] = [x\_{i},x\_{j}]^{-1}.$ 


The only remaining issue is check that $H(A)$ has center $\langle z \rangle$ (but no larger). This requires that $x\_{1}^{b\_{1}} x\_{2}^{b\_{2}} \ldots x\_{2n}^{b\_{2n}}$ is not central when $0 \leq b\_{i} \leq p-1$ and not all $b\_{i}$ are zero.


Using the above commutator relation $[a,bc] = [a,b]^{c}[b,c]$ repeatedly, and taking the commutator of the above element with each $x\_{k}$, we see that this is equivalent to the non-singularity of $A$. 


