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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 $\mathsf{H}(V, \, \omega)$. 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.

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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 $A$.

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    $\begingroup$ @DerekHolt : Yes, I was implicitly using the fact that $[x_{i},x_{j}]$ is central in this case (from the given relations), but perhaps I should have made this explicit. $\endgroup$ Commented Jan 26, 2020 at 11:14
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    $\begingroup$ @FrancescoPolizzi : You only need to check $[x_{k},x_{1}^{b_{1}} \ldots x_{2n}^{b_{2n}}]$ for each $k$, and this turns out to be $z^{b_{1}a_{k1} + \ldots + b_{2n}a_{k 2n}}$ in each case. $\endgroup$ Commented Jan 26, 2020 at 11:18
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    $\begingroup$ @DerekHolt : I am not sure whether this was addressed to Francesco or me (or both). The question was to give a "direct" proof without using symplectic forms. But of course, extraspeciaal groups and symplecctic forms are inseparable, and it is certainly the case that the symplectic form proof and the above "direct" group-theoretic proof are equivalent, just using a different language. $\endgroup$ Commented Jan 26, 2020 at 11:56
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    $\begingroup$ Geoff and Derek: yes, for the reasons your are giving I suspect that every proof will be equivalent to the original one. However, I would like to comply with the referee's request, possibly learning some math during the process. Thanks again for your help. $\endgroup$ Commented Jan 26, 2020 at 12:10
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    $\begingroup$ For the first question, I think everything does go through with hardly any modification. Calling $G(A)$, the new group, it is clear that $G(A)/\langle z \rangle$ is elementary Abelian of order $p^{2n}$, so $G(A)$ has class at most $2$ and all its commutators are central (and, from general properties of class $2$ groups, the exponent is $p^{2}$). The linear independence of the rows of $A$ is equivalent to the fact that $\langle z \rangle$ is precisely the center of $G(A)$, just as before. As for the last question, yes, one such relation forces the group to have exponent $p^{2}$. $\endgroup$ Commented Jan 30, 2020 at 11:43

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