Let $G$ be an extraspecial $2$-group, i.e. $Z(G)=G'=\Phi(G)$ has order $2$. Then $|G|=2^{2n+1}$ for some $n\geq 1$, $G\cong D_8^{*n}$ or $G\cong Q_8*D_8^{*(n-1)}$, and $G$ has one of the following presentations:
$$\langle x_1,y_1, ..., x_n,y_n\mid [x_i,x_j]=[y_i,y_j]=1, [x_i,y_j]=1 \mbox{ for } i\neq j,$$ $$\hspace{-10mm}[x_i,y_i]=z \mbox{ and } x_i^2=y_i^2=z^2=1 \mbox{ for } z\in Z(G)\rangle$$or
$$\langle x_1,y_1, ..., x_n,y_n\mid [x_i,x_j]=[y_i,y_j]=1, [x_i,y_j]=1 \mbox{ for } i\neq j,$$ $$\hspace{-6mm}[x_i,y_i]=z \mbox{ and } y_i^2=z, x_i^2=z^2=1 \mbox{ for } z\in Z(G)\rangle.$$Also, $G$ possesses elementary abelian subgroups of order at most $2^{n+1}$. Is it known an explicit formula for the number of elementary abelian subgroups of order $2^k$, $k=1,...,n+1$, of $G$ containing $Z(G)$?
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3$\begingroup$ It's the same thing as the number of totally isotropic subspaces of $G/Z(G)$, for the quadratic form defining the extension (= the cohomology class). For this, start with mathoverflow.net/questions/102615/… and follow the references. $\endgroup$– PierreCommented Sep 29, 2018 at 12:10
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$\begingroup$ @Pierre - is this clear? If $H\subset G$ is a subgroup on which the commutator form vanishes, that just makes $H$ abelian, not nec. elem. abelian. E.g. the subgroup of $Q_8$ generated by $i$ corresponds to a subspace of $Q_8/Z(Q_8)$ with triv. commutator form, but it is not elem. ab. $\endgroup$– benblumsmithCommented Sep 29, 2018 at 21:38
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$\begingroup$ (Aside: it seems to me the def. of "totally isotropic" given in the linked question differs from the Wikipedia def: en.wikipedia.org/wiki/Isotropic_quadratic_form. Does it mean all the vectors in the subsp. are orth. to each other, or just themselves?) $\endgroup$– benblumsmithCommented Sep 29, 2018 at 21:40
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$\begingroup$ @benblumsmith a subspace $W$ of a space $V$ with a quadratic form $q$ is totally singular if $q(w)=0$ for all $w \in W$. That implies that it is also totally isotropic i.e. $\beta(v,w)=0\, \forall v,w \in W$, where $\beta$ is the associated bilinear form. As Pierre said, elementary abelian subgroups of $G$ that contain $Z(G)$ correspond exactly to totally isotropic subspaces of $G/Z(G)$. $\endgroup$– Derek HoltCommented Sep 30, 2018 at 11:06
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$\begingroup$ @DerekHolt - I trust you completely, but I am confused by both claims. You and Pierre must be referring to a different quadratic form than I am thinking of? I assumed Pierre meant the commutator form $\mathbb{F}_2^n=G/Z(G)\rightarrow [G,G]=Z(G) = \mathbb{F}_2$ given by $\langle \bar a,\bar b\rangle =[a,b]$, where $a,b\in G$ represent $\bar a, \bar b\in G/Z(G)$. This is a skew form on a field of characteristic 2. All vectors satisfy $q(v)=0$, but many do not satisfy $\beta(v,w)=0$. $\endgroup$– benblumsmithCommented Oct 1, 2018 at 12:20
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