I have a sequence of groups $H_n$ which I know to be extraspecial 2groups of order $2^{2n+1}$. I also know the number of order 4 elements I have in $H_n$ for every $n$. Precisely, the number of order 4 elements is given by $2\sum_{i=0}^{4i \leq 2n1} {2n+1 \choose 2+4i}$. Is there a slick way of determining which extraspecial 2group I have in general. For $H_1$, the explicit paper calculations gave me $Q_8$ and for $H_2$ I got $Q_8*D_8$ (here $*$ denotes the $\mathbb{Z}_2$ central product). It appears knowing the number of order 4 elements should be sufficient to decide which one I have, yet I do not know a nice way of computing the number of order 4 elements for arbitrary $n$ copies of central products of Quaternion or Dihedral groups.
This may be answered as follows: any extraspecial $2$ group of order $2^{2n+1}$ is either the central product of $n$copies of $D_{8}$ or else is the central product of $n1$ copies of $D_{8}$ with one copy of $Q_{8}.$ The first type has all its complex irreducible representations realizable over $\mathbb{R}$, while the second type has $2^{2n}$ linear characters and one irreducible representation of degree $2^{n}$ which has FrobeniusSchur indicator $1$ (and which is not realizable over $\mathbb{R}$).
The number of solutions of $x^{2} = 1_{G}$ in $G$ of the first type is $\sum_{ \chi \in {\rm Irr}(G)} \nu(\chi)\chi(1) = 2^{2n}+2^{n},$ so a group of the first type has $2^{2n}2^{n}$ elements of order $4$, while the number of solutions of $x^{2} = 1_{G}$ in $G$ of the second type is $\sum_{ \chi \in {\rm Irr}(G)} \nu(\chi)\chi(1) = 2^{2n}2^{n},$ so a group of the second type has $2^{2n}+2^{n}$ elements of order $4.$

$\begingroup$ This should be equivalent to computing the Arf invariant of the natural quadratic form on the $\mathbb{Z}_2$vector space $G/Z(G)$, right? $\endgroup$ – Francesco Polizzi Apr 3 '19 at 14:27

1$\begingroup$ @FrancescoPolizzi : I think so, but the FS indicator can be computed entirely from characteristic zero charactertheoretic information. $\endgroup$ – Geoff Robinson Apr 3 '19 at 14:29

1$\begingroup$ @JakePatel The natural quadratic form on $G/Z(G)$ maps $g \in G$ to $c$ such that $g^2 = (1)^c$, where 1 is the unique central element of $G$ different from $1$. Thus counting the elements of order $4$ in $G$ corresponds to counting the vectors in $G/Z(G)$ where the natural quadratic form is not zero. The Arf invariant can be computed that way, see en.wikipedia.org/wiki/Arf_invariant . $\endgroup$ – Martin Seysen Apr 4 '19 at 9:21