I have a sequence of groups $H\_n$ which I know to be extraspecial 2-groups 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 $2\sum\_{i=0}^{4i \leq 2n-1} {2n+1 \choose 2+4i}$. Is there a slick way of determining which extraspecial 2-group 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. 



Classification of the Extraspecial 2-groups $H_n$
This may be answered as follows: any extra-special $2$ group of order $2^{2n+1}$ is either the central product of $n$-copies of $D\_{8}$ or else is the central product of $n-1$ 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 Frobenius-Schur 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 $4.$


