# Classification of the Extraspecial 2-groups $H_n$

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.

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 $$4.$$

• 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? – Francesco Polizzi Apr 3 '19 at 14:27
• @FrancescoPolizzi : I think so, but the F-S indicator can be computed entirely from characteristic zero character-theoretic information. – Geoff Robinson Apr 3 '19 at 14:29
• @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 . – Martin Seysen Apr 4 '19 at 9:21