# A new family of $(q^{4}+1)$-tight sets with an automorphism group $F_{4}(q)$

Tao Feng∗, Weicong Li, Qing Xiang† Tao Feng, School of Mathematical Sciences,
Zhejiang University, Hangzhou, China<EMAIL_ADDRESS>Weicong Li,
Department of Mahtematics, School of Sciences, Great Bay University, Dongguan,
China<EMAIL_ADDRESS>Qing Xiang, Department of Mathematics and
Shenzhen International Center for Mathematics, Southern University of Sicence
and Technolohgy, ShenZhen, China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we construct a new family of $(q^{4}+1)$-tight sets in
$Q(24,q)$ or $Q^{-}(25,q)$ according as $q=3^{f}$ or $q\equiv 2\pmod{3}$. The
novelty of the construction is the use of the action of the exceptional simple
group $F_{4}(q)$ on its minimal module over ${\mathbb{F}}_{q}$.

Keywords: Exceptional simple group, intriguing set, polar space, tight set

MSC(2020): 51E20, 05B25, 51A50

∗Research partially supported by National Natural Science Foundation of China
under Grant No. 12171428, 12225110

†Research partially supported by the National Natural Science Foundation of
China Grant No. 12071206, 12131011, 12150710510, and the Sino-German Mobility
Programme M-0157

##  1\. Introduction

Let ${\mathcal{P}}_{r}$ be a finite classical polar space of rank $r\geq 2$. A
tight set is a subset ${\mathcal{T}}$ of points of ${\mathcal{P}}_{r}$ such
that for all points $P$ of ${\mathcal{P}}_{r}$, the intersection sizes
$|P^{\perp}\cap{\mathcal{T}}|$ take exactly two values involving a parameter
$i\geq 1$, according as $P\in{\mathcal{T}}$ or not, where $P^{\perp}$ is the
set of points in ${\mathcal{P}}_{r}$ that are collinear with $P$. Tight sets
in finite generalized quadrangles were first introduced by Payne [13], and
this notion has been generalized to finite classical polar spaces by Drudge
[6]. Parallel to the notion of a tight set is the concept of an $m$-ovoid. The
notions of tight sets and $m$-ovoids were unified under the umbrella of
intriguing sets in [2]. During the past two decades, intriguing sets have been
extensively investigated because of their close connections with many other
combinatorial/geometric objects such as strongly regular graphs, partial
difference sets, Boolean degree one functions, and Cameron-Liebler line
classes, cf. [2, 7, 8].

In general, most known examples of tight sets are found in finite polar spaces
of low ranks. However, 1-tight sets exist in all finite classical polar space
${\mathcal{P}}_{r}$: every generator $\mathcal{G}$ of ${\mathcal{P}}_{r}$ is a
1-tight set since $|P^{\perp}\cap\mathcal{G}|=|\mathcal{G}|$ if
$P\in\mathcal{G}$, and $|P^{\perp}\cap\mathcal{G}|=$ the size of a hyperplane
of $\mathcal{G}$ if $P\not\in\mathcal{G}$. In fact, Drudge [6] proved that any
1-tight set of ${\mathcal{P}}_{r}$ must be a generator. Tight sets with large
parameters are much more complicated. One of the main problems concerning
tight sets is to determine for which values of $i$ there exist $i$-tight sets
in ${\mathcal{P}}_{r}$. A second problem is to characterize $i$-tight sets in
${\mathcal{P}}_{r}$ for specific parameter $i$.

As we saw before, one can deduce fairly easily that a generator of
${\mathcal{P}}_{r}$ is a 1-tight set. Suppose that $A$ and $B$ are an
$i$-tight set and a $j$-tight set in ${\mathcal{P}}_{r}$, respectively. If
$A,B$ are disjoint, then $A\cup B$ is an $(i+j)$-tight set; if $A\subseteq B$,
then $B\setminus A$ is a $(j-i)$-tight set. Thus, the union of any $i$
pairwise disjoint generators (i.e., the elements of a partial spread) of
${\mathcal{P}}_{r}$ forms an $i$-tight set. For this reason, the first
question has a simple and complete answer if the classical polar space
${\mathcal{P}}_{r}$ admits a spread of generators. Thus, we will only consider
the problem of constructing tight sets for polar spaces which do not admit a
spread; and for such a polar space, it is interesting to construct tight sets
with large parameters $i$, where $i$ is greater than the maximum size of a
partial spread. Some results on the existence of spreads of classical polar
spaces can be found in [10, Table 7.4].

In this paper, we construct a new family of $(q^{4}+1)$-tight sets in
$Q(24,q)$ or $Q^{-}(25,q)$ according as $q=3^{f}$ or $q\equiv 2\pmod{3}$. Our
main theorem in this paper is the following.

######  Theorem 1.1.

Let $q=p^{f}$, where $p$ is prime and $f\geq 1$. If $q\equiv 2\pmod{3}$, then
the classical polar space $Q^{-}(25,q)$ admits a $(q^{4}+1)$-tight set with an
automorphism group isomorphic to the exceptional group $F_{4}(q)$; if
$q=3^{f}$, then the classical polar space $Q(24,q)$ admits a $(q^{4}+1)$-tight
set with an automorphism group isomorphic to the exceptional group $F_{4}(q)$.

The novelty of the construction is the use of the action of the exceptional
simple group $F_{4}(q)$ on its minimal module over ${\mathbb{F}}_{q}$. We
remark that the characteristic $3$ case of Theorem 1.1 is essentially due to
Cohen and Cooperstein. They showed in [5, Table 2] that the group $F_{4}(q)$
has two orbits on the singular points of the polar space $Q(24,q)$ associated
with its $25$-dimensional minimal module, and from this fact we deduce that
each orbit is a tight set.

The paper is organized as follows. In Section 2, we introduce some preliminary
results on intriguing sets, octonions and the minimal module of $F_{4}(q)$. In
particular, we give a more effective way to decide whether a subset
${\mathcal{M}}$ of points of a finite classical polar space
${\mathcal{P}}_{r}$ is an intriguing set, see Lemma 2.1. In Section 3, we
present the proof of Theorem 1.1.

##  2\. Preliminary

###  2.1. Intriguing sets on finite classical polar spaces

Let ${\mathbb{F}}_{q}$ be the finite field of order $q$, where $q=p^{f}$, $p$
is a prime and $f\geq 1$. Let $V$ be a $d$-dimensional vector space over
${\mathbb{F}}_{q}$ equipped with a nondegenerate reflexive sesquilinear form
or quadratic form $\kappa$, and let ${\mathcal{P}}_{r}$ be the associated
polar space. A point of ${\mathcal{P}}_{r}$ is defined as a $1$-dimensional
totally isotropic/singular subspace of $V$. A maximal totally
isotropic/singular subspace of ${\mathcal{P}}_{r}$ is called a generator of
${\mathcal{P}}_{r}$. All generators have the same dimension $r$, which is
called the rank of ${\mathcal{P}}_{r}$. A generator of ${\mathcal{P}}_{r}$ has
$\frac{q^{r}-1}{q-1}$ points. An ovoid of ${\mathcal{P}}_{r}$ is a set of
points which meets each generator in exactly one point. We use $\theta_{r}$ to
denote the size of a putative ovoid, which we call the ovoid number of
${\mathcal{P}}_{r}$. A simple counting arguement shows that
$|{\mathcal{P}}_{r}|=\theta_{r}\cdot\frac{q^{r}-1}{q-1}$. We list the ranks
and the ovoid numbers of the six classical polar spaces in Table 2.1.

Table 2.1. The parameters $r$ and $\theta_{r}$ |  $d$ |  $f$ |  polar space ${\mathcal{P}}_{r}$ |  rank $r$ |  ovoid number $\theta_{r}$
---|---|---|---|---|---
${\mathbf{S}}$ |  even |  - |  $W(d-1,q)$ |  $d/2$ |  $q^{d/2}+1$
${\mathbf{O}}$ |  even |  - |  $Q^{+}(d-1,q)$ |  $d/2$ |  $q^{d/2-1}+1$
even |  - |  $Q^{-}(d-1,q)$ |  $d/2-1$ |  $q^{d/2}+1$
odd |  - |  $Q(d-1,q)$ |  $(d-1)/2$ |  $q^{(d-1)/2}+1$
${\mathbf{U}}$ |  odd |  even |  $H(d-1,q)$ |  $(d-1)/2$ |  $q^{d/2}+1$
even |  even |  $H(d-1,q)$ |  $d/2$ |  $q^{(d-1)/2}+1$

Suppose that $r\geq 2$, and let ${\mathcal{M}}$ be a nonempty set of points of
${\mathcal{P}}_{r}$. The set ${\mathcal{M}}$ is called an intriguing set if
there exist some constants $h_{1}\neq h_{2}$ such that
$|P^{\perp}\cap{\mathcal{M}}|=h_{1}$ or $h_{2}$ according as
$P\in{\mathcal{M}}$ or not, where $P$ ranges over all the points of
${\mathcal{P}}_{r}$, cf. [2]. An intriguing set ${\mathcal{M}}$ is proper if
${\mathcal{M}}\neq{\mathcal{P}}_{r}$. There are exactly two types of
intriguing sets:

  1. (1)

$i$-tight sets: $|{\mathcal{M}}|=i\cdot\frac{q^{r}-1}{q-1}$,
$h_{1}=q^{r-1}+i\cdot\frac{q^{r-1}-1}{q-1}$,
$h_{2}=i\cdot\frac{q^{r-1}-1}{q-1}$, and

  2. (2)

$m$-ovoids: $|{\mathcal{M}}|=m\theta_{r}$, $h_{1}=(m-1)\theta_{r-1}+1$,
$h_{2}=m\theta_{r-1}$.

We refer the reader to [2] for more properties of intriguing sets. In
particular, if $H$ is a subgroup of semisimilarities that has exactly two
orbits $O_{1},\,O_{2}$ on the points of ${\mathcal{P}}_{r}$, then both $O_{1}$
and $O_{2}$ are intriguing sets of the same type.

To prove that a candidate subset ${\mathcal{M}}$ of points of
${\mathcal{P}}_{r}$ is an intriguing set, one needs to show that
${\mathcal{M}}$ is a two-intersection set with respect to the perp of singular
points, cf. [2]. However, there is one shortcut method which seems to have
gone unnoticed (see Leema 2.1 below). We observe that
$\theta_{r}-1=q(\theta_{r-1}-1)$ by Table 2.1.

######  Lemma 2.1.

Let ${\mathcal{M}}$ be a subset of size $i\frac{q^{r}-1}{q-1}$ or
$m\theta_{r}$ in ${\mathcal{P}}_{r}$ for some $i$ or $m$, and let
$h_{1},h_{2}$ be the corresponding parameters determined by $|{\mathcal{M}}|$
(see above). Then the following are equivalent:

  * (1)

${\mathcal{M}}$ is an intriguing set in ${\mathcal{P}}_{r}$;

  * (2)

$|P^{\perp}\cap{\mathcal{M}}|=h_{1}$ for all $P\in{\mathcal{M}}$;

  * (3)

$|P^{\perp}\cap{\mathcal{M}}|=h_{2}$ for all
$P\in{\mathcal{P}}_{r}\setminus{\mathcal{M}}$.

###### Proof.

We will prove the equivalence of (1) and (2). It is clear that (1) implies
(2). So we only need to prove that (2) implies (1). We will use
${\mathcal{P}}_{j}$ to denote the polar space of rank $j$ and of the same type
as ${\mathcal{P}}_{r}$. We compute
$\sum_{P\in{\mathcal{P}}_{r}}|P^{\perp}\cap{\mathcal{M}}|$ and
$\sum_{P\in{\mathcal{P}}_{r}}|P^{\perp}\cap{\mathcal{M}}|^{2}$, and will show
that
$\sum_{P\in{\mathcal{P}}_{r}\setminus{\mathcal{M}}}\left(|P^{\perp}\cap{\mathcal{M}}|-h_{2}\right)^{2}=0$.
First we have

$\displaystyle\sum_{P\in{\mathcal{P}}_{r}}|P^{\perp}\cap{\mathcal{M}}|$
$\displaystyle=|\\{(P,z):P\in{\mathcal{P}}_{r},z\in{\mathcal{M}},P\sim z\\}|$
$\displaystyle=\sum_{z\in{\mathcal{M}}}|z^{\perp}\cap{\mathcal{P}}_{r}|=q|{\mathcal{M}}|\cdot|{\mathcal{P}}_{r-1}|+|{\mathcal{M}}|.$

For the last equality, we observe that for a point $z\in{\mathcal{M}}$ the
intersection $z^{\perp}\cap{\mathcal{P}}_{r}$ is a cone with vertex $z$ and
base ${\mathcal{P}}_{r-1}$. Furthermore we have

$\displaystyle\sum_{P\in{\mathcal{P}}_{r}}|P^{\perp}\cap{\mathcal{M}}|^{2}$
$\displaystyle=\sum_{P\in{\mathcal{P}}_{r}}|\\{(P,z_{1},z_{2}):\,z_{1},z_{2}\in{\mathcal{M}},\,z_{1}\sim
P,\,z_{2}\sim P\\}|$
$\displaystyle=\sum_{z\in{\mathcal{M}}}|z^{\perp}\cap{\mathcal{P}}_{r}|+\sum_{\begin{subarray}{c}z_{1},z_{2}\in{\mathcal{M}},\\\
z_{1}\neq z_{2}\end{subarray}}|\langle
z_{1},z_{2}\rangle^{\perp}\cap{\mathcal{P}}_{r}|.$ (2.1)

For the last summation in (2.1), each summand takes value either
$q^{2}\cdot|{\mathcal{P}}_{r-2}|+q+1$ or $|{\mathcal{P}}_{r-1}|$ depending on
whether $z_{1},z_{2}$ are perpendicular or not: $\langle
z_{1},z_{2}\rangle^{\perp}\cap{\mathcal{P}}_{r}$ is a cone with vertex
$\langle z_{1},z_{2}\rangle$ and base ${\mathcal{P}}_{r-2}$ if $\langle
z_{1},z_{2}\rangle$ is totally singular or isotropic, and it is
${\mathcal{P}}_{r-1}$ otherwise. Next we compute directly

$\displaystyle\sum_{P\in{\mathcal{P}}_{r}}\left(|P^{\perp}\cap{\mathcal{M}}|-h_{2}\right)^{2}$
$\displaystyle=\sum_{P\in{\mathcal{P}}_{r}}\Big{(}|P^{\perp}\cap{\mathcal{M}}|^{2}-2h_{2}|P^{\perp}\cap{\mathcal{M}}|+h_{2}^{2}\Big{)}$
$\displaystyle=|{\mathcal{M}}|\cdot\Big{(}|{\mathcal{P}}_{r-1}|(|{\mathcal{M}}|+q-h_{1}-2h_{2}q)+(qh_{1}-q+h_{1}-2h_{2})$
$\displaystyle\
+q^{2}(h_{1}-1)|{\mathcal{P}}_{r-2}|+\frac{h_{2}^{2}|{\mathcal{P}}_{r}|}{|{\mathcal{M}}|}\Big{)}.$
(2.2)

First consider the case where ${\mathcal{M}}$ is an $i$-tight set, so that
$|{\mathcal{M}}|=i\frac{q^{r}-1}{q-1},h_{1}=i\frac{q^{r-1}-1}{q-1}+q^{r-1},h_{2}=i\frac{q^{r-1}-1}{q-1}$.
Observe that the right hand side of $\eqref{eqn_PcapMh2}$ divided by
$|{\mathcal{M}}|$ can be viewed as a degree-one polynomial in variable $i$.
Then some tedious but routine computations show that the coefficient of $i$
equals $0$. Hence we are only concerned with the constant term of the
aforementioned polynomial, which equals

$\displaystyle|{\mathcal{M}}|\big{(}|{\mathcal{P}}_{r-1}|(q-q^{r-1})+(q+1)q^{r-1}-q+q^{2}(q^{r-1}-1)|{\mathcal{P}}_{r-2}|\big{)}$
$\displaystyle=$ $\displaystyle
q^{2r-2}=|{\mathcal{M}}|\cdot(h_{1}-h_{2})^{2}.$

From
$\sum_{P\in{\mathcal{P}}_{r}}\left(|P^{\perp}\cap{\mathcal{M}}|-h_{2}\right)^{2}=|{\mathcal{M}}|(h_{1}-h_{2})^{2}$
we deduce that
$\sum_{P\in{\mathcal{P}}_{r}\setminus{\mathcal{M}}}\left(|P^{\perp}\cap{\mathcal{M}}|-h_{2}\right)^{2}=0$.
This proves that (2) implies (1) when ${\mathcal{M}}$ is an $i$-tight set.
When ${\mathcal{M}}$ is an $m$-ovoid, we can use the same method to obtain the
result. The proof of (1)$\Leftrightarrow$(3) is similar and we omit the
details. This completes the proof. ∎

###  2.2. Octonions and the minimal module of $F_{4}(q)$

The octonion algebra $\mathbb{O}_{{\mathbb{F}}_{q}}$ is an 8-dimensional non-
commutative and non-associative algebra over ${\mathbb{F}}_{q}$. A detailed
introduction to octonions can be found in the book [15]. Let
$\\{x_{1},\ldots,x_{8}\\}$ be the basis of the octonion algebra $\mathbb{O}$
(here in order to simplify notation we omit the subscript) as defined in [17,
(4.26)]. With respect to this basis, the multiplication of $\mathbb{O}$ is
defined as follows:

| $x_{1}$ | $x_{2}$ | $x_{3}$ | $x_{4}$ | $x_{5}$ | $x_{6}$ | $x_{7}$ | $x_{8}$
---|---|---|---|---|---|---|---|---
$x_{1}$ |  |  |  |  | $x_{1}$ | $x_{2}$ | $-x_{3}$ | $-x_{4}$
$x_{2}$ |  |  | $-x_{1}$ | $x_{2}$ |  |  | $-x_{5}$ | $x_{6}$
$x_{3}$ |  | $x_{1}$ |  | $x_{3}$ |  | $-x_{5}$ |  | $x_{7}$
$x_{4}$ | $x_{1}$ |  |  | $x_{4}$ |  | $x_{6}$ | $x_{7}$ |
$x_{5}$ |  | $x_{2}$ | $x_{3}$ |  | $x_{5}$ |  |  | $x_{8}$
$x_{6}$ | $-x_{2}$ |  | $-x_{4}$ |  | $x_{6}$ |  | $x_{8}$ |
$x_{7}$ | $x_{3}$ | $-x_{4}$ |  |  | $x_{7}$ | $-x_{8}$ |  |
$x_{8}$ | $-x_{5}$ | $-x_{6}$ | $x_{7}$ | $x_{8}$ |  |  |  |
(2.3)

where the blank entries are $0$, cf. [17, (4.27)]. It is clear that
$\mathbf{1}:=x_{4}+x_{5}$ is the identity of $\mathbb{O}$. The octonion
conjugation - swaps $x_{4},\,x_{5}$ and maps the other $x_{i}$’s to their
negations, and it is an anti-isomorphism of $\mathbb{O}$. For an octonion
$x=\sum_{i=1}^{8}\lambda_{i}x_{i}$, we define its trace as
$\textup{Tr}(x)=x+\overline{x}=\lambda_{4}+\lambda_{5}$ and its norm as
$N(x)=x\overline{x}$. It is routine to show that
$N(x)=\sum_{i=1}^{4}\lambda_{i}\lambda_{9-i}$ which takes value in
${\mathbb{F}}_{q}$, and defines a nondegenerate hyperbolic quadric
$Q^{+}(7,q)$. Its associated bilinear form is $B(x,y):=N(x+y)-N(x)-N(y)$, and
we have $B(x,y)=\textup{Tr}(x\overline{y})$ for $x,y\in\mathbb{O}$. In
particular, we have $\textup{Tr}(x)=B(x,1)$. The octonion algebra $\mathbb{O}$
is an alternative ring by [16, Lemma 3], so that the subalgebra generated by
any two elements is associative, cf. [15, Chapter 1.4]. It follows that
$\mathbb{O}$ is a composition algebra, i.e., $N(xy)=N(x)N(y)$ for
$x,y\in\mathbb{O}$. The automorphism group of the octonion algebra
$\mathbb{O}$ is known as $G_{2}(q)$, and each of its element fixes the vector
$\mathbf{1}:=x_{4}+x_{5}$. It holds that $x^{2}-\textup{Tr}(x)x+N(x)=0$ for
$x\in\mathbb{O}$ by [15, Proposition 1.2.3], so $G_{2}(q)$ lies in the
isometry group of the quadratic form $N$. We write $\mathbf{1}^{\perp}$ for
the perp of $\mathbf{1}$ with respect to $B$, which is stabilized by
$G_{2}(q)$.

The $G_{2}(q)$-orbits on the nonzero octonions are implicitly known by the
results in [17, Chapter 4.3]. We list them explicitly in the next lemma for
future use.

######  Lemma 2.2.

Suppose that $q>2$. For $q$ odd, let $\alpha$ be a nonsquare of
${\mathbb{F}}_{q}^{*}$ and $T$ be a complete set of coset representatives of
$\\{1,-1\\}$ in ${\mathbb{F}}_{q}^{*}$; for $q$ even, let $\beta$ be an
element of ${\mathbb{F}}_{q}^{*}$ with absolute trace $1$ and and $S$ be a
complete set of coset representatives of $\\{0,1\\}$ in ${\mathbb{F}}_{q}$.
Then the $G_{2}(q)$-orbits on the octonions are as listed in Table 2.2.

Table 2.2. The $G_{2}(q)$-orbits on the octonions Representative | Orbit size | Stabilizer | Trace | Norm | Condition
---|---|---|---|---|---
$k$, $k\in{\mathbb{F}}_{q}$ | $1$ | $G_{2}(q)$ | $2k$ | $k^{2}$ |
$x_{1}+k$, $k\in{\mathbb{F}}_{q}$ | $q^{6}-1$ | $q^{2+1+2}:\textup{SL}_{2}(q)$ | $2k$ | $k^{2}$
|  $a(x_{4}-x_{5})+k$,
---
$a\in T,k\in{\mathbb{F}}_{q}$
$q^{6}+q^{3}$ | $\textup{SL}_{3}(q)$ | $2k$ | $k^{2}-a^{2}$ | $q$ odd
|  $a(x_{1}-\alpha x_{8})+k$,
---
$a\in T,k\in{\mathbb{F}}_{q}$
$q^{6}-q^{3}$ | $\textup{SU}_{3}(q)$ | $2k$ | $k^{2}-\alpha a^{2}$
|  $a(x_{4}+k)$,
---
$a\in{\mathbb{F}}_{q}^{*},k\in S$
$q^{6}+q^{3}$ | $\textup{SL}_{3}(q)$ | $a$ | $a^{2}(k^{2}+k)$ | $q$ even
|  $a(x_{4}+x_{1}+\beta x_{8}+k)$,
---
$a\in{\mathbb{F}}_{q}^{*}$, $k\in S$
$q^{6}-q^{3}$ | $\textup{SU}_{3}(q)$ | $a$ | $a^{2}(k^{2}+k+\beta)$

###### Proof.

We give a sketch of the proof by quoting the relevant results in [17, Chapter
4.3]. Since each element of $G_{2}(q)$ fixes $\mathbf{1}$, each element
$k\in{\mathbb{F}}_{q}$ is stabilized by $G_{2}(q)$. Take an element
$k\in{\mathbb{F}}_{q}$ and set $v=x_{1}+k$. For an element $g\in G_{2}(q)$, it
stabilizes $v$ if and only if it stabilizes $x_{1}$. The stabilizer of
$\langle x_{1}\rangle$ is a maximal parabolic subgroup
$q^{2+1+2}:\textup{GL}_{2}(q)$, and the stabilizer of $x_{1}$ in the latter
subgroup is $q^{2+1+2}:\textup{SL}_{2}(q)$. Hence the $G_{2}(q)$-orbit of
$x_{1}+k$ has size $\frac{|G_{2}(q)|}{q^{5}\cdot|\textup{SL}_{2}(q)|}=q^{6}-1$
for each $k\in{\mathbb{F}}_{q}$. The trace and norm of $x_{1}+k$ are
respectively $2k,k^{2}$, so  the $(x_{1}+k)$’s  are in different
$G_{2}(q)$-orbits for distinct $k$’s in ${\mathbb{F}}_{q}$. The arguments so
far work for both even and odd $q$.

Suppose that $q$ is odd and set $u=x_{4}-x_{5}$, $v=x_{1}-\alpha x_{8}$, where
$\alpha$ is a nonsquare in ${\mathbb{F}}_{q}^{*}$. The stabilizer of $\langle
u\rangle$ in $G_{2}(q)$ is a maximal subgroup $\textup{SL}_{3}(q):\langle
s\rangle$, where $s$ has order $2$ and swaps $x_{4}$ and $x_{5}$. The
stabilizer of $u$ in the latter subgroup is $\textup{SL}_{3}(q)$. As in the
previous paragraph, we deduce that the $G_{2}(q)$-orbit of $au+k$ has size
$\frac{|G_{2}(q)|}{|\textup{SL}_{3}(q)|}=q^{6}+q^{3}$ for each
$a\in{\mathbb{F}}_{q}^{*}$ and $k\in{\mathbb{F}}_{q}$. The trace and norm of
$au+k$ are respectively $2k,-a^{2}+k^{2}$, so $au+k$ and
$a^{\prime}u+k^{\prime}$ are in the same $G_{2}(q)$-orbit if and only if
$a^{\prime}=\pm a$ and $k^{\prime}=k$. The subalgebra
$C:=\langle\mathbf{1},v\rangle$ is isomorphic to ${\mathbb{F}}_{q^{2}}$, and
the stabilizer of $v$ in $G_{2}(q)$ is $\textup{SU}_{3}(q)$. There is an
involution $r\in G_{2}(q)$ that maps $x_{1},x_{8}$ to $-x_{1},-x_{8}$ and thus
$v$ to $-v$, and $\textup{SU}_{3}(q):\langle r\rangle$ is a maximal subgroup
of $G_{2}(q)$ that stabilizes $\langle v\rangle$. Similarly, we deduce that
the $G_{2}(q)$-orbit of $av+k$ has size
$\frac{|G_{2}(q)|}{|\textup{SU}_{3}(q)|}=q^{6}-q^{3}$ for each
$a\in{\mathbb{F}}_{q}^{*}$ and $k\in{\mathbb{F}}_{q}$. The trace and norm of
$av+k$ are respectively $2k,k^{2}-a^{2}$, so $av+k$ and
$a^{\prime}v+k^{\prime}$ are in the same $G_{2}(q)$-orbit if and only if
$a^{\prime}=\pm a$ and $k^{\prime}=k$. Since

$q+q\cdot(q^{6}-1)+\frac{1}{2}(q-1)q\cdot(q^{6}+q^{3})+\frac{1}{2}(q-1)q\cdot(q^{6}-q^{3})=q^{8},$

we conclude that we have obtained all the $G_{2}(q)$-orbits and their relevant
information is as listed in Table 2.2 for $q$ odd.

Suppose that $q$ is even and set $u=x_{4}$, $v=x_{1}+\beta x_{8}+x_{4}$, where
$\beta$ has absolute trace $1$, i.e., $X^{2}+X+\beta$ is irreducible over
${\mathbb{F}}_{q}$. The element $v$ has trace $1$ and norm $\beta$, and there
are $q^{6}-q^{3}$ octonions with those properties. The stabilizer of $u$ in
$G_{2}(q)$ is $\textup{SL}_{3}(q)$, and there is an order $2$ element $s$ that
swaps $u_{4},u_{5}$ such that $\textup{SL}_{3}(q):\langle s\rangle$ is a
maximal subgroup. We have $v^{2}+v+\beta=0$, so the subalgebra
$\langle\mathbf{1},v\rangle$ is a field with $q^{2}$ element. There is an
involution $r\in G_{2}(q)$ that fixes $x_{1},x_{8}$ and swaps $x_{4}$ and
$x_{5}$, so that $r$ maps $v$ to $v+\mathbf{1}$. The stabilizer of $v$ in
$G_{2}(q)$ is $\textup{SU}_{3}(q)$, and $\textup{SU}_{3}(q):\langle r\rangle$
is a maximal subgroup. The remaining arguments are exactly the same as in the
$q$ odd case, and we omit the details. This completes the proof. ∎

For a nonzero element $a$ in $\mathbb{O}$, we define its left and right
annihilators as follows

$\displaystyle\textup{ann}_{L}(a)=\\{x\in\mathbb{O}:\,xa=0\\},\quad\textup{ann}_{R}(a)=\\{x\in\mathbb{O}:\,ax=0\\}.$

By [15, Lemma 1.3.3], we have $x(\overline{x}a)=N(x)a$ and
$(ax)\overline{x}=N(x)a$, so both $\textup{ann}_{L}(a)$ and
$\textup{ann}_{R}(a)$ are totally singular subspaces for the quadratic form
$N$. If $N(a)\neq 0$, then both of its annihilators are trivial by a similar
argument.

######  Lemma 2.3.

Suppose that $D,E$ are nonzero elements in $\mathbb{O}$ such that $DE=0$. Then
$D\overline{D}=E\overline{E}=0$, $\textup{ann}_{L}(D)$ and
$\textup{ann}_{R}(D)$ are totally isotropic subspaces of dimension $4$, and

  * (1)

$\textup{ann}_{L}(D)\cap\textup{ann}_{R}(D)$ has dimension $3$ or $1$
according as $D\in\mathbf{1}^{\perp}$ or not;

  * (2)

$\textup{ann}_{L}(D)\cap\textup{ann}_{R}(E)$ has dimension $3$.

###### Proof.

By the arguments preceding this lemma, we have
$D\overline{D}=E\overline{E}=0$, and $\textup{ann}_{L}(D)$ and
$\textup{ann}_{R}(D)$ are totally isotropic subspaces. First consider the case
where $D$ is in $\mathbf{1}^{\perp}$. Since $G_{2}(q)$ is transitive on the
nonzero singular vectors in $\mathbf{1}^{\perp}$, cf. [17, p. 125], we assume
that $D=x_{1}$ without loss of generality. It follows from (2.3) that
$\textup{ann}_{L}(x_{1})=\langle x_{1},x_{2},x_{3},x_{5}\rangle$ and
$\textup{ann}_{R}(x_{1})=\langle x_{1},x_{2},x_{3},x_{4}\rangle$. Both
annihilators of $x_{1}$ have dimension $4$ and their intersection $\langle
x_{1},x_{2},x_{3}\rangle$ has dimension $3$. Since $DE=0$, we deduce that
$E=\sum_{i=1}^{4}a_{i}x_{i}$ for some $a_{i}$’s in ${\mathbb{F}}_{q}$. For an
element $y\in\textup{ann}_{L}(D)$, we have
$y=b_{1}x_{1}+b_{2}x_{2}+b_{3}x_{3}+b_{5}x_{5}$ for some $b_{i}$’s in
${\mathbb{F}}_{q}$. If further $y$ is in $\textup{ann}_{R}(E)$, then
$Ey=(a_{1}b_{5}-a_{2}b_{3}+a_{3}b_{2}+a_{4}b_{1})x_{1}=0$. There are $q^{3}$
such $(b_{1},b_{2},b_{3},b_{5})$ tuples, so
$\textup{ann}_{L}(x_{1})\cap\textup{ann}_{R}(E)$ has dimension $3$ as desired.

We next consider the case $D$ is not in $\mathbf{1}^{\perp}$. Since $\langle
x_{4}\rangle$ and $\langle x_{5}\rangle$ are in the same $G_{2}(q)$-orbit and
$D$ is singular, we assume without loss of generality that $D=x_{4}$ up to
scalar by Table 2.2. We deduce from (2.3) that
$\textup{ann}_{L}(x_{4})=\langle x_{1},x_{5},x_{6},x_{7}\rangle$ and
$\textup{ann}_{R}(x_{4})=\langle x_{2},x_{3},x_{5},x_{8}\rangle$. Their
intersection is $\langle x_{5}\rangle$ which has dimension $1$. Since $DE=0$,
we deduce that $E=a_{2}x_{2}+a_{3}x_{3}+a_{5}x_{5}+a_{8}x_{8}$ for some
$a_{i}$’s in ${\mathbb{F}}_{q}$. For an element $y\in\textup{ann}_{L}(D)$, we
have $y=b_{1}x_{1}+b_{5}x_{5}+b_{6}x_{6}+b_{7}x_{7}$ for some $b_{i}$’s in
${\mathbb{F}}_{q}$. If further $y$ is in $\textup{ann}_{R}(E)$, then
$Ey=(-a_{2}b_{7}-a_{3}b_{6}+a_{5}b_{5}-a_{8}b_{1})x_{5}=0$. There are $q^{3}$
such $(b_{1},b_{2},b_{3},b_{5})$ tuples, so
$\textup{ann}_{L}(x_{4})\cap\textup{ann}_{R}(E)$ has dimension $3$ as desired.
This completes the proof. ∎

The following result is well-known, and we include a short proof here. We
refer the reader to [9, Proposition 9.10] for the case where $q$ is odd. For a
property $\mathbf{P}$, we use the Iverson bracket $[\\![\mathbf{P}]\\!]$ which
takes value $1$ or $0$ according as $\mathbf{P}$ holds or not.

######  Lemma 2.4.

For a natural number $k$ and an element $a\in{\mathbb{F}}_{q}$, write
$N_{k}(a)$ for the number of tuples
$(a_{1},\ldots,a_{2k})\in{\mathbb{F}}_{q}^{2k}$ such that
$a_{1}a_{2}+\ldots+a_{2k-1}a_{2k}=a$. Then

$N_{k}(a)=q^{2k-1}-q^{k-1}+q^{k}\cdot[\\![a=0]\\!]\quad\textup{ for
}a\in{\mathbb{F}}_{q}.$

###### Proof.

It is easy to see that the claim holds for $k=1$. We have

$\displaystyle N_{k}(a)$
$\displaystyle=\sum_{a_{2k-1}\in{\mathbb{F}}_{q}}\sum_{a_{2k}\in{\mathbb{F}}_{q}}N_{k-1}(a-a_{2k-1}a_{2k})$
$\displaystyle=q^{2}N_{k-1}(1)+(N_{k-1}(0)-N_{k-1}(1))\cdot|\\{a_{2k-1},a_{2k}\in{\mathbb{F}}_{q}:\,a_{2k-1}a_{2k}=a\\}|$
$\displaystyle=q^{2}N_{k-1}(1)+(N_{k-1}(0)-N_{k-1}(1))\left(q-1+q\cdot[\\![a=0]\\!]\right),$

and the claim follows by induction on $k$. ∎

######  Lemma 2.5.

For $a\in{\mathbb{F}}_{q}$ there are $q^{7}-q^{3}+q^{4}\cdot[\\![a=0]\\!]$
octonions of norm $a$, and among them there are
$q^{6}-q^{3}+q^{4}\cdot[\\![a=0]\\!]$ such $\alpha$’s that
$\textup{Tr}(x_{1}{\overline{\alpha}})=0$.

###### Proof.

The first claim immediately follows by specifying $k=4$ in Lemma 2.4. For an
octonion $\alpha=\sum_{i=1}^{8}\lambda_{i}x_{i}$, we have
$\textup{Tr}(x_{1}\overline{\alpha})=\lambda_{8}$ and
$N(\alpha)=\sum_{i=1}^{4}\lambda_{i}\lambda_{9-i}$. If $\lambda_{8}=0$, then
$N(\alpha)=\sum_{i=2}^{4}\lambda_{i}\lambda_{9-i}$. We deduce that there are
$q\cdot N_{3}(a)$ such $\alpha$’s that satisfy $\lambda_{8}=0$ and have norm
$a$, where $N_{3}(a)$ is as in Lemma 2.4. This completes the proof. ∎

######  Lemma 2.6.

There are $(q^{6}-q^{3}+q^{4}-1)(q^{4}-1)$ of $(D,E)$ pairs of nonzero
octonions such that $DE=0$ and $\textup{Tr}(x_{1}\overline{D})=0$.

###### Proof.

If $DE=0$ for some nonzero $D,E$, then $D\overline{D}=0$ by Lemma 2.3. There
are $q^{6}-q^{3}+q^{4}-1$ nonzero $D$’s such that $D\overline{D}=0$ and
$\textup{Tr}(x_{1}\overline{D})=0$ by Lemma 2.5, and for each such $D$, there
are $q^{4}-1$ nonzero $E$’s such that $DE=0$ by Lemma 2.3. The claim then
follows. ∎

Let ${\mathcal{A}}$ the the algebra of $3\times 3$ Hermitian matrices over the
octonions $\mathbb{O}$. To be specific, a $3\times 3$ matrix over $\mathbb{O}$
is an Hermitian matrix if $x^{\top}=\overline{x}$ and the diagonal entries of
$x$ lie in ${\mathbb{F}}_{q}$. Here, $\overline{x}$ is the matrix obtained by
applying the octonion conjugate to the entries of $x$. For
$d,e,f\in{\mathbb{F}}_{q}$ and $D,E,F\in\mathbb{O}$, we define

$(\lambda_{0},\lambda_{0}^{\prime},\lambda_{0}^{\prime\prime}\mid
D,E,F):=\begin{pmatrix}\lambda_{0}&F&\overline{E}\\\
\overline{F}&\lambda_{0}^{\prime}&D\\\
E&\overline{D}&\lambda_{0}^{\prime\prime}\end{pmatrix}$

which is in ${\mathcal{A}}$. For $u,v\in{\mathcal{A}}$, their product in the
algebra ${\mathcal{A}}$ is $u\circ v:=uv+vu$, where $uv$ and $vu$ are matrix
multiplications. We set $I:=(1,1,1\mid 0,0,0)$, so that $I\circ a=2a$ for any
element $a$ in ${\mathcal{A}}$. The algebra ${\mathcal{A}}$ has close
connections with Albert algebras and Jordan algebras, cf. [12, 14]. We choose
a basis $\\{w_{i},w_{i}^{\prime},w_{i}^{\prime\prime}:\,0\leq i\leq 8\\}$ of
${\mathcal{A}}$ as follows:

$\displaystyle w_{0}=(1,0,0\mid 0,0,0)\textup{ and }w_{i}=(0,0,0\mid
x_{i},0,0)\textup{ for }i>0,$ $\displaystyle w_{0}^{\prime}=(0,1,0\mid
0,0,0)\textup{ and }w_{i}^{\prime}=(0,0,0\mid 0,x_{i},0)\textup{ for }i>0,$
$\displaystyle w_{0}^{\prime\prime}=(0,0,1\mid 0,0,0)\textup{ and
}w_{i}^{\prime\prime}=(0,0,0\mid 0,0,x_{i})\textup{ for }i>0.$

The multiplication table of ${\mathcal{A}}$ with respect to the basis
$\\{w_{i},w_{i}^{\prime},w_{i}^{\prime\prime}:\,0\leq i\leq 8\\}$ can be
written down explicitly as in [17, (4.90)-(4.92)]. The similarity group of the
Dickson-Freudenthal determinant of ${\mathcal{A}}$, which we denote by
$\tilde{E}_{6}(q)$, is the universal covering group of the simple group
$E_{6}(q)$, cf. [16]. It has three orbits on the nonzero vectors of
${\mathcal{A}}$, which are called the white, gray and black vectors
respectively, cf. [1, 5]. They correspond to vectors of rank $1$, rank $2$ and
rank $3$ in [11]. The $1$-dimensional subspace spanned by a white, gray, black
vector is called a white, gray and black point respectively. The stabilizer of
$I$ in $\tilde{E}_{6}(q)$ is $F_{4}(q)$.

For an element
$v=\sum_{t=0}^{8}(\lambda_{t}w_{t}+\lambda_{t}^{\prime}w_{t}^{\prime}+\lambda_{t}^{\prime\prime}w_{t}^{\prime\prime})\in{\mathcal{A}}$,
we define the trace of $v$ as
$\textup{Tr}_{{\mathcal{A}}}(v)=\lambda_{0}+\lambda_{0}^{\prime}+\lambda_{0}^{\prime\prime}$.
We define the $F_{4}(q)$-invariant subspaces
$U=\\{v\in{\mathcal{A}}:\textup{Tr}_{{\mathcal{A}}}(v)=0\\}$,
$U^{\prime}=\langle I\rangle_{{\mathbb{F}}_{q}}$, and set $W:=U/(U\cap
U^{\prime})$. For $v\in U$, we define

$Q_{0}(v)=\lambda_{0}^{2}+\lambda_{0}\lambda_{0}^{\prime}+\lambda_{0}^{\prime
2}+\sum_{t=1}^{4}(\lambda_{t}\lambda_{9-t}+\lambda_{t}^{\prime}\lambda_{9-t}^{\prime}+\lambda_{t}^{\prime\prime}\lambda_{9-t}^{\prime\prime}).$

In particular, if
$v=(\lambda_{0},\lambda_{0}^{\prime},\lambda_{0}^{\prime\prime}\mid D,E,F)$,
then
$Q_{0}(v)=\lambda_{0}^{2}+\lambda_{0}\lambda_{0}^{\prime}+\lambda_{0}^{\prime
2}+D\overline{D}+E\overline{E}+F\overline{F}$. The quadratic form $Q_{0}$ on
$U$ is $F_{4}(q)$-invariant and has $U\cap U^{\prime}$ as its radical, so it
induces a nondegenerate quadratic form $Q$ on $W$. The space $W$ is the
minimal module of $F_{4}(q)$, and $\dim(W)=25$ or $26$ according as the
characteristic $p$ of ${\mathbb{F}}_{q}$ is $3$ or not. The form $Q$ on $W$ is
hyperbolic if $q\equiv 1\pmod{3}$ and is elliptic if $q\equiv 2\pmod{3}$. The
polar spaces in Theorem 1.1 are defined by the quadratic space $(W,Q)$.

##  3\. The proof of Theorem 1.1

In this section we give the proof of Theorem 1.1. Take the same notation as in
the previous section, and we consider the polar space defined by the quadratic
space $(W,Q)$. If the characteristic $p$ of ${\mathbb{F}}_{q}$ is $3$, then
the associated polar space is $Q(24,q)$ and the group $F_{4}(q)$ has two
orbits on the singular points by [5, Table 2]. One orbit ${\mathcal{M}}_{1}$
has size $(q^{4}+1)\frac{q^{12}-1}{q-1}$ which is not divisible by the ovoid
number $q^{12}+1$, so it is a $(q^{4}+1)$-tight set of $Q(24,q)$ as desired.
We suppose that $q\equiv 2\pmod{3}$ in the sequel, so that $W=U$ and the
associated polar space is $Q^{-}(25,q)$. Let $M_{1}$ be the set of white
vectors in $U$, and write ${\mathcal{M}}_{1}$ for the corresponding set of
projective points. The set $M_{1}$ has size $(q^{4}+1)(q^{12}-1)$ and forms a
single $F_{4}(q)$-orbit by [5, (W.3)]. By [16, Lemma 5], we enumerate the
white vectors in $U$ as follows:

  1. (I)

$v=f\cdot(A\overline{A},B\overline{B},1\mid\overline{B},A,\overline{A}B)$ for
some $f\in{\mathbb{F}}_{q}^{*}$ and $A,B\in\mathbb{O}$ such that
$A\overline{A}+B\overline{B}+1=0$,

  2. (II)

$v=e\cdot(C\overline{C},1,0\mid A,\overline{CA},C)$ for some
$e\in{\mathbb{F}}_{q}^{*}$ and octonions $A,C$ such that $A\overline{A}=0$,
$C\overline{C}+1=0$,

  3. (III)

$v=(0,0,0\mid D,E,F)$, where $D,E,F$ are octonions such that

$D\overline{D}=E\overline{E}=F\overline{F}=0,\quad DE=EF=FD=0.$ (3.1)

In particular, $M_{1}$ is ${\mathbb{F}}_{q}^{*}$-invariant and it contains the
singular vector $(0,0,0\mid x_{1},0,0)$. It follows that ${\mathcal{M}}_{1}$
has size $(q^{4}+1)\frac{q^{12}-1}{q-1}$ and consists of singular points of
$Q^{-}(25,q)$.

We claim that ${\mathcal{M}}_{1}$ is a $(q^{4}+1)$-tight set of $Q^{-}(25,q)$.
By Lemma 2.1 and the fact ${\mathcal{M}}_{1}$ is single $F_{4}(q)$-orbit, it
suffices to show that there are $q^{11}+(q^{4}+1)\frac{q^{11}-1}{q-1}$ points
in ${\mathcal{M}}_{1}$ that is perpendicular to $\langle(0,0,0\mid
x_{1},0,0)\rangle$. Suppose that $\langle v\rangle$ is such a point in
${\mathcal{M}}_{1}$, and we examine the three types (I)-(III) one by one. If
$v=(\lambda,\lambda^{\prime},\lambda^{\prime\prime}\mid D,E,F)$, then the
condition that $\langle v\rangle$ is perpendicular to $\langle(0,0,0\mid
x_{1},0,0)\rangle$ translates to $\textup{Tr}(x_{1}\overline{D})=0$.

First consider the case where $\langle v\rangle$ is of type (I). The number of
such points in ${\mathcal{M}}_{1}$ that are perpendicular to
$\langle(0,0,0\mid x_{1},0,0)\rangle$ equals the number of $(A,B)$ pairs such
that $A\overline{A}+B\overline{B}+1=0$, $\textup{Tr}(x_{1}B)=0$. If we write
$B=\sum_{i=1}^{8}b_{i}x_{i}$, then $\textup{Tr}(x_{1}B)=-b_{8}=0$ and
$B\overline{B}=b_{2}b_{7}+b_{3}b_{6}+b_{4}b_{5}$. By Lemma 2.4, we deduce that
the number of such $(A,B)$ pairs is $q\cdot N_{7}(-1)=q^{7}(q^{7}-1)$. Here,
$q$ is contributed by $b_{1}$ and $N_{7}(-1)$ is as defined in Lemma 2.4.

Next consider the case where $\langle v\rangle$ is of type (II). The number of
such points in ${\mathcal{M}}_{1}$ perpendicular to $\langle(0,0,0\mid
x_{1},0,0)\rangle$ equals the number of $(A,C)$ pairs such that
$C\overline{C}+1=0$, $A\overline{A}=0$ and $\textup{Tr}(x_{1}\overline{A})=0$.
By Lemma 2.5, it equals $(q^{7}-q^{3})(q^{6}-q^{3}+q^{4})$.

Finally, consider the case where $\langle v\rangle$ is of type (III), i.e.,
$v=(0,0,0\mid D,E,F)$ such that (3.1) holds and
$\textup{Tr}(x_{1}\overline{D})=0$. We observe that the conditions in (3.1)
are invariant under the cyclic shift of $(D,E,F)$. We divide into the
following three cases.

  1. (1)

Suppose that $D,E,F$ are all nonzero. There are
$(q^{6}-q^{3}+q^{4}-1)(q^{4}-1)$ of $(D,E)$ pairs of nonzero octonions such
that $DE=0$, $\textup{Tr}(x_{1}\overline{D})=0$ by Lemma 2.6. For a pair
$(D,E)$ of nonzero octonions such that $DE=0$, there are $q^{3}-1$ nonzero
$F$’s such that $EF=FD=0$ by Lemma 2.5, and we deduce that
$D\overline{D}=E\overline{E}=F\overline{F}=0$ for such $D,E,F$ by Lemma 2.3.
This case contributes $\frac{q^{4}-1}{q-1}(q^{6}-q^{3}+q^{4}-1)(q^{3}-1)$.

  2. (2)

Suppose that exactly one of $D,E,F$ is zero. If $F=0$, then (3.1) reduces to
$DE=0$ by Lemma 2.3. There are $(q^{6}-q^{3}+q^{4}-1)(q^{4}-1)$ of $(D,E)$
pairs of nonzero octonions such that $DE=0$,
$\textup{Tr}(x_{1}\overline{D})=0$ by Lemma 2.6. By symmetry, we obtain the
same number if $E=0$. If $D=0$, then (3.1) reduces to $EF=0$ and similarly
there are $(q^{3}+1)(q^{4}-1)^{2}$ of such $(E,F)$ pairs of nonzero octonions
that $EF=0$ by Lemmas 2.3 and 2.5. To sum up, this case contributes
$\frac{q^{4}-1}{q-1}(2(q^{6}-q^{3}+q^{4}-1)+(q^{3}+1)(q^{4}-1))$.

  3. (3)

Suppose that exactly one of $D,E,F$ is nonzero. If $E=F=0$, then (3.1) reduces
to $D\overline{D}=0$. There are $q^{6}-q^{3}+q^{4}-1$ nonzero $D$’s such that
$D\overline{D}=0$ and $\textup{Tr}(x_{1}\overline{D})=0$ by Lemma 2.5. If
$D=E=0$, then (3.1) reduces to $F\overline{F}=0$ and there are
$(q^{3}+1)(q^{4}-1)$ such nonzero $F$’s. By symmetry we obtain the same number
if $D=F=0$. To sum up, this case contributes
$\frac{1}{q-1}\left(q^{6}-q^{3}+q^{4}-1+2(q^{3}+1)(q^{4}-1)\right)$.

By adding up all the numbers, we deduce that there are exactly
$q^{11}+(q^{4}+1)\frac{q^{11}-1}{q-1}$ points in ${\mathcal{M}}_{1}$ that is
perpendicular to $\langle(0,0,0\mid x_{1},0,0)\rangle$. This completes the
proof of Theorem 1.1.

######  Remark 3.1.

Suppose that the characteristic $p$ of ${\mathbb{F}}_{q}$ is not $3$. Take the
same notation as in Section 2, and let $K$ be the normalizer of $F_{4}(q)$ in
the normalizer of the quadratic form $Q_{0}$ on $U$. The vectors $x=w_{1}$,
$y=w_{4}+w_{4}^{\prime}$, $z=w_{0}-w_{0}^{\prime\prime}+w_{4}-w_{5}$ are
singular vectors of $Q_{0}$ in $U$, and they are stabilized by the Frobenius
automorphism that raises the coefficients of a vector with respect to the
basis $\\{w_{i},w_{i}^{\prime},w_{i}^{\prime\prime}:\,0\leq i\leq 8\\}$ to
their $p$-th powers. Moreover, they are white, gray and black vectors
respectively and lie in distinct $\tilde{E}_{6}(q)$-orbits. We conclude that
$K$ has at least three orbits on the singular points of the polar space
associated with $(U,Q_{0})$. When $p=2$, we verify with Magma [3] that there
are exactly three $F_{4}(q)$-orbits on the singular point and each orbit form
a tight set. It is not clear whether this holds for further values of $q$ such
that $q\equiv 2\pmod{3}$, and we leave it as an open problem.

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# Proof of a Conjecture of Wiegold

Alexander Skutin

###### Key words and phrases:

Group theory, Finite groups

###### 2010 Mathematics Subject Classification:

20D15

##  1\. Introduction

In this short note we confirm a conjecture of James Wiegold [1, 4.69]. The
breadth $b(x)$ of an element $x$ of a finite $p$-group $G$ is defined by the
equation $|G:C_{G}(x)|=p^{b(x)}$, where $C_{G}(x)$ is the centralizer of $x$
in $G$. We prove the following:

######  Theorem 1.1.

Let $G$ be a finite $p$-group and let $|G^{\prime}|>p^{n(n-1)/2}$ for some
non-negative integer $n$. Then the group $G$ can be generated by the elements
of breadth at least $n$.

An overview of this problem can be found in [2]. Also M.R.Vaughan-Lee in [3]
proved that if in a $p$-group $G$ we have $|G^{\prime}|>p^{n(n-1)/2}$, then
there exists an element in $G$ of breadth at least $n$.

In this article we prove that in the case $p\not=2$ a more general result is
true:

######  Theorem 1.2.

Let $p\not=2$ be a prime number. Let $G$ be a finite $p$-group and let
$|G^{\prime}|>p^{n(n-1)/2}$ for some non-negative integer $n$. Then the set of
elements of breadth at least $n$ cannot be covered by two proper subgroups in
$G$.

In the particular case $p=2$ we also get a more general result:

######  Theorem 1.3.

Let $G$ be a finite $2$-group and let $|G^{\prime}|>2^{n(n-1)/2}$ for some
non-negative integer $n$. Then the set of elements of breadth at least $n$
cannot be covered by two proper subgroups in $G$, one of which has index at
least $4$ in $G$.

So Theorem 1.1 is a consequence of the two Theorems 1.2 and 1.3.

##  2\. Proofs of Theorems 1.2 and 1.3

The breadth $b_{H}(g)$ of an element $g$ of a finite $p$-group $G$ with
respect to a subgroup $H\subseteq G$ is defined by the equation
$|H:C_{H}(g)|=p^{b_{H}(g)}$, where $C_{H}(g)=\\{h\in H|hg=gh\\}$ is the
centralizer of $g$ in $H$. By definition, $b(g)=b_{G}(g)$.

First we formulate Lemmas 2.1, 2.2, 2.3, which will be useful in the proofs of
the Theorems 1.2 and 1.3.

######  Lemma 2.1.

Let $G$ be a finite $p$-group and let $C$ be a finite subgroup of index $p$.
Then for any element $g$ from the set $G\setminus C$ we get
$\log_{p}|G^{\prime}|\leq b(g)+\log_{p}|C^{\prime}|$.

###### Proof.

The cardinality of the set $X=\\{[g,c]|c\in C\\}$ is not bigger than
$p^{b(g)}$. So it is enough to prove that $G^{\prime}=XC^{\prime}$. This
follows from the following properties of the set $XC^{\prime}$ :

  1. (1)

$XC^{\prime}$ is a subgroup in $G$ :
$[g,c_{1}][g,c_{2}]C^{\prime}=[g,c_{1}c_{2}]C^{\prime}$;

  2. (2)

$XC^{\prime}$ is a normal subgroup;

  3. (3)

Group $G/XC^{\prime}$ is abelian.

These facts imply Lemma 2.1.

∎

The next two lemmas are the well-known facts in theory of $p$-groups, so we
state them without proof.

######  Lemma 2.2.

Let $G$ be a finite $p$-group and let $|G:Z(G)|\leq p^{2}$. Then
$\log_{p}|G^{\prime}|\leq 1$.

######  Lemma 2.3.

Let $G$ be a finite $p$-group and let $G=H_{1}\cup H_{2}\cup H_{3}$ for some
three proper subgroups of $G$. Then $p=2$ and $H_{i}$ has index $2$ in $G$ for
$i=1,2,3$.

###  2.1. Proof of Theorem 1.2

Assume the converse. Let the proper subgroups $H_{1}$ and $H_{2}$ cover all
the elements of breadth at least $n$. We will prove that then
$|G^{\prime}|\leq p^{n(n-1)/2}$. The proof is by induction on $|G|$. We can
assume that $|G:H_{i}|=p$ (because every proper subgroup is contained in a
subgroup of index $p$) and $H_{1}\not=H_{2}$ (because in every non-cyclic
$p$-group, there are at least two different maximal proper subgroups).
Consider any subgroup $C$ of $G$ such that $C$ has index $p$ in $G$ and $C\cap
H_{1}=C\cap H_{2}=H_{1}\cap H_{2}$. Notice that the set $\\{c\in
C|b_{C}(c)=b(c)\\}$ is contained in the subgroup $Y=\\{c\in C|[c,g]\in
C^{\prime},\forall g\in G\\}$. This is because if $b_{C}(c)=b(c)$, then for
any $g\in G$ there exists $c^{\prime}\in C$ such that $[c,g]=[c,c^{\prime}]$.
Consider the case when $Y=C$. In this case the central subgroup $C/C^{\prime}$
has a prime index in $G/C^{\prime}$, so $G/C^{\prime}$ is abelian and
$G^{\prime}=C^{\prime}$. The rest follows from the induction hypothesis : the
group $C$ has smaller order and all its elements of breadth at least $n$ are
contained in the subgroup $C\cap H_{1}=C\cap H_{2}$ (because $b_{C}(c)\leq
b(c)$). Now consider the case $Y\not=C$. Notice that the set $\\{c\in
C|b_{C}(c)\geq n-1\\}$ is contained in $(C\cap H_{1})\cup Y$ (because
$b_{C}(c)<b(c)$, when $c\notin Y$). Apply the induction hypothesis to the
group $C$ and its proper subgroups $C\cap H_{1}$ and $Y$. We conclude that
$\log_{p}|C^{\prime}|\leq\frac{(n-2)(n-1)}{2}$. Consider any element $g$ not
lying in $H_{1}\cup H_{2}\cup C$ (this is possible because of Lemma 2.3 and
$p>2$). Its breadth is less than $n$ (because $g\notin H_{1}\cup H_{2}$), so
from Lemma 2.1 we get that $\log_{p}|G^{\prime}|\leq
b(g)+\log_{p}|C^{\prime}|\leq\frac{n(n-1)}{2}$. $\Box$

######  Theorem 2.1.

Let $G$ be a finite $p$-group. Let given that for some integers $n\leq k+1$

  1. (1)

The set of all elements of the breadth at least $n$ can be covered by two
proper subgroups of $G$.

  2. (2)

The set of elements of the breadth at most $k$ generates $G$.

Then $\log_{p}|G^{\prime}|\leq\frac{(n-1)(n-2)}{2}+k$.

###### Proof.

The proof is by induction on $|G|$. Let the subgroups $H_{1}$ and $H_{2}$
cover all the elements of the breadth at least $n$. We can assume that
$|G:H_{i}|=p$ (because any proper subgroup is contained in the subgroup of
index $p$) and $H_{1}\not=H_{2}$ (because in every non-cyclic $p$-group, there
are at least two different maximal proper subgroups). Consider any subgroup
$C$ of $G$ such that $C$ has index $p$ in $G$ and $C\cap H_{1}=C\cap
H_{2}=H_{1}\cap H_{2}$. Notice that the set $\\{c\in C|b_{C}(c)=b(c)\\}$ is
contained in the subgroup $Y=\\{c\in C|[c,g]\in C^{\prime},\forall g\in G\\}$.
It is because if $b_{C}(c)=b(c)$, then for any $g\in G$ there exists
$c^{\prime}\in C$ such that $[c,g]=[c,c^{\prime}]$. So we can conclude that
the set $\\{c\in C|b_{C}(c)\geq n-1\\}$ is contained in $(C\cap H_{1})\cup Y$
(because $b_{C}(c)<b(c)$ if $c\notin Y$). Consider the case $Y=C$. In this
case the central subgroup $C/C^{\prime}$ has a prime index in $G/C^{\prime}$,
so $G/C^{\prime}$ is abelian and $G^{\prime}=C^{\prime}$. The rest follows
from the induction hypothesis : group $C$ has smaller order and all its
elements of the breadth at least $n$ are contained in the subgroup $C\cap
H_{1}=H_{1}\cap H_{2}$ (because $b_{C}(c)\leq b(c)$), also the set $C\setminus
H_{1}$ is contained in the set $\\{c\in C|b_{C}(c)\leq k\\}$ (because
$b_{C}(c)\leq b(c)\leq n-1\leq k$ for $c\in C\setminus H_{1}$) and so $C$ is
generated by the elements of the breadth at most $k$ in subgroup $C$.

So we can assume that $Y$ is a proper subgroup of $C$. Consider the case
$|C:Y|=p$ and consider a homomorphism $\pi:G\to G/C^{\prime}$. It is clear
that $\pi(Y)$ is the central subgroup of $\pi(G)$ and $|\pi(G):\pi(Y)|\leq
p^{2}$. Apply Lemma 2.2 to $\pi(G)$, so we conclude that
$\log_{p}|\pi(G)^{\prime}|\leq 1$ and
$\log_{p}|G^{\prime}|\leq\log_{p}|C^{\prime}|+1$. Apply the induction
hypothesis to the group $C$ and to its two proper subgroups $H_{1}\cap H_{2}$,
$Y$ so we get $\log_{p}|C^{\prime}|\leq\frac{(n-3)(n-2)}{2}+n-1$ (every
element from the set $C\setminus H_{1}$ is of the breadth at most $n-1$ in $C$
and this set generates $C$, also for any element $g$ from the set
$C\setminus(H_{1}\cup Y)$, we have $b_{C}(g)\leq n-2$). So if $k\geq 2$ we get
$\log_{p}|G^{\prime}|\leq\log_{p}|C^{\prime}|+1\leq\frac{(n-3)(n-2)}{2}+n=\frac{(n-2)(n-1)}{2}+2\leq\frac{(n-2)(n-1)}{2}+k$
and the induction step is clear. In the case $k\leq 1$, we get $n=2$ (the case
$n\leq 1$ is trivial) and the set $C\setminus(H_{1}\cup Y)$ is contained in
the center of $C$ (because $b_{C}(g)\leq b(g)-1\leq 0$ for $g\in
C\setminus(H_{1}\cup Y)$). It is clear that the set $C\setminus(H_{1}\cup Y)$
generates the central subgroup of index at most $p$ in $C$, so $C$ is abelian.
From the conditions of Theorem 2.1 there exists an element $g\notin C$ such
that $b(g)\leq k\leq 1$. So the group $C_{G}(g)$ has index at most $p$ in $G$.
Also the subgroup $C_{G}(g)\cap C$ has index at most $p^{2}$ in $G$ and is
central in $G$. Apply Lemma 2.2 to the group $G$, so we conclude that
$\log_{p}|G^{\prime}|\leq 1\leq\frac{(n-1)(n-2)}{2}+k$.

Eventually, we can assume that $|C:Y|\geq p^{2}$. So the set
$C\setminus(H_{1}\cup Y)$ generates $C$, because otherwise if it generates
proper subgroup $H$ of $C$, then $C=(H_{1}\cap C)\cup Y\cup H$ and from Lemma
2.3 we conclude that $|C:Y|\leq 2$, contradicting our assumption that
$|C:Y|\geq p^{2}$. Also the set $C\setminus(H_{1}\cup Y)$ is contained in the
set $\\{g\in C|b_{C}(g)\leq n-2\\}$, so $C$ is generated by the elements of
the breadth at most $n-2$. And we can apply the induction hypothesis to $C$
and its two subgroups $H_{1}\cap H_{2}$, $Y$ and conclude that
$\log_{p}|C^{\prime}|\leq\frac{(n-3)(n-2)}{2}+n-2=\frac{(n-2)(n-1)}{2}$. Also
from the conditions of the Theorem 2.1 there exists an element $a\notin C$
such that $b(a)\leq k$. Apply Lemma 2.1 to $C$ and $a$, so
$\log_{p}|G^{\prime}|\leq
b(a)+\log_{p}|C^{\prime}|\leq\frac{(n-2)(n-1)}{2}+k$.

∎

###  2.2. Proof of Theorem 1.3

Theorem 1.3 is a consequence of Theorem 2.1. Assume the converse. Let the
proper subgroups $H_{1}$ and $H_{2}$ cover all the elements of breadth at
least $n$ and $|G:H_{2}|\geq 4$. We will prove that then $|G^{\prime}|\leq
2^{n(n-1)/2}$. The set $G\setminus(H_{1}\cup H_{2})$ generates $G$, because
otherwise if it generates proper subgroup $H$ of $G$, then $G=H_{1}\cap
H_{2}\cup H$ and from Lemma 2.3 we conclude that $|G:H_{2}|\leq 2$,
contradicting our assumption that $|G:H_{2}|\geq 4$. Now, notice that all the
elements from $G\setminus(H_{1}\cup H_{2})$ have breadth at most $n-1$ in $G$.
So denote $k=n-1$ and apply Theorem 2.1 to $G$, $H_{1}$, $H_{2}$ and $n\leq
k+1$. We conclude that
$\log_{2}|G^{\prime}|\leq\frac{(n-1)(n-2)}{2}+k=\frac{n(n-1)}{2}$.

######  Remark 2.1.

In fact, Theorem 1.2 is also a consequence of Theorem 2.1.

The following Theorems 2.2, 2.3 are generalizations of Theorems 1.2 and 1.3,
respectively. Their proofs are similar to the proofs of Theorems 1.2 and 1.3.

######  Theorem 2.2.

Let $p$ be a prime number. Let $G$ be a finite $p$-group and let
$|G^{\prime}|>p^{n(n-1)/2}$ for some non-negative integer $n$. Then the set of
elements of breadth at least $n$ cannot be covered by $p-1$ proper subgroups
in $G$.

######  Theorem 2.3.

Let $p$ be a prime number. Let $G$ be a finite $p$-group and let
$|G^{\prime}|>p^{n(n-1)/2}$ for some non-negative integer $n$. Then the set of
elements of breadth at least $n$ cannot be covered by $p$ proper subgroups in
$G$, one of which has index at least $p^{2}$ in $G$.

##  3\. Acknowledgements

I am grateful to Anton A. Klyachko and to Michael Vaughan-Lee for useful
discussions and the improvement of an earlier version of the manuscript.

## References

  * [1] V.D. Mazurov and E.I. Khukhro (Editors), Unsolved problems in group theory - The Kourovka notebook, No. 18, arXiv:1401.0300v10, 8 Sep 2017.
  * [2] Wiegold. J, Commutator subgroups of finite p-groups, J. Australian Math, Soc., 10 (1969), 480-484.
  * [3] M.R. Vaughan-Lee, Breadth and commutator subgroups of p-groups, J. Algebra 32 (1974), 278-285.