# Schur rings over ${\bf{\rm Sp}(n,2)}$ and multiplicity one subgroups

Stephen Humphries  Department of Mathematics, Brigham Young University, Provo,
UT 84602, U.S.A. E-mail<EMAIL_ADDRESS>
###### Abstract.

We study commutative Schur rings over the symplectic groups Sp$(n,2)$
containing the class $\mathcal{C}$ of symplectic transvections. We find the
possible partitions of $\mathcal{C}$ determined by the Schur ring. We show how
this restricts the possibilities for multiplicity one subgroups of Sp$(n,2)$.

Keywords: Symplectic group; characteristic $2$; Schur ring; Orthogonal group;
symmetric group; $3$-transposition group.

###### 2010 Mathematics Subject Classification:

Primary 05B10. Secondary: 20C05.

##  1\. Introduction

Let $G$ be a finite group. For $X\subseteq G$ we let $\overline{X}=\sum_{x\in
X}x\in\mathbb{Q}G$ and $X^{(-1)}=\\{x^{-1}:x\in X\\}$. Then a Schur ring [21,
24, 8, 20] over $G$ is a subalgebra $\mathfrak{S}$ of $\mathbb{Q}G$ determined
by a partition $C_{1}=\\{1\\},C_{2},\cdots,C_{r}$ of $G$ where (i) for $1\leq
i\leq r$ there is $1\leq i^{\prime}\leq r$ with $C_{i}^{(-1)}=C_{i^{\prime}}$;
(ii) there are $\lambda_{ijk}\in\mathbb{Z}^{\geq 0}$ with
$\overline{C_{i}}\,\overline{C_{j}}=\sum_{k=1}^{r}\lambda_{ijk}\overline{C_{k}}.$

The $C_{i}$ are called principal sets (or p-sets) and an $\mathfrak{S}$-set is
a union of principal sets. One way to construct a Schur ring over $G$ is to
take a subgroup $H\leq G$ and let the $C_{i}$ be the $H$-classes
$g^{H}=\\{g^{h}:h\in H\\}$. This Schur ring is denoted $\mathfrak{S}(G,H)$.

With this definition we can propose the following:

Problem: Suppose that we have a finite group $G$ that is generated by a
conjugacy class $\mathcal{C}$. Determine all partitions of $\mathcal{C}$ that
come from a commutative Schur ring $\mathfrak{S}$ over $G$ that contains
${\mathcal{C}}$ as an $\mathfrak{S}$-set.

This problem is particularly relevant for groups containing a canonical class
$\mathcal{C}$ of generators, and more specifically when those generators are
involutions. Some examples of such a class of groups are the 3-transposition
groups: a group $G$ that has a conjugacy class $\mathcal{C}$ of involutary
generators satisfying: for $a,b\in\mathcal{C}$ the order of $ab$ is $1,2$ or
$3$ [13, 4, 15]. The most-studied class of 3-transposition groups are the
symmetric groups $S_{n}$ where $\mathcal{C}$ is the set of transpositions.
Other examples of 3-transposition groups are the symplectic groups Sp$(n,2)$
and the orthogonal groups ${\rm SO}^{\pm}(n,2)\leq{\rm Sp}(n,2)$. In [16] we
addressed the above problem for $S_{n}$ obtaining:

######  Theorem 1.1.

Let $n\geq 6$ and let $\mathfrak{S}$ be a commutative S-ring over $S_{n}$
where ${\mathcal{C}}=(1,2)^{S_{n}}$ is an $\mathfrak{S}$-set. Suppose that
$C_{1},\dots,C_{r}$ are the principal sets of $\mathfrak{S}$ that are
contained in $\mathcal{C}$. Then $r\leq 3$.

More information about the partition of $(1,2)^{S_{n}}$ in the $S_{n}$ case is
given in [16].

In this paper we consider the 3-transposition group $G=G_{n}:={\rm
Sp}(n,2)={\rm Sp}(V)$ where $V$ is a non-degenerate symplectic space over
$\mathbb{F}_{2}=\\{0,1\\}$ of even dimension $n\geq 2$. Here the conjugacy
class $\mathcal{C}$ is the class of symplectic transvections
$\mathcal{T}=\mathcal{T}_{n}$ that determines ${\rm Sp}(n,2)$ as a
3-transposition group.

The main result in this paper is

######  Theorem 1.2.

Let $n\geq 14$ and let $\mathfrak{S}$ be a commutative S-ring over Sp$(n,2)$
where ${\mathcal{T}}$ is an $\mathfrak{S}$-set. Suppose that
$C_{1},\dots,C_{r}$ are the principal sets of $\mathfrak{S}$ that are
contained in $\mathcal{T}$. Then $r=1$ and $C_{1}=\mathcal{T}$.

We also consider the cases $n=2,4,6$. For a group $G$ and $H\leq G$ we say
that $(G,H)$ is a strong Gelfand pair if for every
$\chi\in\hat{G},\psi\in\hat{H}$ we have $\langle\psi,\chi_{H}\rangle\leq 1$.
We then say that $H$ is a strong Gelfand subgroup of $G$. Here $\hat{G}$ is
the set of irreducible characters of $G$. Note that $G$ is always a strong
Gelfand subgroup of $G$, and if $H\leq G$ is one, then so is $H^{g},g\in G$.
We note that the concept of a strong Gelfand pair is what is sometimes
referred to as the ‘multiplicity one’ property [1, 2].

Now by [23, Corollary 1’] or [17, Corollary 2.2] $(G,H)$ is a strong Gelfand
pair exactly when $\mathfrak{S}(G,H)$ is commutative, and in this case, if we
have a 3-transposition group $G$ with conjugacy class $\mathcal{C}$, then
$\mathcal{C}$ will be an $\mathcal{S}$-set of $\mathfrak{S}(G,H)$. We show
that for $n=2,4$ the only partitions of $\mathcal{T}$ are those determined by
strong Gelfand pairs. This result is included in [16] since $G_{2}={\rm
Sp}(2,2)\cong S_{3},G_{4}={\rm Sp}(4,2)\cong S_{6}$:

######  Theorem 1.3.

(A) There are three strong Gelfand subgroups of $G_{2}\cong S_{3}$, namely
$\langle(1,2)\rangle,\langle(1,2,3)\rangle,S_{3}$. Their sizes and the sizes
of their p-sets in $\mathcal{T}$ are

(i) $2,[1,2]$; (ii) $3,[3]$; (iii) $6,[3]$.

(B) There are seven strong Gelfand subgroups of $G_{4}\cong S_{6},$ each of
which is either $G_{4}$ or a maximal subgroup. Their sizes and the sizes of
their p-sets in $\mathcal{T}$ are

(i) $48,[3,12]$; (ii) $48,[1,6,8]$; (iii) $72,[6,9]$; (vi) $120,[5,10]$; (v)
$120,[15]$; (vi) $360,[15]$; (vii) $720,[15]$.

A Magma [18] calculation allows us to find the only strong Gelfand subgroups
of $G_{6},G_{8},G_{10},G_{12}$:

######  Theorem 1.4.

There are two strong Gelfand subgroups of $G_{6}$. Their sizes and the sizes
of their p-sets in $\mathcal{T}$ are

(i) $25920,[27,36]$; (ii) $1451520,[63]$.

For $n=8,10,12$ there are no proper strong Gelfand subgroups of $G_{n}$. ∎

To prove this it suffices to check the maximal subgroups of each such $G_{n}$,
since if we have $H\leq K\leq G$ where $H$ is a strong Gelfand subgroup of
$G$, then $K$ is also a strong Gelfand subgroup of $G$. In general, our main
theorem gives:

######  Corollary 1.5.

Let $H$ be a proper strong Gelfand subgroup of $G_{n},n\geq 8$. Then

(i) $H$ acts transitively on $\mathcal{T}$; and

(ii) $H$ does not contain any symplectic transvections.

Proof (i) By Theorem 1.4 we can assume $n\geq 14$. If $H$ does not act
transitively on $\mathcal{T}$, then the commutative Schur ring
$\mathfrak{S}(G_{n},H)$ partitions $\mathcal{T}$ into at least two subsets,
contrary to Theorem 1.2. (ii) If $H$ contains a transvection, then by (i) it
will contain all transvections. But then $H=G_{n}$, a contradiction. ∎

Here, as indicated above, we could restrict ourselves to the situation where
$(G,H)$ is a strong Gelfand pair and $H$ is maximal. Then conditions (i) and
(ii) of Corollary 1.5 are satisfied by the maximal subgroups $H<G_{n}$ studied
by Dye in [11, 12]. These only occur when $n=2m$ with $m$ prime, and they are
the stabilizers of a set $\\{U_{i}\\}_{i}$ (called a spread) of subspaces
where $\\{U_{i}\setminus\\{0\\}\\}_{i}$ partitions $V\setminus\\{0\\}$. We
denote this $H$ by ${\rm Dye}_{n}$ and we show later that ${\rm Dye}_{n}$ is
not a strong Gelfand subgroup.

In §§2-4 we give definitions relating to the symplectic structure of
$V=\mathbb{F}_{2}^{n}$ and results regarding products of symplectic
transvections. We also define functions $f_{1}(C),\cdots,f_{4}(C)$ depending
on a set $C\subset\mathcal{T}$ and with domain $C$. We show that these
functions are constant on p-sets of our Schur ring. We also show that if
$f_{2}(C)=0$, then $\langle C\rangle$ acts on $C$ by conjugation. Then in §5
we give an overview of the remainder of the proof, showing how we reduce to
the case $r=1$.

Acknowledgement All calculations made in the preparation of this paper were
accomplished using Magma [18].

##  2\. Definitions and Preliminary results

If $V=\mathbb{F}_{2}^{n}$ is a non-degenerate symplectic space with symplectic
form $\cdot$ then there is a (standard) symplectic basis $e_{1},\cdots,e_{n}$,
where $e_{i}\cdot e_{j}=1$ if and only if $i+j=n+1$. We will also sometimes
write a symplectic basis as
$a_{1}=e_{1},b_{1}=e_{n},a_{2}=e_{2},b_{2}=e_{n-1},\cdots,a_{n/2}=e_{n/2},b_{n/2}=e_{n/2+1}$.
We let $V^{\sharp}:=V\setminus\\{0\\}$. Then $G_{n}={\rm Sp}(n,2)$ is the
group of linear maps of $V$ that preserve this form [3, 10].

To each $v\in V^{\sharp}$ we associate the symplectic transvection $t_{v}\in
G_{n}$, where

$(w)t_{v}=w+(v\cdot w)v\text{ for }w\in V.$

If $v=0\in V$, then we naturally define $t_{0}=id_{G}$. Let
$\mathcal{T}=\mathcal{T}_{n}$ denote the set of symplectic transvections in
$G_{n}$. Then $\mathcal{T}$ is a conjugacy class of involutions in $G$. For
$1\leq i,j\leq n$ we let $t_{i}=t_{e_{i}},t_{i,j}=t_{e_{i}+e_{j}}$ etc.

In what follows we will identify $v\in V^{\sharp}$ with $t_{v}\in\mathcal{T}$
and so identify any $U\subseteq V^{\sharp}$ with
$\mathcal{T}(U):=\\{t_{u}:u\in U\\}$.

We recall that a subset $X\subset V$ is isotropic if $u\cdot v=0$ for all
$u,v\in X.$ For $X\subseteq V$ we let $X^{\perp}=\\{u\in V:u\cdot x=0\text{
for all }x\in X\\}$. For $X,Y\subseteq V$ we write $X\perp Y$ if $x\cdot y=0$
for all $x\in X,y\in Y$.

We will have need to use the Schur-Wielandt principle: if $\alpha=\sum_{g\in
G}f(g)g\in\mathfrak{S},$ where $f:G\to\mathbb{R}$ is the coefficient function,
then for each $\lambda\in\mathbb{R}$ we have $\sum_{g\in
G:f(g)=\lambda}g\in\mathfrak{S}$. We will also use the following without
reference [3, 10]:

######  Theorem 2.1 (Witt’s Theorem).

Let $U,W$ be subspaces of $V$ and let $\varphi:U\to W$ be an isometry. Then
$\varphi$ can be extended to an isometry $\psi:V\to V.$

We have the following fundamental properties of transvections::

######  Lemma 2.2.

Let $t_{a},t_{b}\in\mathcal{T}$. Then

(i) $t_{a}=t_{b}$ if and only if $a=b$ if and only if the order of
$t_{a}t_{b}$ is $1$.

(ii) For $a\neq b$ we have: $a\cdot b=0$ if and only if the order of
$t_{a}t_{b}$ is $2$ if and only if $t_{a}$ and $t_{b}$ commute.

(iii) $a\cdot b=1$ if and only if the order of $t_{a}t_{b}$ is $3$ if and only
if $t_{a}$ and $t_{b}$ do not commute if and only if
$t_{a}^{t_{b}}=t_{a+b}=t_{b}^{t_{a}}$.

(iv) For $\alpha\in G,\alpha\in V^{\sharp},$ we have
$\alpha^{-1}t_{a}\alpha=t_{(a)\alpha}$.

(v) For pairs $t_{a},t_{b}\in\mathcal{T},a\neq b,$ and
$t_{c},t_{d}\in\mathcal{T},c\neq d,$ the elements $t_{a}t_{b}$ and
$t_{c}t_{d}$ are conjugate if and only if $a\cdot b=c\cdot d$.

(vi) If $\alpha=t_{v_{1}}t_{v_{2}}\cdots t_{v_{s}},v\in V$, then
$(v)\alpha=v+w$, where $w\in{\rm Span}(v_{1},\cdots,v_{s})$. ∎

######  Lemma 2.3.

Let $a,b\in V^{\sharp}$ with $a\cdot b=1.$ If $t_{c},t_{d}\in\mathcal{T}$,
then $t_{c}t_{d}=t_{a}t_{b}$, if and only if we have one of:

(i) $c=a,d=b$;

(ii) $c=b,d=a+b$; or

(iii) $c=a+b,d=a$.

Here the pair $[t_{c},t_{d}]$ has the form $[t_{a}^{g^{i}},t_{b}^{g^{i}}],$
for some $i=0,1,2,$ where $g=t_{a}t_{b}$.

Proof Note that $t_{a}^{t_{b}}=t_{a+b}$. For $c,d$ as in (i)-(iii) above, one
checks that $t_{c}t_{d}=t_{a}t_{b}$.

Conversely: assume $t_{c}t_{d}=t_{a}t_{b}$. If $c\cdot d=0$, then $t_{a}t_{b}$
and $t_{c}t_{d}$ are not conjugate and so not equal. Thus $c\cdot d=1$, and
from $(x)t_{a}t_{b}=(x)t_{c}t_{d},x\in V,$ we get

(2.1) $\displaystyle(b\cdot x)b+(a\cdot x)[a+b]=(d\cdot x)d+(c\cdot x)[c+d].$

Putting $x=a$ and $x=b$ in Eq. (2.1) we get $b,a+b\in{\rm{\rm Span}}(c,d)$
(respectively), so that ${\rm Span}(a,b)$= Span$(c,d)$ and we write:

$c=\alpha a+\beta b,\qquad d=\gamma a+\delta b\text{ where
}\alpha,\beta,\gamma,\delta\in\mathbb{F}_{2}.$

Putting $x=a$ in Eq. (2.1) we get

$b=\delta(\gamma a+\delta b)+\beta(\alpha a+\beta b+\gamma a+\delta b)$

Since $a\cdot b=1$, $a,b$ are linearly independent, and equating coefficients
of $a,b$ we get

(2.2) $\displaystyle 0=\delta\gamma+\beta\alpha+\beta\gamma,\qquad
1=\delta+\gamma+\delta\gamma.$

Similarly, putting $x=b$ in Eq. (2.1), we get

(2.3) $\displaystyle 1=\gamma+\alpha+\alpha\gamma,\qquad
1=\gamma\delta+\alpha\beta+\alpha\delta.$

Solving Eqs (2.2), (2.3) over $\mathbb{F}_{2}$ gives
$[c,d]\in\\{[a,b],[b,a+b],[a+b,a]\\}$. ∎

######  Lemma 2.4.

(i) If $a,b\in V^{\sharp}$ with $a\cdot b=0,a\neq b$, then the only $c,d\in
V^{\sharp}$ with $t_{c}t_{d}=t_{a}t_{b}$ are where
$[c,d]\in\\{[a,b],[b,a]\\}$.

(ii) For $a,b,c\in V^{\sharp}$ we cannot have $t_{a}t_{b}=t_{c}$.

Proof (i) This is similar to the proof of Lemma 2.3. (ii) If $a=b$, then
$t_{c}=1$, a contradiction. If $a\neq b$, then one shows that the image of
$t_{a}t_{b}-1$ has dimension $2$, a contradiction, since the image of
$t_{c}-1$ has dimension $1$. ∎

The next result tells us when $\overline{C}$ and $\overline{D}$ commute, for
disjoint $C,D\subseteq\mathcal{T}$.

######  Lemma 2.5.

For disjoint and non-empty $C,D\subseteq\mathcal{T}$ we have:
$\overline{C},\overline{D}$ commute if and only if for all $t_{a}\in
C,t_{b}\in D$ with $a\cdot b=1$ we have $t_{a+b}\in C\cup D$.

Proof First note that if $C\perp D$, then $\overline{C},\overline{D}$ commute,
so we may now assume that $\overline{C},\overline{D}$ commute and there are
$t_{a}\in C,t_{b}\in D$ with $a\cdot b=1$. Then
$t_{a}t_{b}\in\overline{C}\cdot\overline{D}$, and so there are $t_{c}\in
D,t_{d}\in C$ with $t_{c}t_{d}=t_{a}t_{b}$. By Lemma 2.3 we have one of:

(i) $c=a,d=b$;

(ii) $c=b,d=a+b$; or

(iii) $c=a+b,d=a$.

If (i), then $t_{a}\in C\cap D=\emptyset$, a contradiction; if (ii) or (iii),
then $t_{a+b}\in C\cup D$.

For the converse: assume that if $t_{a}\in C,t_{b}\in D,a\cdot b=1$, then
$t_{a+b}=t_{a}^{t_{b}}=t_{b}^{t_{a}}\in C\cup D$. We show that
$\overline{C},\overline{D}$ commute. Let $t_{a}\in C,t_{b}\in D$ so that
$t_{a}t_{b}\in\overline{C}\,\overline{D}$. We first show that $t_{a}t_{b}$ is
a summand of $\overline{D}\,\overline{C}$. If $a\cdot b=0$, then
$t_{a}t_{b}=t_{b}t_{a}\in\overline{D}\cdot\overline{C}$ and we are done. So
assume $a\cdot b=1$. Then $t_{a}^{t_{b}}=t_{a+b}\in C\cup D$. We have two
cases:

(a) If $t_{a}^{t_{b}}\in C$, then $\overline{D}\,\overline{C}$ contains
$t_{b}t_{a}^{t_{b}}=t_{a}t_{b}$ as a summand, and we are done.

(b) If $t_{a}^{t_{b}}\in D$, then $\overline{D}\,\overline{C}$ contains
$t_{a}^{t_{b}}t_{a}=t_{b}^{t_{a}}t_{a}=t_{a}t_{b}$, as required.

From Lemma 2.3, Lemma 2.4 and the above enumeration of cases we see that, for
fixed $t_{a}\in C,t_{b}\in D,$ there is only one way to find $t_{c}\in
D,t_{d}\in C$ with $t_{a}t_{b}=t_{c}t_{d}$. Thus the coefficient of
$t_{a}t_{b}$ in $\overline{C}\,\overline{D}$ and in
$\overline{D}\,\overline{C}$ is $1$, and so
$\overline{C}\,\overline{D}=\overline{D}\,\overline{C}$.∎

######  Corollary 2.6.

Let $C,D\subset\mathcal{T}$ be disjoint and assume that
$\overline{C},\,\overline{D}$ commute. Let $W\leq V$ be a subspace. Then
$\overline{C\cap W},\overline{D\cap W}$ also commute.

Proof We use Lemma 2.5: assume that $t_{a}\in C\cap\mathcal{T}(W),t_{b}\in
D\cap\mathcal{T}(W),a\cdot b=1.$ Then $a,b\in W$ and by Lemma 2.5 applied to
the pair $t_{a}\in C,t_{b}\in D$, we have $t_{a+b}\in C\cup D$. But $W$ is a
subspace and so $a+b\in W$; so either $t_{a+b}\in C\cap\mathcal{T}(W)$ or
$t_{a+b}\in D\cap\mathcal{T}(W)$. ∎

######  Lemma 2.7.

Let $C\subseteq\mathcal{T}$. Choose $a,b\in V^{\sharp},a\cdot b=1$. Then
$t_{a}t_{b},t_{b}t_{a}$ occur in $\overline{C}^{2}$ with the same non-zero
coefficient as follows:

  1. 1\.

the coefficient is zero if and only if $|C\cap\\{t_{a},t_{b},t_{a+b}\\}|\leq
1.$

  2. 2\.

the coefficient is $1$ if and only if $|C\cap\\{t_{a},t_{b},t_{a+b}\\}|=2.$

  3. 3\.

the coefficient is $3$ if and only if $|C\cap\\{t_{a},t_{b},t_{a+b}\\}|=3.$

Proof By Lemma 2.3 we need only check the cases where $C$ is one of
$\\{t_{a}\\},\\{t_{a},t_{b}\\},$
$\\{t_{a},t_{a+b}\\},\\{t_{b},t_{a+b}\\},\\{t_{a},t_{b},t_{a+b}\\}$. Checking
each case gives the result. ∎

######  Corollary 2.8.

(i) Let $a,b\in V^{\sharp},a\neq b,a\cdot b=0$. Then the coefficient of
$t_{a}t_{b}$ in $\overline{\mathcal{T}}^{2}$ is $2$.

(ii) Let $a,b\in V^{\sharp},a\neq b,a\cdot b=1$. Then the coefficient of
$t_{a}t_{b}$ in $\overline{\mathcal{T}}^{2}$ is $3$.

In particular,
$\overline{\mathcal{T}}^{2}=|\mathcal{T}|+2(t_{1}t_{2})^{G}+3(t_{1}t_{n})^{G}$
and $(t_{1}t_{2})^{G},(t_{1}t_{n})^{G}\in\mathfrak{S}$.∎

Definition Let $\overline{C}\in\mathfrak{S},C\subseteq\mathcal{T}$. Define
${\rm Tr}^{2}(C)$ to be the set of triples $\\{a,b,a+b\\},a,b\in
V^{\sharp},a\cdot b=1,$ that have exactly two of $t_{a},t_{b},t_{a+b}$ in $C$.
Let ${\rm Tr}^{3}(C)$ denote the set of all $\\{a,b,a+b\\},a,b\in
V^{\sharp},a\cdot b=1$ where $t_{a},t_{b},t_{a+b}\in C$.

Each set $U=\\{a,b,a+b\\},a\cdot b=1,$ (or $U=\\{a,b\\},a\cdot b=1,$)
determines a sum:

$U^{*}:=t_{a}t_{b}+t_{b}t_{a}=t_{a}t_{a+b}+t_{a+b}t_{a}=t_{b}t_{a+b}+t_{a+b}t_{b}\in\mathbb{Q}G_{n}.$

Note that $U^{*}$ is well-defined. The point of this definition is that if
$a\cdot b=1$, then

$(t_{a}+t_{b})^{2}=2+\\{t_{a},t_{b}\\}^{*},\text{ and
}(t_{a}+t_{b}+t_{a+b})^{2}=3+3\\{t_{a},t_{b},t_{a+b}\\}^{*}.$

Further, if $C\subseteq\mathcal{T}$, then the only way that
$\\{t_{a},t_{b}\\}^{*}$ can occur in $\overline{C}^{2}$ is if $C$ contains at
least two of $\\{t_{a},t_{b},t_{a+b}\\}$. We have:

######  Lemma 2.9.

Let $\overline{C}\in\mathfrak{S},C\subseteq\mathcal{T}$. Then

(i) $\sum_{T\in{\rm Tr}^{2}(C)}T^{*}\in\mathfrak{S}$;

(ii) $\sum_{T\in{\rm Tr}^{3}(C)}T^{*}\in\mathfrak{S}$.

Proof By Corollary 2.8 $(t_{1}t_{n})^{G}\in\mathfrak{S}$. By Lemma 2.7 we have

$\overline{C}^{2}=\left(\sum_{T\in{\rm
Tr}^{2}(C)}T^{*}\right)+3\left(\sum_{T\in{\rm Tr}^{3}(C)}T^{*}\right)+Z,$

where no summand of $Z$ is in $(t_{1}t_{n})^{G}$. Now Corollary 2.8 and the
Schur-Wielandt principle give the result. ∎

######  Lemma 2.10.

Let $t_{a},t_{x},t_{y},t_{z}\in\mathcal{T}$, where $t_{x}t_{y}t_{z}=t_{a}$.
Then $[x,y,z]$ is one of:

(i) $[a,d,d]$ for some $d\in V^{\sharp}$;

(ii) $[d,a,d]$ for some $d\in V^{\sharp}$ with $a\cdot d=0$;

(iii) $[d,d,a]$ for some $d\in V^{\sharp}$;

(iv) $[b,a,a+b]$ for some $b\in V^{\sharp}$ with $a\cdot b=1$;

(v) $[b,a+b,b]$ for some $b\in V^{\sharp}$ with $a\cdot b=1$;

Proof First, it is easy to check that each of (i)-(v) gives a solution to
$t_{x}t_{y}t_{z}=t_{a}$.

If $x=y$, then $z=a$ and we have (iii). So assume $x\neq y$. If $y=z$, then
$x=a$ and we have (i), so assume $y\neq z$. If $x=z$, then
$t_{a}=t_{x}t_{y}t_{x}$. If $x\cdot y=0$, then $t_{a}=t_{x}t_{y}t_{x}=t_{y}$
and so $y=a$, so we have (ii); if $x\cdot y=1$, then
$t_{a}=t_{x}t_{y}t_{x}=t_{x+y}$, so $a=x+y$, and we have case (v). Now we
assume that $x,y,z$ are distinct.

If $x\cdot y=0$, then $t_{x}t_{y}=t_{a}t_{z}$ and by Lemma 2.4
$\\{x,y\\}=\\{a,z\\}$. If $x=a$, then $y=z$, a contradiction. If $y=a,$ then
$t_{a}t_{x}=t_{x}t_{a}=t_{a}t_{z}$, so that $x=z$, a contradiction.

If $x\cdot y=1$, then by Lemma 2.3 applied to $t_{x}t_{y}=t_{a}t_{z}$ we have
$[x,y]\in\\{[a,z],[z,a+z],[a+z,a]\\}$. If $x=a,y=z$, then we have a
contradiction. Also $x=z,y=a+z$, gives a contradiction. Lastly, $x=a+z,y=a$,
gives case (iv). ∎

##  3\. The functions $f_{1},f_{2},f_{3},f_{4}$

Given $C\subseteq\mathcal{T}$ we define four functions
$f_{i}=f_{i}(C):C\to\mathbb{Z}^{\geq 0},i=1,2,3,4$:

For $a\in C$ we let $f_{1}(C,a)$ be the number of $b\in C$ such that $a\cdot
b=1$ and $a+b\in C$.

For $a\in C$ we let $f_{2}(C,a)$ be the number of $b\in C$ such that $a\cdot
b=1$ and $a+b\notin C$.

For $a\in C$ we let $f_{3}(C,a)=|C|$.

For $a\in C$ we let $f_{4}(C,a)$ be the number of $b\in C,b\neq a,$ such that
$a\cdot b=0$.

For $t_{a},t_{b}\in C$ we will say that the pair $t_{a},t_{b}$ has type
$f_{1}$ if $a\cdot b=1$ and $a+b\in C$. Similarly for types $f_{2}$ and
$f_{4}$.

Note that for fixed $t_{a}\in C$ and $t_{b}\in C,b\neq a,$ either $a\cdot b=1$
or $a\cdot b=0$. Thus

(3.1) $\displaystyle
2f_{1}(C,t_{a})+f_{2}(C,t_{a})+f_{4}(C,t_{a})+1=f_{3}(C,t_{a}).$

One of our goals will be to show that if $C\subseteq\mathcal{T}$ is a p-set of
a commutative Schur ring $\mathfrak{S}$ over $G_{n}$ containing $\mathcal{T}$,
then the functions $f_{i}(C)$ are constant on $C$.

For $i=1,2,4$ and a p-set $C$ we let

(3.2) $\displaystyle(C^{2})_{f_{i}}=\\{t_{a}t_{b}:t_{a},t_{b}\in C\text{ where
}t_{a},t_{b}\text{ has type }f_{i}\\}.$

######  Lemma 3.1.

For any $C\subseteq\mathcal{T}$ each $(C^{2})_{f_{i}},i=1,2,4,$ is in
$\mathfrak{S}$.

Proof By Lemma 2.7
$\overline{C}^{2}=|C|+3(C^{2})_{f_{1}}+1(C^{2})_{f_{2}}+2(C^{2})_{f_{4}}$ and
we are done. ∎

Now let $n\geq 4,C\subseteq\mathcal{T},|C|>0,C^{\prime}=\mathcal{T}\setminus
C,\overline{C}\in\mathfrak{S}$, so that
$\overline{C^{\prime}}\in\mathfrak{S}$. Let

$A={\rm Tr}^{3}(C),\,\,B={\rm Tr}^{2}(C),\,\,A^{\prime}={\rm
Tr}^{3}(C^{\prime}),\,\,B^{\prime}={\rm Tr}^{2}(C^{\prime}).$

Then by Lemma 2.9

$\sum_{T\in A}T^{*}\in\mathfrak{S},\,\,\sum_{T\in
B}T^{*}\in\mathfrak{S},\,\,\sum_{T\in
A^{\prime}}T^{*}\in\mathfrak{S},\,\,\sum_{T\in
B^{\prime}}T^{*}\in\mathfrak{S}.$

Note that if $U=\\{a,b,a+b\\}\in A,a\cdot b=1$ so that
$U^{*}=t_{a}t_{b}+t_{b}t_{a}$, then

(3.3) $\displaystyle U^{*}(t_{a}+t_{b}+t_{a+b})=2\cdot(t_{a}+t_{b}+t_{a+b})$

while if $U=\\{a,b\\}\in B,$ where $a,b\in C,a\cdot b=1$, then

(3.4) $\displaystyle U^{*}(t_{a}+t_{b})=2t_{a+b}+t_{a}+t_{b};$ (3.5)
$\displaystyle U^{*}(t_{a}+t_{a+b})=2t_{b}+t_{a}+t_{a+b};$ (3.6)
$\displaystyle U^{*}(t_{b}+t_{a+b})=2t_{a}+t_{b}+t_{a+b}.$

If $a\cdot b=1$, then we define $\sigma(\\{a,b,a+b\\}):=t_{a}+t_{b}+t_{a+b}$.

######  Lemma 3.2.

For $A,A^{\prime}$ as above we have:

(i) $\sum_{\\{a,b,a+b\\}\in A}\sigma(\\{a,b,a+b\\})\in\mathfrak{S};$

(ii) $\sum_{\\{a,b,a+b\\}\in A^{\prime}}\sigma(\\{a,b,a+b\\})\in\mathfrak{S}.$

Proof (i) From $\sum_{T\in A}T^{*},\overline{(t_{1}t_{n})^{G}}\in\mathfrak{S}$
we get $\sum_{T\in A}T^{*}\cdot\overline{(t_{1}t_{n})^{G}}\in\mathfrak{S}$. By
Eq. (3.3) the coefficient of $t_{a}$ in this product is twice the number of
$\\{a,b,a+b\\}_{b\in V^{\sharp}}$ that are in $A$. So (i) follows and (ii) is
similar, using Eqs. (3.4), (3.5), (3.6) .∎

######  Corollary 3.3.

If $C\subset\mathcal{T}$ is a p-set, then for $t_{a}\in C$ the number of
$t_{b}\in C,a\cdot b=1,$ where $t_{a+b}\in C$, is independent of $a$, and the
function $f_{1}(C)$ is constant on $C$.

Proof Let $C$ be a p-set. Let $q:=\sum_{\\{a,b,a+b\\}\in
A}\sigma(\\{a,b,a+b\\})\in\mathfrak{S}.$ The transvection $t_{a}\in C$ occurs
with coefficient $\lambda$ in $q$ if and only if the number of
$\\{a,b,a+b\\}\in A$ with $a$ constant and $b$ varying is $\lambda$. Since $C$
is a p-set and $q\in\mathfrak{S}$, this number must be constant over $C$. But
this number is $f_{1}(C,t_{a})$.∎

We now consider $f_{2}(C,t_{a})$.

######  Lemma 3.4.

Let $C\subseteq\mathcal{T}$ and $t_{a}\in C$. Then the coefficient of $t_{a}$
in $\overline{C}\cdot(t_{1}t_{2})^{G}$ is $f_{4}(C,t_{a})$.

Proof This follows directly from Lemma 2.10 as an element of such a product
has the form $t_{u}\cdot t_{v}t_{w}$, where $v\cdot w=0$ and $t_{u}\in C$.∎

######  Proposition 3.5.

Given a p-set $C\subseteq\mathcal{T}$ the functions
$f_{1}(C),f_{2}(C),f_{3}(C),f_{4}(C)$ are constant.

Proof From Lemma 3.4 $f_{4}(C)$ is constant on $C$; also $f_{3}(C)=|C|$. By
Corollary 3.3 $f_{1}(C)$ is constant on $C$. But then Eq. (3.1) shows that
$f_{2}(C)$ is constant on $C$.∎

Since $f_{i}(C_{j},\\_\\_)$ is constant on each p-set $C_{j}$ we will write
$f_{i}(C_{j})$ for this constant value. Define

(3.7) $\displaystyle D_{1}(C):=\sum_{\begin{matrix}T\in B\text{ where}\\\
T\cap C=\\{u,v\\}\end{matrix}}t_{u}+t_{v},\quad
D_{2}(C):=\sum_{\begin{matrix}T\in B\text{ where}\\\ T\cap
C=\\{u,v\\}\end{matrix}}t_{u+v}.$

######  Lemma 3.6.

Let $C\subseteq\mathcal{T}$ be a p-set of $\mathfrak{S}$. Then
$D_{1}=D_{1}(C),D_{2}=D_{2}(C)\in\mathfrak{S}$.

Proof From $\sum_{T\in B}T^{*},\overline{C}\in\mathfrak{S}$ we get
$\left(\sum_{T\in B}T^{*}\right)\cdot\overline{C}\in\mathfrak{S}.$ From Eq.
(3.4) we get

$\left(\sum_{T\in B}T^{*}\right)\cdot\overline{C}=D_{1}+2D_{2}+X,$

where $X$ does not contain any elements of $\mathcal{T}$. It follows that
$D_{1}+2D_{2}\in\mathfrak{S}$.

Now from the definition of $D_{1}$ we see that $D_{1}\subseteq C$, and since
$C$ is a p-set it follows that $C\subseteq D_{1}$. Thus $D_{1}=C$ and the
result follows.∎

##  4\. $f_{2}(C)=0$ for some p-set $C$

Given $\varepsilon\in\\{0,1\\}=\mathbb{F}_{2}$ and any $C\subseteq\mathcal{T}$
we define the graph $\Gamma^{\varepsilon}(C)$ to have vertices $C$ and an edge
$\\{a,b\\}\subseteq C,a\neq b,$ if $a\cdot b=\varepsilon$. If $a,b\in
V^{\sharp},a\cdot b=1$, then we will call the set $\\{a,b,a+b\\}$ a zero-
triangle.

######  Proposition 4.1.

Let $C\subseteq\mathcal{T}$ be a p-set where $f_{2}(C)=0$ and let $t_{a}\in
C$. Then $C^{t_{a}}=C.$ In particular, $\langle C\rangle$ acts by conjugation
on $C$.

Proof So assume that $f_{2}(C)=0$. If we also have $f_{1}(C)=0$, then any
$t_{a}\in C$ commutes with every $t_{b}\in C$. Thus $C^{t_{a}}=C$ in this
case.

Now we assume $f_{1}(C)\neq 0$, so that each $t_{a+b}\in C$ is in a typical
zero-triangle $\\{t_{a},t_{b},t_{a+b}\\}\subseteq C$ (as $f_{1}$ is constant
on $C$). Consider without loss of generality the conjugation action of
$t_{a+b}=t_{b}^{t_{a}}=t_{a}^{t_{b}}\in C$ on three types of vertices of
$\Gamma^{1}(C)$:

(i) those in $\\{t_{a},t_{b},t_{a+b}\\}$;

(ii) those not in $\\{t_{a},t_{b},t_{a+b}\\}$, but which are adjacent in
$\Gamma^{1}(C)$ to any of the vertices in $\\{t_{a},t_{b},t_{a+b}\\}$;

(iii) those not adjacent in $\Gamma^{1}(C)$ to any of the vertices
$t_{a},t_{b},t_{a+b}$.

For (i) the three choices give: $t_{a}^{t_{a+b}}=t_{b}\in
C,t_{b}^{t_{a+b}}=t_{a}\in C,t_{a+b}^{t_{a+b}}=t_{a+b}\in C.$

For (ii) there are three cases for such a vertex $t_{c}\in C$:

  1. 1\.

$t_{c}\in C$ is adjacent to $t_{a}$ in $\Gamma^{1}(C)$;

  2. 2\.

$t_{c}\in C$ is adjacent to $t_{b}$ in $\Gamma^{1}(C)$;

  3. 3\.

$t_{c}\in C$ is adjacent to $t_{a+b}$ in $\Gamma^{1}(C)$.

CASE (ii) 1: Here $c\cdot a=1$. Since $f_{1}(C)>0,t_{a},t_{c}\in C,$ and
$f_{2}(C)=0$ we have $\\{t_{a},t_{c},t_{a+c}\\}\subseteq C$. Then $c\cdot a=1$
shows that either (a) $c\cdot b=1$ or (b) $c\cdot(a+b)=1$.

If (a), then $f_{2}(C)=0$ gives $\\{t_{b},t_{c},t_{b+c}\\}\subseteq C$ is a
zero triangle. Then $c\cdot(a+b)=0,$ and so $t_{c}^{t_{a+b}}=t_{c}\in C.$

If (b), then $c\cdot b=0$ and so $f_{2}(C)=0$ shows that
$\\{t_{c},t_{a+b},t_{a+b+c}\\}\subseteq C$ is a zero triangle. Then
$c\cdot(a+b)=1$ gives $t_{c}^{t_{a+b}}=t_{a+b+c}\in C.$

CASE (ii) 2: This case is analogous to CASE 1 (just switch $a$ and $b$).

CASE (ii) 3: Here $c\cdot(a+b)=1$. Then, since $t_{c},t_{a+b}\in
C,c\cdot(a+b)=1$, again we have $t_{c}^{t_{a+b}}=t_{a+b+c}\in C.$

For (iii): here $c\cdot a=c\cdot b=0$ and so $t_{c}^{t_{a+b}}=t_{c}$ . Thus in
all cases we have $t_{c}^{t_{a+b}}\in C$. This gives the result. ∎

######  Lemma 4.2.

(i) The class of $t_{1}t_{2}$ has size $(2^{n}-1)(2^{n-2}-1)$.

(ii) The class of $t_{1}t_{n}$ has size $(2^{n}-1)2^{n-1}/3$.

(iii) The number of dimension $2$ symplectic subspaces of $V$ is equal to the
number of zero triangles $\\{a,b,a+b\\}$ in $V$, which is equal to
$(2^{n}-1)2^{n-2}/3$.

(iv) The order of $G_{n},n=2m,$ is
$2^{m^{2}}(2^{n}-1)(2^{n-2}-1)(2^{n-4}-1)\cdots(2^{2}-1).$

Proof (i) This is equal to the number of subsets $\\{a,c\\}\subset V^{\sharp}$
of size $2$ where $a\cdot c=0$. The rest are similar.∎

##  5\. An overview of the proof

Given any subsets $C_{1},\cdots,C_{r}\subseteq\mathcal{T}$ we define the graph
$\Gamma^{1}(C_{1},\cdots,C_{r})$ to have vertices $C_{1},\cdots,C_{r}$ and an
edge between $C_{i},C_{j},i\neq j,$ if there are $t_{a}\in C_{i},t_{b}\in
C_{j}$ that do not commute. If a partition $\mathcal{T}=\cup_{i=1}^{r}C_{i}$
is understood (as given by the p-sets of a Schur ring for example), then we
will denote $\Gamma^{1}(C_{1},\cdots,C_{r})$ by $\Gamma^{1}(\mathcal{T})$. The
following are steps in the proof:

  1. Step 1:

Show that $\Gamma^{1}(\mathcal{T})$ is connected and bipartite: Corollary 6.2.

  2. Step 2:

Show that we cannot have an edge $\\{C_{i},C_{j}\\}$ of
$\Gamma^{1}(\mathcal{T})$ where $f_{2}(C_{i}),f_{2}(C_{j})>0$: Theorem 7.1.

  3. Step 3:

Show that we can assign an orientation to each edge of
$\Gamma^{1}(\mathcal{T})$, so that $\Gamma^{1}(\mathcal{T})$ is a directed
graph with an edge $C_{i}\to C_{j}$ if $f_{2}(C_{i})>0,f_{2}(C_{j})=0$: Lemma
7.5.

  4. Step 4:

If we have $i\neq j$ with either $f_{2}(C_{i}),f_{2}(C_{j})>0$ or
$f_{2}(C_{i})=f_{2}(C_{j})=0$, then $C_{i}\perp C_{j}:$ Corollary 7.7 and
Proposition 7.4.

  5. Step 5:

We show that we cannot have distinct vertices $C_{i},C_{j},C_{k}$ of
$\Gamma^{1}(\mathcal{T})$ with directed edges $C_{i}\to C_{j}$ and $C_{k}\to
C_{j}$ of $\Gamma^{1}(\mathcal{T})$: Proposition 8.1.

  6. Step 6:

The graph $\Gamma^{1}(\mathcal{T})$ is a connected star graph with one source
vertex. More specifically this gives what we call the standard situation: the
p-sets $C_{k}$ can be numbered so that $C_{1}$ is the source and so that
$f_{2}(C_{1})\geq 1,f_{2}(C_{i})=0,1<i\leq r$. Further we also have:
$C_{i}\perp C_{j}$ for distinct $i,j>1$ and there is a directed edge
$C_{1}\rightarrow C_{i}$ for each $1<i\leq r$: Corollary 8.2.

  7. Step 7:

Each $C_{i},1\leq i\leq r,$ is invariant under the action by conjugation of
the group $H:=\langle C_{2},\cdots,C_{r}\rangle$: Corollary 8.3.

  8. Step 8:

No more than one of the $C_{i},i>1,$ has $f_{1}(C_{i})>0$: Proposition 9.1.

  9. Step 9:

For $n\geq 6,r>1,$ the case where each $C_{i},1<i\leq r,$ is isotropic does
not occur: Propositions 10.1, 10.2, 10.3.

By Step 9 we can assume that $f_{1}(C_{2})>0$ and $C_{3},\cdots,C_{r}$ are
isotropic (the up-dated standard situation).

  10. Step 10:

For $n\geq 6$ the standard situation is only possible if either (A) $r=2$; or
(B) $r=3$ and $|C_{3}|=1$: Corollary 11.9.

  11. Step 11:

The graph $\Gamma^{1}(C_{2})$ is connected and $H=\langle C_{2}\rangle$ is an
irreducible $3$-transposition subgroup of $G$: Propositions 12.1 and 12.2.

  12. Step 12:

Here we obtain relations among the ‘variables’
$a_{i}=f_{i}(C_{1}),b_{i}=f_{i}(C_{2}),1\leq i\leq 4,\lambda_{1},\lambda_{2}$:
see $\S 13$.

Now the results of McLaughlin [19] show that any proper irreducible subgroup
of ${\rm Sp}(n,2)$ generated by transvections is one of:

$(i)\,\,S_{n+1};\quad(ii)\,\,S_{n+2};\quad(iii)\,\,{\rm
SO}^{+}(n,2);\quad(iv)\,\,{\rm SO}^{-}(n,2).$

Here irreducible means that there are no invariant subspaces. This gives the
possibilities for $H.$ The next steps are to consider these four cases
separately.

  13. Step 13:

If we have (A), and $H$ is a symmetric group (cases (i) and (ii) above), then
we obtain a contradiction from the set of relations that we have obtained (if
$n\geq 14$).

We also similarly show that (B) does not happen for any of (i)-(iv).

  14. Step 14:

We then show what happens when we have (A) and $H={\rm SO}^{\pm}(n,2)$.

##  6\.  Step 1: $\Gamma^{1}(\mathcal{T})$ is connected

Let $\mathcal{T}=\bigcup_{i=1}^{r}C_{i}$, a union of p-sets. We first show
that $\Gamma^{1}(\mathcal{T})$ is connected:

######  Lemma 6.1.

Define $v_{1},v_{2},\cdots,v_{n+1}\in V^{\sharp}$ as follows:

$\displaystyle v_{1}=e_{1},\,\,\,v_{2}=e_{n}$ $\displaystyle
v_{3}=(e_{1}+e_{2})+(e_{n}),\,\,\,v_{4}=(e_{1})+(e_{n-1}+e_{n}),$
$\displaystyle
v_{5}=(e_{1}+e_{2}+e_{3})+(e_{n-1}+e_{n}),\,\,\,v_{6}=(e_{1}+e_{2})+(e_{n-2}+e_{n-1}+e_{n}),$
$\displaystyle v_{7}=(e_{1}+e_{2}+e_{3}+e_{4})+(e_{n-2}+e_{n-1}+e_{n}),$
$\displaystyle v_{8}=(e_{1}+e_{2}+e_{3})+(e_{n-3}+e_{n-2}+e_{n-1}+e_{n}),$
$\displaystyle\qquad\vdots$ $\displaystyle v_{n+1}=e_{1}+e_{2}+\cdots+e_{n}.$

Then $v_{i}\cdot v_{j}=1$ for all $1\leq i\neq j\leq n+1$. Further
$\\{v_{1},\cdots,v_{n}\\}$ is a basis.

Proof Just check that $v_{i}\cdot v_{j}=1$ for $1\leq i\neq j\leq n+1$. ∎

######  Corollary 6.2.

The graph $\Gamma^{1}(\mathcal{T})$ is connected.

Proof Let $v_{1},\cdots,v_{n+1}$ be as in Lemma 6.1. Then
$\Gamma^{1}(\\{v_{1}\\},\\{v_{2}\\},\cdots,\\{v_{n+1}\\})$ is a complete graph
and so is connected. Since $\\{v_{1},\cdots,v_{n}\\}$ is a basis for $V$, if
$v\in V,v\neq 0$, then there is some $1\leq i\leq n$ such that $v_{i}\cdot
v=1$. Thus $\Gamma^{1}(\mathcal{T})$ is connected.∎

######  Proposition 6.3.

Let $\mathcal{T}=\bigcup_{i=1}^{r}C_{i}$ as a union of p-sets. Then we cannot
have $f_{2}(C_{i})=0$ for all $1\leq i\leq r$.

Proof If $f_{2}(C_{i})=0$ for all $1\leq i\leq r$, then the graph
$\Gamma^{1}(\mathcal{T})$ is not connected.∎

##  7\. Steps 2, 3, 4: $f_{2}(C_{i})f_{2}(C_{j})\neq 0$ cannot happen

######  Theorem 7.1.

If $\mathfrak{S}$ is a commutative Schur ring, then we cannot have an edge
$\\{C_{i},C_{j}\\}$ of $\Gamma^{1}(\mathcal{T})$ where
$f_{2}(C_{i}),f_{2}(C_{j})>0$.

Proof So assume that we have distinct p-sets $C_{1},C_{2}$ giving an edge of
$\Gamma^{1}(\mathcal{T})$ where $f_{2}(C_{1}),f_{2}(C_{2})>0,$ so that there
are $t_{a},t_{b}\in C_{1},a\cdot b=1,$ with $t_{a+b}\in C_{2}$.

Since $C_{1},C_{2}$ are p-sets, $f_{2}$ is constant on $C_{1}$ and on $C_{2}$.
Since $t_{a+b}\in C_{2}$ and $f_{2}(C_{2})\neq 0$ there is $c\in C_{2}$ such
that $t_{a+b},t_{c}\in C_{2}$ has type $f_{2}$, so that $c\cdot(a+b)=1$ and
$t_{a+b+c}\notin C_{2}$. There are two cases: either

(A): $a+b+c\in C_{1}$; or

(B): there is $m>2$ such that $a+b+c\in C_{m}$. We may assume that $m=3$.

In either of these cases we have the following: If
$a\cdot(a+b+c)=0=b\cdot(a+b+c)$, then $a\cdot c=1=b\cdot c$, showing
$c\cdot(a+b)=0$, a contradiction. So either $a\cdot(a+b+c)=1$ or
$b\cdot(a+b+c)=1$. Since we can interchange $a$ and $b$ in what we have done
so far, we may, without loss of generality, assume that $b\cdot(a+b+c)=1$.
Then:

$a\cdot b=1,\,\,\,b\cdot c=0,\,\,\,a\cdot c=1.$

Since $b\cdot(a+b+c)=1$ the pair $b,a+b+c$ is either type $f_{1}$ or type
$f_{2}$. There are now four cases to be considered:

Case A1: (A) and $b,a+b+c$ has type $f_{1}$. Here
$\\{a,b,a+b+c,a+c\\}\subseteq C_{1},\\{a+b,c\\}\subseteq C_{2}$.

Case A2: (A) and $b,a+b+c$ has type $f_{2}$. Here $(a+b+c)+b=a+c\notin C_{1}$
and so there are two cases that we can index as follows:

Case A21: $a+c\in C_{2}$, so that $\\{a,b,a+b+c\\}\subseteq
C_{1},\\{a+b,c,a+c\\}\subseteq C_{2}$.

Case A22: $a+b\in C_{3}$, so that $\\{a,b,a+b+c\\}\subseteq
C_{1},\\{a+b,c\\}\subseteq C_{2},a+c\in C_{3}$. Here $a\in C_{1},c\in
C_{2},a+c\in C_{3},a\cdot c=1,$ which violates Lemma 2.5, so this case does
not happen.

Case B1: (B) and the pair $b,a+b+c$ has type $f_{1}$. This case cannot happen
since $b\in C_{1},a+b+c\in C_{3}\neq C_{1}$.

Case B2: (B) and the pair $b\in C_{1},a+b+c\in C_{3}$ has type $f_{2}$. Here
there are two cases (by Lemma 2.5):

Case B21: we have (B2) and $a+c\in C_{1}$;

Case B22: we have (B2) and $a+c\in C_{3}$. Here $t_{a}\in C_{1},t_{c}\in
C_{2},t_{a+c}\in C_{3}$ and $a\cdot c=1$, violating Lemma 2.5.

We thus need to consider the three cases A1, A21, B21, as follows:

First, from $a\cdot b=1,b\cdot c=0,a\cdot c=1$ we can assume:
$a=e_{1},b=e_{n},c=e_{2}+e_{n}$. Let $W={\rm Span}(e_{1},e_{2},e_{n})$. Note
that

$a\cdot(b+c)=b\cdot(b+c)=c\cdot(b+c)=0,$

so that $t_{b+c}$ commutes with each of
$t_{a},t_{b},t_{c},t_{a+b},t_{a+c},t_{a+b+c}$.

Case A1 Here

(7.1)
$\displaystyle\\{a,b,a+b+c,a+c\\}=\\{e_{1},e_{n},e_{1}+e_{2},e_{1}+e_{2}+e_{n}\\}\subseteq
C_{1},$ $\displaystyle\\{a+b,c\\}=\\{e_{1}+e_{n},e_{2}+e_{n}\\}\subset C_{2}.$

Lemma 2.9 gives $(C_{1}^{2})_{f_{2}}\in\mathfrak{S}$. Now $t_{2,n}\in
C_{2},t_{1}t_{n}\in(C_{1}^{2})_{f_{2}}$ and so $t_{2,n}t_{1}t_{n}\in
C_{2}\cdot(C_{1}^{2})_{f_{2}}\subseteq C_{2}C_{1}^{2}$; we will show that
$t_{2,n}t_{1}t_{n}\notin(C_{1}^{2})_{f_{2}}C_{2}\subseteq C_{1}^{2}\cdot
C_{2}$. We need:

######  Lemma 7.2.

If $t_{u},t_{v},t_{w}\in\mathcal{T}$,
$t_{u}t_{v}t_{w}=\alpha:=t_{2,n}t_{1}t_{n}$, then $(t_{u},t_{v},t_{w})$ is one
of:

(i) $(t_{1,2},t_{1},t_{1,2,n})$ (ii) $(t_{1,2},t_{1,2,n},t_{2,n})$; (iii)
$(t_{1,2},t_{2,n},t_{1})$;

(iv) $(t_{1},t_{1,2},t_{1,2,n})$, (v) $(t_{1},t_{n},t_{1,2})$; (vi)
$(t_{1},t_{1,2,n},t_{n});$

(vii) $(t_{n},t_{1,2},t_{2,n})$; (viii) $(t_{n},t_{1,n},t_{1,2})$; (ix)
$(t_{n},t_{2,n},t_{1,n})$;

(x) $(t_{1,2,n},t_{n},t_{2,n})$; (xi) $(t_{1,2,n},t_{2,n},t_{n})$;

(xii) $(t_{1,n},t_{1,2},t_{1});$ (xiii) $(t_{1,n},t_{1},t_{1,2})$;

(xiv) $(t_{2,n},t_{1},t_{n})$; (xv) $(t_{2,n},t_{n},t_{1,n})$; (xvi)
$(t_{2,n},t_{1,n},t_{1})$.

In particular $u,v,w\in\\{t_{1},t_{n},t_{1,n},t_{1,2},t_{2,n},t_{1,2,n}\\}$.

Proof First one checks that each of these cases gives a solution to
$t_{u}t_{v}t_{w}=\alpha$. Next we note that
$(e_{n})\alpha=e_{n}+(e_{1}+e_{n}),(e_{1})\alpha=e_{1}+(e_{1}+e_{2}+e_{n})$
and $(e_{n-1})\alpha=e_{n-1}+(e_{1}+e_{2})$, showing that ${\rm
Span}(e_{1},e_{2},e_{n})={\rm
Span}(e_{1}+e_{n},e_{1}+e_{2}+e_{n},e_{1}+e_{2})\subseteq{\rm Span}(u,v,w)$.
Thus Span$(u,v,w)={\rm Span}(e_{1},e_{2},e_{n})$. Thus $u,v,w$ is a basis for
Span$(e_{1},e_{2},e_{n})$, and we check all such sets to find the $16$ cases
listed.∎

Now in case A1 we have $t_{2+n}t_{1}t_{n}\in C_{2}\cdot(C_{1}^{2})_{f_{2}}$.
Since $\mathfrak{S}$ is commutative there are $t_{u},t_{v}\in C_{1}$ of type
$f_{2}$ and $t_{w}\in C_{2}$ such that $t_{u}t_{v}t_{w}=t_{2,n}t_{1}t_{n}$. By
Lemma 7.2

$t_{u},t_{v},t_{w}\in S:=\\{t_{1},t_{n},t_{1,n},t_{2,n},t_{1,2,n},t_{1,2}\\}.$

In fact from Eq. (7.1) we know which $C_{i}$ each of the elements of $S$ is
contained in. Using Lemma 7.2 we now show that there are no triples
$t_{u},t_{v}\in C_{1}$ of type $f_{2}$ and $t_{w}\in C_{2}$ satisfying these
conditions. Specifically, in Lemma 7.2:

options (i) and (iv) are not possible, since $t_{w}=t_{1,2,n}\notin C_{2}$;

options (ii), (vii), (x) and (xiii) are not possible, since $t_{u},t_{v}$ is
type $f_{1}$;

options (iii), (xii) and (xvi) are not possible, since $t_{w}=t_{1}\notin
C_{2}$;

options (v) and (viii) are not possible, since $t_{w}=t_{1,2}\notin C_{2}$;

options (vi), (xi) and (xiv) are not possible, since $t_{w}=t_{n}\notin
C_{2}$;

options (ix) and (xv) are not possible since $t_{u}\cdot t_{v}=0$.

Thus Case A1 does not happen.

Case A21: Here

(7.2) $\displaystyle\\{a,b,a+b+c\\}=\\{e_{1},e_{n},e_{1}+e_{2}\\}\subseteq
C_{1},$
$\displaystyle\\{a+b,c,a+c\\}=\\{e_{1}+e_{n},e_{2}+e_{n},e_{1}+e_{2}+e_{n}\\}\subseteq
C_{2},$

and we have
$(C_{1}^{2})_{f_{2}}\cap\mathcal{T}(W)^{2}=\\{t_{1}t_{n},t_{n}t_{12n}\\}.$

Now we consider $\beta:=t_{1,n}\cdot t_{n}t_{1,2}\in
C_{2}\cdot[(C_{1}^{2})_{f_{2}}\cap\mathcal{T}(W)]$ where commutativity means
that there are $t_{u},t_{v}\in C_{1}$ of type $f_{2}$ and $t_{w}\in C_{2}$
such that $t_{u}t_{v}t_{w}=\beta.$ Now we show that
$\beta\notin[(C_{1}^{2})_{f_{2}}\cap\mathcal{T}(W)]\cdot C_{2}$ using a result
similar to Lemma 7.2 to conclude that there are no $u,v,w$ satisfying the
required conditions: $t_{u},t_{v}\in C_{1}$ of type $f_{2}$, $t_{w}\in C_{2}$
and $t_{u}t_{v}t_{w}=\beta.$ This shows that Case A21 does not occur.

Case B21: Here $\\{a,b,a+c\\}=\\{t_{1},t_{n},t_{1,2,n}\\}\subseteq
C_{1},\\{a+b,c\\}=\\{t_{1,n},t_{2,n}\\}\subseteq
C_{2},\\{a+b+c\\}=\\{t_{1,2}\\}\subseteq C_{3}$.

Now let $U:=C_{1}C_{2}\cap(t_{1}t_{2})^{G}\cap{\rm
Sp}(W)=\\{t_{n}t_{2+n},t_{1+2+n}t_{1+n}\\}$. We will need

######  Lemma 7.3.

Let $\alpha=t_{n}t_{2,n}t_{1,n}$. Then the only
$t_{u},t_{v},t_{w}\in\mathcal{T}$ such that $v\cdot w=0$ and
$t_{u}t_{v}t_{w}=\alpha$ are

(i) $t_{1,2,n},t_{n},t_{2,n}$;

(ii) $t_{1,n},t_{1},t_{1,2}$;

(iii) $t_{1,2,n},t_{2,n},t_{n}$;

(iv) $t_{1,n},t_{1,2},t_{1}.$

Proof This is similar to the proof of Lemma 7.2. First note that

$e_{1}\alpha=e_{1}+(e_{1}+e_{2}+e_{n}),\quad
e_{n}\alpha=e_{n}+(e_{1}+e_{n}),\quad e_{n-1}\alpha=e_{n-1}+(e_{1}+e_{2}).$

Thus Span$(e_{1}+e_{2}+e_{n},e_{1}+e_{n},e_{1}+e_{2}\\}\subseteq{\rm
Span}(u,v,w)$, which shows that $u,v,w$ is a basis for
Span$(\\{e_{1}+e_{2}+e_{n},e_{1}+e_{n},e_{1}+e_{2}\\})={\rm
Span}(\\{e_{1},e_{2},e_{n}\\})$. This again reduces the proof to a short
enumeration. ∎

Let $\alpha:=t_{n}t_{2,n}t_{1,n}$ and note that $t_{n}t_{2,n}\in
C_{1}C_{2}\cap(t_{1}t_{2})^{G}$, where $E:=C_{1}C_{2}\cap(t_{1}t_{2})^{G}$ is
an $\mathfrak{S}$-set. We show that $C_{2}$ does not commute with $E$; in
particular $t_{1,n}\in C_{2}$ and so $(t_{n}t_{2,n})\cdot t_{1,n}\in EC_{2}$.
We show that $(t_{n}t_{2,n})\cdot t_{1,n}\notin C_{2}E$.

Since $E$ and $C_{2}$ commute, there are $t_{u}\in C_{2}$ and $t_{v}t_{w}\in
E$, so that $v\cdot w=0$ and $t_{u}(t_{v}t_{w})=\alpha$. But the possibilities
for such $u,v,w$ are given in Lemma 7.3 (i)-(iv). Now we cannot have (i) or
(iii) since we would then have $t_{u}=t_{1,2,n}\in C_{1}$, contradicting
$t_{u}\in C_{2}$. We cannot have (ii) or (iv) since $t_{1,2}\in C_{3}$. This
concludes the proof of Case B21 and so completes the proof of Theorem 7.1.∎

######  Proposition 7.4.

If $C_{i},C_{j},i\neq j,$ are commuting p-sets with
$f_{2}(C_{i})=f_{2}(C_{j})=0$, then $C_{i}\perp C_{j}$.

Proof Suppose that $C_{i},C_{j},i\neq j,$ are p-sets with
$f_{2}(C_{i})=f_{2}(C_{j})=0$ and we have $u\in C_{i},v\in C_{j}$ with $u\cdot
v=1$. Since $C_{i},C_{i}$ are p-sets that commute, we must have $u+v\in
C_{i}\cup C_{j}$ by Lemma 2.5. So suppose without loss of generality that
$u,u+v\in C_{i},v\in C_{j}$; then this contradicts $f_{2}(C_{i})=0$.∎

We will show how to make $\Gamma^{1}(C_{1},\cdots,C_{r})$ into a directed
graph i.e. we need to assign an orientation to each edge. So suppose that
there is an edge $\\{C_{i},C_{j}\\}$. Then we cannot have
$f_{2}(C_{i})=f_{2}(C_{j})=0$ as then $C_{i}\perp C_{j}$ by Proposition 7.4.

So assume without loss of generality that $f_{2}(C_{i})>0.$ Then Theorem 7.1
shows that $f_{2}(C_{j})=0$ and we orient the edge $\\{C_{i},C_{j}\\}$ from
$C_{i}$ to $C_{j}$. Thus

######  Lemma 7.5.

As defined above, $\Gamma^{1}(C_{1},\cdots,C_{r})$ is a directed graph. ∎

######  Lemma 7.6.

The graph $\Gamma^{1}(C_{1},\cdots,C_{r})$, considered as an un-directed
graph, is bipartite and connected.

Proof Whenever we have an edge $\\{C_{i},C_{j}\\}$ exactly one of
$C_{i},C_{j}$ has non-zero $f_{2}$. This determines the bipartite partition of
$\Gamma^{1}(C_{1},\cdots,C_{r})$. We have already noted that it is connected.
∎

If $\\{C_{i},C_{j}\\}$ is an edge of $\Gamma^{1}(\mathcal{T})$ where the
orientation is from $C_{i}$ to $C_{j}$, then we notate this as $C_{i}\to
C_{j}$.

######  Corollary 7.7.

If $C_{i},C_{j},i\neq j,$ are p-sets with $f_{2}(C_{i}),f_{2}(C_{j})>0,$ then
$C_{i}\perp C_{j}$.

Proof The situation where $f_{2}(C_{i}),f_{2}(C_{j})>0$ and there are
$t_{a}\in C_{i},t_{b}\in C_{j}$ with $a\cdot b=1$ is prohibited by Theorem
7.1. ∎

This concludes the proofs of Steps 2, 3, 4.

##  8\. Steps 5, 6, 7: Star graph

######  Proposition 8.1.

The situation where there are distinct $i,j,k$ such that $C_{i}\rightarrow
C_{j}$ and $C_{k}\rightarrow C_{j}$ are edges of
$\Gamma^{1}(C_{1},\cdots,C_{r})$ does not occur.

Proof Since we have $C_{i}\rightarrow C_{j}$ there are $a\in C_{i},a+b\in
C_{j}$ where $a\cdot(a+b)=1$. Since $f_{2}(C_{j})=0$ we see that $b=a+(a+b)\in
C_{i}$ and the pair $a,b\in C_{i}$ has type $f_{2}$. Now define
$D_{2}(C_{i})\subset\mathcal{T}$ as in Eq. (3.7). Then we have $t_{a+b}\in
D_{2}(C_{i})\cap C_{j}\neq\emptyset$. Applying Lemma 3.6 we see that
$C_{j}\subseteq D_{2}(C_{i})$ since $C_{j}$ is a p-set. We similarly obtain
$C_{j}\subseteq D_{2}(C_{k})$. So there are $t_{c},t_{d}\in C_{k}$, of type
$f_{2}$, with $c+d=a+b$. Now

$c\cdot(a+b)=c\cdot(c+d)=1,\text{ and so }1\in\\{c\cdot a,c\cdot b\\};$

but $c\in C_{k},a,b\in C_{i}$, and so there is an edge $\\{C_{i},C_{k}\\}$ in
$\Gamma^{1}(\mathcal{T})$, which contradicts the fact that
$\Gamma^{1}(C_{1},\cdots,C_{r})$ is bipartite.∎

Let $\Gamma^{1}(C_{1},\cdots,C_{r})^{+}$ be the set of vertices $C_{i}$ of
$\Gamma^{1}(C_{1},\cdots,C_{r})$ where $f_{2}(C_{i})>0$, and let
$\Gamma^{1}(C_{1},\cdots,C_{r})^{0}$ denote the rest of the vertices. A
directed star graph with one source is a directed graph $\Gamma$ where there
is a vertex $v$ (the source) such that for all vertices $w\neq v$ we have an
edge from $v$ to $w$, these being the only edges of $\Gamma$.

######  Corollary 8.2.

The set $\Gamma^{1}(C_{1},\cdots,C_{r})^{+}$ has exactly one element. In
particular, $\Gamma^{1}(C_{1},\cdots,C_{r})$ is a directed star graph with one
source.

Proof Recall that $\Gamma^{1}(C_{1},\cdots,C_{r})$ is connected. Thus if
$\Gamma^{1}(C)_{1},\cdots,C_{r})^{+}$ has two vertices $C_{i},C_{k},i\neq k$,
then there is a path between them in $\Gamma^{1}(C_{1},\cdots,C_{r})$. That
path in $\Gamma^{1}(C_{1},\cdots,C_{r})$ has to have vertices $C_{u},C_{v}$ in
$\Gamma^{1}(C_{1},\cdots,C_{r})^{+}$ at distance $2$. This then gives us the
situation described in Lemma 8.1, and so we have a contradiction. The rest
follows. ∎

So we now have the standard situation defined in Step 6.

Now let $t_{u}\in C_{k},1<k\leq r$. Then, by Proposition 4.1,
$C_{k}^{t_{u}}=C_{k}$. We also have $t_{u}\in C_{k}$ and $C_{k}\perp
C_{j},1<j\neq k\leq r$ which then gives $C_{j}^{t_{u}}=C_{j}$ for all $1<j\neq
k\leq r$. But $\mathcal{T}^{t_{u}}=\mathcal{T},\mathcal{T}=C_{1}\cup
C_{2}\cup\cdots\cup C_{r}$, so that $C_{1}^{t_{u}}=C_{1}$. Thus:

######  Lemma 8.3.

Let $\mathcal{T}=C_{1}\cup C_{2}\cup\cdots\cup C_{r}$ be in the standard
situation and let $t_{u}\in C_{i},1<i\leq r$. Then $C_{i}^{t_{u}}=C_{i}$ for
all $1\leq i\leq r$.

In particular, if $H=\langle C_{2}\rangle\times\langle
C_{3}\rangle\times\cdots\times\langle C_{r}\rangle\leq G,$ then for $h\in H$
we have $C_{i}^{h}=C_{i}$ for $1\leq i\leq r$, and the $C_{i}$ are unions of
orbits of the action of $H$ on $\mathcal{T}$. ∎

##  9\. Step 8: $f_{1}(C_{i}),i>1$

For $a,b\in\mathcal{T}$ we will say that $a$ meets $b$ if $a\cdot b=1$.

######  Proposition 9.1.

Assume the standard situation. Then no more than one of the $C_{i},i>1,$ can
have $f_{1}(C_{i})>0$.

Proof So assume that $f_{1}(C_{2}),f_{1}(C_{3})>0$. Without loss we can assume
that $e_{1},e_{n}\in C_{2}$, which then gives
$\\{e_{1},e_{n},e_{1}+e_{n}\\}\subseteq C_{2}$ as
$f_{1}(C_{2})>0,f_{2}(C_{2})=0$. Since $f_{1}(C_{3})>0$ and $C_{2}\perp C_{3}$
we may similarly assume that $e_{2},e_{n-1},e_{2}+e_{n-1}\in C_{3}$. Let
$W:={\rm Span}(e_{1},e_{n},e_{2},e_{n-1})$. Now if $a\in C_{2},b\in C_{3}$,
then $a$ meets elements of $C_{2}$ and $b$ meets elements of $C_{3}$, so that
$a+b$ meets elements of $C_{2}$ and of $C_{3}$. Since $C_{i}\perp C_{j}$ for
all $i\neq j\geq 2$ we must have $a+b\in C_{1}$. It follows that

$\displaystyle\\{e_{1}+e_{2},e_{1}+e_{n-1},e_{1}+e_{2}+e_{n-1},e_{n}+e_{2},e_{n}+e_{n-1},e_{n}+e_{2}+e_{n-1},$
$\displaystyle\qquad\qquad
e_{1}+e_{n}+e_{2},e_{1}+e_{n}+e_{n-1},e_{1}+e_{n}+e_{2}+e_{n-1}\\}\subseteq
C_{1}.$

Note that we now have: $(C_{1}\cap W)\cup(C_{2}\cap W)\cup(C_{3}\cap
W)=W^{\sharp}$ (a disjoint union).

Define the multisets

$X=C_{1}C_{2}\cap(t_{1}t_{2})^{G},\quad Y=C_{1}^{2}\cap(t_{1}t_{n})^{G},\quad
Z=C_{1}\cdot Y.$

First note that $X,Y,Z\in\mathfrak{S}$ and that

$\alpha_{1}:=t_{1,2,n-1,n}t_{1,n}\in X,\quad t_{n-1,n}t_{2,n}\in Y,$

so that $\alpha_{2}:=t_{1,n-1,n}t_{n-1,n}t_{2,n}\in Z$. Let
$\alpha:=\alpha_{1}\alpha_{2}\in XZ$.

######  Lemma 9.2.

Assume that $n\geq 6$. Then

(i) As a product of elements of $\mathcal{T}$ $\alpha$ has minimal length $5$.

(ii) If we write $\alpha=t_{a}t_{b}t_{u}t_{v}t_{w}$, then $a,b,u,v,w\in W$.

(iii) $\alpha$ occurs in $XZ$ with multiplicity $5$.

(iv) $\alpha$ occurs in $ZX$ with multiplicity $3$.

Proof Note that $\alpha=\alpha_{1}\alpha_{2}\in XZ\subset C_{1}C_{2}\cdot
C_{1}^{3}$ and so $\alpha$ has minimal length at most $5$ as a product of
elements of $\mathcal{T}$.

(i) Write $\alpha=t_{a}t_{b}t_{u}t_{v}t_{w}$, where we allow some of the
$a,b,u,v,w$ to be zero, with the convention that $t_{0}=id$. Now note that
$\\{v\alpha-v:v\in W\\}=W$ and so $W\subseteq{\rm Span}(a,b,u,v,w)$, where
$\dim{\rm Span}(a,b,u,v,w)\leq 5$. If $\dim{\rm Span}(a,b,u,v,w)=4$, then
${\rm Span}(a,b,u,v,w)=W$, and $a,b,u,v,w\in W$. If $\dim{\rm
Span}(a,b,u,v,w)=5$, then ${\rm Span}(a,b,u,v,w)={\rm
Span}(e_{1},e_{2},e_{n-1},e_{n},c)$ for some $c\notin W$; further we can
change $c$ to $c+u,u\in W$ so that $c\perp W$. By a change of basis we can
then assume that $c=e_{3}$ (since $n\geq 6$). There are $31$ transvections in
$\mathcal{T}({\rm Span}(e_{1},e_{2},e_{n-1},e_{n},e_{3}))$ and so we can just
find all $k$-tuples, $1\leq k\leq 5$, of such transvections whose product is
$\alpha$. Doing this shows that we only have $k=5$, giving (i), and also
showing that $a,b,u,v,w\in W$, giving (ii).

(iii) The above has reduced the situation to where we are able to write
$\alpha=t_{a}t_{b}t_{u}t_{v}t_{w}$, with
$t_{a},t_{b},t_{u},t_{v},t_{w}\in\mathcal{T}(W)$. If we choose
$\beta_{1}=t_{a}t_{b}\in X,\beta_{2}=t_{u}t_{v}t_{w}\in Z$, then another short
calculation shows that there are $3$ such pairs $[\beta_{1},\beta_{2}]$ with
$\beta_{1}=t_{a}t_{b}\in X\cap\mathcal{T}(W)^{2},\beta_{2}=t_{u}t_{v}t_{w}\in
Z\cap\mathcal{T}(W)^{3}$ and $\beta_{1}\beta_{2}=\alpha$.

For pairs $[\beta_{1},\beta_{2}]$ with $\beta_{1}=t_{u}t_{v}t_{w}\in
Z\cap\mathcal{T}(W)^{3},\beta_{2}=t_{a}t_{b}\in X\cap\mathcal{T}(W)^{2}$ and
$\beta_{1}\beta_{2}=\alpha$ we have $5$ such products. This gives (iii) and
(iv). What we have shown is that $X,Z\in\mathfrak{S}$, but $XZ\neq ZX$ as
multisets; Lemma 9.2 follows.∎

The contradiction to commutativity obtained from Lemma 9.2 concludes the proof
of Proposition 9.1.∎

By Proposition 9.1 there is at most one of the $C_{i},1<i\leq r,$ that is not
isotropic. We now reorder the $C_{i}$ so that if there is one non-isotropic
$C_{i},1<i\leq r$, then it is $C_{2}$. We now consider this condition as being
a part of the standard situation.

##  10\. Step 9: the Isotropic cases

There are three cases for Step $10$. We first prove:

######  Proposition 10.1.

Suppose that we have the standard situation where $C_{2},\cdots,C_{r},$ $r\geq
2,$ are all isotropic. Let $X:={\rm Span}(C_{2},\cdots,C_{r})^{\sharp}$, and
assume that $X\neq C_{2}\cup\cdots\cup C_{r}$. Then $C_{1},\cdots,C_{r}$ are
not the p-sets of a commutative Schur ring.

Proof So assume that $C_{1},\cdots,C_{r}$ are the p-sets of a commutative
Schur ring, that $C_{2},\cdots,C_{r}$ are all isotropic and let $X={\rm
Span}(C_{2},\cdots,C_{r})^{\sharp}$. Let $a\in X\setminus(C_{2}\cup\cdots\cup
C_{r})$. Then $a\in C_{1}$. But $\Gamma^{1}(C_{1},\cdots,C_{r})$ is connected
(as an undirected graph), so $f_{2}(C_{1})>0$. Since $a\in C_{1}$ there is
$b\in C_{1}$ where $a,b$ has type $f_{2}$, so that $t_{a+b}\in
C_{2}\cup\cdots\cup C_{r}$ and $a\cdot b=1$. But $a\in X,a+b\in X$ and $X$
isotropic, so $a\cdot b=1$ is a contradiction.∎

Thus we assume $X={\rm
Span}(C_{2},\cdots,C_{r})^{\sharp}=\bigcup_{i=2}^{r}C_{i},r\geq 2,$ so that
$\dim X\leq n/2$ as $X$ is isotropic. The next case is:

######  Proposition 10.2.

Suppose that we have the standard situation where $C_{2},\cdots,C_{r},$ $r\geq
2,$ are all isotropic and $X={\rm
Span}(C_{2},\cdots,C_{r})^{\sharp}=\bigcup_{i=2}^{r}C_{i}$. Assume that $\dim
X<n/2$. Then $C_{1},\cdots,C_{r}$ are not the p-sets of a commutative Schur
ring.

Proof So let $p=\dim X<n/2$ and let $a_{1},\cdots,a_{p}$ be a basis for $X$.
Then we can extend this basis to a symplectic basis
$a_{1},b_{1},a_{2},b_{2},\cdots,a_{n/2},b_{n/2}$ for $V$. Here we have
$b_{i}\in C_{1},1\leq i\leq n/2$. Now we find that

$|\\{x\in X:u\cdot b_{1}=0\\}|=2^{p-1}-1,\qquad|\\{x\in X:u\cdot
b_{n/2}=0\\}|=2^{p}-1.$

It follows that $f_{4}(C_{1})$ is not constant on $C_{1}$, a contradiction.∎

######  Proposition 10.3.

Let $n\geq 6$. Suppose that we have the standard situation where
$C_{2},\cdots,C_{r}$ are all isotropic, $r\geq 2$. Assume that $A={\rm
Span}(C_{2},\cdots,C_{r})^{\sharp}=\bigcup_{i=2}^{r}C_{i}$ and $\dim A=n/2$.
Then $C_{1},\cdots,C_{r}$ are not the p-sets of a commutative Schur ring.

Proof So let $e_{1},\cdots,e_{n/2}$ be a basis for $A$ where
$e_{1},\cdots,e_{n}$ is a symplectic basis for $V$. Let $B={\rm
Span}(\\{e_{n/2+1},\cdots,e_{n}\\})^{\sharp}$ so that
$C_{1}=B\cup\left(\cup_{b\in B}(b+A)\right)$. Define the multisets

$X:=C_{1}^{2}\cap(t_{1}t_{2})^{G},\quad Y:=C_{1}A\cap(t_{1}t_{n})^{G},\quad
Z:=YA,$

so that $X,Y,Z\in\mathfrak{S}$. Now $Z$ has multiplicities $1,2,4$ and we let
$Z_{1}$ be the set of elements of $Z$ that have multiplicity $1$. We will show
that $XZ_{1}\neq Z_{1}X$.

Note that $\alpha_{1}:=t_{1,2,3,n}t_{1,n-1}\in
X,\alpha_{2}:=t_{2,n-1}t_{1,2,3}\in Y,\alpha_{3}=\alpha_{2}t_{1,2}\in Z_{1}$.
Let $\alpha=\alpha_{1}\cdot\alpha_{3}\in X\cdot Z_{1}.$

Let $V_{6}:={\rm Span}(e_{1},e_{2},e_{3},e_{n-2},e_{n-1},e_{n}),W:={\rm
Span}(e_{1}+e_{2}+e_{3}+e_{n},e_{1}+e_{n-1},e_{2}+e_{n-1},e_{1}+e_{2}+e_{3},e_{1}+e_{3})={\rm
Span}(e_{1},e_{2},e_{3},e_{n-1},e_{n})\leq V_{6}$ Then $\\{u\alpha-u:u\in
V_{6}\\}=W$, showing that (i) $\alpha$ is not the product of less than five
transvections, and (ii) if $t_{a}t_{b}t_{u}t_{v}t_{w}=\alpha$, then
$a,b,u,v,w$ is a basis for $W$.

Thus to consider whether $\alpha$ is in $XZ_{1}$ or $Z_{1}X$ we can now
restrict ourselves to considering products of transvections in
$\mathcal{T}(W))$. This now reduces the problem to checking the result in
Sp$(W)\leq G$. Specifically, one finds that $\alpha$ has multiplicity $4$ in
$XZ_{1}$, but multiplicity $0$ in $Z_{1}X$. This gives the result. ∎

Thus we can now assume: $\Gamma^{1}(\mathcal{T})$ is a star graph with $C_{1}$
the source, $f_{1}(C_{2})>0,f_{2}(C_{i})=0,i>1,$ and that
$\bigcup_{i=3}^{r}C_{i}$ is isotropic.

##  11\.  Step 10: The $f_{1}(C_{2})>0$ case

######  Proposition 11.1.

Consider the standard situation where $f_{1}(C_{2})>0$. Let
$X:=C_{3}\cup\cdots\cup C_{r}$. Then $X$ is isotropic and $X={\rm
Span}(X)^{\sharp}$.

Proof Each $C_{i},i>2,$ is isotropic and $C_{i}\perp C_{j}$ for $2\leq i\neq
j\leq r$. So $X$ is isotropic.

Let $x\in{\rm Span}(X)^{\sharp}\setminus X$, so that $x\in C_{1}\cup C_{2}$
also $x$ meets no elements of $C_{2}\cup\cdots\cup C_{r}$. If $x\in C_{1}$,
then $x$ meet $2f_{1}(C_{1})+f_{2}(C_{1})$ elements of $C_{1}$ and meets no
elements of $C_{2}\cup\cdots\cup C_{r}$. Thus
$2f_{1}(C_{1})+f_{2}(C_{1})=2^{n-1}$. However there are $y\in C_{1}$ that meet
elements of $C_{2}$ since $C_{1}\to C_{2}$ is an edge of
$\Gamma^{1}(\mathcal{T})$. Thus $y$ meets $2f_{1}(C_{1})+f_{2}(C_{1})=2^{n-1}$
elements of $C_{1}$ and meets at least one element of $C_{2}$. Thus $y$ meets
more than $2^{n-1}$ elements of $\mathcal{T}$, a contradiction. Thus $x\notin
C_{1}$ and so $x\in C_{2}$. Now $x\in{\rm Span}(X)$ gives $x\perp C_{2}$. Thus
we cannot have $x\in C_{2}$ since $f_{1}(C_{2})>0$. So we have a
contradiction. ∎

######  Proposition 11.2.

The standard situation where
$C_{1}\cap\left(\bigcap_{j=2}^{r}C_{j}^{\perp}\right)\neq\emptyset$ cannot
happen.

Proof Assume to the contrary that $b\in
C_{1}\cap\left(\bigcap_{j=2}^{r}C_{j}^{\perp}\right)$. Then the only elements
of $\mathcal{T}$ that $b$ meets are in $C_{1}$, and there are
$2f_{1}(C_{1})+f_{2}(C_{1})$ of them; thus
$2f_{1}(C_{1})+f_{2}(C_{1})=2^{n-1}$. However there is an edge of
$\Gamma^{1}(C_{1},\cdots,C_{r})$ from $C_{1}$ to $C_{2}$. Thus there is some
$c\in C_{1}$ that meets some element of $C_{2}$. Since the $f_{i}$ are
constant on $C_{1}$ we see that $c$ meets $2f_{1}(C_{1})+f_{2}(C_{1})$
elements of $C_{1}$; but it meets at least one element of $C_{2}$. Thus $b$
and $c$ meet a different number of elements of $\mathcal{T}$. This
contradiction gives the result.∎

Thus we may now assume

(11.1) $\displaystyle
C_{1}\cap\left(\bigcap_{j=2}^{r}C_{j}^{\perp}\right)=\emptyset.$

Note that if none of the $C_{i},1\leq i\leq r$ is isotropic, then we must have
$r=2$. So we now consider the situation where $C_{3},\cdots,C_{r}$ are all
isotropic and $r\geq 3$.

Since $f_{1}(C_{2})>0$ we may assume that $e_{1},e_{n}\in C_{2},$ so that
$e_{1}+e_{n}\in C_{2}$. We may also assume that $e_{2}\in C_{3}$. Let $a\in
V,a\cdot e_{2}=1$. Then we must have $a\in C_{1}$, so that
$e_{n-1}+e_{2}^{\perp}\subset C_{1}$. Let $W:={\rm
Span}(e_{1},e_{n},e_{2},e_{n-1})$. Since $\Gamma^{1}(C_{1},\cdots,C_{r})$ is
connected there is $a\in C_{1},a\cdot e_{2}=1$. Then $b:=a+e_{2}\in C_{1}$ as
$f_{2}(C_{3})=0$.

We recall that $\langle t_{1},t_{n},t_{2}\rangle\leq\langle
C_{2},C_{3}\rangle$ acts on each $C_{i},1\leq i\leq r$ and also acts on
$\mathcal{T}(W)$. Let $N_{12n}=N_{G}(\langle t_{1},t_{n},t_{2}\rangle).$ One
checks that there are two orbits under the action of $N_{12n}$ on
$\mathcal{T}$ for elements $a\in W$ where $a\cdot e_{2}=1$. Representatives
for these orbits are $a\in\\{e_{n-1},e_{n-1}+e_{n}\\}$. Thus we can, without
loss of generality, assume that $a\in\\{e_{n-1},e_{n-1}+e_{n}\\}$. For such an
$a$ we then take $b:=a+e_{2}$, so that $a,b\in C_{1},a+b=e_{2},a\cdot b=1$.

Then we have $a+e_{2}^{\perp}\subset C_{1},e_{1},e_{n},e_{1}+e_{n}\in
C_{2},e_{2}\in C_{3}$. We also note that $a+e_{2}^{\perp}$ does not depend on
the choice of $a$ if $a\cdot e_{2}=1$. This leaves three elements of
$W^{\sharp}$ that are not known to be in $C_{1}\cup C_{2}\cup C_{3}$, namely
the elements of $R:=\\{e_{1}+e_{2},e_{1}+e_{2}+e_{n},e_{2}+e_{n}\\}$. This
leads us to do the following: Choose $R_{1}\subseteq R$ and $R_{2}\subseteq
R\setminus R_{1}$ and let $D_{1}=C_{1}\cup R_{1},D_{2}=C_{2}\cup
R_{2},D_{3}=C_{3}.$ We show

######  Lemma 11.3.

Assume the above and $a\in\\{e_{n-1},e_{n-1}+e_{n}\\}.$ Then the only
possibility for having $D_{1},D_{2},$ be the restriction to $W$ of p-sets
$E_{1},E_{2}$ of a commutative Schur ring is to have

$D_{1}=a+e_{2}^{\perp},\quad D_{2}=\\{e_{1},e_{n},e_{1}+e_{n}\\}\cup R.$

Furthermore, the set $D_{2}$ is
$\\{u+v:u\in\\{e_{1},e_{n},e_{1}+e_{n}\\},v\in\\{0,e_{2}\\}\\}$.

Proof This is similar to Proposition 9.1 and Lemma 9.2: let
$X:=D_{1}D_{2}\cap(t_{1}t_{2})^{G},Y=D_{1}^{2}\cap(t_{1}t_{n})^{G},Z:=D_{1}Y$
and show that the only situation where $XZ$ is equal to $ZX$ is the situation
described.∎

If $a\in\mathcal{T}$, then by $y_{a}$ we mean any element of
$\mathcal{T}\setminus a^{\perp}$.

######  Corollary 11.4.

Assume that we have the standard situation where $f_{1}(C_{2})>0$ and $r>2$.
Then $x+C_{2}=C_{2}$ for all $x\in X=\bigcup_{i=3}^{r}C_{i}$.

Proof Since $f_{1}(C_{2})>0$, if $a\in C_{2}$, then there is $b\in C_{2}$ with
$a\cdot b=1$ and $a,b,a+b\in C_{2}$. Now let $X:=C_{3}\cup\cdots\cup C_{r}$,
an isotropic set orthogonal to $C_{2}$. By Lemma 11.3 in the above set-up we
see that for $x\in X$ we have $\\{a+x,b+x,a+b+x\\}\subseteq C_{2}$. In
particular $a+x\in C_{2}$ and so $x+C_{2}=C_{2}$ for all $x\in X$. ∎

Thus $\bigcup_{x\in X\cup\\{0\\}}\\{x+e_{1},x+e_{n},x+e_{1}+e_{n}\\}\subseteq
C_{2},\text{ and }\bigcup_{x\in X}y_{x}+x^{\perp}\subseteq C_{1}$

######  Lemma 11.5.

Let $X=C_{3}\cup\cdots\cup C_{r}$ (an isotropic set). Let $x\in{\rm
Span}(C_{3}\cup\cdots\cup C_{r})$. Then $y_{x}+x^{\perp}\subseteq C_{1}$ and
${\rm Span}(X)^{\sharp}=X.$

Proof Let $x\in{\rm Span}(X)$ and $c\in y_{x}+x^{\perp}$, so that $c\cdot
x=1$. Then $c\notin X$ as $X$ is isotropic, so that $c\in C_{1}\cup C_{2}$. If
$c\in C_{2}$, then $c\cdot x\in\\{c\cdot u:u\in X\\}\neq\\{0\\}$; but $c\perp
X$ gives a contradiction. Thus $c\in C_{1}$ and so $y_{x}+x^{\perp}\subseteq
C_{1}$.

Now if $x\in{\rm Span}(X)\setminus X$, then $x\in C_{1}\cup C_{2}$ and again
$x\notin C_{2}$, so that $x\in C_{1}$. Now $f_{2}(C_{1})>0$ means that there
is some $c\in y_{x}+x^{\perp}\subseteq C_{1},c\cdot x=1$, and $c+x\in C_{2}$.
But $x\cdot(c+x)=1$, which contradicts $X\perp C_{2}$. Thus ${\rm
Span}(X)^{\sharp}=X$. ∎

######  Proposition 11.6.

Suppose that we have the standard situation where $f_{1}(C_{2})>0$ and
$C_{3},\cdots,C_{r}$ are isotropic, $r\geq 3$. Let $X={\rm
Span}(C_{3},\cdots,C_{r}).$ If $C_{1},\cdots,C_{r}$ are the p-sets of a
commutative Schur ring, then either

(i) $C_{1}\cap C_{2}^{\perp}\cap X^{\perp}\neq\emptyset$; or

(ii) $C_{2}^{\perp}=X.$

Proof So assume that (i) and (ii) are false, so that $C_{1}\cap
C_{2}^{\perp}\cap X^{\perp}=\emptyset$ and $C_{2}^{\perp}\neq X.$ But
$X\subseteq C_{2}^{\perp}$ and $C_{2}^{\perp}\neq X$ so we can choose $a\in
C_{2}^{\perp}\setminus X$. Now $a\in C_{2}^{\perp}$ and $f_{1}(C_{2})>0$ means
that $a\notin C_{2}$ and then $a\notin X$ gives $a\in C_{1}$.

Now we have $a\in C_{1}\cap C_{2}^{\perp}$, so that $C_{1}\cap
C_{2}^{\perp}\cap X^{\perp}=\emptyset$ tells us that $a\notin X^{\perp}$. Thus
there is some $b\in X$ such that $a\cdot b=1$.

Let $c\in C_{2}$. Then $a\in C_{2}^{\perp}$ gives $a\cdot c=0$. By Corollary
11.4 we have $X+C_{2}=C_{2}$, so that $b+c\in C_{2}$. However $a\cdot(b+c)=1$,
contradicting $a\in C_{2}^{\perp}$. This contradiction completes the proof. ∎

By Lemma 11.2 we do not have Lemma 11.6 (i), so we now assume

######  Corollary 11.7.

Suppose that we have the standard situation where $f_{1}(C_{2})>0$ and
$C_{3},\cdots,C_{r}$ are isotropic, $r\geq 3$. Then $C_{2}^{\perp}=X$. ∎

######  Proposition 11.8.

Suppose that we have the standard situation where $f_{1}(C_{2})>0,r\geq
3,n\geq 6,X={\rm Span}(X)^{\sharp}$ where $|X|>1$ and $C_{2}^{\perp}=X$. Then
the $C_{i}$ cannot be the p-sets of a commutative Schur ring.

Proof Since $X={\rm Span}(X)^{\sharp}$ where $|X|>1$ we can assume that
$X={\rm Span}(a_{1},a_{2},\cdots,$ $a_{k}),1<k<n/2,$ where
$a_{1},b_{1},\cdots,a_{n/2},b_{n/2}$ is a symplectic basis. Since
$f_{1}(C_{2})>0$ and $f_{2}(C_{2})=0$ we can assume:
${a_{k+1}},{b_{k+1}},{a_{k+1}+b_{k+1}}\in C_{2}$. Since $C_{2}+X=C_{2}$ we
then see that ${a_{k+1}+u},{b_{k+1}+u},{a_{k+1}+b_{k+1}+u}\in C_{2}$ for $u\in
X$. We also know that
$(y_{a_{1}}+a_{1}^{\perp})\cup(y_{a_{2}}+a_{2}^{\perp})\subseteq C_{1}$.

Let $W:={\rm
Span}(a_{1},b_{1},a_{2},b_{2},a_{k+1},b_{k+1}),D_{i}=C_{i}\cap\mathcal{T}(W),i=1,2,$
and $D_{X}=\\{a_{1},a_{2},a_{1}+a_{2}\\}$. Then $|D_{1}|=48$. Now there are
$12$ elements of $D_{2}$ of the form $u+v$ where
$u\in\\{a_{k+1},b_{k+1},a_{k+1}+b_{k+1}\\},v\in D_{X}\cup\\{0\\}$. Denote the
set of these $12$ elements by $E_{2}$. Now let
$E_{1}:=(b_{1}+a_{1}^{\perp})\cup(b_{2}+a_{2}^{\perp}).$ Then
$E_{1},E_{2},D_{X}$ are disjoint and we have $E_{1}\subset D_{1},E_{2}\subset
D_{2}$. So we obtain $|E_{1}|+|E_{2}|+|E_{X}|\geq 48+12+3=63=|W^{\sharp}|$,
showing that $D_{1}=E_{1},D_{2}=E_{2},D_{X}$ is a partition of $W^{\sharp}$.

Now let $A:=(C_{1}^{3})_{1},$ the set of elements of $C_{1}^{3}$ that have
multiplicity $1$ in $C_{1}^{3}$, and let $B:=(C_{1}C_{2}C_{1})_{1}$ the set of
elements of $C_{1}C_{2}C_{1}$ that have multiplicity $1$ in $C_{1}C_{2}C_{1}$.
Then $A,B\in\mathfrak{S}$. Now let

$\displaystyle\alpha_{1}:=t_{b_{2}+b_{1}}t_{a_{1}+a_{k+1}+b_{2}+b_{1}}t_{a_{1}+b_{k+1}+b_{2}+b_{1}},$
$\displaystyle\alpha_{2}=t_{a_{2}+a_{k+1}+b_{k+1}+b_{2}}t_{a_{2}+a_{k+1}+b_{k+1}},t_{a_{1}+a_{2}+b_{2}},$

and note that $\alpha_{1}\in A,\alpha_{2}\in B$. Let
$\alpha=\alpha_{1}\alpha_{2}\in AB$. We note that $\\{v\alpha-v:v\in W\\}=W$,
showing that the minimal number of transvections giving $\alpha$ as a product
is six, and that any such transvections must be in $\mathcal{T}(W)$, and must
span $W$. Thus to obtain $\alpha$ as a product
$\beta_{1}\beta_{2},\beta_{1}\in A,\beta_{2}\in B$ (or where $\beta_{1}\in
B,\beta_{2}\in A)$ we only need to consider products of transvections in
$\mathcal{T}(W)$.

Now the multiplicity of $\alpha$ in the multi-set $AB$ is $6$, while the
multiplicity of $\alpha$ in $BA$ is $3$, so that $AB\neq BA$.∎

######  Corollary 11.9.

Suppose that we have the standard situation where $f_{1}(C_{2})>0,r\geq
3,n\geq 6,X={\rm Span}(X)^{\sharp}$. Then $r=3$ and we can assume that
$|X|=|C_{3}|=1$ and $C_{2}^{\perp}=X$. ∎

##  12\.  Step 11

###  12.1. The $r=2$ case

Assume that we have the standard situation with $r=2.$

######  Proposition 12.1.

If $f_{1}(C_{2})>0,f_{2}(C_{2})=0,f_{2}(C_{1})>0$, then

(i) $\Gamma^{1}(C_{2})$ is connected, and

(ii) $\langle C_{2}\rangle$ is a 3-transposition subgroup of ${\rm Sp}(n,2)$
(where $C_{2}$ is the set of involutions of $\langle C_{2}\rangle$ determing
the $3$-transposition group structure) that acts transitively on $C_{2}$ by
conjugation and also acts on $C_{1}$ by conjugation.

Proof (i) Suppose that $\Gamma^{1}(C_{2})$ is not connected and let
$\Gamma_{2i},1\leq i\leq s,$ be the components of $\Gamma^{1}(C_{2})$. Then
$f_{1}(\Gamma_{2i})=f_{1}(C_{2})>0,1\leq i\leq s$.

CASE 1: $s\geq 3$. Here we choose $b_{i}\in\Gamma_{2i}$. Then $b_{i}$ does not
meet any elements of $\Gamma_{2j},j\neq i$. Then
$b_{1}+b_{2},b_{1}+b_{2}+b_{3}\in C_{1}$ (as these element meet $\Gamma_{21}$
and $\Gamma_{22}$). But they meet the same number of elements of
$C_{1},\Gamma_{21}$ and $\Gamma_{22}$, but not the same number of elements of
$\Gamma_{23}$ (and they meet no elements of $\Gamma_{2i},i>3$). This
contradiction gives the result in this case. So we are left to consider:

CASE 2: $s=2$. We first claim that ${\rm
Span}(\Gamma_{2i})^{\sharp}=\Gamma_{2i},i=1,2$. To this end assume that ${\rm
Span}(\Gamma_{21})^{\sharp}\neq\Gamma_{21}$ and let $b_{0}\in{\rm
Span}(\Gamma_{21})^{\sharp}\setminus\Gamma_{21}$. Now choose
$b_{2}\in\Gamma_{22}.$ Then $b_{0},b_{0}+b_{2}\in C_{1}$ and they both meet
the same number of elements of $C_{1}$ and of $\Gamma_{21}$, however $b_{0}$
meets zero elements of $\Gamma_{22}$, while $b_{0}+b_{2}$ meets
$2f_{1}(C_{2})>0$ elements of $\Gamma_{22}$. This contradiction gives the
claim. It then follows that each $\Gamma_{2i}$ is a (non-degenerate)
symplectic subspace of $V$, since $0<f_{1}(C_{2})=f_{1}(\Gamma_{2i})$ and so
there are no isotropic elements of $\Gamma_{2i},i=1,2$.

So we now have $\Gamma_{21}=V_{u}^{\sharp},\Gamma_{22}=V_{v}^{\sharp},u+v\leq
n$, where $V_{k}$ is a symplectic space of dimension $k$ and $V_{u}\perp
V_{v}$. Also $V_{u}\cap V_{v}=\\{0\\}$.

We next claim that ${\rm Span}(C_{2})=V$. If not, then from the above there is
a symplectic subspace $V_{k},k>1,$ such that $V=V_{u}\oplus V_{v}\oplus
V_{k}$. Let $b_{0}\in V_{k}^{\sharp},b_{i}\in\Gamma_{2i},i=1,2$. Then
$b_{0}+b_{1},b_{0}+b_{1}+b_{2}\in C_{1}$ again meet different numbers of
elements of $V$. Thus $V=V_{u}\oplus V_{v}.$ [This is the end of the proof of
Proposition 12.3 \- see below.]

Since $n\geq 6$ and $u,v>0$ we can assume that $u\geq 4,v\geq 2$. Let

$W:={\rm Span}(e_{1},e_{2},e_{3},e_{n-2},e_{n-1},e_{n}),U:={\rm
Span}(e_{1}+e_{6},e_{2},e_{3}+e_{6},e_{4}+e_{6},e_{5}+e_{6})$

and assume that ${\rm Span}(e_{1},e_{2},e_{n-1},e_{n})\subseteq V_{u},{\rm
Span}(e_{3},e_{n-2})\subseteq V_{v}$. Then $W\setminus({\rm
Span}(e_{1},e_{2},e_{n-1},e_{n})\cup{\rm Span}(e_{3},e_{n-2}))\subset C_{1}$.
Define the following multisets:

$X:=(C_{1}\cdot C_{2})\cap(t_{1}t_{2})^{G},\quad
Y:=C_{1}^{2}\cap(t_{1}t_{n})^{G},\quad Z:=C_{1}\cdot Y.$

Then $X,Y,Z\in\mathfrak{S}$. Let $Z_{i}$ be the set of those elements of the
set $Z$ that have multiplicity $i\in\mathbb{Z}$ in the multiset $Z$. We will
show that $XZ_{1}\neq Z_{1}X$. Let

$z:=t_{n-2,n}t_{2,n-2,n-1}t_{n-2,n-1}\in Z_{1},\quad
x:=t_{2,3,n}t_{1,2,n-1}\in X,\quad\alpha:=zx\in Z_{1}X.$

Then $\alpha$ is a product of $5$ transvections each in $\mathcal{T}(U)$, and
that if $\alpha=t_{u}t_{v}t_{w}t_{x}t_{y}$, then ${\rm Span}(u,v,w,x,y)=U$.
Thus to find such $u,v,w,x,y$ we need only look in $U$.

Now $\mathcal{T}(U)$ is partitioned as
$(C_{1}\cap\mathcal{T}(U))\cup(C_{2}\cap\mathcal{T}(U))$ and it is easy to
check that $\alpha$ occurs in $Z_{1}X$ with multiplicity $5$ while $\alpha$
occurs with multiplicity $3$ in $XZ_{1}$. Thus $XZ_{1}\neq Z_{1}X$ and this
concludes the proof of Proposition 12.1 (i).

By (i) $\Gamma^{1}(C_{2})$ is connected, and if $t_{a},t_{b}\in C_{2},a\cdot
b=1$, then $t_{a}^{t_{b}t_{a}}=t_{b}.$ Thus elements of $C_{2}$ are conjugate
by elements of $\langle C_{2}\rangle$. Now $C_{2}$ is fixed by the conjugation
action of any element of $\langle C_{2}\rangle$. Thus $C_{2}$ is a conjugacy
class in $\langle C_{2}\rangle.$ This shows that $\langle C_{2}\rangle$ is a
3-transposition subgroup of ${\rm Sp}(n,2)$ and the rest follows. ∎

######  Lemma 12.2.

Assume: $r=2,f_{2}(C_{2})=0,f_{1}(C_{2})>0,f_{2}(C_{1})>0,$ and
$\Gamma^{1}(C_{2})$ is connected. Then ${\rm Span}(C_{2})=V$ and $\langle
C_{2}\rangle$ is an irreducible $3$-transposition group.

Proof So assume that $0\neq x\in{\rm Span}(C_{2})^{\perp}$. Since
$f_{1}(C_{2})>0$ we have $x\notin C_{2}$, so that $x\in C_{1}$. Thus $x\in
C_{1}\cap(C_{2}^{\perp})=\\{0\\}$, contradicting $f_{2}(C_{1})>0$. Thus ${\rm
Span}(C_{2})$ is a symplectic subspace and if ${\rm Span}(C_{2})\neq V$, then
there is $0\neq x\in C_{2}^{\perp}$ and again we have $x\in
C_{1}\cap(C_{2}^{\perp})$.∎

###  12.2. The $r=3$ case

Assume the standard situation where $\mathcal{T}=C_{1}\cup C_{2}\cup
C_{3},C_{3}=\\{e_{1}\\},f_{1}(C_{2})>0,f_{2}(C_{2})=0,f_{2}(C_{1})>0$.

######  Proposition 12.3.

Then

(i) $\Gamma^{1}(C_{2})$ is connected, and

(ii) $\langle C_{2}\rangle$ is a 3-transposition subgroup of $Sp(n,2)$ (where
$C_{2}$ is the set of involutions) that acts transitively on $C_{2}$ by
conjugation and also acts on $C_{1}$ by conjugation.

Proof (i) The proof follows the proof of Proposition 12.1 down to where it is
shown that $V_{u}\oplus V_{v}=V$. This gives a contradiction as then
$C_{3}\perp C_{2}$ and ${\rm Span}(C_{2})=V$. (ii) follows as in Proposition
12.1. ∎

##  13\. Step 12: Relations

###  13.1. Relations when $r=2$

So assume that $r=2,n\geq 6,\mathcal{T}=C_{1}\cup C_{2}$ where
$f_{2}(C_{2})=0,f_{2}(C_{1})>0$. Let
$a_{i}=f_{i}(C_{1}),b_{i}=f_{i}(C_{2}),1\leq i\leq 4$. The goal will be to
find polynomial relations among $a_{1},\cdots,a_{4},b_{1},\cdots,b_{4}$ and
some other indeterminates. We certainly have (see Eq. (3.1))

(13.1) $\displaystyle a_{3}=2a_{1}+a_{2}+a_{4}+1;$ (13.2) $\displaystyle
b_{3}=2b_{1}+b_{2}+b_{4}+1;$ (13.3) $\displaystyle a_{3}+b_{3}=2^{n}-1;$
(13.4) $\displaystyle b_{2}=0.$

For $0\neq a\in V,e\in\mathbb{F}_{2},$ and $C\subseteq\mathcal{T}$ we define
$B_{a}^{e}(C)=\sum_{t_{b}\in C,a\cdot b=e}t_{b}.$

######  Lemma 13.1.

Let $C_{1},C_{2}\subseteq\mathcal{T}$ be disjoint $\mathfrak{S}$-sets. Then

(i) $\sum_{t_{a}\in C_{1}}B_{a}^{0}(C_{2})\in\mathfrak{S}$.

(ii) If $C_{2}$ is a p-set, then $\sum_{t_{a}\in
C_{1}}B_{a}^{0}(C_{2})=\lambda C_{2}$ for some integer $\lambda\geq 0$.

(iii) If $\mathcal{T}=C_{1}\cup C_{2}$ and $t_{a}\in C_{1}$, then
$|B_{a}^{0}(C_{2})|=2^{n-1}-2-f_{4}(C_{1}).$

Proof Now the $(t_{1}t_{2})^{G}$ part of $\overline{C_{1}}\,\overline{C_{2}}$
is $U:=\sum_{t_{a}\in C_{1}}t_{a}B_{a}^{0}(C_{2})$ since $C_{1}\cap
C_{2}=\emptyset$. So

$\overline{C_{1}}\cdot U=\sum_{t_{a^{\prime}},t_{a}\in
C_{1}}t_{a^{\prime}}t_{a}B_{a}^{0}(C_{2}).$

Let $t_{a^{\prime}}t_{a}t_{b}\in\overline{C_{1}}\cdot U$, where
$t_{a^{\prime}},t_{a}\in C_{1},t_{b}\in B_{a}^{0}(C_{2})$, so that $a\cdot
b=0$. Since $a,a^{\prime}\neq b$ Lemma 2.10 shows that the only way that
$t_{a^{\prime}}t_{a}t_{b}\in\mathcal{T}$ is if $a=a^{\prime}$.

Thus we see that the $\mathcal{T}$ part of this sum is $\sum_{T_{a}\in
C_{1}}B_{a}^{0}(C_{2})$, which is thus in $\mathfrak{S}$. This gives (i) and
(ii) follows.

(iii) Note that $|a^{\perp}|=2^{n-1}$. Since there are $f_{4}(C_{1})$ elements
$t_{b}\in C_{1},b\neq a,$ with $b\cdot a=0$, there are (since
$\mathcal{T}=C_{1}\cup C_{2}$) $2^{n-1}-2-f_{4}$ elements $t_{c}\in C_{2}$
with $c\cdot a=0.$ This gives $|B_{a}^{0}(C_{2})|=2^{n-1}-2-f_{4}(C_{1})$.∎

######  Corollary 13.2.

If $\mathcal{T}=C_{1}\cup C_{2}$ where $C_{1},C_{2}$, then there are
$\lambda_{1},\lambda_{2}\in\mathbb{Z}^{\geq 0}$ with

(13.5) $\displaystyle a_{3}(2^{n-1}-2-a_{4})=\lambda_{1}b_{3};$ (13.6)
$\displaystyle b_{3}(2^{n-1}-2-b_{4})=\lambda_{2}a_{3}.\qed$

######  Lemma 13.3.

As a function of the $a_{i},b_{j}$ the number of zero-triangles in
$\mathcal{T}$ is

(13.7)
$\displaystyle\frac{1}{3}a_{1}a_{3}+\frac{1}{2}a_{2}a_{3}+\frac{1}{3}b_{1}b_{3}+\frac{1}{2}b_{2}b_{3}=\frac{1}{3}(2^{n}-1)2^{n-2}.$

Proof Any $t_{a}\in C_{1}$ is in $f_{1}(C_{1})=a_{1}$ zero-triangles contained
in $C_{1}$. Summing over all $a_{3}$ elements of $C_{1}$ we get a contribution
of $a_{1}a_{3}/3$ to the number of zero-triangles in $\mathcal{T}$. Further
$t_{a}\in C_{1}$ is in $a_{2}$ zero triangles $\\{t_{a},t_{b},t_{a+b}\\}$ for
some $t_{b}\in C_{1}$ and $t_{a+b}\in C_{2}$. Summing over all $a_{3}$
elements of $C_{1}$ we obtain a contribution of $a_{2}a_{3}/2$ to the number
of zero-triangles in $\mathcal{T}$. Now we do the same argument for $t_{a}\in
C_{2}$, noting that counting in this way we never count any zero-triangle
twice. Thus the total number is
$a_{1}a_{3}/3+a_{2}a_{3}/2+b_{1}b_{3}/3+b_{2}b_{3}/2.$ We now use Lemma 4.2. ∎

Assume $\mathcal{T}=C_{1}\cup C_{2}\cup\cdots\cup C_{r}$. Let
$\Sigma(C_{2},C_{1},a),a\in C_{2},$ be the graph with vertices $C_{1}$ and
edges $\\{v_{1},v_{2}\\}\subset C_{1}$ if $v_{1}\cdot v_{2}=1$ and
$v_{1}+v_{2}=a$. For $S\subseteq C_{2}$ let

$\Sigma(C_{2},C_{1},S)=\bigcup_{a\in S}\Sigma(C_{2},C_{1},a)\text{ and
}\Sigma(C_{2},C_{1})=\Sigma(C_{2},C_{1},C_{2})$

We first note: if we have an edge in $\Sigma(C_{2},C_{1},a)$, then
$f_{2}(C_{1})\neq 0$. Suppose $\mathcal{T}=C_{1}\cup C_{2}$ and
$f_{2}(C_{1})\neq 0$. Then for $b\in C_{1}$ there are $f_{2}(C_{1})$ elements
$a+b\in C_{1}$ with $a\cdot b=1$ and $a\in C_{2}$. This gives $f_{2}(C_{1})$
elements $a\in C_{2}$ for each $b\in C_{1}$.

Conversely, for $a\in C_{2}$, if there is $b\in C_{1}$ with $a\cdot b=1,a+b\in
C_{1}$, then this $b\in C_{1}$ is accounted for as in the above paragraph.

Fix $a\in C_{2}$. Now the number of $b\in\mathcal{T}$ with $a\cdot b=1$ is
$2^{n-1}$ of which $2f_{1}+f_{2}$ are in $C_{2}$. Thus there are
$2^{n-1}-2f_{1}-f_{2}$ such elements $b$ in $C_{1}$. There are $a_{2}$ of the
$b\in C_{1}$ with $a+b\in C_{2}$. This leaves $2^{n-1}-2a_{1}-2a_{2}$ elements
$b\in C_{2}$ where $a+b\in C_{1}$. These $b$s come in pairs $b,a+b$, each such
pair giving an edge of $\Sigma(C_{2},C_{1},a)$. Thus the number of edges of
$\Sigma(C_{2},C_{1},a)$ is $b_{3}(2^{n-1}-2b_{1}-2b_{2})/2$. But the number of
such edges is also $f_{3}(C_{1})f_{2}(C_{1})/2=a_{2}a_{3}/2$. Thus we have

(13.8) $\displaystyle a_{2}a_{3}=b_{3}(2^{n-1}-2b_{1}-2b_{2})$

###  13.2. Relations for $r=3$

So assume that $r=3,n\geq 6$ and that $\mathcal{T}=C_{1}\cup C_{2}\cup
C_{3},C_{3}=\\{e_{1}\\}$ where $f_{2}(C_{2})=0,f_{2}(C_{1})>0$. We also have
$e_{n}+e_{1}^{\perp}\subseteq C_{1},C_{2}+e_{1}=C_{2}.$ Let
$a_{i}=f_{i}(C_{1}),b_{i}=f_{i}(C_{2}),1\leq i\leq 4$. We again find
polynomial relations among
$a_{1},\cdots,a_{4},b_{1},\cdots,b_{4},\lambda_{1},\lambda_{2}$. We certainly
have

(13.9) $\displaystyle a_{3}=2a_{1}+a_{2}+a_{4}+1;$ (13.10) $\displaystyle
b_{3}=2b_{1}+b_{2}+b_{4}+1;$ (13.11) $\displaystyle a_{3}+b_{3}=2^{n}-2;$
(13.12) $\displaystyle b_{2}=0.$

An argument similar to Corollary 13.2 gives:

######  Lemma 13.4.

If $\mathcal{T}=C_{1}\cup C_{2}$ where $C_{1},C_{2}$ are p-sets, then
$f_{3}(C_{1})(2^{n-1}-2-f_{4}(C_{1}))=\lambda f_{3}(C_{2})$ for some integral
$\lambda\geq 0$. Thus

(13.13) $\displaystyle a_{3}(2^{n-1}-2-a_{4})=\lambda_{1}b_{3};$ (13.14)
$\displaystyle b_{3}(2^{n-1}-3-b_{4})=\lambda_{2}a_{3}.\qed$

##  14\. Symmetric group cases

We now consider the situations where $\langle C_{2}\rangle$ is a symmetric
group $S_{n+1}$ or $S_{n+2}$.

Note that for $r=2,3$: if we know $a_{3}$ or $b_{3}$ as a function of $n$,
then we know $a_{3}$ and $b_{3}$ as a function of $n$ (since
$a_{3}+b_{3}=2^{n}-1$ or $a_{3}+b_{3}=2^{n}-2$). Then the quadratic equations
Eqs. (13.1)-(13.8) in the $a_{i},b_{i},1\leq i\leq 4,\lambda_{1},\lambda_{2}$
(or Eqs. (13.9)-(13.14)) become linear equations with coefficients that are
functions of $n$.

###  14.1. Symmetric group $S_{n+1}$ for $r=2$

So assume that $\mathcal{T}=C_{1}\cup C_{2}$ and note that if $\langle
C_{2}\rangle=S_{n+1}$, then $b_{3}=\binom{n+1}{2}\text{ and
}b_{4}=\binom{n-1}{2}.$ Using Eq (13.3) gives
$a_{3}=2^{n}-1-\frac{1}{2}n^{2}-\frac{1}{2}n.$ Using $b_{2}=0$ and these
values for $a_{3},b_{3},b_{4}$ Eqs. (13.1) -(13.8) become

$\displaystyle
2a_{1}+a_{2}+a_{4}-2^{n}+\frac{1}{2}n^{2}+\frac{1}{2}n+2=0;\quad
2b_{1}-2n+2=0,$
$\displaystyle\left(\frac{1}{3}2^{n}-\frac{1}{6}n^{2}-\frac{1}{6}n-\frac{1}{3}\right)a_{1}+\left(2^{n-1}-\frac{1}{4}n^{2}-\frac{1}{4}n-\frac{1}{2}\right)a_{2}+\left(\frac{1}{6}n^{2}+\frac{1}{6}n\right)b_{1}-\frac{1}{12}2^{2n}+\frac{1}{12}2^{n}=0,$
$\displaystyle\left(-2^{n}+\frac{1}{2}n^{2}+\frac{1}{2}n+1\right)a_{4}+\left(-\frac{1}{2}n^{2}-\frac{1}{2}n\right)\lambda_{1}+\frac{1}{2}2^{2n}-\frac{1}{4}2^{n}n^{2}-\frac{1}{4}2^{n}n-\frac{5}{2}2^{n}+n^{2}+n+2=0,$
$\displaystyle\left(-2^{n}+\frac{1}{2}n^{2}+\frac{1}{2}n+1\right)\lambda_{2}+\frac{1}{4}2^{n}n^{2}+\frac{1}{4}2^{n}n-\frac{1}{4}n^{4}+\frac{1}{2}n^{3}-\frac{3}{4}n^{2}-\frac{3}{2}n=0,$
$\displaystyle\left(-2^{n+1}+n^{2}+n+2\right)a_{1}+\left(-2^{n+1}+n^{2}+n+2\right)a_{2}+2^{2n-1}-\frac{1}{4}2^{n}n^{2}-\frac{1}{4}2^{n}n-\frac{1}{2}2^{n}=0,$
$\displaystyle\left(2^{n}-\frac{1}{2}n^{2}-\frac{1}{2}n-1\right)a_{2}+\left(-\frac{1}{2}n^{2}-\frac{1}{2}n\right)\mu=0.$

We solve these linear equations in the variables
$a_{1},a_{2},a_{4},b_{1},\lambda_{1},\lambda_{2},\mu$ to get $a_{1}=N/D$ where

$N=2^{2n-2}-3\cdot 2^{n-3}n^{2}-3\cdot 2^{n-3}n-2^{n-2}+n^{3}-n;\quad
D=2^{n}-\frac{1}{2}n^{2}-\frac{1}{2}n-1.$

One finds that $N/D\notin\mathbb{Z}$ if $n\geq 8$, a contradiction.∎

###  14.2. Symmetric group $S_{n+2}$ for $r=2$

The method is the same as for the $S_{n+1}$ case above, except that
$b_{3}=\binom{n+2}{2},b_{4}=\binom{n}{2}$. One finds that $a_{1}=N/D$ where

$N=2^{2n-2}-3\cdot 2^{n-3}n^{2}-9\cdot 2^{n-3}n-2^{n}+n^{3}+3\cdot
n^{2}+2n,\quad D=2^{n}-\frac{1}{2}n^{2}-\frac{3}{2}n-2.$

One then checks that $N/D$ is not an integer for all $n\geq 14.$

##  15\. Orthogonal Groups for $r=2$

We recall the definitions of quadratic forms, their Arf invariants and their
Orthogonal groups. A quadratic form on $V$ (a vector space over
$\mathbb{F}_{2}$) is a function $Q:V\to\mathbb{F}_{2}$ satisfying
$Q(ax+by)=aQ(x)+bQ(y)+(ab)(x\cdot y)$ for $a,b\in\mathbb{F}_{2},x,y\in V$. We
note that then $x\cdot y=Q(x+y)+Q(x)+Q(y)$. Put $n=2m$. Then [9, pp. 197-199]
the quadratic form $Q$ is equivalent to $Q_{0}$ or $Q_{1}$ where

$Q_{\alpha}\left(\sum_{i=1}^{n}x_{i}e_{i}\right)=(\alpha
x_{1}+x_{1}x_{n}+\alpha x_{n})+x_{2}x_{n-1}+x_{3}x_{n-2}+\cdots+x_{m}x_{m+1}.$

Here $\alpha\in\mathbb{F}_{2}$. Further $Q_{0}$ and $Q_{1}$ are not equivalent
and these give the two types of orthogonal group.

The Arf invariant of the quadratic form (which takes values in
$\mathbb{F}_{2}$) has a number of definitions: it is $0$ if the quadratic form
is equivalent to a direct sum of copies of the binary form $xy$, and it is $1$
if the form is a direct sum of $x^{2}+xy+y^{2}$ with a number of copies of
$xy.$

The Arf invariant is also the value which is assumed most often by the
quadratic form [5]. Another characterization: $Q$ has Arf invariant $0$ if and
only if the underlying $n$-dimensional vector space over the field
$\mathbb{F}_{2}$ has a $n/2$-dimensional subspace on which $Q$ is identically
$0$.

If $a_{1},b_{1},\cdots,a_{n/2},b_{n/2}$ is a symplectic basis then also: ${\rm
Arf}(Q)=\sum_{i=1}^{n/2}Q(a_{i})Q(b_{i})$; here it is known that this is
independent of the choice of symplectic basis.

Thus there are two types of quadratic form, as determined by their Arf
invariant. These give rise to two Orthogonal groups ${\rm SO}^{\pm}(n,2)$. The
form $Q_{0}$ corresponds to the Orthogonal group ${\rm SO}^{-}(2m,1)$ while
the form $Q_{1}$ corresponds to ${\rm SO}^{+}(2m,1)$.

Given a basis $B=\\{b_{1},\cdots,b_{n}\\}$ for $V$ a quadratic form $Q_{B}$
can be defined using $B$: just take $Q(b_{i})=1$ for all $i\leq n$ and extend
using the above equation. If $H=\langle B\rangle$ and the graph
$\Gamma^{1}(B)$ is connected, then $H$ acts transitively on $B$ and so on the
orbit $(B)H$, this being the conjugacy class of symplectic transvections in
$H$. Further, the quadratic form $Q_{B}$ is invariant under the action of $H$:
$Q((a)h)=Q(a)$ for all $a\in V,h\in H$, and $H$ is the corresponding
orthogonal group.

######  Proposition 15.1.

The group ${\rm SO}^{+}(2m,2)$ has $2^{m-1}(2^{m}-1)$ symplectic
transvections, while ${\rm SO}^{-}(2m,2)$ has $2^{m-1}(2^{m}+1)$ symplectic
transvections.

Proof This will be by simultaneous induction on $m\geq 2$, the case $m=2$
being easy to check. Thus we assume that the group ${\rm SO}^{+}(2m,2)$ has
$2^{m-1}(2^{m}-1)$ symplectic transvections, while the group ${\rm
SO}^{-}(2m,2)$ has $2^{m-1}(2^{m}+1)$ symplectic transvections. We then
consider the case $m+1$.

We write the form $\sum_{i=1}^{m+1}x_{i}y_{i}$ as
$\left(\sum_{i=1}^{m}x_{i}y_{i}\right)+(x_{m+1}y_{m+1})$. We note that
$x_{m+1}y_{m+1}$ is zero for exactly three choices of
$x_{m+1},y_{m+1}\in\mathbb{F}_{2}$ (and is $1$ otherwise).

By induction, the number of $(x_{1},y_{1},\cdots,x_{m},y_{m})$ with
$\sum_{i}x_{i}y_{i}=0$ is $2^{m-1}(2^{m}-1)$ while the number with
$\sum_{i}x_{i}y_{i}=1$ is $2^{m-1}(2^{m}-1)$, so the number of
$(x_{1},y_{1},\cdots,\break x_{m+1},y_{m+1})$ with $\sum_{i}x_{i}y_{i}=0$ is

$3\cdot 2^{m-1}(2^{m}-1)+1\cdot 2^{m-1}(2^{m}+1)=2^{m+1-1}(2^{m+1}-1),$

as required. For the other case the sum that we get is $3\cdot
2^{m-1}(2^{m}+1)+1\cdot 2^{m-1}(2^{m}-1)=2^{m+1-1}(2^{m+1}+1),$ as required. ∎

######  Proposition 15.2.

For $H={\rm SO}^{\pm}(n,2),n\geq 8,$ we have $f_{4}(C_{2})=2^{n-2}-1$.

Proof We induct on $n\geq 8$ We use a set $B=B_{n}$ of transvection generators
for $H=H_{n}\in\\{{\rm SO}^{+}(n,2),{\rm SO}^{-}(n,2)\\}$ given by [22]. See
Figure 1, where there are two graphs $J_{1},J_{2}$ shown. We let $J_{n,i}$ be
the subgraph of $J_{i},i=1,2,$ on vertices labeled $1,2,\cdots,n$. The
relationship between $B$ and the graphs is that $\Gamma^{1}(B)$ is one of the
graphs $J_{n,1},J_{n,2}$; in each case $B$ is a basis for $V$. For example,
the graph for ${\rm SO}^{+}(n,2)$ is the graph $J_{n,i}$, where vertex $n$ is
labelled ${\rm O}^{+}$ in $J_{i}$. Vertex $j$ of the graph $J_{n,i}$
corresponds to $b_{j}\in B$ if $B=\\{b_{1},\cdots,b_{n}\\}$. Each graph
$J_{i},i=1,2,$ is extended to the right with vertex labelings of period $8$.
Let $Q_{B}$ be the corresponding quadratic form.

Figure 1.

For $H\leq G$ and any $t\in\mathcal{T}(H)$ let $B(t)=(B(t)_{i})_{i=1}^{n}$ be
the coordinates of $t$ relative to the basis $B=\\{b_{1},\cdots,b_{n}\\}$. Now
for $\varepsilon,\delta\in\mathbb{F}_{2}$ we define:

$n_{\varepsilon,\delta}=|\\{t\in\mathcal{T}(H):Q_{B}(t)=\varepsilon,B(t)_{n}=\delta,b_{1}\cdot
t=0\\}|.$

Note that the condition $b_{1}\cdot t=0$ is equivalent to $B(t)_{2}=0$. We
also note that for each $v\in V$ the value $Q_{B}(v)$ is just the number of
components (mod $2$) of the subgraph of $\Gamma^{1}(B)$ determined by $v$ i.e.
determined by the non-zero coordinates of $B(v)$. Also $t\in\mathcal{T}(H)$
determines a pair $\pi(t):=[\varepsilon,\delta]$ where
$Q_{B}(t)=\varepsilon,B(t)_{n}=\delta.$

One checks the $n=8$ case to start the induction [18]. Now to get from the $n$
case to the $n+2$ case we just add two vertices ($n+1$ and $n+2$) to the right
of vertex $n$. We want to find linear equations relating the
$(n+2)_{\varepsilon,\delta}$ to the
$n_{\varepsilon,^{\prime}\delta^{\prime}}$. Now any $u\in V_{n+2}$ can be
written as $u=v+w$ where $v\in V_{n}$ and
$w=\lambda_{1}b_{n+1}+\lambda_{2}b_{n+2}$. So, for fixed $v,$ there are $4$
cases. For example, if $t_{v}\in\mathcal{T}(H_{n})$ has $\pi(t_{v})=[0,1]$ and
we add $0b_{n+1}+1b_{n+2}$, then $\pi(t_{v+b_{n+2}})=[1,1]$. Considering all
such possibilities gives the following system of equations:

$\displaystyle(n+2)_{0,0}=n_{0,0}+n_{1,0}+2n_{0,1}+1;$
$\displaystyle(n+2)_{1,0}=n_{0,0}+n_{1,0}+2n_{1,1}+1;$
$\displaystyle(n+2)_{0,1}=2n_{1,0}+n_{0,1}+n_{1,1}+2;$
$\displaystyle(n+2)_{1,1}=2n_{0,0}+n_{0,1}+n_{1,1}+2.$

As a matrix equation this is:

$\displaystyle\begin{bmatrix}1&1&2&0&1\\\ 1&1&0&2&1\\\ 0&2&1&1&2\\\
2&0&1&1&2\\\ 0&0&0&0&1\end{bmatrix}\begin{bmatrix}n_{0,0}\\\ n_{1,0}\\\
n_{0,1}\\\ n_{1,1}\\\ 1\end{bmatrix}=\begin{bmatrix}(n+2)_{0,0}\\\
(n+2)_{1,0}\\\ (n+2)_{0,1}\\\ (n+2)_{1,1}\\\ 1\end{bmatrix}$

The extra $1,1,2,2$ (in the last column of the $5\times 5$ matrix) that are
added come from considering the cases $t_{v}=t_{0}=id$ and $t_{v}=t_{b_{1}}$.
Denote the above $5\times 5$ matrix by $K$. Then $K$ has eigen-values
$0,4,1,2i,-2i,i=\sqrt{-1}$, and so is diagonalizable with diagonalizing matrix
[18]

$Q:=\begin{bmatrix}4&4&-4&-4&8\\\ 60&60&60&60&120\\\ 0&0&0&0&15\\\
20i&-20i&20&-20&0\\\ -20i&20i&20&-20&0\end{bmatrix}$

Here $QKQ^{-1}={\rm diag}[0,4,1,2i,-2i]$. One then shows that for $m\geq 0$ we
have:

$K^{m}\begin{bmatrix}31,31,32,32,1\end{bmatrix}^{T}=\begin{bmatrix}2^{2m+5}-1,2^{2m+5}-1,2^{2m+5},2^{2m+5},1\end{bmatrix}^{T}.$

With this, Proposition 15.2 follows as the number required is either
$n_{0,0}+n_{0,1}$ or $n_{1,0}+n_{1,1}$, which in either case is
$2^{n-2}-1$.\̇qed

###  15.1. Orthogonal group ${\rm SO}^{+}(n,2)$ for $r=2$

Let $H={\rm SO}^{+}(n,2),n=2m$. We now have $b_{4}=f_{4}(C_{2})=2^{n-2}-1$
(Proposition 15.2) and $b_{3}=2^{m-1}(2^{m}-1)$ (Proposition 15.1). We also
have Eqs (13.1), (13.2) and (13.3). These five equations are linear and we
solve them to get values for
$a_{1},a_{3},b_{1},b_{2}(=0),b_{3},b_{4},\lambda_{1},$ $\lambda_{2}$.
Substituting these values into the remaining equations we obtain:

######  Theorem 15.3.

Let $H={\rm SO}^{+}(n,2),n=2m$. Then
$a_{1},\cdots,b_{4},\lambda_{1},\lambda_{2}$ are

$\displaystyle a_{1}=0,$ $\displaystyle a_{2}=2^{n-2},$ $\displaystyle
a_{3}=2^{n-1}+2^{m-1}-1,$ $\displaystyle a_{4}=2^{n-2}+2^{m-1}-2,$
$\displaystyle b_{1}=2^{n-3}-2^{m-2},$ $\displaystyle b_{2}=0,$ $\displaystyle
b_{3}=2^{n-1}-2^{m-1},$ $\displaystyle b_{4}=2^{n-2}-1,$
$\displaystyle\lambda_{1}=2^{n-2}-1,$
$\displaystyle\lambda_{2}=2^{n-2}-2^{m-1}.\qed$

###  15.2. Orthogonal group ${\rm SO}^{-}(n,2)$ for $r=2$

A similar calculation, again using Propositions 15.2 and 15.1, deals with the
case $H={\rm SO}^{-}(n,2)$:

######  Theorem 15.4.

Let $H={\rm SO}^{-}(n,2),n=2m$. Then
$a_{1},\cdots,b_{4},\lambda_{1},\lambda_{2}$ are

$\displaystyle a_{1}=0,$ $\displaystyle a_{2}=2^{n-2},$ $\displaystyle
a_{3}=2^{n-1}-2^{m-1}-1,$ $\displaystyle a_{4}=2^{n-2}-2^{m-1}-2,$
$\displaystyle b_{1}=2^{n-3}+2^{m-2},$ $\displaystyle b_{2}=0,$ $\displaystyle
b_{3}=2^{n-1}+2^{m-1},$ $\displaystyle b_{4}=2^{n-2}-1,$
$\displaystyle\lambda_{1}=2^{n-2}-1,$
$\displaystyle\lambda_{2}=2^{n-2}+2^{m-1}.\qed$

##  16\.  Relations for the $r=3$ cases and their groups

###  16.1. The $S_{n+1}$ case for $r=3$

Here we have $b_{3}=\binom{n+1}{2},b_{4}=\binom{n-1}{2}.$ Using these values
for $b_{3},b_{4}$ we solve equations (13.9)-(13.14) to obtain
$\lambda_{2}=N/D$, where

$N:=\frac{1}{4}2^{n}n^{2}+\frac{1}{4}2^{n}n-\frac{1}{4}n^{4}+\frac{1}{2}n^{3}-\frac{5}{4}n^{2}-2n;\quad
D=2^{n}-\frac{1}{2}n^{2}-\frac{1}{2}n-3.$

Now

$N-\left(\frac{1}{4}n^{2}+\frac{1}{4}n\right)D=-\frac{1}{8}n^{4}+\frac{3}{4}n^{3}-\frac{5}{8}n^{2}-\frac{3}{4}n.$

But $N,D,N/D\in\mathbb{Z}$ and
$\frac{1}{4}n^{2}+\frac{1}{4}n\in\frac{1}{2}\mathbb{Z}$ gives

$A:=\frac{-\frac{1}{8}n^{4}+\frac{3}{4}n^{3}-\frac{5}{8}n^{2}-\frac{3}{4}n}{2^{n}-\frac{1}{2}n^{2}-\frac{1}{2}n-2}\in\frac{1}{2}\mathbb{Z},$

and one checks that $A$ is not a half-integer for $n\geq 6$, a contradiction.

###  16.2. The $S_{n+2}$ case for $r=3$

Here we have $b_{3}=\binom{n+2}{2},b_{4}=\binom{n}{2}.$ Using these values for
$b_{3},b_{4}$ we solve equations (13.9)-(13.14) to obtain $\lambda_{2}=N/D$,
where

$N=2^{n-2}n^{2}+\frac{3}{4}2^{n}n+2^{n-1}-\frac{1}{4}n^{4}-\frac{1}{2}n^{3}-\frac{5}{4}n^{2}-4n-3;\quad
D=2^{n}-\frac{1}{2}n^{2}-\frac{3}{2}n-3.$

Then

$N-\left(\frac{1}{4}n^{2}+\frac{3}{4}n+\frac{1}{2}\right)D=-\frac{1}{8}n^{4}+\frac{1}{4}n^{3}+\frac{7}{8}n^{2}-n-\frac{3}{2}.$

Then $N,D,N/D\in\mathbb{Z}$ and
$\frac{1}{4}n^{2}+\frac{3}{4}n+\frac{1}{2}\in\frac{1}{2}\mathbb{Z}$ gives

$\frac{-\frac{1}{8}n^{4}+\frac{1}{4}n^{3}+\frac{7}{8}n^{2}-n-\frac{3}{2}}{2^{n}-\frac{1}{2}n^{2}-\frac{3}{2}n-4}\in\frac{1}{2}\mathbb{Z}.$

One checks that the left-hand-side is not a half-integer for $n\geq 6$, a
contradiction.

###  16.3. The ${\rm SO}^{+}(n,2)$ case for $r=3$

Let $n=2m$. Here we have $b_{3}=2^{m-1}(2^{m}-1),b_{4}=2^{n-2}-1$. Using these
values for $b_{3},b_{4}$ we solve equations (13.9)-(13.14) to obtain
$\lambda_{2}=N/D$, where

$N=\frac{1}{4}2^{n}-\frac{1}{4}2^{n+m}-2^{n+1}+2^{m+1};\quad D=2^{n}+2^{m}-4.$

Then $N-(2^{n-2}-2^{m-1}-1/2)D=2^{m-1}-2$ and one checks that
$\frac{2^{m-1}-2}{2^{n}+2^{m}-4}$ is not a half-integer if $n\geq 4$.

###  16.4. The ${\rm SO}^{-}(n,2)$ case for $r=3$

Let $n=2m$. Here we have $b_{3}=2^{m-1}(2^{m}+1),b_{4}=2^{n-2}-1$. Using these
values for $b_{3},b_{4}$ we solve equations (13.9)-(13.14) to obtain
$\lambda_{2}=N/D$, where

$N=\frac{1}{4}2^{2n}+\frac{1}{4}2^{n+m}-2^{n+1}-2^{m+1};\quad
D=2^{n}-2^{m}-4.$

Then $N-(2^{n-2}+2^{m-1}-1/2)D=-2^{m-1}-2$ and one checks that
$\frac{-2^{m-1}-2}{2^{n}-2^{m}-4}$ is not a half-integer if $n\geq 4$.

Having checked these four cases we conclude:

######  Proposition 16.1.

The case $r=3$ cannot happen if $n\geq 6$.∎

So we now assume that $r=2$ and $H=\langle C_{2}\rangle={\rm
SO^{\varepsilon}}(n,2),\varepsilon=\pm$.

##  17\.  The $r=2,H={\rm SO}^{\pm}(n,2)\leq{\rm Sp}(n,2)$ cases

The following result will allow us to reduce to the $H={\rm SO}^{+}(6,2)$
case. Let $P_{k}$ denote the path graph with $k$ vertices.

######  Proposition 17.1.

For $n\geq 10$ even and $\varepsilon\in\\{+,-\\}$, the subgroup ${\rm
SO}^{\varepsilon}(n,2)\leq{\rm Sp}(n,2)$ contains a subgroup isomorphic to
${\rm SO}^{+}(6,2)\cong S_{8}$.

Proof Let $n\geq 10$ be even and let $\varepsilon\in\\{+,-\\}$. We choose a
basis $B$ using Figure 1 so that $\langle B\rangle={\rm
SO}^{\varepsilon}(n,2)$ and $B$ contains vertices labeled $1,\cdots,n$, where
vertex $n$ is labeled $O^{\varepsilon}$ in Figure 1. Then referring to Figure
1 we see that there is a subset
$C=\\{b_{1},b_{2},b_{3},b_{4},b_{5},b_{6}\\}\subset B$ such that
$\Gamma^{1}(\\{b_{1}\\},\\{b_{2}\\},\\{b_{3}\\},\\{b_{4}\\},\\{b_{5}\\},\\{b_{6}\\})$
is the path graph $P_{6}$. Then
$\Gamma^{1}(\\{b_{1}\\},\\{b_{2}\\},\\{b_{3}\\},\\{b_{4}\\},\\{b_{5}\\},\\{b_{6}\\},\\{b_{1}+b_{3}+b_{5}\\}\cong
P_{7}$ and $\langle C\cup\\{b_{1}+b_{3}+b_{5}\\}\rangle\cong S_{8}={\rm
SO}^{+}(6,2)$. ∎

Now consider the case where $r=2,H={\rm SO}^{\pm}(n,2)$ and let
$b_{1},b_{2},\cdots,b_{6}$ be a part of a basis $B$ as in Proposition 17.1 and
Figure 1 where $\langle B\rangle=H$ and where
$t_{b_{1}},t_{b_{2}},\cdots,t_{b_{6}},t_{b_{1}+b_{3}+b_{5}}\in H$ generate
$S_{8}\cong{\rm SO}^{+}(6,2)\leq{\rm SO}^{\pm}(n,2)$. Let $W:={\rm
Span}(b_{1},\cdots,b_{6})$. Then
$t_{b_{1}},t_{b_{2}},\cdots,t_{b_{6}},t_{b_{1}+b_{3}+b_{5}}\in
C_{2}\cap\mathcal{T}(W)$ and the elements of $C_{1}\cap\mathcal{T}(W)$ include
those elements $e\in W$ where $Q_{B}(e)=0$. We look at
$(C_{1}\cap\mathcal{T}(W))^{3}$, noting that it has multiplicities
$1,3,4,6,20,87$ and we let $X_{i}$ be those elements of the set
$(C_{1}\cap\mathcal{T}(W))^{3}$ that have multiplicity $i$. We show
$X_{1}X_{3}\neq X_{3}X_{1}$.

Define $\alpha=\alpha_{1}\alpha_{2}$ where

$\alpha_{1}=t_{b_{1}+b_{2}+b_{4}}t_{b_{2}+b_{5}}t_{b_{1}+b_{5}}\in
X_{1},\,\alpha_{2}=t_{b_{1}+b_{2}+b_{3}+b_{4}+b_{5}}t_{b_{2}+b_{3}+b_{5}+b_{6}}t_{b_{1}+b_{2}+b_{3}+b_{5}+b_{6}}\in
X_{3},$

and note that the span of
$b_{1}+b_{2}+b_{4},b_{2}+b_{5},b_{1}+b_{5},b_{1}+b_{2}+b_{3}+b_{4}+b_{5},b_{2}+b_{3}+b_{5}+b_{6},b_{1}+b_{2}+b_{3}+b_{5}+b_{6}$
is $W$. This shows (i) that the minimal number of transvections needed to give
$\alpha$ is six, and (ii) that if $\alpha=t_{u}t_{v}t_{w}t_{x}t_{y}t_{z}$,
then $u,v,w,x,y,z\in W$. Now the multiplicities of $\alpha$ in
$(X_{1}\cap\mathcal{T}(W)^{3})(X_{3}\cap\mathcal{T}(W)^{3})$ and
$(X_{3}\cap\mathcal{T}(W)^{3})(X_{1}\cap\mathcal{T}(W)^{3})$ are $1$ and $3$
(respectively) and so
$(X_{1}\cap\mathcal{T}(W)^{3})(X_{3}\cap\mathcal{T}(W)^{3})\neq(X_{3}\cap\mathcal{T}(W)^{3})(X_{1}\cap\mathcal{T}(W)^{3})$.
It follows from (i) and (ii) that $X_{1}X_{3}\neq X_{3}X_{1}.$

This concludes consideration of all cases and so the proof of Theorem 1.2.∎

Proof that ${\rm Dye}_{n},n>6,$ is not a strong Gelfand subgroup: We note from
[11],[12] that if $n=2m$ with $m$ prime, then ${\rm Dye}_{n}\cong{\rm
SL}(2,2^{m})\rtimes\mathcal{C}_{m}$. We use a result from [14, Lemma 3.3]: For
a group $H$ with irreducible character degrees $d_{1},\cdots,d_{t}$ the total
character degree of $H$ is $\tau_{H}:=\sum_{i=1}^{t}d_{i}$; the result is: if
$H\leq G$ and $\chi\in\hat{G},$ where $\tau_{H}<\chi(1)$, then $(G,H)$ is not
a strong Gelfand pair. We apply this to ${\rm Dye}_{n}\leq G_{n}$, where
$\tau_{{\rm Dye}_{n}}<|{\rm Dye}_{n}|=(2^{n}-1)2^{m}m$ and where
$\chi\in\hat{G_{n}}$ is the Steinberg character so that $\chi(1)=2^{m^{2}}$,
the size of a Sylow $2$-subgroup of $G_{n}$. Since
$(2^{2m}-1)2^{m}m<2^{m^{2}}$ for $m>3$, the result follows.∎

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