$\mathbb{H}=F^{2\times 2}$ and $w^{2}=1$.

  1. ((a))

If $\dim{A}=1$ then $A=F$.

  2. ((b))

If $\dim{A}=2$ then we have one of the following cases:

    1. ((i))

$A$ is a separable quadratic extension field over $F$.

    2. ((ii))

$A$ is an inseparable quadratic extension field over $F$ (and
$\operatorname{char}{F}=2$).

    3. ((iii))

$A\cong F[X]/(X^{2}-X)\cong F\times F$ is a split composition algebra.

    4. ((iv))

$A\cong F[X]/(X^{2})$ is a local algebra.

  3. ((c))

If $\dim{A}=3$ then the radical $R:=A^{\perp}\cap A$ is not trivial, and one
of the following occurs.

    1. ((i))

$\dim{R}=1$, and $A$ is in the $\operatorname{Aut}_{F}(\mathbb{O})$-orbit of
the subalgebra $F+Fp_{0}+Fn_{0}$ of upper triangular matrices in $\mathbb{H}$
(see 5.7).

    2. ((ii))

$\dim{R}\geq 2$, and $A$ is in the orbit of the subalgebra
$F+(Fp_{0}+Fn_{0})w$ (see 5.9).

  4. ((d))

If $\dim{A}=4$ then one of the following occurs:

    1. ((i))

$A$ is a quaternion algebra (split or not).

    2. ((ii))

$A$ is in the orbit of $B+(Fp_{0}+Fn_{0})w$, where $B\subseteq\mathbb{H}$ is
either a two-dimensional composition algebra, or an inseparable quadratic
field extension.

    3. ((iii))

The radical $Q:=\left\\{{x\in A^{\perp}\cap A}\left|\vphantom{x\in
A^{\perp}\cap A{\mathrm{N}}(x)=0}\right.\,{{\mathrm{N}}(x)=0}\right\\}$ of the
restriction ${\mathrm{N}}|_{A}$ of the quadratic form has dimension $3$. Then
$Q$ is isomorphic to the Heisenberg algebra, and $A$ is in the orbit of
$F+(n_{0}\mathbb{O}\cap\mathbb{O}n_{0})$.

    4. ((iv))

$A=A^{\perp}$ is a totally inseparable field extension of degree $4$ and
exponent $1$ (and $\operatorname{char}{F}=2$).
Of course, this case can only occur if $\operatorname{char}{F}=2$. Two such
subfields are in the same orbit under $\operatorname{Aut}_{F}(\mathbb{O})$ if,
and only if, they are isomorphic as $F$-algebras, see [7, 5.7].

In each of these cases, subalgebras are in the same orbit under
$\operatorname{Aut}_{F}(\mathbb{O})$ precisely if they are isomorphic as
$F$-algebras (and contain $1$, as assumed here).

Each subalgebra $A$ with $1\in A$ and $\dim{A}\leq 4$ is associative.

###### Proof.

If $\dim{A}=2$, pick $a\in A\smallsetminus F$. Then $1,a$ is a basis for $A$,
and the minimal polynomial of $a$ is
$m_{a}(X)=X^{2}-{\mathrm{T}}(a)X+{\mathrm{N}}(a)$. The four cases in assertion
(b) now belong to the cases where $m_{a}(X)$ is irreducible and separable
(viz., $\operatorname{char}{F}\neq 2$ or ${\mathrm{T}}(a)\neq 0$), or
$m_{a}(X)$ is irreducible and inseparable (viz., $\operatorname{char}{F}=2$
and ${\mathrm{T}}(a)=0$), or $m_{a}(X)$ is reducible with two different roots
in $F$, or $m_{a}(X)$ has a double root in $F$, respectively. Two such
algebras $A$ and $A^{\prime}$ are in the same orbit under
$\operatorname{Aut}_{F}(\mathbb{O})$ if, and only if, there are elements $a\in
A\smallsetminus F$ and $a^{\prime}\in A^{\prime}\smallsetminus F$ such that
$a$ and $a^{\prime}$ have the same norm and trace (see 4.4). In that case, the
algebras $A$ and $A^{\prime}$ are both isomorphic to $F[X]/(m_{a}(X))$.

If $\dim{A}=3$ then $A$ is not a composition algebra. Therefore, the
restriction of the polar form is degenerate, and the radical $R:=A^{\perp}\cap
A$ is not trivial. From 6.1(a) we know $AR=R=RA$.

Assume that $\dim{R}=1$, and choose a vector space complement $B$ for $R$ in
$A$ such that $1\in B$. Then $B$ is a subalgebra, and the restriction of the
polar form to $B$ is not degenerate. So $B$ is a composition algebra, and $R$
is a $B$-module. As $\dim{R}=1$, the annihilator $\left\\{{a\in
B}\left|\vphantom{a\in BaR=\\{0\\}}\right.\,{aR=\\{0\\}}\right\\}$ is not
trivial. Therefore, the algebra $B$ is not a field but a split composition
algebra, and $R$ does not contain invertible elements (see also 6.1). Then
$B\cong F\times F$, and each non-trivial element of $1^{\perp}\cap B$ is
invertible. Choose $a\neq 0$ in the annihilator of $R$ in $B$. Then
${\mathrm{N}}(a)=0\neq{\mathrm{T}}(a)$, and $M:=Fa+R$ is a totally singular
subalgebra containing an idempotent. From 5.11 we now infer that $M$ is in the
orbit of either $Fp_{0}+Fn_{0}$ or $\kappa(Fp_{0}+Fn_{0})$. In any case, the
algebra $A=F+M$ is in the orbit of the algebra
$F+Fp_{0}+Fn_{0}=F+\kappa(Fp_{0}+Fn_{0})$ of upper triangular matrices, as
claimed.

Now assume $\dim{R}\geq 2$ (and still $\dim{A}=3$). Then the radical $Q$ of
${\mathrm{N}}|_{A}$ has dimension $2$ by 6.3, and forms a totally singular
subalgebra, see 6.1. From 5.11 we conclude that $Q$ is in the orbit of
$(Fp_{0}+Fn_{0})w$. So $A=F+Q$ is in the orbit of $F+(Fp_{0}+Fn_{0})w$, as
claimed.

Finally, consider the case where $\dim{A}=4$. From 6.1 we know that $A/R$ is a
composition algebra if $R\neq A$. So $\dim{R}=4-\dim(A/R)\in\\{0,2,3,4\\}$. If
$\dim{R}=0$ then $A$ is a composition algebra, and then a quaternion algebra.

If $\dim{R}=2$ then the totally singular subalgebra $R$ is in the orbit of
$(Fp_{0}+Fn_{0})w$, see 5.11, and we assume $R=(Fp_{0}+Fn_{0})w$ without loss
of generality. We have $A\subseteq\\{p_{0}w,n_{0}w\\}^{\perp}=\mathbb{H}+R$,
and $B:=A\cap\mathbb{H}$ has dimension $2$. As $A$ contains invertible
elements, we have $1\in B$. So $B$ is a subalgebra. From $B\cap R=\\{0\\}$ we
infer that the restriction of the polar form to $B$ is not degenerate. So $B$
is a composition algebra (either a separable quadratic extension field of $F$,
or isomorphic to $F\times F$), and $A=B+(Fp_{0}+Fn_{0})w$.

Consider a four-dimensional subalgebra $A^{\prime}$ with $R\subseteq
A^{\prime}\subseteq\mathbb{H}+R$. Then $A$ and $A^{\prime}$ are isomorphic as
$F$-algebras if, and only, the subalgebras $B$ and
$B^{\prime}:=A^{\prime}\cap\mathbb{H}$ are isomorphic. In that case, we find
$b\in B\smallsetminus F$ and $b^{\prime}\in B^{\prime}\smallsetminus F$ with
the same norm (i.e., the same determinant) and the same trace. So there exists
$s\in\mathrm{GL}_{2}{(F)}=\mathbb{H}^{\times}$ with $sbs^{-1}=b^{\prime}$.
Pick an upper triangular matrix $t\in\mathbb{H}$ (${}=F^{2\times 2}$) with
$\det t=\det s$. The map $\alpha_{s,t}\colon a+xw\mapsto sas^{-1}+(txs^{-1})w$
is an $F$-linear automorphism of $\mathbb{O}$, see 4.2. Now
$\alpha_{s,t}(b)=b^{\prime}$, and it is easy to verify $\alpha_{s,t}(R)=R$. So
$\alpha_{s,t}(A)=A^{\prime}$. This settles those cases in assertion (d)(ii)
where $\dim{R}=2$. The remaining cases that belong to assertion (d)(ii) will
be discussed below; those are cases where $R=A$ but $\dim{Q}=2$.

Now assume $\dim{R}\geq 3$, and consider the radical $Q:=\left\\{{x\in
A^{\perp}\cap A}\left|\vphantom{x\in A^{\perp}\cap
A{\mathrm{N}}(x)=0}\right.\,{{\mathrm{N}}(x)=0}\right\\}$ of the restriction
${\mathrm{N}}|_{A}$ of the quadratic form. From 6.1 we know that $Q$ is a
subalgebra. As $1\notin Q$, we have $\dim{Q}\leq 3$.

If $\dim{Q}=3$ then the totally singular subalgebra $Q$ is in the orbit of the
Heisenberg algebra $n_{0}\mathbb{O}\cap\mathbb{O}n_{0}$ (see 5.11), and
assertion (d)(iii) follows.

If $\dim{Q}<3$ then $Q\neq R$, and $\operatorname{char}{F}=2$. In particular,
the polar form is alternating, and $R=A$ because the co-dimension of $R$ in
$A$ is even. From 6.1 we know that $A$ is commutative.

If $\dim{Q}=0$, we pick $b\in A\smallsetminus F$, and note that $B:=F+Fb$ is
an inseparable quadratic field extension of $F$. For any $c\in A\smallsetminus
B$, we then have $B\cap Bc=\\{0\\}$, and $A=B+Bc$ shows that the algebra $A$
is generated by $\\{1,b,c\\}$. So $A$ is associative, commutative, and thus a
totally inseparable field extension of $F$ (of degree $4$ and exponent $1$),
as claimed in (d)(iv).

If $\dim{Q}\neq 0$ then there are $u,v\in A\smallsetminus Q$ such that
$F+Fu+Fv$ contains a vector space complement of $Q$ in $A$. Then the
subalgebra spanned by $\\{1,x,y\\}$ is associative, and the quotient $A/Q$ is
a totally inseparable field extension of $F$. In particular, the degree of
that extension is a power of $2$, and we are left with the case that
$\dim{Q}=2$. Now 5.11 yields that $Q$ is in the orbit of $(Fp_{0}+Fn_{0})w$,
and we may assume $Q=(Fp_{0}+Fn_{0})w$. Then $A\subseteq\mathbb{H}+Q$, and
$B:=A\cap\mathbb{H}$ is a two-dimensional subalgebra, again (albeit not a
composition algebra, but an inseparable quadratic extension field).

If $A^{\prime}$ is another such algebra (contained in $\mathbb{H}+Q$), we note
that $A$ is isomorphic to $A^{\prime}$ precisely if $A/Q$ is isomorphic to
$A^{\prime}/Q$. So $A\cong A^{\prime}$ means $B\cong
B^{\prime}:=A^{\prime}\cap\mathbb{H}$. As above, any algebra isomorphism
between $B$ and $B^{\prime}$ extends to an algebra automorphism $a\mapsto
sas^{-1}$ of $\mathbb{H}$, any such automorphism extends to an automorphism
$\alpha_{s,t}$ of $\mathbb{O}$ such that $\alpha_{s,t}(Q)=Q$, and
$\alpha_{s,t}(A)=A^{\prime}$ follows. We have thus also completed the proof of
assertion (d)(ii).

It remains to verify that each subalgebra $A$ with $1\in A$ and $\dim{A}\leq
4$ is associative. For the cases in (a) and in (b), this is obvious. For the
cases in (c), we note that $A$ is of the form $A=F+X$, where $X$ is a
subalgebra generated by two elements. So $X$ is associative by Artin’s theorem
(see 1.4(e)), and $A$ is associative, as well.

In (d), the quaternion algebras and the field extensions are associative. For
the algebra $F+(n_{0}\mathbb{O}\cap\mathbb{O}n_{0})$ we use the fact that the
subalgebra $n_{0}\mathbb{O}\cap\mathbb{O}n_{0}$ is associative (it is a
Heisenberg algebra, generated by $n_{0}w$ and
${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}w$, see
3.7).

In the last remaining case, we have to consider $A=B+X$, where
$B=F+Fb\subseteq\mathbb{H}$ is a two-dimensional (associative, and
commutative) subalgebra of $\mathbb{H}=F^{2\times 2}$, and
$X=(Fp_{0}+Fn_{0})w$. Since $({\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{yw}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{yw}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{yw}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{yw}\hfil$\crcr}}}})(xw)=0$
holds for all $x,y\in X$, the multiplication in $A$ is given by
$(a+xw)(b+yw)=ab+(ya+x{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{b}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{b}\hfil$\crcr}}}})w$, where
$a,b\in B$ and $x,y\in Fp_{0}+Fn_{0}$. Using the associative and commutative
laws in $B$, we find that $A$ is associative. ∎

######  6.5 Remark.

The examples in 6.4(d)(ii) show that not every associative subalgebra of a
split octonion algebra is contained in a quaternion algebra.

##  7\. Associative and commutative subalgebras

######  7.1 Associativity of subalgebras.

Let $\mathbb{O}$ be an octonion algebra.

  1. ((a))

Every subalgebra of dimension less than $4$ is associative.

  2. ((b))

If $\mathbb{O}$ is split then there do exist subalgebras of dimension $4$ that
are not associative, namely, the totally singular ones (of the form
$n\mathbb{O}$ or $\mathbb{O}n$ with ${n^{2}=0\neq n\in\mathbb{O}}$). Every
other subalgebra of dimension $4$ is associative.

  3. ((c))

Every subalgebra of dimension greater than $4$ contains a non-associative
subalgebra of dimension $4$.

  4. ((d))

If $\mathbb{O}$ is split then a given subalgebra is not associative if, and
only if, it contains a maximal totally singular subspace.

  5. ((e))

If $\mathbb{O}$ is not split then every proper subalgebra is associative.

###### Proof.

If a subalgebra is not totally singular then it contains $1$. We have seen in
6.4 that each subalgebra $A$ with $1\in A$ and $\dim{A}\leq 4$ is associative.
A totally singular subalgebra is associative if, and only if, it has dimension
less than $4$; see 5.11. By 5.11.(d), the totally singular subalgebras of
dimension $4$ are those of the form $n\mathbb{O}$ and $\mathbb{O}n$,
repsectively, with $n^{2}=0\neq n$. So assertions (a) and (b) are established.

Each proper subalgebra of dimension greater than $4$ contains a totally
singular subalgebra of dimension $4$ (of the form $n\mathbb{O}$ for some
nilpotent element $n$, see 6.2). These subalgebras are not associative. This
gives assertions (c) and (d).

If $\mathbb{O}$ is not split then every proper subalgebra has dimension at
most $4$, and there are no nilpotent elements apart from $0$. So assertions
(a) and (b) yield assertion (e). ∎

######  7.2 Corollary.

A $4$-dimensional subalgebra of $\mathbb{O}$ is associative if, and only if,
it contains a neutral element for its multiplication. By 5.5, such a neutral
element in a $4$-dimensional subalgebra necessarily equals $1$. ∎

######  7.3 Commutativity of subalgebras.

Let $\mathbb{O}$ be an octonion algebra, and let $A$ be a subalgebra. If
$\mathbb{O}$ is split, we write $\mathbb{O}=\mathbb{H}+\mathbb{H}w$, with
$\mathbb{H}=F^{2\times 2}$ and $w^{2}=1$. The subalgebra $A$ is commutative in
the following cases:

  1. ((a))

$\dim{A}=1$.

  2. ((b))

$\dim{A}=2$, and $A=F+Fa\cong F[X]/(X^{2}-{\mathrm{T}}(a)X+{\mathrm{N}}(a))$.

  3. ((c))

$\dim{A}=2$, and $A$ is in the $\operatorname{Aut}_{F}(\mathbb{O})$-orbit of
$(Fp_{0}+Fn_{0})w$.

  4. ((d))

$\dim{A}=3$, and $A$ is in the $\operatorname{Aut}_{F}(\mathbb{O})$-orbit of
$F+(Fp_{0}+Fn_{0})w$.

If $\operatorname{char}{F}\neq 2$ then these are the only commutative
subalgebras.
If $\operatorname{char}{F}=2$ we have the following additional cases (and no
others):

  1. ((e))

$\dim{A}=3$, and $A$ is in the $\operatorname{Aut}_{F}(\mathbb{O})$-orbit of
the Heisenberg algebra ${Fn_{0}+(Fp_{0}+Fn_{0})w}$.

  2. ((f))

$\dim{A}=4$, and $A$ is in the $\operatorname{Aut}_{F}(\mathbb{O})$-orbit of
$B+(Fp_{0}+Fn_{0})w$, where $B=F+Fb$ with
$b\in(1^{\perp}\cap\mathbb{H})\smallsetminus F$.

  3. ((g))

$\dim{A}=4$, and $A$ is a totally inseparable field extension of exponent $1$.

In any case, every commutative subalgebra of $\mathbb{O}$ is associative, as
well.

###### Proof.

We use our classification of subalgebras, see 5.11, 6.4, and 6.2. We note that
$Fp_{0}+Fn_{0}$ is not commutative, in fact $p_{0}n_{0}=n_{0}\neq
0=n_{0}p_{0}$. So we can exclude from the list of subalgebras each algebra
that contains a subalgebra isomorphic to $Fp_{0}+Fn_{0}$, or isomorphic to
$\kappa(Fp_{0}+Fn_{0})=F{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}+Fn_{0}$.
In particular, we note that $n_{0}\mathbb{O}$ is excluded by the observation
$p_{0}=\left(\begin{smallmatrix}1&0\\\
0&0\end{smallmatrix}\right)=\left(\begin{smallmatrix}0&1\\\
0&0\end{smallmatrix}\right)\left(\begin{smallmatrix}0&0\\\
1&0\end{smallmatrix}\right)\in n_{0}\mathbb{H}\subseteq n_{0}\mathbb{O}$. This
leaves us with the following list:

  * •

Subalgebras of dimension $1$: these are clearly commutative.

  * •

Subalgebras in the orbit of $Q:=(Fp_{0}+Fn_{0})w$: here the multiplication is
trivial, and thus commutative.

  * •

Subalgebras in the orbit of $F+Q$: these algebras are commutative because $Q$
is.

  * •

Heisenberg algebras (in the orbit of
$Fn_{0}+Q=(n_{0}w)\mathbb{O}\cap\mathbb{O}(n_{0}w)$, see 2.7(b)); the only
nontrivial products of elements in $\\{n_{0},p_{0}w,n_{0}w\\}$ are
$n_{0}(p_{0}w)=n_{0}w$ and $(p_{0}w)n_{0}=-n_{0}w$. So the Heisenberg algebras
are commutative if, and only if, the field $F$ has characteristic two.

  * •

Subalgebras in the orbit of $F+Fn_{0}+Q$: such an algebra is commutative if,
and only if, the Heisenberg algebra $Fn_{0}+Q$ is commutative, and we already
know that this happens precisely if $\operatorname{char}{F}=2$.

  * •

Totally inseparable extensions fields of degree $4$ (only possible if
$\operatorname{char}{F}=2$) are commutative.

  * •

Subalgebras in the orbit of $B+Q$, where $B\subseteq\mathbb{H}$ is either a
two-dimensional composition algebra, or an inseparable quadratic field
extension.

For $b\in B$ and $x\in Fp_{0}+Fn_{0}$ we compute $b(xw)=(xb)w$ and
$(xw)b=(x{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{b}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{b}\hfil$\crcr}}}})w$. If
$B\subseteq 1^{\perp}$ then ${\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{b}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{b}\hfil$\crcr}}}}=b$ holds
for each $b\in B$, and $B+Q$ is commutative. If $B\not\subseteq 1^{\perp}$
then $B$ is a composition algebra. So either $B\cong F\times F$, or $B$ is a
separable quadratic extension field. In both cases, we find $b\in B$ such
$b-{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{b}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{b}\hfil$\crcr}}}}$ is
invertible, and $b(xw)=(xw)b$ implies $x=0$. So the algebra $B+Q$ is not
commutative in these cases.

Our discussion yields that the commutative subalgebras of $\mathbb{O}$ are
exactly those given in the statement of the theorem. From 7.1 we infer that
each one of these commutative subalgebras is in fact associative. ∎

##  8\. Maximal subalgebras

######  8.1 The lattice of isomorphism types of subalgebras.

Let $p,n,m$ be elements of a (necessarily split) octonion algebra $\mathbb{O}$
with $p^{2}=p\notin\\{0,1\\}$, $n^{2}=0=m^{2}$, and such that $m,n$ are
linearly independent. Moreover, assume that $pn=n$ and $np=0=nm$ (then
$mn=0$). We abbreviate $S:=F+Fp\cong F\times F$, and $Q:=Fm+Fn$. Note that $S$
is a split two-dimensional composition algebra, while the multiplication on
$Q$ is trivial. Note also that $T:={F+Fp+Fn}$ is isomorphic to the algebra of
upper triangular matrices in $F^{2\times 2}$, and that
$n\mathbb{O}\cap\mathbb{O}n$ is a Heisenberg algebra (see 2.7).

Let $X$ be any subalgebra of dimension $d$ in $\mathbb{O}$. Our classification
of subalgebras (see 5.11, 6.2, and 6.4) can be rephrased, as follows.

$d=1$::

Then $X$ is in the orbit of $Y\in\\{F,Fp,Fn\\}$ under
$\operatorname{Aut}_{F}(\mathbb{O})$.

$d=2$::

Either $X$ is in the $\operatorname{Aut}_{F}(\mathbb{O})$-orbit of
$Y\in\\{S,{F+Fn},{Fn+Fp},{Fn+F{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p}\hfil$\crcr}}}}},Q\\}\,,$
or $X$ is a separable extension field $E$ over $F$, or
$\operatorname{char}{F}=2$ and $X$ is an inseparable extension field $D$ of
$F$. Moreover, any subalgebra isomorphic to $E$ or $D$ is in the
$\operatorname{Aut}_{F}(\mathbb{O})$-orbit of $E$ or $D$, respectively.

$d=3$::

Then $X$ is in the $\operatorname{Aut}_{F}(\mathbb{O})$-orbit of
$Y\in\\{T,F+Q,{m\mathbb{O}\cap\mathbb{O}n},{n\mathbb{O}\cap\mathbb{O}n}\\}$.

$d=4$::

Then $X$ is in the orbit of $Y$ such that either
$Y\in\\{{F+(n\mathbb{O}\cap\mathbb{O}n)},{n\mathbb{O}},{\mathbb{O}n}\\}$, or
one of the following holds:

  * •:

$Y\cong F^{2\times 2}$ is a split quaternion algebra,

  * •:

there is a subalgebra $S\cong F\times F$ contained in $Q^{\perp}$ such that
$Y={S+Q}$,

  * •:

$Y$ is a quaternion field (denoted by $H$ in Figure 1),

  * •:

there exists a separable extension field $E$ of degree $2$ contained in
$Q^{\perp}$ such that $Y=E+Q$,

  * •:

$\operatorname{char}{F}=2$ and there exists an inseparable extension field $D$
of degree $2$ contained in $Q^{\perp}$ such that $Y=D+Q$,

  * •:

$\operatorname{char}{F}=2$ and $Y$ is a totally inseparable extension field
(denoted by $K$ in Figure 1) of degree $4$ and exponent $1$.

$d=5$::

Then $X$ is in the $\operatorname{Aut}_{F}(\mathbb{O})$-orbit of
$n\mathbb{O}+\mathbb{O}n$.

$d=6$::

Then $X$ is in the $\operatorname{Aut}_{F}(\mathbb{O})$-orbit of
$Q^{\perp}=m\mathbb{O}+n\mathbb{O}=\mathbb{O}m+\mathbb{O}n$.

The algebras $F$ and $Fp$ are isomorphic but not in the same orbit under
$\operatorname{Aut}_{F}(\mathbb{O})$. In any other case, two subalgebras of
$\mathbb{O}$ are in the same orbit under $\operatorname{Aut}_{F}(\mathbb{O})$
if, and only if, the subalgebras are isomorphic (as abstract $F$-algebras).

Consider subalgebras $X\neq F$ and $Y^{*}$ with $\dim{Y^{*}}>1$. Since every
subalgebra isomorphic to $Y^{*}$ is in the
$\operatorname{Aut}_{F}(\mathbb{O})$-orbit of $Y^{*}$, there exists a
subalgebra $X^{*}$ isomorphic to $Y^{*}$ with $X\subseteq X^{*}$ if, and only
if, the algebra $Y^{*}$ has a subalgebra $Y$ isomorphic to $X$. We can thus
describe inclusions between subalgebras by a graph with the
$\operatorname{Aut}_{F}(\mathbb{O})$-orbits of subalgebras as vertices, and
edges indicating embeddings. In Figure 1 we have further simplified that graph
by identifying orbits of subalgebras that contain extension fields;
corresponding vertices are indicated by doubled boundaries at their labels
(namely, $E$, $H$, $E+Q$, $D$, $K$, $D+Q$). Recall that the existence of such
extensions heavily depends on the structure of $F$ (e.g., no such extension
exists if $F$ is algebraically closed); the corresponding embeddings
(indicated by dotted lines in Figure 1) exist only if the subalgebras in
question exist. The inseparable extensions only exist if
$\operatorname{char}{F}=2$, this is indicated by dashed boundaries (and dashed
edges indicating embeddings).

$F\vphantom{p}$$Fp$$Fn\vphantom{p}$$Fn+Fp$$Q\vphantom{p}$$Fn+F{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p}\hfil$\crcr}}}}$$m\mathbb{O}\cap\mathbb{O}n\vphantom{p}$$n\mathbb{O}\cap\mathbb{O}n\vphantom{p}$
$n\mathbb{O}\cap\mathbb{O}n\vphantom{p}$$n\mathbb{O}\vphantom{p}$$\mathbb{O}n\vphantom{p}$$n\mathbb{O}+\mathbb{O}n\vphantom{p}$$\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}Q^{\perp}$$D\vphantom{p}$$E\vphantom{p}$$\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}K\vphantom{p}$$\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}H\vphantom{p}$$S\vphantom{p}$$F+Fn\vphantom{p}$$T\vphantom{p}$$F+Q\vphantom{p}$$F^{2\times
2}$$\,F+nO\cap On\,$ $F+(n\mathbb{O}\cap\mathbb{O}n)\vphantom{p}$
$D+Q\vphantom{p}$ $E+Q\vphantom{p}$ $S+Q\vphantom{p}$ Figure 1. The lattice of
orbits of subalgebras, see 8.1.

In Figure 1, the maximal element $\mathbb{O}$ is omitted. The maximal proper
subalgebras are in the orbits with black labels, i.e., those represented by
non-split quaternion subalgebras, totally inseparable extensions of degree $4$
and exponent $1$, and $Q^{\perp}$. Grey labels indicate subalgebras that do
not contain $1$ (i.e., totally singular ones), and rectangular labels (as
opposed to elliptical ones) indicate associative algebras. Rectangular labels
with rounded corners indicate commutative algebras (all of them are
associative). Double lining is used for subalgebras whose existence depends on
the ground field, and dashed lines indicate subalgebras that only exist if
$\operatorname{char}{F}=2$. Finally, the algebras $n\mathbb{O}\cap\mathbb{O}n$
and $F+(n\mathbb{O}\cap\mathbb{O}n)$ are commutative precisely if
$\operatorname{char}{F}=2$.

Figures 2 and 3 give the simpler graphs for the cases where
$\operatorname{char}{F}\neq 2$, and where $\operatorname{char}{F}\neq 2$ and
$F$ allows no quadratic field extension. If $F$ is finite then the vertices
labeled by “$K$” or by “$H$” have to be omitted from the graph in Figure 1.

######  8.2 Remarks.

Let $\mathbb{O}$ be the split octonion algebra over $F$; we know that
$\mathbb{O}$ is unique up to isomorphism.

If there does exist a quadratic extension field (separable or not) over $F$
then the split quaternion algebra $F^{2\times 2}$ over $F$ contains
subalgebras isomorphic to that extension field. Doubling $F^{2\times 2}$ we
obtain a split octonion algebra, and thus see that $\mathbb{O}$ contains
subalgebras isomorphic to any quadratic extension field of $F$.

If there exists a quaternion field $H$ over $F$ then a suitable double of $H$
is a split octonion algebra, and we obtain that $\mathbb{O}$ contains
subalgebras isomorphic to $H$. However, using the Kaplansky radical (cp. [14,
Ch. XII, Prop. 6.1, pp. 450 f]) one sees that there do exist fields $F$ with
quadratic extensions of $F$ that are not contained in any quaternion field
over $F$ — although there do exist quaternion fields over $F$ (see the
examples in [14, p. 461] or [2, 2.4]). If $F$ is such a field, then the split
octonion algebra over $F$ contains quaternion fields and also quadratic
extensions that are not contained in any quaternion field, so the inclusion
between $E$ and $H$ in Figure 1 has to be interpreted with due care.

Finally, assume that there exists a totally inseparable extension field $K$ of
degree $4$ and exponent $1$ over $F$. Then [7, 4.3, 4.6] shows that there is a
split octonion algebra containing a subalgebra isomorphic to $K$.

$F\vphantom{p}$$Fp$$Fn\vphantom{p}$$Fn+Fp$$Q\vphantom{p}$$Fn+F{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p}\hfil$\crcr}}}}$$m\mathbb{O}\cap\mathbb{O}n\vphantom{p}$$n\mathbb{O}\cap\mathbb{O}n\vphantom{p}$$n\mathbb{O}\cap\mathbb{O}n\vphantom{p}$$n\mathbb{O}\vphantom{p}$$\mathbb{O}n\vphantom{p}$$n\mathbb{O}+\mathbb{O}n\vphantom{p}$$\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}Q^{\perp}$$E\vphantom{p}$$\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}H\vphantom{p}$$S\vphantom{p}$$F+Fn\vphantom{p}$$T\vphantom{p}$$F+Q\vphantom{p}$$F^{2\times
2}$$F+(n\mathbb{O}\cap\mathbb{O}n)\vphantom{p}$ $E+Q\vphantom{p}$
$S+Q\vphantom{p}$ Figure 2. The lattice of orbits of subalgebras for the case
where $\operatorname{char}{F}\neq 2$, see 8.1.
$F\vphantom{p}$$Fp$$Fn\vphantom{p}$$Fn+Fp$$Q\vphantom{p}$$Fn+F{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p}\hfil$\crcr}}}}$$m\mathbb{O}\cap\mathbb{O}n\vphantom{p}$$n\mathbb{O}\cap\mathbb{O}n\vphantom{p}$$n\mathbb{O}\cap\mathbb{O}n\vphantom{p}$$n\mathbb{O}\vphantom{p}$$\mathbb{O}n\vphantom{p}$$n\mathbb{O}+\mathbb{O}n\vphantom{p}$$\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}Q^{\perp}$$S\vphantom{p}$$F+Fn\vphantom{p}$$T\vphantom{p}$$F+Q\vphantom{p}$$F^{2\times
2}$$F+(n\mathbb{O}\cap\mathbb{O}n)\vphantom{p}$ $S+Q\vphantom{p}$ Figure 3.
The lattice of orbits of subalgebras for the case where $F$ is quadratically
closed and $\operatorname{char}{F}\neq 2$, see 8.1.

######  8.3 Open Problems.

For each subalgebra $X$ in a given octonion algebra $\mathbb{O}$ over $F$:

  1. ((a))

Determine the full group $\operatorname{Aut}(X)$ of all $\mathbb{Z}$-linear
automorphisms of $X$, and the group $\operatorname{Aut}_{F}(X)$ of all algebra
automorphisms of $X$.

  2. ((b))

Determine the stabilizers $\operatorname{Aut}(\mathbb{O})_{X}$ and
$\operatorname{Aut}_{F}(\mathbb{O})_{X}$ of $X$ in
$\operatorname{Aut}(\mathbb{O})$ and in $\operatorname{Aut}_{F}(\mathbb{O})$,
repsectively.

  3. ((c))

Determine the subgroups of $\operatorname{Aut}(X)$ induced by
$\operatorname{Aut}(\mathbb{O})_{X}$ and by
$\operatorname{Aut}_{F}(\mathbb{O})_{X}$, respectively.

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Norbert Knarr,

Markus J. Stroppel

LExMath

Fakultät 8

Universität Stuttgart

70550 Stuttgart

<EMAIL_ADDRESS>


MnLargeSymbols’164 MnLargeSymbols’171

# flag transitive geometries with trialities and no dualities coming from
Suzuki Groups

Dimitri Leemans  Dimitri Leemans, Université Libre de Bruxelles, Département
de Mathématique, C.P.216 - Algèbre et Combinatoire, Boulevard du Triomphe,
1050 Brussels, Belgium, Orcid number 0000-0002-4439-502X.
<EMAIL_ADDRESS>,  Klara Stokes  Klara Stokes, Department of
Mathematics and Mathematical Statistics, Umeå University, 901 87 Umeå, Sweden,
Orcid number 0000-0002-5040-2089<EMAIL_ADDRESS>and  Philippe Tranchida
Philippe Tranchida, Université Libre de Bruxelles, Département de
Mathématique, C.P.216 - Algèbre et Combinatoire, Boulevard du Triomphe, 1050
Brussels, Belgium, Orcid number 0000-0003-0744-4934.
<EMAIL_ADDRESS>
###### Abstract.

Recently, Leemans and Stokes constructed an infinite family of incidence
geometries admitting trialities but no dualities from the groups
$\operatorname{PSL}(2,q)$ (where $q=p^{3n}$ with $p$ a prime and $n>0$ a
positive integer). Unfortunately these geometries are not flag transitive. In
this paper, we construct the first infinite family of incidence geometries of
rank three that are flag transitive and have trialities but no dualities.
These geometries are constructed using chamber systems of Suzuki groups
$\operatorname{Sz}(q)$ (where $q=2^{2e+1}$ with $e$ a positive integer and
$2e+1$ is divisible by 3) and the trialities come from field automorphisms. We
also construct an infinite family of regular hypermaps with automorphism group
$\operatorname{Sz}(q)$ that admit trialities but no dualities.

###### Key words and phrases:

Incidence geometry, triality, Suzuki groups

###### 1991 Mathematics Subject Classification:

51A10, 51E24, 20C33

This research was made possible thanks to an Action de Recherche Concertée
grant from the Communauté Française Wallonie-Bruxelles.

##  1\. Introduction

In [9], Leemans and Stokes showed how to construct coset geometries having
trialities by using a group $G$ that has outer automorphisms of order three.
They gave examples for the smallest simple Suzuki group
$\operatorname{Sz}(8)$, some small $\operatorname{PSL}(2,q)$ groups (with
$q\in\\{8,27,64,125,512\\}$), and the Mathieu group $M_{11}$. They used this
technique in [10] to construct the first infinite family of coset geometries
admitting trialities but no dualities. These geometries are of rank four, and
constructed using regular maps of Wilson class III for the groups
$\operatorname{PSL}(2,q)$, where $q=p^{3n}$ for a prime $p$ and a positive
integer $n>0$. These geometries are not flag transitive. The group
$\operatorname{PSL}(2,q)$ has two orbits on their chambers.

The geometries for $\operatorname{Sz}(8)$ mentioned in [9] are known to be
flag transitive thanks to [8] where Leemans classified all thin flag
transitive geometries having a group $\operatorname{Sz}(8)$ as type-preserving
automorphism group. Among the 183 thin geometries found for
$\operatorname{Sz}(8)$, four of them admit trialities but no dualities.
Nevertheless, in [9], Leemans and Stokes point out that they do not know if,
in the general case, the regular hypermaps they construct are flag transitive
coset geometries. The goal of this article is to prove that at least some of
these hypermaps are indeed flag transitive coset geometries, therefore
providing the first infinite family of flag transitive geometries having
trialities but no dualities.

Using chamber systems, we give a geometric description of rank three flag
transitive coset geometries $\Gamma(G,(G_{i})_{i\in\\{0,1,2\\}})$ admitting
trialities but no dualities. Here $G=\operatorname{Sz}(q)$ is the Suzuki group
over the finite field of order $q$ (where $q=2^{2e+1}$ for some positive
integer $e$ such that $2e+1$ is divisible by 3) and the $G_{i}$’s are suitably
chosen dihedral subgroups of $G$. These geometries all have the following
diagram.

$\;\;\;\;\;5\;\;\;\;\;$$5$$5\ $

Most of the work in this article generalizes to constructions of similar
geometries with diagrams with labels any integer $m$ dividing one of $q-1$,
$q+\sqrt{2q}+1$ or $q-\sqrt{2q}+1$ instead of 5. At the moment, we can only
prove that these geometries are flag transitive for $m=5$. For any integer $m$
dividing one of $q-1$, $q+\sqrt{2q}+1$ or $q-\sqrt{2q}+1$ and such that
$m\equiv 2(\bmod\;3)$, we nevertheless show that they are regular hypermaps,
see Corollary 3.7. These hypermaps also admit trialities but no dualities, in
the sense of operations on hypermaps (see [6]).

Hypermaps in the context of Suzuki groups were already investigated in [3],
where the authors give a formula to count the number of regular and orientably
regular maps and hypermaps with automorphism group $\operatorname{Sz}(q)$.

The paper is organised as follows. In Section 2, we recall the basic
properties of the Suzuki groups and the notions in incidence geometry needed
to understand this paper. We also recall the definition of an hypermap. In
Section 3, we construct the geometries mentioned above and show that they are
residually connected, flag transitive, and that they admit trialities and no
dualities. We conclude the paper with a series of remarks in Section 4.

##  2\. Preliminaries

We start by an introduction to the Suzuki groups $\operatorname{Sz}(q)$ and
some of their properties. We then define incidence geometries, chamber systems
and hypermaps.

###  2.1. Suzuki groups

Here we introduce the Suzuki groups, one of the 18 infinite families of finite
simple groups. We mainly follow the exposition in [11], where many more
details can be found.

Let $\operatorname{\mathbb{F}}=\operatorname{\mathbb{F}}(q)$ be the field of
order $q=2^{2e+1}$ for some integer $e\geq 1$. Let
$\Phi\colon\operatorname{\mathbb{F}}\to\operatorname{\mathbb{F}}$ be the
Frobenius automorphism sending $x$ to $x^{2}$ and let $\theta$ be the
automorphism $x\to x^{r}$ with $r=2^{e+1}$, so that $\theta^{2}=\Phi$. Let
${\sf P}=\operatorname{\mathbf{P}\mathbf{G}}(3,\operatorname{\mathbb{F}})$ be
the $3$-dimensional projective space over $\operatorname{\mathbb{F}}$ and let
$(X_{0}:X_{1}:X_{2}:X_{3})$ be homogeneous coordinates for $\sf P$.

Let $E$ be the plane of $\sf P$ defined by the equation $X_{0}=0$ and set
$U=(0:1:0:0)$. Let

$x=X_{2}X_{0}^{-1}\quad\quad\quad\quad y=X_{3}X_{0}^{-1}\quad\quad\quad\quad
z=X_{1}X_{0}^{-1}$

be affine coordinates on the affine space ${\sf P}_{E}=\sf P$ \$E$. Finally,
define $\operatorname{\mathcal{O}}$ to be the set of points of $\sf P$
consisting of $U$ together with all the points of ${\sf P}_{E}$ that satisfy
the equation

$z=xy+x^{\theta+2}+y^{\theta}$

This set $\operatorname{\mathcal{O}}$ is called a Suzuki-Tits ovoid. The
Suzuki group $G:=\operatorname{Sz}(q)$ is then the group of projectivities of
$\sf P$ that preserve $\operatorname{\mathcal{O}}$.

The definition of ovoid is due to Tits [14]. Let $\operatorname{\mathcal{O}}$
be a set of points of some projective space $\Sigma$. Then
$\operatorname{\mathcal{O}}$ is called an ovoid if every line of $\Sigma$
intersects $\operatorname{\mathcal{O}}$ in at most two points, and if, for
each point $p$, the union of the set of lines of $\Sigma$ intersecting
$\operatorname{\mathcal{O}}$ in precisely $p$ is a hyperplane. It was proven
by Tits that the set $\operatorname{\mathcal{O}}$ we defined here is indeed an
ovoid under this definition.

We will mostly understand the Suzuki group $G$ by its action on
$\operatorname{\mathcal{O}}$. We very briefly state some of the properties of
this action. The group $G$ acts doubly transitively on
$\operatorname{\mathcal{O}}$ and only the identity element fixes $3$ points of
$\operatorname{\mathcal{O}}$. Let $S$ be a Sylow $2$-subgroup of $G$. Then $S$
is contained in its normalizer $N_{G}(S)$, a Frobenius group which is the
stabilizer $G_{P}$ of some point $P$ of $\operatorname{\mathcal{O}}$ and has
cardinality $q^{2}(q-1)$. The center $Z(S)$ of $S$ is $Z(S)=\\{\gamma\in
S\mid\gamma^{2}=1\\}$ and is of cardinality $q$. All elements of even order of
$G$ are conjugate to some element in $S$, and therefore have a unique fixed
point. The stabilizer $G_{P,Q}$ of two points $P$ and $Q$ of
$\operatorname{\mathcal{O}}$ is a cyclic group of order $q-1$, which is a
Frobenius complement of $G_{P}$.

The cardinality of $G$ is $(q^{2}+1)q^{2}(q-1)$, which can be deduced from the
double transitivity together with $|G_{P,Q}|=q-1$.

The Frobenius automorphism $\Phi$ also acts on $\operatorname{\mathcal{O}}$.
Conjugating elements of $G$ by the action of $\Phi$ gives outer automorphisms
of $G$. In fact, $\operatorname{Out}(G)$ is cyclic of order $2e+1$ and is
generated by the action of $\Phi$.

The structure of the maximal subgroups of $G$ is very well understood. For two
groups $H$ and $K$, the notation $H:K$ denotes a semi-direct product of $H$ by
$K$. Here are the facts we will need about maximal subgroups, the proofs can
be found in [12].

######  Theorem 2.1.

Let $\alpha_{q}=2^{2e+1}+2^{e+1}+1$ and $\beta_{q}=2^{2e+1}-2^{e+1}+1$. The
maximal subgroups of $\operatorname{Sz}(q)$ are all conjugate to one of the
following.

  1. (1)

A Suzuki subgroup $\operatorname{Sz}(q_{0})$ where
$\operatorname{\mathbb{F}}_{q_{0}}\subset\operatorname{\mathbb{F}}_{q}$ is a
field extension with $q_{0}=2^{2e_{0}+1}$ for some positive integer $e_{0}$
and $q_{0}$ is maximal for this property,

  2. (2)

A dihedral group $D_{2(q-1)}$,

  3. (3)

A group isomorphic to $C_{\alpha_{q}}:C_{4}$,

  4. (4)

A group isomorphic to $C_{\beta_{q}}:C_{4}$.

Moreover, the cyclic subgroups $C_{q-1},C_{\alpha_{q}}$ and $C_{\beta_{q}}$
all intersect trivially.

We conclude this section by a technical lemma that will be of crucial
importance later.

######  Lemma 2.2.

[11, Lemma 24.3] Let $A$ be the set of involutions of $G$ fixing a point
$P\in\operatorname{\mathcal{O}}$ and let $B$ be the set of involutions of $G$
fixing $Q\in\operatorname{\mathcal{O}}$. Then, for any $\omega\in
A,\zeta,\zeta^{\prime}\in B$ we have that $\omega\zeta$ is never conjugated to
$\omega\zeta^{\prime}$.

###  2.2. Incidence and coset geometries

To their core, most of the geometric objects of interest to mathematicians are
composed of elements together with some relation between them. This very
general notion is made precise by the notion of an incidence system, or an
incidence geometry. For a more detailed introduction to incidence geometry, we
refer to [1].

A triple $\Gamma=(X,*,t)$ is called an incidence system over $I$ if

  1. (1)

$X$ is a set whose elements are called the elements of $\Gamma$,

  2. (2)

$*$ is a symmetric and reflexive relation (called the incidence relation) on
$X$, and

  3. (3)

$t$ is a map from $X$ to $I$, called the type map of $\Gamma$, such that
distinct elements $x,y\in X$ with $x*y$ satisfy $t(x)\neq t(y)$.

Elements of $t^{-1}(i)$ are called the elements of type $i$. The rank of
$\Gamma$ is the cardinality of the type set $I$. A flag in an incidence system
$\Gamma$ over $I$ is a set of pairwise incident elements. The type of a flag
$F$ is $t(F)$, that is the set of types of the elements of $F.$ A chamber is a
flag of type $I$. An incidence system $\Gamma$ is an incidence geometry if all
its maximal flags are chambers.

Let $F$ be a flag of $\Gamma$. An element $x\in X$ is incident to $F$ if $x*y$
for all $y\in F$. The residue of $\Gamma$ with respect to $F$, denoted by
$\Gamma_{F}$, is the incidence system formed by all the elements of $\Gamma$
incident to $F$ but not in $F$. The rank of the residue $\Gamma_{F}$ is equal
to rank$(\Gamma)$ \- $|F|$.

The incidence graph of $\Gamma$ is a graph with vertex set $X$ and where two
elements $x$ and $y$ are connected by an edge if and only if $x*y$. Whenever
we talk about the distance between two elements $x$ and $y$ of a geometry
$\Gamma$, we mean the distance in the incidence graph of $\Gamma$ and simply
denote it by $d_{\Gamma}(x,y)$, or even $d(x,y)$ if the context allows.

Let $\Gamma=\Gamma(X,*,t)$ be an incidence geometry over the type set $I$. A
correlation of $\Gamma$ is a bijection $\phi$ of $X$ respecting the incidence
relation $*$ and such that, for every $x,y\in X$, if $t(x)=t(y)$ then
$t(\phi(x))=t(\phi(y))$. If, moreover, $\phi$ fixes the types of every element
(i.e $t(\phi(x))=t(x)$ for all $x\in X$), then $\phi$ is said to be an
automorphism of $\Gamma$. The _type_ of a correlation $\phi$ is the
permutation it induces on the type set $I$. A correlation of type $(i,j)$ is
called a duality if it has order $2$ and a correlation of type $(i,j,k)$ is
called a triality if it has order $3$. The group of all correlations of
$\Gamma$ is denoted by $\operatorname{Cor}(\Gamma)$ and the automorphism group
of $\Gamma$ is denoted by $\operatorname{Aut}(\Gamma)$. Remark that
$\operatorname{Aut}(\Gamma)$ is a normal subgroup of
$\operatorname{Cor}(\Gamma)$ since it is the kernel of the action of
$\operatorname{Cor}(\Gamma)$ on $I$.

If $\operatorname{Aut}(\Gamma)$ is transitive on the set of chambers of
$\Gamma$ then we say that $\Gamma$ is flag transitive. If moreover, the
stabilizer of a chamber in $\operatorname{Aut}(\Gamma)$ is reduced to the
identity, we say that $\Gamma$ is simply transitive or regular.

Incidence geometries can be obtained from a group $G$ together with a set
$(G_{i})_{i\in I}$ of subgroups of $G$ as described in [13]. The _coset
geometry_ $\Gamma(G,(G_{i})_{i\in I})$ is the incidence geometry over the type
set $I$ where:

  1. (1)

The elements of type $i\in I$ are right cosets of the form $G_{i}\cdot g$,
$g\in G$.

  2. (2)

The incidence relation is given by non empty intersection. More precisely, the
element $G_{i}\cdot g$ is incident to the element $G_{j}\cdot k$ if and only
if $i\neq j$ and $G_{i}\cdot g\cap G_{j}\cdot k\neq\emptyset$.

###  2.3. Chamber systems

The concept of chamber system was invented by Tits in [16]. A chamber system
over $I$ is a pair $\operatorname{\mathcal{C}}=(C,\\{\sim_{i},i\in I\\})$
consisting of a set $C$, whose members are called chambers, and a collection
of equivalence relations $\sim_{i}$ on $C$, indexed by $i\in I$. Two chambers
$c$ and $d$ are called $i-$adjacent if $c\sim_{i}d$. For $i\in I$, each
$\sim_{i}$-equivalence class is called an i-panel. The rank of
$\operatorname{\mathcal{C}}$ is $|I|$. The chamber system
$\operatorname{\mathcal{C}}$ is called thin if every $i-$panel is of size
exactly $2$.

A weak homomorphism $\varphi\colon(C,\\{\sim_{i},i\in
I\\})\to(C^{\prime},\\{\sim_{i^{\prime}},i^{\prime}\in I\\})$ of chamber
systems over $I$ is a map $\varphi\colon C\to C^{\prime}$ for which a
permutation $\pi$ of $I$ can be found such that, for all $c,d\in C$, the
relation $c\sim_{i}d$ implies $\varphi(c)\sim_{\pi(i)}\varphi(d)$. If
$\pi=\operatorname{Id}$, the weak homomorphism $\varphi$ is said to be a
homomorphism. A bijective homomorphism whose inverse is an homomorphism is
called an isomorphism and an isomorphism from $\operatorname{\mathcal{C}}$ to
$\operatorname{\mathcal{C}}$ is called an automorphism of
$\operatorname{\mathcal{C}}$. We denote by
$\operatorname{Aut}(\operatorname{\mathcal{C}})$ the group of all
automorphisms of $\operatorname{\mathcal{C}}$

The graph of $\operatorname{\mathcal{C}}$ is the graph whose vertices are the
chambers of $\operatorname{\mathcal{C}}$ and where two chambers $c,d$ are
connected by an edge if there exists an $i\in I$ such that $c\sim_{i}d$. The
chamber system $\operatorname{\mathcal{C}}$ is connected if the graph of
$\operatorname{\mathcal{C}}$ is connected.

For $J\subset I$, we denote by $\sim_{J}$ the union of all $\sim_{j}$ with
$j\in J$. A connected component of $(\operatorname{\mathcal{C}},\sim_{J})$ is
called a J-cell of $\operatorname{\mathcal{C}}$. For $i\in I$, the $(I$
\$\\{i\\})$-cells are called $i$-objects of $\operatorname{\mathcal{C}}$.

If $\operatorname{\mathcal{C}}$ is a chamber system, the incidence system of
$\operatorname{\mathcal{C}}$, denoted by $\Gamma(\operatorname{\mathcal{C}})$,
is the incidence system over $I$ determined as follows. Its $i$-elements, for
$i\in I$ are the pairs $(x,i)$ with $x$ an $i$-object of
$\operatorname{\mathcal{C}}$; two elements $(x,k)$ and $(y,l)$ of
$\Gamma(\operatorname{\mathcal{C}})$ are incident if and only if $x\cap
y\neq\emptyset$ in $\operatorname{\mathcal{C}}$, i.e., $x$ and $y$ have a
chamber in common.

A chamber system $\operatorname{\mathcal{C}}$ over $I$ is residually connected
if, for every subset $J$ of $I$ and every system of $j$-objects $Z_{j}$, one
for each $j\in J$, with the property that any two have a non-empty
intersection, it follows that $\bigcap_{j\in J}Z_{j}$ is an $(I$ \$J$)-cell.

Let $\psi\colon G\to\operatorname{Aut}(\operatorname{\mathcal{C}})$ be a
representation of a group $G$ on a chamber system
$\operatorname{\mathcal{C}}$. When $G$ is transitive on the chambers of
$\operatorname{\mathcal{C}}$, we say that $G$ is chamber transitive on
$\operatorname{\mathcal{C}}$. We also say that $\operatorname{\mathcal{C}}$ is
chamber transitive if $\operatorname{Aut}(\operatorname{\mathcal{C}})$ is
chamber transitive.

Let $G$ be a group, $B$ a subgroup, $(G^{(i)})_{i\in I}$ a system of subgroups
of $G$ with $B\subset G^{(i)}$. The coset chamber system of $G$ on $B$ with
respect to $(G^{(i)})_{i\in I}$, denote by
$\operatorname{\mathcal{C}}(G,B,(G^{(i)})_{i\in I})$, has the chamber set
consisting of all cosets $gB,g\in G$, and $i$-adjacency determined by
$gB\sim_{i}hB$ if and only if $gG^{(i)}=hG^{(i)}$.

For $i\in I$, the group $G^{(i)}$ is called the standard parabolic subgroup of
type $I\textbackslash\\{i\\}$. The subgroup $B$ is called the Borel subgroup
of $G$. In the cases we will explore in this article, $B$ will always be the
identity subgroup.

######  Proposition 2.3.

[1, Proposition 3.6.4] If $\psi\colon
G\to\operatorname{Aut}(\operatorname{\mathcal{C}})$ is a chamber transitive
representation of $G$ on a chamber system $\operatorname{\mathcal{C}}$ over
$I$, then, for every chamber $c$ of $\operatorname{\mathcal{C}}$, the
canonical representation of $G$ on
$\operatorname{\mathcal{C}}(G,B,(G^{(i)})_{i\in I})$ is equivalent to $\psi$,
where $G^{(i)}$ is the stabilizer of the $i$-cell containing $c$.

This proposition allows us to work with chamber transitive chamber systems in
a group theoretical way. The notion of connectedness and residual
connectedness can then also be translated into group theoretical language.

######  Lemma 2.4.

[1, Lemma 3.6.7] Let $G$ be a group, $B$ a subgroup of $G$, $(G^{(i)})_{i\in
I}$ a system of subgroups of $G$ with $B\leq G^{(i)}$. The chamber system
$\operatorname{\mathcal{C}}(G,B,(G^{(i)})_{i\in I})$ is connected if and only
if $G$ is generated by the subgroups $G^{(i)}$, $i\in I$.

For $G$ a group with a system of subgroups $(G^{(i)})_{i\in I}$ and for
$J\subset I$, we write $G^{(J)}=\langle G^{(j)}\mid j\in J\rangle$. We use the
following theorem. Observe that in [1], the hypothesis that the chamber system
is connected is missing.

######  Theorem 2.5.

[1, Theorem 3.6.9] Let $I$ be a finite index set. Suppose that $G$ is a group
with a system of subgroups $(G^{(i)})_{i\in I}$ and that $B$ is a subgroup of
$\bigcap_{i\in I}G^{(i)}$. If the chamber system
$\operatorname{\mathcal{C}}(G,B,(G^{(i)})_{i\in I})$ is connected, the
following statements are equivalent.

  1. (1)

The chamber system $\operatorname{\mathcal{C}}(G,B,(G^{(i)})_{i\in I})$ is
residually connected.

  2. (2)

For all $J,K,L\subseteq I$, we have $G^{(L)}\cap G^{(J)}G^{(K)}=G^{(L\cap
J)}G^{(L\cap K)}$.

  3. (3)

For all $J,K,L\subseteq I$, we have $G^{(J)}G^{(L)}\cap
G^{(K)}G^{(L)}=G^{(J\cap K)}G^{(L)}$.

Finally, if we restrict ourselves to residually connected incidence geometries
and residually connected chamber system, we can go back and forth between the
two notions freely.

######  Theorem 2.6.

[1, Theorem 3.4.6] Let $\mathcal{G}(I)$ be the collection of all residually
connected geometries over $I$ and let $\operatorname{\mathcal{C}}(I)$ be the
collection of all residually connected chamber systems over $I$. There is a
bijective homomorphism preserving structures between the collections
$\mathcal{G}(I)$ and $\operatorname{\mathcal{C}}(I)$. Thus, each residually
connected geometry over $I$ corresponds to a unique residually connected
chamber system over $I$ (up to isomorphism), with same automorphism group, and
vice versa.

###  2.4. Hypermaps

Hypermaps were introduced in [2], see also [7, 5]. A hypermap $\mathcal{M}$ is
a transitive permutation representation
$\Delta\rightarrow\textup{Sym}(\Sigma)$ of the group

$\Delta=\left\langle
r_{0},r_{1},r_{2}~{}|~{}r_{0}^{2}=r_{1}^{2}=r_{2}^{2}=1\right\rangle\cong
C_{2}*C_{2}*C_{2},$

where $\Sigma$ is a set, the elements of which are called flags. One can think
of each flag as consisting of a hypervertex, a hyperedge and a hyperface. If
the automorphism group $\textup{Aut}(M)$ of the hypermap acts transitively on
the flags, the hypermap is called regular. In that case, the hypervertices,
the hyperedges and the hyperfaces are the orbits of the dihedral subgroups
$\langle r_{1},r_{2}\rangle$, $\langle r_{2},r_{0}\rangle$ and $\langle
r_{0},r_{1}\rangle$, respectively and the hypermap is said to be of type
$(p,q,r)$ where $p$ is the order of $r_{1}r_{2}$, $q$ is the order of
$r_{2}r_{0}$ and $r$ is the order of $r_{0}r_{1}$.

The coset geometry constructed from the automorphism group $\textup{Aut}(M)$
and the three dihedral subgroups is a rank 3 incidence geometry on the type
set $\\{\mbox{hypervertex, hyperedge, hyperface}\\}$. A maximal flag (a
chamber) of this incidence geometry is a triple consisting of one element of
each type such that they are pairwise incident.

It is important to note that the notion of a flag is different for hypermaps
and incidence geometries; in a hypermap flag the incidence between the objects
is a ternary relation, while the incidence relation in the coset geometry is
binary. This implies that the coset geometry constructed from a regular
hypermap can have more flags than the hypermap. In particular, the transitive
action of the automorphism group on the flags of the hypermap does not imply
the transitive action of the automorphism group on the chambers of the
corresponding coset geometry. See for instance [4, Example 4.4].

##  3\. The Chamber System approach

Let $e>1$ be a positive integer such that $3\mid 2e+1$ and let $q=2^{2e+1}$.
Let $G\cong\operatorname{Sz}(q)$ be the Suzuki group acting on the $q^{2}+1$
points of the ovoid $\mathcal{O}$ (see section 2.1). The aim of this section
is to construct a chamber system over $I:=\\{0,1,2\\}$ from $G$. We first fix
a few conventions. Regular letters like $q,p$ and $m$ designate integers.
Capital letters like $P,Q,R$ designate points of $\operatorname{\mathcal{O}}$.
Greek letters like $\rho,\gamma$ and $\tau$ designate group elements, with
$\tau$ restricted to outer automorphisms of $G$ of order $3$.

Since $3\mid 2e+1$, the underlying field $\operatorname{\mathbb{F}}_{q}$ is a
extension of degree 3 of a subfield $\operatorname{\mathbb{F}}_{q_{0}}$ and,
therefore, there are outer automorphisms of $G$ of order $3$. We will call any
such automorphism a triality of $G$. Let $\tau$ be a triality of $G$ and let
$P,Q,R$ be a triple of points of $\mathcal{O}$ permuted cyclically by $\tau$.
Since the triality $\tau$ comes from the action of a power of the Frobenius
automorphism $\Phi$ on the ovoid $\operatorname{\mathcal{O}}$, we can consider
$\tau$ as a permutation of the points $\operatorname{\mathcal{O}}$. With that
in mind, for any point $P\in\operatorname{\mathcal{O}}$, $\tau(P)$ is just the
image of $P$ by that permutation, while for an element $\gamma\in G$, we will
denote by $\tau(\gamma):=\tau\gamma\tau^{-1}=\gamma^{\tau}$, the element of
$G$ that sends any $X\in\operatorname{\mathcal{O}}$ to
$\tau\circ\gamma\circ\tau^{-1}(X)$. As such, if $\gamma$ fixes some point
$P\in\operatorname{\mathcal{O}}$, then $\tau(\gamma)$ fixes $\tau(P)$.

######  Lemma 3.1.

Let $P$ be a point of $\operatorname{\mathcal{O}}$ and let $\tau$ be a
triality of $G$ such that $\tau(P)\neq P$. Let $A$ be the set of involutions
of $G$ with fixed point $P$. Then the set $\\{\rho\cdot\tau(\rho)\mid\rho\in
A\\}$ contains only elements of odd order, no two of which are conjugate.

###### Proof.

All elements of even order of $G$ have exactly one fixed point. Since $\rho$
and $\tau(\rho)$ are always involutions with different fixed points, their
product has either $0$ or $2$ fixed points, and must thus have odd order.

It remains to show that, if $\rho\neq\rho^{\prime}\in A$, then
$\rho\tau(\rho)$ cannot be conjugated to $\rho^{\prime}\tau(\rho^{\prime})$.

Let $H:=G_{P,\tau^{2}(P)}$ be the stabilizer in $G$ of $p$ and $\tau^{2}(P)$.
Hence, $H$ is a cyclic group of order $q-1$. Let $\zeta$ be a generator of
$H$. For $\omega:=\zeta^{i}$, $i=1,2,\cdots,q-2$, we have that

$(\rho\tau(\rho))^{\omega}=\rho^{\omega}\tau(\rho)^{\omega}=\rho^{\omega}\omega\tau(\rho)\omega^{-1}=\rho^{\omega}\tau(\tau^{-1}(\omega)\rho\tau^{-1}(\omega^{-1}))=\rho^{\omega}\tau(\rho^{\tau^{-1}(\omega)})$

We claim that $\rho^{\omega}\neq\rho^{\tau^{-1}(\omega)}$. Indeed, suppose
that $\rho^{\omega}=\rho^{\tau^{-1}(\omega)}$. Conjugating both sides by
$\omega^{-1}$, we obtain that $\rho=\rho^{\tau^{-1}(\omega)\omega^{-1}}$. Set
$\beta=\tau^{-1}(\omega)\omega^{-1}=\tau^{2}(\omega)\omega^{-1}$. We have that
$\rho^{\beta}=\rho$ thus $\rho^{\beta}$ fixes $P$ and $\beta$ fixes $P$ as
well. This means that $\tau^{-1}(\omega)$ must fix $P$ since $\omega^{-1}$ and
$\beta$ fix $P$. But $\tau^{-1}(\omega)$ also fixes $\tau^{-1}(P)=\tau^{2}(P)$
and $\tau^{-1}(\tau^{2}(P))=\tau(P)$ so $\tau^{-1}(\omega)$ would have three
fixed points, a contradiction.

We conclude using Lemma 2.2. Indeed we showed that $\rho\tau(\rho)$ is
conjugated to $\rho^{\omega}\tau(\rho^{\tau^{-1}(\omega)})$ which in turn
cannot be conjugated to $\rho^{\omega}\tau(\rho^{\omega})$ by Lemma 2.2,
unless $\rho^{\omega}=\rho^{\tau^{-1}(\omega)}$, which we ruled out. Since $H$
acts transitively on $A$, this concludes the lemma. ∎

######  Theorem 3.2 (Existence of base Chamber).

Let $m$ be any odd integer that arises as the order of some element of $G$.
For any point $P$ of $\operatorname{\mathcal{O}}$ and for any triality $\tau$
of $G$ such that $\tau(P)\neq P$, there exists an involution $\rho_{0}$ fixing
$P$ such that $\rho_{0}\cdot\tau(\rho_{0})$ has order $m$. Moreover, if
$\langle\rho_{0}\tau(\rho_{0})\rangle\neq\langle\rho_{0}\tau(\rho_{0})\rangle^{\tau}$
then $G$ is generated by $\rho_{0},\tau(\rho_{0})$ and $\tau^{2}(\rho_{0})$.

###### Proof.

By [12, §11, Theorem 5], the Suzuki group $G$ has exactly $q-1$ conjugacy
classes of elements of odd order. The first part of the statement is thus a
direct consequence of Lemma 3.1 as there are exactly $q-1$ involutions fixing
a given point $P$.

Let $m$ be the order of $\rho_{0}\tau(\rho_{0})$, let
$H=\langle\rho_{0},\tau(\rho_{0}),\tau^{2}(\rho_{0})\rangle$, and suppose that
$H$ is a proper subgroup of $G$. Then $H$ must be contained in some maximal
subgroup of $G$. We use the classification of maximal subgroups provided by
Theorem 2.1. Suppose first that $H$ is contained in a maximal subgroup $M$,
isomorphic to either $D_{2(q-1)}$, $C_{\alpha_{q}}:C_{4}$. or
$\operatorname{C}_{\beta_{q}}:C_{4}$. By the last sentence of Theorem 2.1,
there is a unique cyclic subgroup $K$ of order $m$ in $M$. Hence, the three
subgroups
$\langle\rho_{0}\tau(\rho_{0})\rangle,\langle\tau(\rho_{0})\tau^{2}(\rho_{0})\rangle$
and $\langle\tau^{2}(\rho_{0})\rho_{0}\rangle$ all contain $K$. But any two of
these subgroups also have an involution in common, so they must then all be
equal, a contradiction.

Suppose instead that $M$ is isomorphic to a maximal Suzuki subgroup
$\operatorname{Sz}(q_{0})$. If $H=M$, since $H$ is always stabilized by
$\tau$, it would imply that $M$ is also stabilized by $\tau$. This means that
$\tau$ restricts to an automorphism of $\operatorname{Sz}(q_{0})$. This
implies that $q_{0}=2^{2e_{0}+1}$ with $2e_{0}+1$ divisible by $3$ and
therefore, by maximality, $q_{0}=q^{1/3}$. But the only Suzuki subgroup
$\operatorname{Sz}(q^{1/3})$ stabilized by $\tau$ is precisely the one fixed
element-wise by $\tau$. This is a contradiction since $\tau$ clearly moves
$\rho_{0}$.

If instead $H$ is again a proper subgroup of $M$, we can repeat the same
argument starting with $\operatorname{Sz}(q_{0})$ this time. At some point, we
arrive at a contradiction, hence showing that
$G=\langle\rho_{0},\tau(\rho_{0}),\tau^{2}(\rho_{0})\rangle$. ∎

Using Theorem 3.2, we can find the base chamber of the chamber system we want
to construct. We first investigate some geometric properties of such base
chambers and show their algebraic implications.

Let $\rho_{0}$ be an involution with fixed point $P$ such that the order of
$(\rho_{0}\tau(\rho_{0}))$ is equal to $m$ for some integer $m$. Let
$\rho_{1}:=\tau(\rho_{0})$ and $\rho_{2}:=\tau^{2}(\rho_{0})$. Then $\rho_{1}$
has fixed point $Q=\tau(P)$ and $\rho_{2}$ has fixed point $R=\tau^{2}(P)$.
For $i,j,k$ such that $I=\\{i,j,k\\},$ the subgroups generated by
$\langle\rho_{j},\rho_{k}\rangle$ are dihedral groups
$H_{i}=H_{i}(\rho_{0},\tau)$ of order $2m$. The existence of such a
configuration is guaranteed by Corollary 3.2.

Let $\operatorname{\mathcal{O}}_{i}=\operatorname{\mathcal{O}}(H_{i})$ be the
set of points of $\operatorname{\mathcal{O}}$ that are fixed by some
involution in $H_{i}$. Thus $\operatorname{\mathcal{O}}_{i}$ contains exactly
$m$ points for each $i\in I$.

######  Definition 3.3.

Let $\operatorname{\mathcal{O}}_{i}$ be set of points of
$\operatorname{\mathcal{O}}$ for $i\in I$. We say that an ordered triple
$(\operatorname{\mathcal{O}}_{0},\operatorname{\mathcal{O}}_{1},\operatorname{\mathcal{O}}_{2})$
forms a triangle if
$\operatorname{\mathcal{O}}_{i}\cap\operatorname{\mathcal{O}}_{j}\neq\emptyset$
for all $i\neq j=0,1,2$. We say that the triangle
$(\operatorname{\mathcal{O}}_{0},\operatorname{\mathcal{O}}_{1},\operatorname{\mathcal{O}}_{2})$
is non-degenerate if the $\operatorname{\mathcal{O}}_{i}$’s are all distinct
for $i=0,1,2$. We say that the triangle
$(\operatorname{\mathcal{O}}_{0},\operatorname{\mathcal{O}}_{1},\operatorname{\mathcal{O}}_{2})$
is proper if
$\operatorname{\mathcal{O}}_{0}\cap\operatorname{\mathcal{O}}_{1}\cap\operatorname{\mathcal{O}}_{2}=\emptyset$.
Finally, if $\rho_{0}$ is an involution of $G$ and $\tau$ is a triality as
above, we say that
$T_{\rho_{0},\tau}:=(\operatorname{\mathcal{O}}_{0}=\operatorname{\mathcal{O}}(H_{0}(\rho_{0},\tau)),\operatorname{\mathcal{O}}_{1}=\operatorname{\mathcal{O}}(H_{1}(\rho_{0},\tau)),\operatorname{\mathcal{O}}_{2}=\operatorname{\mathcal{O}}(H_{2}(\rho_{0},\tau)))$
is the triangle associated to $(\rho_{0},\tau)$.

We now show that non-degeneracy of $T_{\rho_{0},\tau}$ implies that $G$ is
generated by $\rho_{0},\tau(\rho_{0})$ and $\tau^{2}(\rho_{0})$.

######  Lemma 3.4.

If $T_{\rho_{0},\tau}$ is degenerate, then
$H_{0}(\rho_{0},\tau)=H_{1}(\rho_{0},\tau)=H_{2}(\rho_{0},\tau)$

###### Proof.

Suppose that
$\operatorname{\mathcal{O}}_{0}=\operatorname{\mathcal{O}}_{1}=\operatorname{\mathcal{O}}_{2}$
but that the three subgroups $H_{0},H_{1}$ and $H_{2}$ are distinct. By the
second part of Theorem 3.2, we can then deduce that
$G=\langle\rho_{0},\tau(\rho_{0}),\tau^{2}(\rho_{0})\rangle$. But all three of
$\rho_{0},\tau(\rho_{0})$ and $\tau^{2}(\rho_{0})$ stabilize
$\operatorname{\mathcal{O}}_{0}$. Therefore,
$\langle\rho_{0},\tau(\rho_{0}),\tau^{2}(\rho_{0})\rangle$ is a subgroup of
the stabilizer in $G$ of $\operatorname{\mathcal{O}}_{0}$, which is a proper
subgroup since $G$ acts transitively on the whole ovoid
$\operatorname{\mathcal{O}}$. This is a contradiction. ∎

######  Corollary 3.5.

If $T_{\rho_{0},\tau}$ is non-degenerate, then
$G=\langle\rho_{0},\tau(\rho_{0}),\tau^{2}(\rho_{0})\rangle$.

###### Proof.

By Lemma 3.4, we know that the subgroups $H_{0},H_{1},H_{2}$ are distinct. We
can then conclude by using the second part of Theorem 3.2. ∎

$P$$\tau(P)$$\tau^{2}(P)$$O_{2}$$O_{1}$$O_{0}$ Figure 1. The base chamber
$T_{\rho_{0},\tau}$ of the chamber system

The next Lemma gives criterion for $T_{\rho_{0},\tau}$ to be non-degenerate or
proper.

######  Lemma 3.6.

If $\rho_{0}$ in an involution and $\tau$ is a triality such that
$\rho_{0}\tau(\rho_{0})$ is of order $m$ with $m\equiv 2(\bmod\;3)$, then the
triangle
$T_{\rho_{0},\tau}=(\operatorname{\mathcal{O}}_{0},\operatorname{\mathcal{O}}_{1},\operatorname{\mathcal{O}}_{2})$
associated to $(\rho_{0},\tau)$ is non-degenerate. Furthermore, if $m=5$, the
triangle
$T_{\rho_{0},\tau}=(\operatorname{\mathcal{O}}_{0},\operatorname{\mathcal{O}}_{1},\operatorname{\mathcal{O}}_{2})$
is also proper.

###### Proof.

Since $\operatorname{\mathcal{O}}_{i}\cap\operatorname{\mathcal{O}}_{j}$
always contains the fixed point of $\rho_{k}$, the triangle associated to any
pair $(\rho_{0},\tau)$ is always a triangle in the sense of Definition 3.3.
Suppose now that
$\operatorname{\mathcal{O}}_{0}=\operatorname{\mathcal{O}}_{1}=\operatorname{\mathcal{O}}_{2}$.
Then the triality $\tau$ acts on $\operatorname{\mathcal{O}}_{0}$ by permuting
its points. Then, since $|\operatorname{\mathcal{O}}_{0}|=m=2(\bmod\;3)$, the
triality $\tau$ must fix at least $2$ points $X$ and $Y$ of
$\operatorname{\mathcal{O}}_{0}$. Since $\tau$ also sends $H_{0}$ to itself by
Lemma 3.4, we conclude that $\tau$ must fix $\rho_{X}$ and $\rho_{Y}$, the two
involutions of $H_{0}$ having respectively $X$ and $Y$ as fixed points. But
$H_{0}$ is generated by $\rho_{X}$ and $\rho_{Y}$, so $\tau$ should act
trivially on $H_{0}$, a contradiction. This shows that the triangle is non-
degenerate.

Suppose now that $m=5$ and that
$\operatorname{\mathcal{O}}_{0}\cap\operatorname{\mathcal{O}}_{1}\cap\operatorname{\mathcal{O}}_{2}$
is not empty and contains some point $X\in\operatorname{\mathcal{O}}$. Let $P$
be a point in
$\operatorname{\mathcal{O}}_{0}\cap\operatorname{\mathcal{O}}_{1}$ which is
not $X$. Let $\rho_{X}^{0}$ be the involution fixing $X$ in $H_{0}$,
$\rho_{X}^{1}$ be the involution fixing $X$ in $H_{1}$ and $\rho$ be the
involution fixing $P$ in $H_{0}\cap H_{1}$. Then both $\rho\rho_{X}^{0}$ and
$\rho\rho_{X}^{1}$ have order $5$. By Lemma 2.2 $\rho\rho_{X}^{0}$ and
$\rho\rho_{X}^{1}$ cannot be conjugated. But there is only one conjugacy class
of elements of order $5$ in $G$, a contradiction. This shows that the triangle
is proper, and concludes the proof. ∎

We are now ready to show that for some well chosen value of
$m=O(\rho_{0}\tau(\rho_{0}))$, we can construct regular hypermaps of type
$(m,m,m)$ by letting $G$ acts on $T_{\rho_{0},\tau}$.

######  Corollary 3.7.

Any triangle $T_{\rho_{0},\tau}$ with $\rho_{0}\tau(\rho_{0})$ of order $m$
and $m\equiv 2(\bmod\;3)$ defines a hypermap of type $(m,m,m)$ with
automorphism group $G$. In particular, for any $q=2^{2e+1}$ with $2e+1$
divisble by 3, there exists an hypermap of type
$(\alpha_{q},\alpha_{q},\alpha_{q})$ where $\alpha_{q}=2^{2e+1}+2^{e+1}+1$ or
of type $(\beta_{q},\beta_{q},\beta_{q})$ where
$\beta_{q}=2^{2e+1}-2^{e+1}+1$, depending on the parity of $e$.

###### Proof.

Lemma 3.6 tells us that
$G=\langle\rho_{0},\tau(\rho_{0}),\tau^{2}(\rho_{0})\rangle$ whenever the
order of $\rho_{0}\tau(\rho_{0})$ is equal to $2$ modulo $3$. Notice that
$2^{2e+1}+2^{e+1}+1\equiv 2\pm 2+1(\bmod\;3)$ where the sign depends on
whether $e$ is odd or even. So for every value of $e$, exactly one of
$\alpha_{q}$ and $\beta_{q}$ is congruent to $2\bmod 3$ while the other is
congruent to $1\bmod 3$. Choosing the right one then yields a regular hypermap
of the wanted type. ∎

Let $\mathcal{M}$ be one of the hypermaps obtained by Corollary 3.7. The
triality $\tau$ then sends $\mathcal{M}$ to a map $\tau(\mathcal{M})$ which is
isomorphic to $\mathcal{M}$, but where the roles of vertices, hyperedges and
hyperfaces are permuted cyclically. Such operation of the finite order on
hypermaps have been studied in [6]. An operation of order three is called a
triality and an operation of order two is called a duality. Since
$\operatorname{Out}(G)$ has no elements of order $2$, the map $\mathcal{M}$
cannot have dualities. The hypermaps of Corollary 3.7 thus admit trialities
but no dualities.

We now proceed, after a few results, to define the chamber system
$C=\\{\gamma(T_{\rho_{0},\tau})\mid\gamma\in G\\}$. From here on, we suppose
that we have a pair $(\rho_{0},\tau)$ such that the associated triangle
$T_{\rho_{0},\tau}$ is proper and non-degenerate.

######  Lemma 3.8.

Let $(P,Q,R)$ be a triple of points permuted by a triality $\tau$. Then,
$G_{\\{P,Q,R\\}}$ is trivial.

###### Proof.

Suppose there is a $\gamma\in G$ preserving the set $\\{P,Q,R\\}$. If $\gamma$
fixes $\\{P,Q,R\\}$ pointwise, it must be the identity as the only element of
$G$ that fixes three points of $\operatorname{\mathcal{O}}$ is the identity
element. Also, $\gamma$ cannot permute $P,Q$ and $R$ cyclically, since $3$
does not divide the order of $G$. Suppose, without loss of generality, that
$\gamma(P)=P$ and that $\gamma$ exchanges $Q$ and $R$. Then $\tau(\gamma)$
fixes $Q$ and exchanges $P$ and $R$. But then one easily checks that
$\gamma\cdot\tau(\gamma)$ permutes cyclically $P,Q$ and $R$, a contradiction.
∎

For a triangle
$(\operatorname{\mathcal{O}}_{0},\operatorname{\mathcal{O}}_{1},\operatorname{\mathcal{O}}_{2})$,
its image
$\gamma(\operatorname{\mathcal{O}}_{0},\operatorname{\mathcal{O}}_{1},\operatorname{\mathcal{O}}_{2})$
by the action of $\gamma\in G$ is the triangle
$(\gamma(\operatorname{\mathcal{O}}_{0}),\gamma(\operatorname{\mathcal{O}}_{1}),\gamma(\operatorname{\mathcal{O}}_{2}))$.

######  Lemma 3.9.

The image $\gamma(T_{\rho_{0},\tau})$ by $\gamma\in G$ of the triangle
associated to $(\rho_{0},\tau)$ is $T_{\rho_{0}^{\gamma},\tau^{\gamma}}$, the
triangle associated to $(\rho_{0}^{\gamma},\tau^{\gamma})$.

###### Proof.

In other words, we need to show that
$\operatorname{\mathcal{O}}_{i}(\rho_{0}^{\gamma},\tau^{\gamma})=\gamma(\operatorname{\mathcal{O}}_{i}(\rho_{0},\tau))$.
By definition,
$\operatorname{\mathcal{O}}_{2}(\rho_{0}^{\gamma},\tau^{\gamma})$ is the set
of fixed points of the involutions of
$\langle\rho_{0}^{\gamma},\tau^{\gamma}\rho_{0}^{\gamma}(\tau^{\gamma})^{-1}\rangle=\langle\rho_{0},\tau\rho_{0}\tau^{-1}\rangle^{\gamma}$.
Since $\operatorname{\mathcal{O}}_{2}(\rho_{0},\tau)$ is the set of fixed
points of involutions in $\langle\rho_{0},\tau\rho_{0}\tau^{-1}\rangle$, this
concludes the Lemma for $\operatorname{\mathcal{O}}_{2}$. The cases of
$\operatorname{\mathcal{O}}_{0}$ and $\operatorname{\mathcal{O}}_{1}$ are
identical. ∎

######  Corollary 3.10 (Chamber-transitivity).

Let $C=\\{\gamma(T_{\rho_{0},\tau})\mid\gamma\in
G\\}=\\{(\gamma(\operatorname{\mathcal{O}}_{0}),\gamma(\operatorname{\mathcal{O}}_{1}),\gamma(\operatorname{\mathcal{O}}_{2}))\mid\gamma\in
G\\}$. Then $|C|=|G|$. Moreover, if
$UC:=\\{\\{\gamma(\operatorname{\mathcal{O}}_{0}),\gamma(\operatorname{\mathcal{O}}_{1}),\gamma(\operatorname{\mathcal{O}}_{2})\\}\mid\gamma\in
G\\}$ is the set of unordered triples, we also have that $|UC|=|G|$.

###### Proof.

We only prove the second statement as it implies the first. Let
$c=T_{\rho_{0},\tau}$ and $c^{\prime}\in C$. Then $c^{\prime}=\gamma(c)$ for
some $\gamma\in G$. Suppose that $c^{\prime}=c$. Then $\gamma$ must send the
triple
$\\{\operatorname{\mathcal{O}}_{0},\operatorname{\mathcal{O}}_{1},\operatorname{\mathcal{O}}_{2}\\}$
to itself. This also means that $\gamma$ must conjugate the triple
$\\{H_{0},H_{1},H_{2}\\}$ to itself. Since $H_{i}\cap
H_{j}=\langle\rho_{k}\rangle$ for $\\{i,j,k\\}=I$, we conclude that $\gamma$
must conjugate the triple $\\{\rho_{0},\rho_{1},\rho_{2}\\}$ to itself.
Therefore, if $P,Q,R$ are the fixed points of $\rho_{i}$, we can conclude that
$P,Q,R$ must be fixed setwise by $\gamma$, and thus that $\gamma=1_{G}$, by
Lemma 3.8. ∎

Let $C=\\{\gamma(T_{\rho_{0},\tau})\mid\gamma\in G\\}$ and let $d\in C$. We
say that $d$ is a chamber of $\operatorname{C}$. Then
$d=T_{\rho^{\gamma},\tau^{\gamma}}$ for some $\gamma\in G$ by Lemma 3.9. Let
$H_{i}(d)$ be the unique dihedral subgroup of order $2m$ inside of the
stabilizer of $\operatorname{\mathcal{O}}_{i}(\rho^{\gamma},\tau^{\gamma})$.
Then $\rho_{i}(d)$ is defined to be the unique involution in $H_{j}(d)\cap
H_{k}(d)$, whenever $\\{i,j,k\\}=I$.

######  Definition 3.11 (The chamber system).

Let $C=\\{\gamma(T_{\rho_{0},\tau})\mid\gamma\in G\\}$ as before and define
equivalence relations $\sim_{i},i\in I$ on $C$ as follows: for $c_{1},c_{2}\in
C$, $c_{1}\sim_{i}c_{2}$ if and only if $c_{1}=\rho_{i}(c_{1})(c_{2})$ or
$c_{1}=c_{2}$.

Note that if $c_{1}=\rho_{i}(c_{1})(c_{2})$, then
$\rho_{i}(c_{1})=\rho_{i}(c_{2})$ so that $\sim_{i}$ is indeed symmetric.

######  Theorem 3.12.

$G$ acts chamber transitively by automorphisms on the chamber system
$(C,\sim_{i})$, $i\in I$.

###### Proof.

Chamber-transitivity is clear by Corollary 3.10. To check that $G$ acts by
automorphisms, it suffices to check that the following diagram commutes:

${C}$${C}$${C}$${C}$$\scriptstyle{\rho_{i}}$$\scriptstyle{\gamma}$$\scriptstyle{\gamma}$$\scriptstyle{\rho_{i}}$

To check that, let $c\in C$. Then we need to compare
$(\gamma\circ\rho_{i}(c))(c)$ to $(\rho_{i}(\gamma(c))\circ\gamma)(c)$. But
since $\rho_{i}(\gamma(c))=\rho_{i}(c)^{\gamma}$, the two expressions are
indeed equal. ∎

We have thus constructed so far a chamber system $(C,\sim_{i})$, $i\in I$ on
which $G$ acts chamber transitively by automorphisms. It remains to show that
$(C,\sim_{i})$, $i\in I$ is residually connected. For that we first need a
technical result.

######  Lemma 3.13.

Let $K$ be a cyclic subgroup of $G=\operatorname{Sz}(q)$ of prime odd order
$p$. Let $N_{G}(K)$ be the normalizer in $G$ of $K$ and let $H$ be a dihedral
subgroup of $G$ of order $2p$. Then $H$ contains $K$ if and only if $H\leq
N_{G}(K)$.

###### Proof.

Let $H$ be a dihedral subgroup containing $K$. Then $H$ normalizes $K$ as $K$
is of odd order. So $H\leq N_{G}(K)$.

Suppose that $H\leq N_{G}(K)$ and $K\not\leq H$. Then $N_{G}(K)$ contains two
cyclic subgroups of order $p$, namely $K$ and the subgroup of order $p$ of
$H$. Also $N_{G}(K)=D_{2(q-1)}$ if $p|q-1$ and $N_{G}(K)=C_{x}:C_{4}$
otherwise (where $x=q+\sqrt{2q}+1$ or $q-\sqrt{2q}+1$). In any case,
$N_{G}(K)$ has a unique cyclic subgroup of odd order $p$, a contradiction. ∎

We are now ready to show that $(C,\sim_{i})$, $i\in I$ is residually
connected, under the assumption that $\gamma(T_{\rho_{0},\tau})$ is proper and
non-degenerate.

######  Theorem 3.14.

Let $C=\\{\gamma(T_{\rho_{0},\tau})\mid\gamma\in G\\}$ for a proper and non-
degenerate triangle $\gamma(T_{\rho_{0},\tau})$. Suppose moreover that
$\rho_{0}\tau(\rho_{0})$ is an element of prime order $p$. The chamber system
$(C,\sim_{i}),i\in I$ is residually connected.

###### Proof.

Since $G$ acts chamber transitively by automorphisms on $(C,\sim_{i})$, by
Proposition 2.3, the coset chamber system $(G,G^{i})$ is equivalent to
$(C,\sim_{i})$, where $G^{i}=\langle\rho_{i}\rangle$ where $\rho_{0}$ is the
involution used to define the triangle $T_{\rho_{0},\tau}$ and
$\rho_{1}=\tau(\rho_{0})$, $\rho_{2}=\tau^{2}(\rho_{0})$. Therefore, we can
use Theorem 2.5. We will check that for all $J,K,L\subseteq I$, we have

(1) $G^{(L)}\cap G^{(J)}G^{(K)}=G^{(L\cap J)}G^{(L\cap K)}.$

Note that $G^{(L)}\cap G^{(K)}=G^{(L\cap K)}$ for all $L,K\subset I$ and that
$G^{(K)}G^{(L)}=G^{(L)}$ whenever $K\subset L$. Most of the cases hold in a
quite straightforward manner. For example, if $J=I$, equation (1) becomes

$G^{(L)}\cap G^{(K)}=G^{(L)}G^{(L\cap K)}=G^{(L\cap K)}.$

The case of $J=I$ or $K=I$ is similar.

Suppose now that $J=\emptyset$. Then equation (1) becomes

$G^{(L)}\cap G^{(K)}=G^{(L\cap K)}.$

The same happens if $K=\emptyset$ or $L=\emptyset$.

If $J=\\{0\\},K=\\{1\\}$ and $L=\\{2\\}$ then equation (1) becomes

$\langle\rho_{2}\rangle\cap\langle\rho_{0}\rangle\langle\rho_{1}\rangle=\\{\operatorname{Id}\\}$

which holds since $\rho_{2}$ is not contained in $H_{2}$ since
$(\rho_{0},\tau)$ is non-degenerated.

The harder case arises when $|J|=|K|=|L|=2$. So suppose that
$J=\\{0,1\\},K=\\{1,2\\}$ and $L=\\{0,2\\}$. Equation (1) then becomes

(2)
$\langle\rho_{0},\rho_{2}\rangle\cap\langle\rho_{0},\rho_{1}\rangle\langle\rho_{1},\rho_{2}\rangle=\langle\rho_{0}\rangle\langle\rho_{2}\rangle=\\{e,\rho_{0},\rho_{2},\rho_{0}\rho_{2}\\}$

which is the same as

$H_{1}\cap H_{2}H_{0}=\\{e,\rho_{0},\rho_{2},\rho_{0}\rho_{2}\\}\subset
H_{1}.$

Note that $\\{e,\rho_{0},\rho_{2},\rho_{0}\rho_{2}\\}\subseteq H_{1}\cap
H_{2}H_{0}$ is evidently true, so it suffices to show the opposite inclusion.
To show this, let $\gamma_{2}\in H_{2}$ and $\gamma_{0}\in H_{0}$.

Suppose first that $\gamma_{2}$ and $\gamma_{0}$ are both involutions. If they
have the same fixed point $x$, then $\gamma_{2}\gamma_{0}$ is also an
involution with fixed point $x$ which can then not be an element of $H_{1}$
since $(\rho_{0},\tau)$ is non-degenerated. So suppose that $\gamma_{2}$ and
$\gamma_{0}$ have different fixed points. If $\gamma_{2}=\rho_{0}$ and
$\gamma_{0}=\rho_{2}$, their product is $\rho_{0}\rho_{2}$ so we are done. So
we can suppose that either $\gamma_{2}\neq\rho_{0}$ or
$\gamma_{0}\neq\rho_{2}$. Suppose that $\gamma_{2}\neq\rho_{0}$. The product
$\gamma_{2}\gamma_{0}$ cannot be an involution since it has $0$ or $2$ fixed
points. Hence, either $\gamma_{2}\gamma_{0}$ is not in $H_{1}$ or it is of
order $p$. If it is of order $p$, then the subgroup generated by $\gamma_{2}$
and $\gamma_{0}$ is a dihedral subgroup $H$ of order $2p$ such that $H\cap
H_{1}=\langle\rho_{0}\rho_{2}\rangle$. Therefore, by Lemma 3.13, $H$ must be
contained in $N=N_{G}(\langle\rho_{0}\rho_{2}\rangle)$. By the same Lemma, we
know that $H_{2}$ cannot be in $N$. But that means that $\gamma_{2}\notin N$
since $H_{2}=\langle\rho_{0},\gamma_{2}\rangle$ and $\rho_{0}$ clearly is in
$N$. This is a contradiction, which shows that $\gamma_{2}\gamma_{0}\notin
H_{1}$ as long as $\gamma_{2}\neq\rho_{0}$, or $\gamma_{0}\neq\rho_{2}$.

Suppose now that both $\gamma_{2}\in H_{2}$ and $\gamma_{0}\in H_{0}$ have
order $p$. Then $\gamma_{2}=(\rho_{0}\rho_{1})^{k}$ for some $k=1,\ldots,p-1$
and $\gamma_{0}=(\rho_{1}\rho_{2})^{m}$ for $m=1,\ldots,p-1$. Therefore,

$\gamma_{2}\gamma_{0}=(\rho_{0}\rho_{1})^{k}(\rho_{1}\rho_{2})^{m}=(\rho_{0}\rho_{1})^{k-1}(\rho_{0}\rho_{2})(\rho_{1}\rho_{2})^{m-1}$

If $k=m=1$ then $\gamma_{2}\gamma_{0}=\rho_{0}\rho_{2}\in H_{1}$. Else, note
that $(\rho_{0}\rho_{1})^{k-1}\rho_{0}=\rho_{0}^{h}$ with $h\in H_{2}$ if $k$
is odd and $(\rho_{0}\rho_{1})^{k-1}\rho_{0}=\rho_{1}^{h}$ with $h$ in $H_{2}$
if $k$ is even. The same thing holds for $\rho_{2}(\rho_{1}\rho_{2})^{m-1}$.
This means that if at least one of $k$ and $m$ is not $1$,
$\gamma_{2}\gamma_{0}$ becomes equal to the product of two involutions, one in
$H_{2}$ and one in $H_{0}$. We have already shown that such a product cannot
be in $H_{1}$. We note that, from the above argument, it can be deduced that
if $h_{i}$ is an element of order $m$ in $H_{i}$ and $h_{j}$ is an element of
order $m$ in $H_{j}$, then if $h_{i}h_{j}$ is in $H_{k}$, it must be of order
$p$.

Finally suppose without loss of generality that $\gamma_{2}$ has order $p$ and
$\gamma_{0}$ has order $2$. Suppose that $\gamma_{2}\gamma_{0}\in H_{1}$. If
$\gamma_{0}=\rho_{2}$ then $\gamma_{0}\in H_{1}$ and
$\gamma_{2}=\gamma_{2}(\gamma_{0})^{2}\in H_{1}$ as well, a contradiction with
Lemma 3.13. Suppose thus that $\gamma_{0}\neq\rho_{0}$. If the order of
$\gamma_{2}\gamma_{0}$ is $2$, then
$\langle\gamma_{2},\gamma_{2}\gamma_{0}\rangle$ is a dihedral group of order
$2p$ and we can use the same method as the first case. If the order of
$\gamma_{2}\gamma_{0}$ is $p$, then $\gamma_{2}$ is of order $p$ in $H_{2}$,
$\gamma_{2}\gamma_{0}$ is of order $p$ is $H_{1}$ so, by the comment at the
end of the previous case, $\gamma_{0}=\gamma_{2}^{-1}(\gamma_{2}\gamma_{0})$
should be of order $p$, a contradiction. ∎

Finally, let $C=\\{\gamma(T_{\rho_{0},\tau})\mid\gamma\in G\\}$ be a chamber
system obtained by Theorem 3.14, and let $\Gamma=(G,(G_{i})_{i\in I}$ be the
associated coset geometry. We want to show that $\Gamma$ admits trialities but
no dualities. The triality $\tau$ naturally induces a triality of $\Gamma$
since it permutes the three generators $\rho_{0},\rho_{1}$ and $\rho_{3}$. To
prove that $\Gamma$ admits no dualities, we show a results that associates to
any correlation of $\Gamma$ an automorphism of $G$.

Let $\Gamma$ be a thin, simply flag transitive incidence geometry, let
$G=\operatorname{Aut}(\Gamma)$ and let $c$ be a chamber of $\Gamma$. Let
$\\{c_{0},\cdots,c_{n}\\}$ be the set of chambers of $\Gamma$ incident to $c$.
For each $c_{i}$, there is a unique element $\gamma_{i}\in G$ such that
$c_{i}=\gamma_{i}(c)$. Let $S=\\{\gamma_{0},\cdots,\gamma_{n}\\}$. We then
have that each $\gamma_{i}$ has order $2$ and that
$G=\langle\gamma_{0},\cdots,\gamma_{n}\rangle$. For any correlation $\alpha$
of $\Gamma$ we can definite a map $\phi_{\alpha}\colon S\to S$ as follows. For
each $i$, there is exactly one chamber $c_{i}$ which is $i$-adjacent to $c$.
The image $\alpha(c_{i})$ is a chamber which is $j-$adjacent to $\alpha(c)$
for some $j\in I$. We then define $\phi_{\alpha}(\gamma_{i}):=\gamma_{j}$ and
note $j=\alpha(i)$. In other words,
$\phi_{\alpha}(\gamma_{i})=\gamma_{\alpha(i)}$ where we let $\alpha$ naturally
act on the type set $I$.

######  Proposition 3.15.

Let $\Gamma$ be a thin, simply flag transitive incidence geometry and let
$\alpha$ be a correlation of $\Gamma$. Then the map $\phi_{\alpha}\colon S\to
S$ naturally extends to an automorphism $\Phi_{\alpha}$ of $G$.

###### Proof.

To show that $\Phi_{\alpha}$ is a homomorphism, it suffices to show that any
word $w$ in the alphabet $S$ representing the identity in $G$ is sent by
$\alpha$ to a word $\alpha(w)$ which also represents the identity in $G$. Let
$w$ be such a word. Then it uniquely defines a loop $l$ with base point $c$ in
the chamber graph $CG(\Gamma)$. This loop can be recorded by a sequence
$i_{1},i_{2},\cdots,i_{n}$, $i_{j}\in I$, of adjacencies. The image
$\alpha(l)$ of this loop is once again a loop, now with base point
$\alpha(c)$, which is recorded by the sequence
$\alpha(i_{1}),\alpha(i_{2}),\cdots,\alpha(i_{n})$. This loop $\alpha(l)$ also
uniquely defines a word which by definition is $\alpha(w)$. Thus $\alpha(w)$
represents the identity and $\Phi_{\alpha}$ is a homomorphism.

The homomorphism $\Phi_{\alpha}$ is surjective since any element $\gamma\in G$
can be represented by a path $p$ in $CG(\Gamma)$. Any path $p^{\prime}$ such
that $\alpha(p^{\prime})=p$ represents an elements of $\gamma^{\prime}\in G$
such that $\alpha(\gamma^{\prime})=\gamma$.

The homomorphism $\Phi_{\alpha}$ is injective since $\alpha$ is injective on
the vertices of $CG(\Gamma)$. Therefore, if $p$ is a path in $CG(\Gamma)$
which is not a loop, the image $\alpha(p)$ cannot be a loop so that any word
not representing the identity in $G$ cannot be sent to the identity by
$\alpha$.

∎

######  Theorem 3.16.

Let $G=\operatorname{Sz}(q)$ with $q=2^{2e+1}$ where $2e+1$ is a multiple of
$3$. There exists a flag transitive, residually connected, thin incidence
geometry $\Gamma$ over $I=\\{0,1,2\\}$ with diagram

$\;\;\;\;\;5\;\;\;\;\;$$5$$5\ $

such that $\operatorname{Aut}(\Gamma)\cong\operatorname{Sz}(q)$ has index 3 in
$Cor(\Gamma)\cong\operatorname{Sz}(q):C_{3}$. Moreover, $\Gamma$ admits
trialities but no dualities.

###### Proof.

By Lemma 3.6, we know that the triangle $T_{\rho_{0},\tau}$ associated to any
pair $(\rho_{0},\tau)$ with $\rho_{0}\tau(\rho_{0})$ of order $5$ is always
proper and non-degenerate. Theorems 3.12 and 3.14 then show that we can
construct a chamber transitive and residually connected chamber system $C$
with automorphism group $G$. Theorem 2.6 implies the existence of a coset
geometry $\Gamma$ with automorphism group $G$ and parabolic subgroups
$H_{i}=\langle\rho_{j},\rho_{k}\rangle,$ for $\\{i,j,k\\}=\\{0,1,2\\}$. The
triality $\tau$ then acts on $\Gamma$ as a correlation of order three.
Proposition 3.15 guarantees that $\Gamma$ cannot have dualities. Indeed, if a
duality were to exist, the map $\Phi_{\alpha}$ fixing $\rho_{i}$ and
exchanging $\rho_{j}$ and $\rho_{k}$ should be an automorphism of $G$. This
automorphism cannot be inner, by Lemma 3.8. It also cannot be outer since $G$
has no outer automorphism of order $2$. ∎

##  4\. Concluding remarks

We conclude this article with a few remarks. In this article, we decided to
first construct a chamber system with the desired properties and then to use
Theorem 2.6 to obtain the desired incidence geometries. There are two main
reasons for that decision. Indeed, the chamber system approach not only
simplifies some technical difficulties, as explained in more details below,
but it also allows the construction to remain more concrete and geometric.

Suppose we were to try to construct directly and geometrically the incidence
geometries of Theorem 3.16. For that, we need to find a subset $S$ of points
of $\operatorname{\mathcal{O}}$ such that the stabilizer of $S$ under the
action of $\operatorname{Sz}(q)$ is a dihedral group of order $10$. This means
we cannot simply set $S=\operatorname{\mathcal{O}}_{5}$, a sub-ovoid of $5$
points, as we did in the chamber system approach. Indeed the stabilizer in
$\operatorname{Sz}(q)$ of $S$ would then be a subgroup isomorphic to
$\operatorname{Sz}(2)$. The group $\operatorname{Sz}(2)$ nonetheless contains
a dihedral group of order $10$ as an index $2$ subgroup, so it is most likely
possible to fix this problem, but it requires additional effort.

On the other hand, if we accept to loose the geometric interpretation, the
incidence geometries of Theorem 3.16 can be constructed directly as coset
incidence geometries in the following way. Let $T_{\rho_{0},\tau}$ is a proper
and non-degenerate triangle and let $H_{0},H_{1}$ and $H_{2}$ be the $3$
dihedral subgroups associated to $T_{\rho_{0},\tau}$. We can then construct
the coset geometry $\Gamma(G,(H_{0},H_{1},H_{2}))$. Equation (2) is satisfied,
meaning that this coset geometry is flag transitive. Indeed, by a result of
Tits [15, Section 1.4], this equation implies that any triple of cosets that
are pairwise incident have a common element, and therefore that $G$ acts
transitively on the chambers of $\Gamma$. Residual connectedness is
straightforward.

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