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# Simpler foundations for the hyperbolic plane

John Bamberg  Department of Mathematics and Statistics
The University of Western Australia
Crawley, WA 6009, Australia<EMAIL_ADDRESS>and  Tim Penttila
School of Mathematical Sciences
The University of Adelaide
Adelaide, SA 5005, Australia<EMAIL_ADDRESS>
###### Abstract.

H. L. Skala (1992) gave the first elegant first-order axiom system for
hyperbolic geometry by replacing Menger’s axiom involving projectivities with
the theorems of Pappus and Desargues for the hyperbolic plane. In so doing,
Skala showed that hyperbolic geometry is incidence geometry. We improve upon
Skala’s formulation by doing away with Pappus and Desargues altogether, by
substituting for them two simpler axioms.

###### 2010 Mathematics Subject Classification:

51A20, 51A30, 51A45, 51G05, 51M10

##  1\. Introduction

The independent discovery by Lobachevsky and Bolyai of hyperbolic geometry in
the 1830’s was followed by slow acceptance of the subject from the 1860’s on,
with the publications of relevant parts of the correspondence of Gauss. A new
phase was entered from 1903, when Hilbert [12], in his work introducing the
_calculus of ends_ , introduced an axiomatisation for hyperbolic plane
geometry by adding a hyperbolic parallel axiom to the axioms for plane
absolute geometry. (A detailed survey of axiomatisations for hyperbolic
geometry is given in [24]).

In 1938, Menger [18] made the important discovery that in hyperbolic geometry
the concepts of betweenness and equidistance can be defined in terms of point-
line incidence. Since an axiom system obtained by replacing all occurrences of
betweenness and equidistance with their definitions in terms of incidence
would look highly unnatural, Menger and his students looked for a more natural
axiom system.

> _“This research, which is still in progress, will result, I believe, in a
> system of fairly simple assumptions in terms of the projective operations
> for the whole Bolyai-Lobachevsky geometry.”_ Karl Menger, 1940 [19].

This task was carried out by Menger [18, 19, 20, 21], and by his students
Abbott [1, 2, 4], DeBaggis [8, 9] and Jenks [14, 15, 16], but even their most
polished axiom system contained a statement on projectivities that was not
reducible to a first-order statement. In 1992, Menger’s student Skala [29]
showed that the statement on projectivities can be replaced by the axioms of
Pappus and Desargues for the hyperbolic plane, thereby producing the first
elegant first-order axiom system for hyperbolic geometry based on incidence
alone. Since one of the Menger school’s other axioms was ‘Pascal’s Theorem’ on
the rimpoints, this axiom system looks a little excessive. In 1905, Hessenberg
[11] had shown that, in a projective plane, the axiom of Pappus implies the
axiom of Desargues. In 1966, Buekenhout [7] had shown that, in a projective
plane, the axiom of Pascal for an oval implies the axiom of Pappus for the
plane. This context suggests that the axiom of Pascal for the rimpoints should
be sufficient without the need for the axioms of Pappus and Desargues. This
paper establishes that suggestion to be valid, fulfilling Menger’s dream, but
for two additional axioms that ensures that midpoints exist, and that
perpendicularity of lines is well-defined.

We will be considering an incidence structure that fulfils the axioms (A1)
through (A7) plus (A10) and (A11) below. The first four axioms allow us to
have a ternary relation of _betweenness_. For three collinear points $A,B,C$,
the point $B$ is said to be between $A$ and $C$, if every line through $B$
intersects at least one line of each pair of intersecting lines which pass
through $A$ and $C$, respectively. Using the concept of betweenness one can
define _ray_ and _segment_ in the usual manner. Two rays $r$ and $s$ with
endpoints $R$ and $S$, respectively, are said to be (strictly) parallel if $r$
and $s$ are noncollinear and if every line that meets one of the rays meets
either the other ray or the segment $\overline{RS}$. Two lines, or a ray and a
line, are said to be parallel if they contain rays that are parallel. Then it
can be shown that there are exactly two lines through a given point which are
parallel to a given line [9, Corollary 2.4]. By (A6) below, two non-collinear
rays have a common parallel line.

A pair $(a,b)$ of parallel lines is called a rimpoint if there is a line
$\ell$ parallel to both $a$ and $b$, and there is a line which intersects $a$,
$b$ and $\ell$. A rimpoint $(a,b)$ is equal to another rimpoint $(c,d)$ if
every pair of lines from $\\{a,b,c,d\\}$ are identical or parallel. A rimpoint
$(a,b)$ is incident with a line $\ell$ if $\ell=a$, $\ell=b$ or $\ell$ is
parallel to both $a$ and $b$, and there exists a line that intersects $a$,
$b$, and $\ell$. From (A1) – (A7) below, every line of our incidence structure
has two rimpoints [1, Theorem 1.7]. It is useful to the reader to have in mind
the examples of hyperbolic planes characterised by our main result: the
Cayley-Klein hyperbolic planes over Euclidean fields. Thus, one constructs
such a geometry by first considering the unit circle $\mathcal{O}$ centred at
the origin of the Cartesian plane $\mathbb{E}^{2}$, where $\mathbb{E}$ is a
Euclidean field. The finite points of the hyperbolic plane are the points of
$\mathbb{E}$ lying within the interior of circle, the lines of the hyperbolic
plane are the chords of the circle, and the rimpoints are (essentially) the
points of the circle. Two lines are parallel if they intersect in a point on
the circle, and two lines are perpendicular if they are conjugate with respect
to the circle.

We will use the term ‘point’ for both a point and a rimpoint, and we will
refer to points of our incidence structure as ‘finite points’. If $A$ and $B$
are points, then we write $AB$ for the unique line incident with both $A$ and
$B$. We will often write a polygon in an order that distinguishes sides of the
polygon. For example, $ABCDEF$ is a hexagon with distinguished sides $AB$,
$BC$, $CD$, and so forth.

######  Theorem 1.1 (Skala [29]).

Every model of axioms (A1) through (A9) is a Cayley-Klein hyperbolic plane
over a Euclidean field.

  1. (A1)

Any two distinct points are incident with a unique line.

  2. (A2)

Each line is on at least one point.

  3. (A3)

There exist three non-collinear points, and three collinear points.

  4. (A4)

(Betweenness) Of three collinear points, at least one has the property that
every line through it intersects at least one of each pair of intersecting
lines through the other two.

  5. (A5)

(Hyperbolicity) If $P$ is not on $\ell$, then there exist two distinct lines
on $P$ not meeting $\ell$ and such that each line meeting $\ell$ meets at
least one of those two lines.

  6. (A6)

(Parallelism) Any two non-collinear rays have a common parallel line.

  7. (A7)

(Pascal’s Theorem on rimpoints) If $A$, $B$, $C$, $D$, $E$, $F$ are rimpoints
such that the three diagonal points $AB\cap DE$, $BC\cap EF$, $CD\cap FA$ of
the hexagon $ABCDEF$ exist, then the diagonal points are collinear.

  8. (A8)

(Pappus’ Theorem) Let $a$ and $b$ be lines containing points $A_{1}$, $A_{2}$,
$A_{3}$, and $B_{1}$, $B_{2}$, $B_{3}$, respectively. If the points
$A_{1}B_{2}\cap A_{2}B_{1}$, $A_{1}B_{3}\cap A_{3}B_{1}$, $A_{2}B_{3}\cap
A_{3}B_{2}$ exist, then they are collinear.

  9. (A9)

(Desargues’ Theorem) Let the lines $a$, $b$, $c$ be concurrent in a point and
contain pairs of points $\\{A_{1},A_{2}\\}$, $\\{B_{1},B_{2}\\}$ and
$\\{C_{1},C_{2}\\}$, respectively. If the points $A_{1}B_{1}\cap A_{2}B_{2}$,
$A_{1}C_{1}\cap A_{2}C_{2}$, $B_{1}C_{1}\cap B_{2}C_{2}$ exist, then they are
collinear.

As Pambuccian observed in footnote 7 of [22], Skala’s original Axiom 4 follows
from her Axiom 6. That is why we have presented only nine of her ten axioms
here. We should also point out that Skala’s statement of (A9) was incorrect,
but that her correct statement of the ‘Dual of Desargues’ Theorem’ [29, p.
259] remedied this anomaly. We will be considering the following axiom in our
formulation:

  1. (A10)

Given distinct finite points $P$, $Q$ there exist lines $\ell$ on $P$ and $m$
on $Q$ such that two common parallels to $\ell$ and $m$ meet on $PQ$.

This axiom allows us to define (unique) midpoints of line segments in our
incidence structure: the _midpoint_ of the segment $\overline{PQ}$ is defined
to be the point of intersection $\ell\cap m$ in the statement of (A10).
Moreover (A10) leads us to a new notion of perpendicularity of lines (see
Definition 4.2), that is independent of a notion of congruence, and based
purely on the properties of _axial maps_. This approach was motivated by the
work of Rigby [27]. Our relation on lines has the usual properties of
perpendicularity that holds for Cayley-Klein hyperbolic planes: (a) two non-
intersecting non-parallel lines have a unique common perpendicular line, (b)
for each point $P$ and line $\ell$ there exists a unique line through $P$
perpendicular to $\ell$, (c) there are no rectilaterals. To ensure that our
definition (in a later section) of perpendicularity is well-defined, we need
another axiom. In the following, if $X$ is a rimpoint and $A$ is a finite
point, then $X^{A}$ is the other rimpoint on the line $XA$ joining $X$ and
$A$.

  1. (A11)

Let $X$ be a rimpoint not on a line $\ell$. Then there is a unique point $R$
on $\ell$ such that for every finite point $S$ on $\ell$, two common parallels
to $\ell$ and $X^{SR}X^{RS}$ meet on $XR$.

We show that we can substitute $\mathrm{(A8)}\wedge\mathrm{(A9)}$ for the
weaker $(\mathrm{A10})\wedge(\mathrm{A11})$, in order to completely
characterise hyperbolic planes over Euclidean fields.

######  Theorem 1.2.

Every model of axioms (A1) through (A7), plus (A10) and (A11), is a hyperbolic
plane and conversely. Moreover, these models are precisely the Cayley-Klein
hyperbolic planes over Euclidean fields.

The proof of this result appears in Section 6 and it draws together the ideas
of Bachmann [5], Buekenhout [7] and Rigby [27] to develop a theory of half-
turns and reflections for an incidence structure satisfying the above axioms.
Thus Sections 2, 3, 4, 5, 6 develop this extra structure on our incidence
structure. We show that axioms (A1) – (A7) plus (A10) and (A11) enable us to
affirm the axioms of a _metric plane_ , and then with a bit more work, we use
Bachmann’s characterisation theorem for hyperbolic planes (Theorem 6.5) to
establish an isomorphism to a Cayley-Klein hyperbolic plane over a Euclidean
field. An alternative proof will be given in Section 7 that uses the theory of
abstract ovals.

### Some notation and definitions

In this paper, we will often encounter standard concepts in incidence geometry
and group theory. An incidence structure is a triple
$(\mathcal{P},\mathcal{B},\mathrm{I})$ with $\mathcal{P}$ and $\mathcal{B}$
disjoint, non-empty sets, and $\mathrm{I}$ a relation between them; the
_incidence relation_. An isomorphism from the incidence structure
$(\mathcal{P},\mathcal{B},\mathrm{I})$ to the incidence structure
$(\mathcal{P^{\prime}},\mathcal{B^{\prime}},\mathrm{I^{\prime}})$ is a
bijection
$\phi\colon\mathcal{P}\cup\mathcal{B}\to\mathcal{P^{\prime}}\cup\mathcal{B^{\prime}}$
such that $\mathcal{P}^{\phi}=\mathcal{P^{\prime}}$,
$\mathcal{B}^{\phi}=\mathcal{B^{\prime}}$, and for all $P\in\mathcal{P}$ and
$B\in\mathcal{B}$:

$P\,\mathrm{I}\,B\Leftrightarrow P^{\phi}\,\mathrm{I^{\prime}}\,B^{\phi}.$

An automorphism of an incidence structure
$\mathcal{S}=(\mathcal{P},\mathcal{B},I)$ is an isomorphism to itself. We
denote the set of all automorphisms of $\mathcal{S}$ by
$\mathsf{Aut}(\mathcal{S})$. Moreover, $\mathsf{Aut}(\mathcal{S})$ is a group
under composition, called the automorphism group of $\mathcal{S}$. If $G$ is a
group acting on a set $\Omega$, and $H$ is a group acting on a set $\Gamma$,
we say that these two actions are permutationally isomorphic, and we write
$(\Omega,G)\cong(\Gamma,H)$, if there exists a group isomorphism $\varphi:G\to
H$ and a bijection $\beta:\Omega\to\Gamma$ such that
$\beta(\omega^{g})=\beta(\omega)^{\varphi(g)}$ for all $\omega\in\Omega$ and
$g\in G$. The pair $(\beta,\varphi)$ is a permutational isomorphism from
$(\Omega,G)$ to $(\Gamma,H)$, and permutational isomorphism is an equivalence
relation on group actions.

##  2\. Cyclic order and separation

A nonempty set $\Omega$ is said to be _cyclically ordered_ if it is equipped
with a ternary relation $\mathcal{C}$ that possesses the following properties
(see [13]):

  1. (C1)

if $\mathcal{C}(x,y,z)$, then $x\neq y\neq z\neq x$;

  2. (C2)

if $x\neq y\neq z\neq x$, then exactly one of the relations
$\mathcal{C}(x,y,z)$ or $\mathcal{C}(x,z,y)$ holds;

  3. (C3)

if $\mathcal{C}(x,y,z)$, then $\mathcal{C}(y,z,x)$;

  4. (C4)

if $\mathcal{C}(x,y,z)$ and $\mathcal{C}(x,z,t)$, then $\mathcal{C}(x,y,t)$.

The _reverse_ $\bar{\mathcal{C}}$ of $\mathcal{C}$ is the order
$\bar{\mathcal{C}}(x,y,z)$ if and only if $\mathcal{C}(z,y,x)$.

The relation of _separation_ is a (nonempty) quaternary relation on a set
$\Omega$ if the following properties hold on five elements $A$, $B$, $C$, $D$,
$X$ of $\Omega$ (see [13] and [26]):

  1. (S1)

if $A$, $B$ separates $C$, $D$, then $A$, $B$, $C$, $D$ are distinct;

  2. (S2)

if $A$, $B$ separates $C$, $D$, then $A$, $B$ separates $D$, $C$;

  3. (S3)

if $A$, $B$ separates $C$, $D$, then it is not the case that $A$, $C$
separates $B$, $D$;

  4. (S4)

$A$, $B$ separates $C$, $D$ or $A$, $C$ separates $B$, $D$ or $A$, $D$
separates $B$, $C$;

  5. (S5)

If $A$, $B$ separates $C$, $D$, and $A$, $C$ separates $B$, $X$, then $A$, $B$
separates $D$, $X$.

We will often use the notation $AB//CD$ for “$A$, $B$ separates $C$, $D$”, and
it will soon become clear why this notation suits our purposes. Given a cyclic
order $\mathcal{C}$ on $\Omega$, we can construct a separation relation on
$\Omega$ by the following rule:

$AC//BD\iff\left(\mathcal{C}(A,B,C)\wedge\mathcal{C}(C,D,A)\right)\vee\left(\mathcal{C}(A,D,C)\wedge\mathcal{C}(C,B,A)\right).$

Conversely, given a separation relation on a set $\Omega$, we can associate
two cyclic orders on $\Omega$, each of which is the reverse of the other.

######  Definition 2.1.

Let $\mathcal{I}$ be an incidence structure of points and lines satisfying
(A1) through to (A7). Let $\mathcal{O}$ be the set of rimpoints of
$\mathcal{I}$. For four rimpoints $A$, $B$, $C$, $D$, we write $AB//CD$ if the
line $AB$ meets $CD$ in a finite point.

We will show that the above quaternary relation on rimpoints satisfies the
axioms for a separation relation. It is clear that Axioms (S1) and (S2) hold,
and Theorem 3.5 shows that (S3), (S4) and (S5) hold. The _Pasch property_
follows from the first five axioms (A1)–(A5) [25, p. 168], and is the
culmination of the work of Menger’s students, from DeBaggis [8] and Jenks
[14], to Abbott [1] (for the _closed_ plane; the finite points and the
rimpoints).

######  Theorem 2.2 (Pasch property ([1, Theorem 1.11])).

If $PQR$ is any triangle in $\mathcal{I}$ and if $\ell$ is any line which
intersects $PQ$, then $\ell$ contains $R$ or intersects exactly one other side
in an interior point.

######  Corollary 2.3.

Let $A,B,C,M,N$ be distinct rimpoints. Then $\\{M,N\\}$ separates two or none
of the pairs $\\{A,B\\}$, $\\{B,C\\}$, $\\{C,A\\}$.

###### Proof.

Suppose $\\{M,N\\}$ separates at least one of the pairs. Without loss of
generality, we may assume that $\\{M,N\\}$ separates $\\{A,B\\}$. Therefore,
$MN$ and $AB$ meet in a finite point $P$. Consider the triangle $ABC$. Then
the line $MN$ intersects a side of $ABC$ and so by Theorem 2.2, it intersects
exactly one of $BC$ or $CA$ (n.b., $MN$ does not contain a vertex as $M,N\neq
A,B,C$). So it follows immediately that $\\{M,N\\}$ separates one of
$\\{B,C\\}$ or $\\{C,A\\}$. ∎

Following Abbott [1], given a line $\ell$, we say that two points $A$ and $B$
that are not incident with $\ell$ are _on the same side of $\ell$_ if one of
the following holds:

  1. (i)

$A=B$,

  2. (ii)

$A$ and $B$ are rimpoints and $AB$ does not intersect $\ell$ in a finite
point,

  3. (iii)

$A$ and $B$ are finite points and the segment $\overline{AB}$ does not have a
finite point in common with $\ell$,

  4. (iv)

$A$ is finite and $B$ is a rimpoint, and if $X$ and $X^{\prime}$ are the two
rimpoints not incident with $\ell$ such that $AX$ and $AX^{\prime}$ are the
two lines through $A$ parallel to $\ell$, then $B$ lies on the same side of
$\ell$ as $X,X^{\prime}$.

  5. (v)

$A$ is a rimpoint and $B$ is finite, and we apply (iv) to $B$ and $A$.

The property (iv) is well-defined since the two rimpoints $X$ and $X^{\prime}$
on the lines through $A$ parallel to $\ell$ satisfy the property (ii) [1,
Theorem 1.2]. We write $A\sim_{\ell}B$ for this relation ‘on the same side of
$\ell$’.

######  Lemma 2.4 ([1, Theorem 1.1]).

The relation $\sim_{\ell}$ is an equivalence relation with two classes.

The two equivalence classes of $\sim_{\ell}$ are the two _sides_ of $\ell$.

######  Remark 2.5.

Another way to express this equivalence relation $\sim_{\ell}$ just for the
rimpoints is by using a cyclic order $\mathcal{C}$ associated to the
separation relation $//$. It is not difficult to see that if the rimpoints of
$\ell$ are $L$ and $L^{\prime}$, then rimpoints $A$ and $B$ lie on the same
side of $\ell$ if and only if $\mathcal{C}(L,A,L^{\prime})$ and
$\mathcal{C}(L,B,L^{\prime})$, or $\mathcal{C}(L^{\prime},A,L)$ and
$\mathcal{C}(L^{\prime},B,L)$.

##  3\. Half-turns and axial permutations

Throughout this section, we will suppose that $\mathcal{I}$ is an incidence
structure satisfying (A1) through (A7), and we will also be adopting (A10)
later in this section, but not immediately.

###  3.1. Basic properties of half-turns

Given a point $P$, the _half-turn_ about $P$ is a permutation of $\mathcal{O}$
defined in the following way: If $X\in\mathcal{O}$, then $X^{P}$ is the unique
element of $\mathcal{O}$ incident with $XP$, not equal to $X$. Therefore, if
$A$ and $B$ are rimpoints and $P$ is a finite point on $AB$, then $A^{P}=B$.
We will use the same letter for a finite point and its half-turn, as is
standard in transformation geometry. Each half-turn is determined by two of
its transpositions in its decomposition into disjoint transpositions, for if
$(A\,B)$ and $(C\,D)$ are two such components of the half-turn about $X$, then
the point $X$ is the intersection of $AB$ and $CD$. We use the symbol ‘$1$’ to
denote the identity permutation on the rimpoints. Theorem 3.1 is the analogue
of [7, Proposition 7.2] for abstract ovals.

######  Theorem 3.1.

If $P$, $Q$, $R$ are finite distinct points such that $PQR$ is an involution,
then $P$, $Q$, $R$ are collinear.

###### Proof.

Let $A\in\mathcal{O}$. Let $B:=A^{P}$, $C:=B^{Q}$, $D:=C^{R}$, $E:=D^{P}$,
$F:=E^{Q}$. Then $F^{R}=A$ since $PQRPQR=1$. Thus the points $P=AB\cap DE$,
$Q=BC\cap EF$ and $R=CD\cap FA$ exist, and are collinear by (A7). ∎

######  Lemma 3.2.

Any half-turn about a point $P$ on a line $\ell$ interchanges the sides of
$\ell$.

###### Proof.

Let $Q$ be a rimpoint not incident with $\ell$. By definition, $Q^{P}$ is the
unique element of $\mathcal{O}$ on $PQ$ not equal to $Q$. So $QQ^{P}$ and
$\ell$ intersect in a finite point (namely $P$), and therefore, $Q^{P}$ lies
on the opposite side of $\ell$ as $Q$. ∎

######  Lemma 3.3.

Let $P$ and $Q$ be finite points.

  1. (a)

If $P\neq Q$, then for all rimpoints $X\in\mathcal{O}$ such that $X$ does not
lie on the line $PQ$, we have that $X$ and $X^{PQPQ}$ lie on different sides
of the line joining $X^{P}$ and $X^{PQ}$.

  2. (b)

If the permutations $PQ$ and $QP$ are equal, then $P=Q$.

###### Proof.

For the proof of (a), let $X$ be a rimpoint not incident with $PQ$. Let $\ell$
be the line spanned by $X^{P}$ and $X^{PQ}$. Now the line joining $X^{P}$ and
$X$ is incident with $P$ and so $X$ and $P$ lie on the same side of $\ell$.
Similarly, the line joining $X^{PQP}$ and $X^{PQ}$ is incident with $P$ and so
$X^{PQP}$ and $P$ lie on the same side of $\ell$. Therefore, by Lemma 2.4, $X$
and $X^{PQP}$ lie on the same side of $\ell$. Now the line spanning $X^{PQP}$
and $X^{PQPQ}$ passes through $Q$, and $Q$ is incident with $\ell$, so
$X^{PQP}$ and $X^{PQPQ}$ lie on different sides of $\ell$ (by Lemma 3.2). It
then follows from Lemma 2.4 that $X$ and $X^{PQPQ}$ lie on different sides of
$\ell$.

Now we show that (b) holds. Suppose $P\neq Q$. Then there exists a rimpoint
$X$ not lying on the line joining $P$ and $Q$ (by (A3)). Then by (a), it is
clear that $X\neq X^{PQPQ}$. Therefore, $PQPQ\neq 1$ and hence $PQ\neq QP$. ∎

######  Lemma 3.4.

Let $M$ and $N$ be two rimpoints, and let $\ell$ be the line spanned by them.
Suppose $X$ and $Y$ are two distinct rimpoints lying on the same side of
$\ell$. Then $XM$ intersects $YN$ or $XN$ intersects $YM$, in a finite point.

###### Proof.

Let $P$ be a finite point on $MN$, and suppose $XM$ does not intersect $YN$.
So $Y$ and $N$ are on the same side of $XM$. Now $P$ and $N$ are on the same
side of $XM$ and hence $P$ and $Y$ are on the same side of $XM$ (by Lemma
2.4). So $PY$ does not meet $XM$. By [9, Lemma 2.5], $PY$ meets $XN$ and so
$P$ and $Y$ lie on different sides of $XN$. Now $P$ and $M$ are on the same
side of $XN$, and $P$ and $Y$ are on different sides of $XN$. So by Lemma 2.4,
$M$ and $Y$ are on different sides of $XN$. Therefore, $XN$ intersects $YM$ in
a finite point. ∎

Now we show that the relation $//$ from Definition 2.1 satisfies (S3), (S4)
and (S5).

######  Theorem 3.5.

$//$ is a separation relation.

###### Proof.

Let $A$, $B$, $C$, $D$, $X$ be distinct rimpoints.

(S3) holds:

The following argument is a direct analogue of [30, Theorem 2.5]. Suppose
$AB//CD$; that is, there exists a finite point $P$ on both $AB$ and $CD$. For
a proof by contradiction, suppose $AC$ and $BD$ intersect in a finite point
$Q$. Now $A^{P}=B$, $C^{P}=D$, $A^{Q}=C$ and $B^{Q}=D$, from which we see that
$PQPQ$ fixes $A$, $B$, $C$ and $D$. Suppose $P\neq Q$. Since $C\neq D$, we
have that $A^{Q}\neq A^{P}$ and hence $A$ is not incident with $PQ$. So by
Lemma 3.3, $A$ and $A^{PQPQ}$ lie on different sides of the line joining
$A^{P}$ and $A^{PQ}$; which is a contradiction as $A=A^{PQPQ}$. Therefore,
$P=Q$ and $B=A^{P}=A^{Q}=C$; a contradiction. We have shown that $AB//CD$
implies that $AC//BD$ does not hold.

(S4) holds:

Suppose $AB//CD$ does not hold. Then $A$ and $B$ lie on the same side of the
line $\ell:=CD$, and so by Lemma 3.4, $AC$ intersects $BD$ or $AD$ intersects
$BC$. Therefore, $AC//BD$ or $AD//BC$.

(S5) holds:

Suppose $AB//CD$ and $AC//BX$. In particular, we do not have $AB//CX$, by
(S3). Now $\\{A,B\\}$ separates one of the pairs $\\{C,D\\}$, $\\{D,X\\}$,
$\\{X,C\\}$ (namely $\\{C,D\\}$) and hence, by Corollary 2.3, two of the
pairs. Therefore, $AB//DX$.∎

######  Lemma 3.6.

Consider two finite points $P$ and $Q$, on a line $\ell$, and suppose $A$ and
$D$ are two rimpoints on different sides of $\ell$. Then
$AD^{{P}{Q}}//A^{{P}{Q}}D$.

###### Proof.

Let $X,X^{\prime}$ be the rimpoints incident with $\ell$. Then
$XX^{\prime}//AD$, and by (S3), we may suppose without loss of generality that
$XA//X^{\prime}A^{PQ}$ (otherwise, reverse the roles of $X$ and $X^{\prime}$).
So by (S5), we have111We can readily see how to apply (S5) via the following
mapping of symbols $\left(\begin{smallmatrix}A&B&C&D&X\\\
X&A&X^{\prime}&A^{PQ}&D\end{smallmatrix}\right)$.

$XA//A^{PQ}D,\quad\text{and similarly,}\quad XD//D^{PQ}A.$

Applying (S5) to both the statements ‘$XA//DA^{PQ}$ and $DX//AD^{PQ}$’ and
‘$XD//AD^{PQ}$ and $AX//DA^{PQ}$’ yields222Again, it helps to know how we map
symbols: $\left(\begin{smallmatrix}A&B&C&D&X\\\
X&A&D&A^{PQ}&D^{PQ}\end{smallmatrix}\right)$.:

$AX//A^{PQ}D^{PQ}\quad\text{ and }\quad DX//D^{PQ}A^{PQ}.$

Again, applying (S5) to ‘$AX//D^{PQ}A^{PQ}$ and $AD^{PQ}//XD$’ yields333On
this occasion, we use the mapping $\left(\begin{smallmatrix}A&B&C&D&X\\\
A&D^{PQ}&X&D&A^{PQ}\end{smallmatrix}\right)$.  $AD^{PQ}//A^{PQ}D$ as required.
∎

Inspired by [7, Proposition 7.3], we have the following important result on
half-turns. We remind the reader that only axioms (A1) through (A7) are
necessary for this result to hold.

######  Theorem 3.7.

If $P$, $Q$, $R$ are collinear finite points belonging to a line $d$, then
$PQR$ is an involution, and moreover, $PQR={P^{\prime}}$ for some point
$P^{\prime}$ on $d$.

###### Proof.

Let $A$ be a rimpoint not incident with $PQ$. Let $B=A^{P}$, $C=B^{Q}$,
$D=C^{R}$, $E=D^{P}$, $F=E^{Q}$. Then $P=AB\cap DE$, $Q=BC\cap EF$. By Lemma
3.6, the lines $AD^{PQ}$ and $DA^{PQ}$ meet, or in other words, $AF$ meets
$CD$. Then by (A7), $P$, $Q$ and $AF\cap CD$ are collinear, and hence $AF\cap
CD=PQ\cap CD=R$. So $A=F^{R}$, and $A$ and $D$ are interchanged by $PQR$:

$A^{PQR}=B^{QR}=C^{R}=D,\quad D^{PQR}=E^{QR}=F^{R}=A.$

Since $A$ is arbitrary subject to not being on $PQ$, we have that $PQR$ is an
involution.

Let $P^{\prime}:=PQ\cap AD$. We will show $PQR=P^{\prime}$. Take a rimpoint
$Y$ not incident with $PQ$, and suppose $Y$ lies on the opposite side of $PQ$
to $B$. Let $Z:=Y^{{P^{\prime}}}$, $Y^{\prime}:=Z^{P}$ and
$Z^{\prime}:=(Y^{\prime})^{Q}$. Note that $B$ and $Y^{\prime}$ lie on opposite
sides of $PQ=PP^{\prime}$ (by Lemma 3.2). By Lemma 3.6, $BY$ meets
$DY^{\prime}$ (where $P$, $P^{\prime}$, $Y^{\prime}$, $B$ play the roles of
$P$, $Q$, $A$, $D$), since $A^{P^{\prime}}=D$. Let $T:=BY\cap DY^{\prime}$. By
(A7) applied to the hexagon $ABYZY^{\prime}D$, we have that $P^{\prime}=AD\cap
YZ$, $P=AB\cap YZ^{\prime}$ and $T=BY\cap DY^{\prime}$ are collinear. So $T$
is incident with $PQ$. Note that $D$ and $Y^{\prime}$ lie on opposite sides of
$PQ=TQ$ as $D$ lies on the same side of $PQ$ as $B$ (by Lemma 3.2). By Lemma
3.6, $YZ^{\prime}$ meets $CD$ (where $T$, $Q$, $Y$, $D$ play the roles of $P$,
$Q$, $A$, $D$), since $Y^{TQ}=B^{Q}=C$ and
$D^{TQ}=(Y^{\prime})^{Q}=Z^{\prime}$. By (A7) applied to the hexagon
$CBYZ^{\prime}Y^{\prime}D$, we have that $T=BY\cap DY^{\prime}$,
$Q=Y^{\prime}Z^{\prime}\cap BC$ and $YZ^{\prime}\cap CD$ are collinear. So
$YZ^{\prime}\cap CD$ is incident with $PQ$ and therefore equals $CD\cap PQ=R$.
Thus $(Z^{\prime})^{R}=Y$ and hence $Y$ is fixed by ${P^{\prime}}PQR$. Since
$Y$ was arbitrary subject to being not incident with $PQ$ and lying on the
opposite side to $PQ$ as $B$, we have that ${P^{\prime}}PQR$ fixes every
rimpoint on one side of $PQ$. A similar argument shows that ${P^{\prime}}PQR$
fixes every rimpoint on the other side of $PQ$ as well. Therefore,
${P^{\prime}}PQR=1$ and so $PQR={P^{\prime}}$. ∎

######  Theorem 3.8.

Each half-turn is an automorphism of $\mathcal{I}$.

###### Proof.

Let $S$ be the set of half-turns as permutations of the set $\mathcal{O}$ of
rimpoints. We will first show that the group $G$ generated by the half-turns
is an _$S$ -group_ as defined in [17, §4]: a group $G$ together with a
distinguished generating subset $S$ of the full set of involutions $J$, such
that the following holds: if $a\neq b$ and $abx,aby,abz\in J$, then $xyz\in
S$. Let $A$ and $B$ be distinct elements of $S$, and let $X,Y,Z\in S$ such
that $ABX$, $ABY$, $ABZ$ are involutions. Now $ABX$ is an involution and so by
Theorem 3.1, $A$, $B$, $X$ are collinear. Likewise, $Y$ and $Z$ lie on $AB$.
Therefore, $X$, $Y$ and $Z$ all lie on $AB$, and so by Theorem 3.7, $XYZ\in
S$. Therefore, the group $G$ generated by $S$ is an $S$-group.

Now given an $S$-group $(G,S)$ we can define an incidence structure $D(G,S)$
as follows: the points are the elements of $S$, and a line is a set of points
$S_{a,b}:=\\{x:(abx)^{2}=1\\}$ where $a$ and $b$ are two distinct elements of
$S$. By [17, page 38], the conjugation action of $G$ induces a subgroup of
automorphisms of $D(G,S)$.

Let $\mathcal{P}$ be the set of points of $\mathcal{I}$ and let $\mathcal{L}$
be the lines of $\mathcal{I}$. For each point $X$, let
$\beta_{\mathcal{P}}(X)$ be the half-turn about $X$. Note that
$\beta_{\mathcal{P}}$ is a bijection between $\mathcal{P}$ and $S$. So the
conjugation action of $G$ on $S$ induces a permutationally isomorphic action
of $G$ on $\mathcal{P}$ via the map $\beta_{\mathcal{P}}$: that is, for every
point $X$ and point $P$, the image $X^{P}$ is the unique point fixed by the
half-turn $PXP$. For lines, it is similar. Let $\ell\in\mathcal{L}$ and define
$\beta_{\mathcal{L}}(\ell)$ to be
$S_{\beta_{\mathcal{P}}(M),\beta_{\mathcal{P}}(N)}$ where $M$ and $N$ are
rimpoints of $\ell$. So again, the action of $G$ on pairs of elements of $S$
induces a permutationally isomorphic action of $G$ on $\mathcal{L}$ whereby
$\ell^{P}$ is the line spanned by $M^{P}$ and $N^{P}$. In $D(G,S)$, the half-
turn $\beta_{\mathcal{P}}(X)$ about $X$ is incident with
$S_{\beta_{\mathcal{P}}(M),\beta_{\mathcal{P}}(N)}$ if and only if
$(\beta_{\mathcal{P}}(M)\beta_{\mathcal{P}}(N)\beta_{\mathcal{P}}(X))^{2}=1$.
By Theorem 3.1, $MNX$ are collinear and hence $X$ is incident with $MN$; and
vice-versa (by Theorem 3.7). Therefore, $D(G,S)$ is isomorphic to
$\mathcal{I}$ (as an incidence structure) and the result follows. ∎

######  Corollary 3.9.

The half-turn about $P$ fixes a line $\ell$ if and only if $P$ is incident
with $\ell$.

###### Proof.

Suppose $P$ fixes $\ell$, and let $Q$ be a point incident with $\ell$. Then by
Theorem 3.8, $Q^{P}$ is incident with $\ell^{P}$, which is equal to $\ell$.
Since $P$, $Q$ and $Q^{P}$ are collinear, and $\ell$ is the unique line
joining $Q$ and $Q^{P}$ (by (A1)), it follows that $P$ is incident with
$\ell$. Conversely, if $P$ is incident with $\ell$, then the half-turn about
$P$ fixes $P$ and maps $\ell$ to a line $\ell^{P}$ incident with $P$ (by
Theorem 3.8). The rimpoints $L$ and $L^{\prime}$ of $\ell$ are mapped to the
rimpoints of $\ell^{P}$, which must be interchanged by $P$. It follows that
$\ell^{P}=\ell$. ∎

######  Corollary 3.10.

Every half-turn preserves the separation relation $//$; that is, for every
quadruple of distinct rimpoints $W$, $X$, $Y$, $Z$, and for every finite point
$P$, we have $WX//YZ$ if and only if $W^{P}X^{P}//Y^{P}Z^{P}$.

A pair of transpositions $(A\,B)$ and $(C\,D)$ in the symmetric group on the
rimpoints is _separating_ if $\\{A,B\\}$ and $\\{C,D\\}$ are separating.

######  Lemma 3.11.

Let $g$ be a fixed-point-free involution in the symmetric group on the
rimpoints such that each pair of transpositions of $g$ are separating. Then
$g$ is a half-turn.

###### Proof.

Consider three pairs of transpositions of $g$: $(A\,B)$, $(C\,D)$, $(E\,F)$.
Suppose, for a proof by contradiction, that the lines $AB$, $CD$, $EF$ are not
all incident with the same finite point. Since every pair of these lines
separate, we have three finite points $X:=AB\cap CD$, $Y:=CD\cap EF$,
$Z:=EF\cap AB$ fixed by $g$. Let us now consider the points $A$, $X$, $Z$, $B$
which lie on $AB$. Then $A$, $C^{X}=D$, $C^{Z}$, $B$ are four distinct
rimpoints. Now $g$ is an automorphism of $\mathcal{I}$ (by Theorem 3.8), and
so $C^{Zg}$ is equal to $D^{Z}$ (because $g$ maps the line $CZ$ to $DZ$).
Since $D$ and $C^{Z}$ lie on the same side of $AB$, it follows that $AB$ does
not intersect $C^{Z}D$. So by (S4), $AC^{Z}//BD$ or $AD//C^{Z}B$. Without loss
of generality, we may suppose that $AC^{Z}//BD$. Now $g$ is an automorphism of
$\mathcal{I}$, and so preserves the separation relation (by Corollary 3.10),
and hence $A^{g}C^{Zg}//B^{g}D^{g}$. Computing each image under $g$ yields
$BD^{Z}//AC$. Now we apply $Z$ and reveal that $AD//BC^{Z}$. However, by
assumption $AC^{Z}//BD$, which contradicts (S3). Therefore, all lines
corresponding to transpositions of $g$ are concurrent in a single finite point
$P$, and therefore, $g$ is a half-turn. ∎

######  Remark 3.12.

A consequence of Lemma 3.11 (and Theorem 3.7) is that if $P$ and $Q$ are
finite points, then the map $PQP$ is the half-turn about the point $Q^{P}$.
Indeed, we have the following corollary of Lemma 3.11.

######  Corollary 3.13.

Let $\tau$ be a permutation of $\mathcal{O}$ that preserves the separation
relation $//$. Then for each finite point $P$, the transformation
$\tau^{-1}P\tau$ is a half-turn about the point $P^{\tau}$.

######  Lemma 3.14.

Let $\mathcal{C}$ be one of the two cyclic orders associated to the separation
relation $//$. Then every half-turn preserves $\mathcal{C}$; that is, for
every triple of distinct rimpoints $X$, $Y$, $Z$, and for every finite point
$P$, we have $\mathcal{C}(X,Y,Z)$ if and only if
$\mathcal{C}(X^{P},Y^{P},Z^{P})$.

###### Proof.

Let $P$ be a finite point. The half-turn about $P$ preserves the separation
relation $//$, by Corollary 3.10. So $P$ preserves or reverses the cyclic
order $\mathcal{C}$. Assume, for a proof by contradiction, that $P$ reverses
$\mathcal{C}$. Suppose we have $\mathcal{C}(X,Y,Z)$ for three distinct
rimpoints $X$, $Y$, $Z$. If $Y=X^{P}$, then $Z$ and $Z^{P}$ lie on opposite
sides of $XY$ and we have by Remark 2.5, $\neg\mathcal{C}(X,X^{\prime},Z^{P})$
and hence $\mathcal{C}(X,Z^{P},Y)$ (by (C2)). However, after applying $P$, we
have $\mathcal{C}(Y,X,Z)$ which contradicts $\mathcal{C}(X,Y,Z)$ and (C3). A
similar argument holds for $Y=Z^{P}$ or $X=Z^{P}$. So we can assume that $P$
does not lie on the lines $XY$, $YZ$, $ZX$. If $Y$ and $Z$ lie on the same
side of $XX^{P}$, then $XZ//X^{P}Y$ (because lying on the same side of
$XX^{P}$ means that $\mathcal{C}(X,Y,X^{P})$ and $\mathcal{C}(X,Z,X^{P})$, or
$\mathcal{C}(X^{P},Y,X)$ and $\mathcal{C}(X^{P},Z,X)$), and similarly, $Y^{P}$
and $Z^{P}$ lie on the same side of $XX^{P}$ and hence $XZ^{P}//X^{P}Y^{P}$.
However, $P$ preserves the separation relation $//$ (by Corollary 3.10) and
hence $X^{P}Z//XY$, which contradicts $XZ//X^{P}Y$ (by (S3)). Therefore, $Y$
and $Z$ lie on opposite sides of $XX^{P}$. In this case, $Y^{P}$ and $Z$ lie
on the same side of $XX^{P}$. So by applying the previous argument with
$\\{Y^{P},Z\\}$ in the place of $\\{Y,Z\\}$ (the ordering dependent on whether
$\mathcal{C}(X,Y^{P},Z)$ or $\mathcal{C}(X,Z,Y^{P})$), we obtain again a
contradiction. Therefore, $P$ preserves $\mathcal{C}$. ∎

###  3.2. Adopting (A10) and enabling midpoints

We now suppose that our incidence structure $\mathcal{I}$ satisfies (A10).
Recall from Axiom (A10) that given distinct finite points $P$, $Q$, there
exist lines $\ell$ on $P$ and $m$ on $Q$ such that two common parallels to
$\ell$ and $m$ meet on $PQ$. We show now that this point is indeed the
‘midpoint’ of $\overline{PQ}$ in the language of half-turns.

######  Lemma 3.15 (Midpoints exist).

Given two finite points $P$ and $Q$, there exists a finite point $M$ such that
$PM=MQ$.

###### Proof.

By (A10), there exist lines $\ell$ on $P$ and $m$ on $Q$ such that if $X$, $Y$
are the rimpoints on $\ell$ and $X^{\prime}$, $Y^{\prime}$ are the rimpoints
on $m$, with $X$ and $X^{\prime}$ lying on the same side of $PQ$, then
$X^{\prime}Y$ and $XY^{\prime}$ meet in a point $M$ incident with $PQ$. Then,
$(X^{\prime})^{M}=Y$ and $(Y^{\prime})^{M}=X$ so $m^{M}=\ell$ (by Theorem 3.8)
and $(PQ)^{M}=PQ$, so $Q^{M}=(PQ\cap m)^{M}=PQ\cap\ell=P$ (again by Theorem
3.8). Now the permutation $MQM$ is a half-turn by Lemma 3.11, and it fixes the
point $Q^{M}=P$. Therefore, $MQM=P$ and the result follows. ∎

######  Lemma 3.16 (Midpoints are unique).

Let $P$ and $Q$ be finite points, and let $\ell$ and $m$ be lines through $P$
and $Q$ respectively such that the intersection $M$ of the two common
parallels to $\ell$ and $m$ is incident with $PQ$. Then $PM=MQ$.

###### Proof.

By Theorem 3.7, there exists a finite point $N$ incident with $PQ$ such that
the permutation $PMQ$ is equal to the half-turn about $N$. Let $X$ be a
rimpoint on $\ell$. Then the common parallels to $\ell$ and $m$ are $XX^{PMQ}$
and $X^{P}X^{PM}$, and so $N$ lies on both $\ell$ and $XX^{PMQ}$ (as
$X^{N}=X^{PMQ}$). Therefore, $M=N$ and $PM=MQ$. ∎

The theory for the remainder of this section was motivated by the late John
Frankland Rigby’s 1969 paper ‘Pascal ovals in projective planes’ [27].

A nonidentity permutation $\alpha$ of the set $\mathcal{O}$ of rimpoints,
preserving a cyclic order $\mathcal{C}$ compatible with the separation
relation $//$, is an order-preserving axial permutation if there exists a line
$\ell$ such that for all rimpoints $X$, $Y$ if $XY^{\alpha}$ meets
$YX^{\alpha}$, it does so in a finite point incident with $\ell$. We then say
that $\ell$ is the axis of $\alpha$.

######  Lemma 3.17.

The set of fixed rimpoints of an order-preserving axial mapping $\alpha$ with
axis $\ell$ is the set of rimpoints on $\ell$. Hence the axis is unique, and
an order-preserving axial mapping has exactly two fixed points.

###### Proof.

Let $X$ be a fixed rimpoint, and suppose $Y$ is a rimpoint that is not fixed
by $\alpha$. Then $X=XY^{\alpha}\cap YX^{\alpha}$ lies on $\ell$. Conversely,
suppose that $A$ is a rimpoint on $\ell$, not fixed by $\alpha$, and let $Y$
be a point not incident with $\ell$ such that $AY^{\alpha}$ meets
$A^{\alpha}Y$ in a finite point $Q$. Then $Q$ does not lie on $\ell$; a
contradiction. So the fixed rimpoints of $\alpha$ are exactly those incident
with $\ell$. ∎

Since $\alpha$ preserves $\mathcal{C}$ and fixes the rimpoints of $\ell$, it
follows that if $XY^{\alpha}$ meets $YX^{\alpha}$, then $X$ and $X^{\alpha}$
lie on the same side of $\ell$, and $X$ and $Y$ lie on opposite sides of
$\ell$.

######  Lemma 3.18.

Let $P$, $Q$ be distinct finite points. Then $PQ$ is an order-preserving axial
mapping with axis the line $\ell$ joining $P$ and $Q$.

###### Proof.

Certainly, $PQ$ fixes the rimpoints on $\ell$. If $X$ and $Y$ are rimpoints
not on $\ell$, apply (A7) to the hexagon with vertices

$X,X^{P},X^{PQ},Y,Y^{P},Y^{PQ}.$

We know that $XX^{P}$ and $YY^{P}$ meet at $P$, and $X^{P}X^{PQ}$ and
$Y^{P}Y^{PQ}$ meet at $Q$, so if $XY^{PQ}$ meets $YX^{PQ}$, it does so on
$\ell=PQ$. ∎

######  Lemma 3.19.

  1. (i)

If $\alpha$ is an order-preserving axial map with axis $\ell$, and if $P$ is a
finite point on $\ell$, then there exists a unique finite point $Q$ on $\ell$
such that $\alpha=PQ$.

  2. (ii)

The composition of two order-preserving axial mappings with the same axis is
an order-preserving axial map (with the same axis), or the identity mapping.

###### Proof.

Let $X$ be a rimpoint not on $\ell$. Then $X^{P}$ and $X^{\alpha}$ are on
opposite sides of $\ell$ (by Lemma 3.2) since an order-preserving axial map
fixes the sides of its axis (by Lemma 3.18 and Lemma 3.14). So the line
$X^{P}X^{\alpha}$ meets $\ell$ in a point $Q$. Let $Y$ be a rimpoint not
incident with $\ell$, lying on the other side of $\ell$ to $X$. So
$XY^{\alpha}$ meets $YX^{\alpha}$ in a finite point $T$, and by definition of
an order-preserving axial map, $T$ lies on $\ell$. Consider the hexagon
$XX^{P}X^{\alpha}YY^{P}Y^{\alpha}$. Then the diagonal points are $XX^{P}\cap
YY^{P}=P$, $XY^{\alpha}\cap YX^{\alpha}=T$, and $X^{P}X^{\alpha}\cap
Y^{P}Y^{\alpha}$. Now $X^{P}$ lies on the opposite side of $\ell$ to $X$, and
$X^{\alpha}$ lies on the same side of $\ell$ as $X$. Therefore,
$X^{P}X^{\alpha}\cap Y^{P}Y^{\alpha}$ exists and lies on $\ell$, and hence
$X^{P}X^{\alpha}\cap Y^{P}Y^{\alpha}=Q$. Therefore, $Y^{\alpha}=Y^{PQ}$. So we
have shown that $\alpha$ is equal to $PQ$ on the rimpoints lying on the side
of $\ell$ opposite to $X$. By a similar argument, and consideration of a
rimpoint $X^{\prime}$ on the opposite of $\ell$ to $X$, there exists a finite
point $Q^{\prime}$ incident with $\ell$ such that $\alpha$ is equal to
$PQ^{\prime}$ on the rimpoints lying on the same side of $\ell$ as $X$.
However, we also know, by definition of $Q$, that $X^{\alpha}=X^{PQ}$, and
hence $QQ^{\prime}$ fixes $X^{P}$. So by Lemma 3.17, $QQ^{\prime}$ is the
identity permutation and $Q=Q^{\prime}$. Therefore, $\alpha=PQ$ and (i) holds.

Suppose $\alpha$ and $\beta$ are two order-preserving axial maps with axis
$XY$ (where $X$ and $Y$ are rimpoints). Let $P$ be a point incident with $XY$.
By Lemma 3.19, there exist points $Q$ and $R$ incident with $XY$ such that
$\alpha=PQ$ and $\beta=PR$. Hence, $\alpha\beta=PQPR=Q^{P}R$ which is the
identity if $R=Q^{P}$, or otherwise, an order-preserving axial map, by Remark
3.12 and Lemma 3.18. Therefore, (ii) holds. ∎

######  Remark 3.20.

It is a simple consequence of Lemma 3.19 that if $\alpha$ is an order-
preserving axial map with axis $\ell$ and $A$ is a finite point, then the map
$A\alpha A$ is an order-preserving axial map with axis $\ell^{A}$. In fact, if
$\alpha=PQ$, then $A\alpha A=(APA)(AQA)=P^{A}Q^{A}$, by Theorem 3.8 and Remark
3.12.

######  Lemma 3.21.

An order-preserving axial map cannot interchange two rimpoints.

###### Proof.

Let $\alpha$ be an order-preserving axial map and suppose it interchanges two
rimpoints $X$ and $Y$. By Lemma 3.19, there exist two finite points incident
with the axis of $\alpha$ such that $\alpha=MN$. Then $\\{X,X^{M}\\}$
separates $\\{Y,Y^{M}\\}$ and hence $XY^{M}$ does not meet $YX^{M}$ (by (S3));
a contradiction as $N$ is the point of intersection of these two lines. ∎

######  Lemma 3.22.

Given a line $\ell$ and rimpoints $U$, $V$ on the same side of $\ell$ there is
a unique order-preserving axial mapping $\alpha$ with axis $\ell$ such that
$U^{\alpha}=V$.

###### Proof.

Let $P$ be a finite point on $\ell$ and let $W:=U^{P}$. Then $WV$ intersects
$\ell$, as $V$, $W$ are on opposite sides of $\ell$; let $Q:=WV\cap\ell$. Then
$\alpha:=PQ$ is an order-preserving axial map (by Lemma 3.18) taking $U$ to
$V$. Any other such mapping can be written as $PQ^{\prime}$, for some
$Q^{\prime}$ on $\ell$ (by Lemma 3.19). Since $U^{PQ^{\prime}}=V$, it follows
that $W^{Q^{\prime}}=V$ and hence $Q=Q^{\prime}$. ∎

######  Lemma 3.23.

Let $\mathcal{C}$ be a cyclic order compatible with $//$, and let $\alpha$ be
an order-preserving axial mapping with axis $\ell$. Then for every rimpoint
$X$ not incident with $\ell$, we have either
$\mathcal{C}(X^{\alpha},X,X^{\alpha^{-1}})$ or
$\mathcal{C}(X^{\alpha^{-1}},X,X^{\alpha})$.

###### Proof.

By Lemma 3.18, there exist finite point $P$ and $Q$ incident with $\ell$ such
that $\alpha=PQ$. Let $X_{1}$ and $X_{2}$ be the rimpoints of $\ell$, such
that we have the ordering $X_{1}\,P\,Q\,X_{2}$. Let $X$ be a rimpoint not
incident with $\ell$. If (without loss of generality)
$\mathcal{C}(X_{1},X,X_{2})$, then we have the ordering,
$X_{2}\,X^{Q}\,X^{P}\,X_{1}$ (via projection). Therefore,
$\mathcal{C}(X^{Q},X^{P},X_{1})$ and hence $\mathcal{C}(X,X^{PQ},X_{2})$ (by
Lemma 3.14). By a similar argument, $\mathcal{C}(X_{1},X,X^{QP},X)$.
Therefore, $\mathcal{C}(X^{\alpha^{-1}},X,X^{\alpha})$. If $P$ and $Q$ are
situated such that we have the ordering $X_{1}\,Q\,P\,X_{2}$, then we have the
other case. ∎

######  Lemma 3.24.

Two order-preserving axial maps with the same axis commute.

###### Proof.

Let $\alpha$ and $\alpha^{\prime}$ be two order-preserving axial maps with
common axis $\ell$. By Lemma 3.19, there exist three finite points $P$,
$P^{\prime}$, $Q$ incident with $\ell$ such that $\alpha=PQ$ and
$\alpha^{\prime}=QP^{\prime}$. By Theorem 3.7, there exists a finite point
$Q^{\prime}$ incident with $\ell$ such that $P^{\prime}PQ=Q^{\prime}$ and
hence $P^{\prime}PQ=(P^{\prime}PQ)^{-1}=QPP^{\prime}$. Therefore,

$\alpha^{\prime}\alpha=QP^{\prime}PQ=Q(P^{\prime}PQ)=Q(QPP^{\prime})=PP^{\prime}=\alpha\alpha^{\prime}$

and we have proved our result. ∎

The following two lemmas will be used later to define _perpendicularity_ of
lines of $\mathcal{I}$. Later we will see that Lemma 3.26 allows us to easily
express the unique common perpendicular to two lines (see Lemma 4.7).

######  Lemma 3.25.

Every order-preserving axial mapping is the square of an order-preserving
axial mapping (with the same axis).

###### Proof.

Let $\alpha$ be an order-preserving axial mapping with axis $\ell$. By Lemma
3.19, there exist finite points $P$ and $Q$ incident with $\ell$ such that
$\alpha=PQ$. By Lemma 3.15, there exists a finite point $M$ incident with $PQ$
such that $P^{M}=Q$ (i.e., $M$ is the _midpoint_ of the segment
$\overline{PQ}$). Then $(PM)^{2}=PMPM=PP^{M}=PQ=\alpha$, and we know that $PM$
is an order-preserving axial map by Lemma 3.18, with axis $\ell$. Therefore,
$\alpha$ is the square of an order-preserving axial mapping. ∎

######  Lemma 3.26.

Let $X_{1}$, $X_{2}$, $X_{3}$, $X_{4}$ be rimpoints such that $X_{1}X_{2}$ and
$X_{3}X_{4}$ are non-intersecting non-parallel lines. Then there is a unique
order-preserving axial map $g$ with axis $X_{1}X_{2}$ such that
$X_{4}^{g^{2}}=X_{3}$.

###### Proof.

Let $\ell$ be the line $X_{1}X_{2}$. By Lemma 3.22, there exists an order-
preserving axial map $\alpha$ with axis $\ell$ such that
$X_{4}^{\alpha}=X_{3}$. By Lemma 3.25, there is an order-preserving axial map
$g$ with axis $\ell$ such that $\alpha=g^{2}$. So we know that an order-
preserving axial map $g$ with axis $\ell$ exists such that
$X_{4}^{g^{2}}=X_{3}$. By Lemma 3.24 and Lemma 3.19, the order-preserving
axial mappings with axis $\ell$ form an Abelian group $A(\ell)$, and so the
squaring map $x\mapsto x^{2}$ is a homomorphism on $A(\ell)$. There are no
elements of order $2$ in $A(\ell)$ (by Lemma 3.21) and so the squaring map is
injective. Hence $g$ is the unique element of $A(\ell)$ such that
$X_{4}^{g^{2}}=X_{3}$. ∎

##  4\. Line-reflections and perpendicularity

Throughout this section, we will suppose that $\mathcal{I}$ is an incidence
structure satisfying axioms (A1) through (A7), plus (A10) and (A11). Suppose
$\ell$ and $m$ are non-parallel lines, and suppose $\ell$ has rimpoints
$\\{L_{1},L_{2}\\}$ and $m$ has rimpoints $\\{M_{1},M_{2}\\}$ such that
$L_{1}M_{2}//L_{2}M_{1}$. We define $\ell\bowtie m$ to be the finite point
$L_{1}M_{2}\cap L_{2}M_{1}$. In this notation, Axiom (A11) states that if $X$
is a rimpoint not on a line $\ell$, then there is a unique point $R$ on $\ell$
such that for every finite point $S$ on $\ell$, $\ell\bowtie X^{SR}X^{RS}$ is
incident with $XR$.

######  Lemma 4.1.

Let $\alpha$ and $\beta$ be order-preserving axial maps with a common axis
$\ell$. Let $A$ be a rimpoint not incident with $\ell$. Then $A$, $\ell\bowtie
A^{\alpha}A^{\alpha^{-1}}$, $\ell\bowtie A^{\beta}A^{\beta^{-1}}$ are
collinear.

###### Proof.

By (A11), there is a unique finite point $R$ on $\ell$ such that for every
finite point $S$ on $\ell$, $\ell\bowtie A^{SR}A^{RS}$ is incident with $AR$.
By Lemma 3.19, there exist finite points $M$ and $N$ on $\ell$ such that
$\alpha=RM$ and $\beta=RN$. Therefore, $\ell\bowtie
A^{\alpha}A^{\alpha^{-1}}=\ell\bowtie A^{RM}A^{MR}$ and $\ell\bowtie
A^{RM}A^{MR}$ is incident with $AR$. Likewise, $\ell\bowtie
A^{\beta}A^{\beta^{-1}}$ is incident with $AR$, and so $A$, $\ell\bowtie
A^{\alpha}A^{\alpha^{-1}}$, $\ell\bowtie A^{\beta}A^{\beta^{-1}}$ all lie on
$AR$. ∎

######  Definition 4.2.

Let $\ell$ be a line and let $A$ be a rimpoint not incident with $\ell$. Let
$X_{1}$ and $X_{2}$ be the rimpoints on $\ell$ and let $\alpha$ be an order-
preserving axial map with axis $\ell$. Let $O:=\ell\bowtie
A^{\alpha}A^{\alpha^{-1}}$. Then $AA^{O}$ is the perpendicular line to $\ell$
incident with $A$.

Axiom (A11) (see Lemma 4.1) implies that the definition (in Definition 4.2) of
the perpendicular line to $\ell$ incident with $A$ is independent of the
choice of $\alpha$, and hence well-defined. Therefore, given a line $\ell$,
there is a function $\sigma_{\ell}$ on the rimpoints defined as follows:
$\sigma_{\ell}$ fixes the rimpoints of $\ell$, and for a rimpoint $A$ not
incident with $\ell$, we define $A^{\sigma_{\ell}}$ to be the rimpoint $A^{O}$
given in Definition 4.2. By Lemma 4.4 (below), $\sigma_{\ell}$ is a
permutation of the rimpoints and we will refer to this map as the reflection
in the line $\ell$.

######  Lemma 4.3.

Let $\ell$ be a line and let $A$ be a rimpoint not incident with $\ell$. Then
the line $AA^{\sigma_{\ell}}$ meets $\ell$ in a finite point. Therefore,
$\sigma_{\ell}$ interchanges the sides of $\ell$.

###### Proof.

Let $\alpha$ be an order-preserving axial map with axis $\ell$, and let
$\\{X_{1},X_{2}\\}$ be the rimpoints of $\ell$. Let $\mathcal{C}$ be a cyclic
order on the rimpoints compatible with the separation relation $//$ such that
$\mathcal{C}(X_{1},A,X_{2})$. Now $A^{\alpha}$ and $A^{\alpha^{-1}}$ lie on
the same side of $\ell$ as $A$ (by Lemma 3.2), and either
$\mathcal{C}(A^{\alpha},A,A^{\alpha^{-1}})$ or
$\mathcal{C}(A^{\alpha^{-1}},A,A^{\alpha})$ (by Lemma 3.23). Suppose, without
loss of generality, that $\mathcal{C}(X_{1},A^{\alpha},A)$ and
$\mathcal{C}(A,A^{\alpha^{-1}},X_{2})$. Let $O:=\ell\bowtie
A^{\alpha}A^{\alpha^{-1}}$. In particular, from our ordering, we have
$X_{1}^{O}=A^{\alpha^{-1}}$ and $X_{2}^{O}=A^{\alpha}$. Now the half-turn $O$
preverses the cyclic order $\mathcal{C}$, and hence
$\mathcal{C}(X_{1}^{O},A^{\alpha O},A^{O})$ and
$\mathcal{C}(A^{O},A^{\alpha^{-1}O},X_{2}^{O})$. So
$\mathcal{C}(A^{\alpha^{-1}},X_{2},A^{O})$ and
$\mathcal{C}(A^{O},X_{1},A^{\alpha})$, and it follows that
$\mathcal{C}(X_{2},A^{O},X_{1})$. In particular, $A^{\sigma_{\ell}}=A^{O}$
lies on the opposite side of $\ell$ as $A$. ∎

######  Lemma 4.4.

For any line $\ell$, we have $\sigma_{\ell}^{2}=1$.

###### Proof.

Let $A$ be a rimpoint not incident with $\ell$, and let $\alpha$ be an order-
preserving axial map with axis $\ell$. By definition,
$A^{\sigma_{\ell}}=A^{O}$ where $O=\ell\bowtie A^{\alpha}A^{\alpha^{-1}}$. Let
$O^{\prime}:=\ell\bowtie(A^{O})^{\alpha}(A^{O})^{\alpha^{-1}}$. By Lemma 4.3,
$\ell$ meets $AO$ in a finite point $R$. Write $\alpha=PQ$ for two finite
points $P$ and $Q$ incident with $\ell$ (see Lemma 3.19). Since $P$, $Q$, $R$
are collinear, we have from Theorem 3.7 that $PQR=RQP$ and $QPR=RPQ$. So

$\displaystyle\left((A^{O})^{\alpha}\right)^{R}$
$\displaystyle=A^{O(PQR)}=A^{O(RQP)}=(A^{OR})^{QP}=A^{QP}=A^{\alpha^{-1}},$
$\displaystyle\left((A^{O})^{\alpha^{-1}}\right)^{R}$
$\displaystyle=A^{O(QPR)}=A^{O(RPQ)}=(A^{OR})^{PQ}=A^{PQ}=A^{\alpha}.$

Therefore, since half-turns are automorphisms by Theorem 3.8, we have

$(O^{\prime})^{R}=(\ell\bowtie(A^{O})^{\alpha}(A^{O})^{\alpha^{-1}})^{R}=\ell\bowtie
A^{\alpha^{-1}}A^{\alpha}=O$

as $\ell^{R}=\ell$. So $O^{\prime}$ lies on the line $RO=AO$ and therefore,
$\left(A^{\sigma_{\ell}}\right)^{\sigma_{\ell}}=(A^{\sigma_{\ell}})^{O^{\prime}}=A^{OO^{\prime}}=A$.
Thus we have shown that $\sigma_{\ell}^{2}$ fixes every rimpoint, and so is
the identity. ∎

######  Lemma 4.5.

Let $\ell$ and $m$ be two lines meeting in a finite point. Then the following
are equivalent:

  1. (a)

$\sigma_{\ell}$ interchanges the rimpoints of $m$;

  2. (b)

$\sigma_{m}$ interchanges the rimpoints of $\ell$;

###### Proof.

Suppose $\sigma_{\ell}$ interchanges the rimpoints $\\{M_{1},M_{2}\\}$ of $m$.
Let $\alpha$ be an order-preserving axial map with axis $\ell$. Let
$\\{X_{1},X_{2}\\}$ be the rimpoints of $\ell$, written in such a way that
$\\{X_{1},M_{1}^{\alpha^{-1}}\\}$ separates $\\{X_{2},M_{1}^{\alpha}\\}$. Let
$T:=X_{1}M_{1}^{\alpha^{-1}}\cap X_{2}M_{1}^{\alpha}$ and $U:=\ell\cap m$. By
assumption, $M_{1}^{\sigma_{\ell}}=M_{2}$ and hence $M_{2}=M_{1}^{T}$.
Therefore, $T$ is incident with $m$ and $m=M_{1}T$ is the perpendicular line
to $\ell$ through $M_{1}$. Let $\beta:=UT$. Then

$\displaystyle X_{1}^{\beta}$
$\displaystyle=(X_{1}^{U})^{T}=X_{2}^{T}=M_{1}^{\alpha},$ $\displaystyle
X_{1}^{\beta^{-1}}$ $\displaystyle=(X_{1}^{T})^{U}=M_{1}^{\alpha^{-1}U},$

and so $X_{1}P$ is the perpendicular to $m$ through $X_{1}$ where
$P:=M_{1}^{\alpha^{-1}U}M_{1}\cap M_{2}M_{1}^{\alpha}$. We will show that $P$
is incident with $\ell$. Now $PUTP$ is axial with axis
$m^{P}=M_{1}^{\alpha^{-1}U}M_{1}^{\alpha}$ (see Remark 3.20), and

$M_{1}^{(PUTP)^{2}}=M_{1}^{PUTUTP}=(M_{1}^{\alpha^{-1}})^{TUTP}=X_{1}^{UTP}=X_{2}^{TP}=M_{1}^{\alpha
P}=M_{2}.$

So by Lemma 3.22, $PUTP=T^{\prime}U^{\prime}$ where $U^{\prime}=\ell\cap
M_{1}^{\alpha}M_{1}^{\alpha^{-1}U}$ and $T^{\prime}=X_{1}M_{1}\cap
X_{2}M_{2}^{U^{\prime}}$. (Note:
$M_{1}^{(T^{\prime}U^{\prime})^{2}}=M_{1}^{T^{\prime}U^{\prime}T^{\prime}U^{\prime}}=X_{1}^{U^{\prime}T^{\prime}U^{\prime}}=X_{2}^{T^{\prime}U^{\prime}}=M_{2}$).
Therefore,

$X_{1}^{P}=(M_{1}^{\alpha^{-1}})^{TP}=(M_{1}^{\alpha^{-1}U})^{UTP}=M_{1}^{PUTP}=M_{1}^{T^{\prime}U^{\prime}}=X_{1}^{U^{\prime}}=X_{2}.$

Hence $P$ lies on $\ell$ and $X_{1}P=\ell$. It follows that $\sigma_{m}$
interchanges the rimpoints of $\ell$. Interchanging the roles of $\ell$ and
$m$ in this argument shows that the converse also holds. ∎

Lemma 4.5 allows us to define perpendicularity of lines. We say that a line
$\ell$ and a line $m$ are perpendicular, and we write $\ell\perp m$, if $\ell$
and $m$ are concurrent and $\sigma_{\ell}$ interchanges the rimpoints of $m$.
So Lemma 4.5 implies that $\perp$ is a symmetric relation on lines. Recall
from Theorem 3.8 that half-turns are automorphisms of $\mathcal{I}$. They also
preserve the perpendicularity relation (Definition 4.2).

######  Lemma 4.6.

Half-turns preserve perpendicularity. That is, if $\ell\perp m$ and $P$ is a
finite point, then $\ell^{P}\perp m^{P}$.

###### Proof.

Suppose $\ell\perp m$ and let $P$ be a finite point. Let $M$ and $M^{\prime}$
be the rimpoints of $m$, and let $\alpha$ be an axial map with axis $\ell$.
Then, $m=MO$ where $O=\ell\bowtie M^{\alpha}M^{\alpha^{-1}}$. Let
$\beta:=P\alpha P$. By Remark 3.20, $\beta$ is axial with axis $\ell^{P}$.
Moreover, by Lemma 3.8, we have

$\displaystyle O^{P}$ $\displaystyle=\ell^{P}\bowtie M^{\alpha
P}M^{\alpha^{-1}P}$ $\displaystyle=\ell^{P}\bowtie M^{P\beta}M^{P\beta^{-1}}$
$\displaystyle=\ell^{P}\bowtie(M^{P})^{\beta}(M^{P})^{\beta^{-1}}$

and hence $M^{P}O^{P}$ is perpendicular to $\ell^{P}$. That is,
$m^{P}\perp\ell^{P}$. ∎

######  Lemma 4.7.

Suppose $\ell$ and $m$ are two non-parallel non-intersecting lines. Then there
exists a unique line $n$ perpendicular to both $\ell$ and $m$. Moreover, there
are unique order-preserving axial maps $g$ and $h$ such that (i) $g$ has axis
$\ell$, (ii) $h$ has axis $m$, (iii) $X_{4}^{g^{2}}=X_{3}$, (iv)
$X_{2}^{h^{2}}=X_{1}$. The line $n$ is equal to $X_{4}^{g}X_{2}^{h}$ and is
the common perpendicular of $\ell$ and $m$.

###### Proof.

Let $\\{X_{1},X_{2}\\}$ be the rimpoints of $\ell$ and let $\\{X_{3},X_{4}\\}$
be the rimpoints of $m$ such that $X_{1}X_{4}//X_{2}X_{3}$. By Lemma 3.26,
there is a unique order-preserving axial map $g$ with axis $\ell$ such that
$g^{2}$ maps $X_{4}$ to $X_{3}$. Likewise, there is a unique order-preserving
axial map $h$ with axis $m$ such that $h^{2}$ maps $X_{2}$ to $X_{1}$. Let
$A:=X_{4}^{g}$. Then

$A^{g}=X_{3}\text{ and }A^{g^{-1}}=X_{4}.$

Let $O:=X_{1}X_{4}\cap X_{2}X_{4}$. By definition, the perpendicular line to
$\ell$ through $A$ is $n:=AO$. Now consider the map $Og^{-1}O$. First, by
Lemma 3.18, $Og^{-1}O$ is an order-preserving axial map as it is the product
of two half-turns: if $g=PQ$, then $Og^{-1}O=Q^{O}P^{O}$ (by Lemma 3.11), and
so $Og^{-1}O$ has axis $m$. Also,

$X_{2}^{(Og^{-1}O)^{2}}=X_{2}^{Og^{-2}O}=X_{3}^{g^{-2}O}=X_{4}^{O}=X_{1}.$

Therefore, by Lemma 3.26, $h=Og^{-1}O$. Let $B:=X_{2}^{h}$. Then

$B^{h}=X_{1}\text{ and }B^{h^{-1}}=X_{2},$

and so the perpendicular line to $m$ through $B$ is $BO$. On the other hand,

$X_{2}^{h}=X_{2}^{Og^{-1}O}=X_{3}^{g^{-1}O}=A^{O},$

and therefore $B=A^{O}$ and $BO=n$. So we have shown that $n$ is perpendicular
to $\ell$ and $m$.

It remains to prove uniqueness of the line $n$. Let $n^{\prime}$ be another
line perpendicular to both $\ell$ and $m$. (The following argument is similar
to that given in [3, Theorem 3.8]). Let the rimpoints of $n$ and $n^{\prime}$
be $U$ and $V$, and $U^{\prime}$ and $V^{\prime}$, respectively, arranged so
that $U$ and $U^{\prime}$ lie on the same side of $\ell$ and $m$. So there
exist finite points $Q_{1}$ and $Q_{2}$ such that $Q_{1}=\ell\cap U^{\prime}V$
and let $Q_{2}=m\cap U^{\prime}V$. Let $V_{1}^{\prime}$ and $V_{2}^{\prime}$
be the other rimpoints on $UQ_{1}$ and $UQ_{2}$, respectively. Since $\ell$ is
perpendicular to $n$, and the half-turn about $Q_{1}$ maps $V$ to $U_{1}$ and
$U$ to $V_{1}^{\prime}$, by Lemma 4.6, $U_{1}^{\prime}V_{1}^{\prime}$ is
perpendicular to $\ell$ too. However, the perpendicular line to $\ell$ through
a rimpoint not incident with $\ell$ is unique and hence
$U_{1}^{\prime}V_{1}^{\prime}=n^{\prime}=U^{\prime}V^{\prime}$. Therefore,
$V^{\prime}=V_{1}^{\prime}$. Similarly, using the fact that $m$ is
perpendicular to $n$, we have via the half-turn about $Q_{2}^{\prime}$ that
$U^{\prime}V_{2}^{\prime}$ is the perpendicular line to $m$ through
$U^{\prime}$, and so $V^{\prime}=V_{2}^{\prime}$. Thus $Q_{1}=U^{\prime}V\cap
UV^{\prime}=Q_{2}$ is on $\ell$ and $m$; contrary to the fact that $\ell$ and
$m$ do not intersect. Therefore, $n$ is the unique line perpendicular to
$\ell$ and $m$. ∎

We will denote the unique common perpendicular to non-parallel non-
intersecting lines $\ell$ and $m$ by $\bot(\ell,m)$.

######  Lemma 4.8.

Let $\ell$ and $m$ be non-intersecting non-parallel lines. Then $\ell\bowtie
m$ is incident with $\bot(\ell,m)$. Moreover, $\ell\bowtie m$ is the midpoint
of the segment joining $\ell\cap\bot(\ell,m)$ and $m\cap\bot(\ell,m)$.

###### Proof.

Let $X$ and $X^{\prime}$ be the rimpoints on $\ell$, let $Y$ and $Y^{\prime}$
be the rimpoints on $m$, labelled so that $u=XY^{\prime}$ meets
$v=X^{\prime}Y$ in a finite point $P$. Then the half-turn about $P$
interchanges $X$ and $Y^{\prime}$, and also $Y$ and $X^{\prime}$; and hence
interchanges $\ell=XX^{\prime}$ and $m=Y^{\prime}Y$ (as half-turns are
automorphisms by Theorem 3.8). Let $n:=\bot(\ell,m)$. Then (by Lemma 4.7)
there are unique order-preserving axial maps $g$ and $h$ such that (i) $g$ has
axis $\ell$, (ii) $h$ has axis $m$, (iii) $Y^{g^{2}}=Y^{\prime}$, (iv)
$X^{h^{2}}=X^{\prime}$. By Lemma 3.19 and Theorem 3.8, $PgP$ is order-
preserving axial with axis $m$. Now

$(X^{\prime})^{(PgP)^{2}}=(X^{\prime})^{PgPPgP}=(X^{\prime})^{Pg^{2}P}=Y^{g^{2}P}=(Y^{\prime})^{P}=X$

and so by Lemma 3.26, $PgP=h^{-1}$. Therefore,
$Y^{gP}=(X^{\prime})^{PgP}=(X^{\prime})^{h^{-1}}=X^{h}$ and hence $P$ fixes
$n=Y^{g}X^{h}$. So by Corollary 3.9, $P$ is incident with $n$. Moreover, (by
Theorem 3.8), $(\ell\cap n)^{P}=\ell^{P}\cap n^{P}=m\cap n$ and hence $P$ is
the midpoint of the segment joining $\ell\cap n$ and $m\cap n$. ∎

######  Lemma 4.9.

Two distinct lines $\ell$ and $m$ are parallel if and only if they are not
concurrent (in a finite point) nor share a common perpendicular.

###### Proof.

If $\ell$ and $m$ are not concurrent in a finite point, nor parallel, then
they share a common perpendicular by Lemma 4.7. Conversely, suppose $\ell$ and
$m$ are parallel; that is, $\ell$ and $m$ share a rimpoint $X$. Clearly,
$\ell$ and $m$ cannot meet in a finite point $P$, say, since otherwise,
$\ell=XP=m$. So suppose a line $n$ is perpendicular to $\ell$ and $m$. Let $g$
be an order-preserving axial map with axis $\ell$. Then there is a rimpoint
$A$ incident with $n$, such that if $O:=\ell\bowtie A^{g}A^{g^{-1}}$, then
$n=AO$. Now $\sigma_{n}$ is a permutation of the rimpoints, by Lemma 4.4.
Therefore, $X^{\sigma_{n}}$ cannot be a rimpoint on both $\ell$ and $m$, as
$\ell\neq m$, except possibly if $X=X^{\sigma_{n}}$, but this cannot occur by
Lemma 4.3. Therefore, $\ell$ and $m$ do not share a common perpendicular. ∎

Let $P$ and $\ell$ be a point and line that are not incident. The following
lemma shows that there exists a line incident with $P$ that is perpendicular
to $\ell$.

######  Lemma 4.10.

Let $P$ be a finite point not incident with a line $\ell$. Let $X$ and $Y$ be
the two rimpoints incident with $\ell$. Then $X^{P}Y^{P}$ is a line not
intersecting $\ell$ and not parallel to $\ell$, and $\bot(X^{P}Y^{P},\ell)$ is
incident with $P$.

###### Proof.

Let $m:=X^{P}Y^{P}$. First note that $X^{P}$ and $Y^{P}$ lie on the same side
of $\ell$ and hence $m$ does not intersect $\ell$ in a finite point or
rimpoint. Therefore, $m$ is not parallel to $\ell$. By Lemma 4.8, $XX^{P}\cap
YY^{P}$ is a point and is incident with $\bot(\ell,m)$. The result follows by
noting that $P=XX^{P}\cap YY^{P}$. ∎

Now if $P$ is a finite point not incident with a line $\ell$, then there
cannot be two distinct lines through $P$ perpendicular to $\ell$, since
otherwise $\ell$ would be the common perpendicular of two concurrent lines. So
Lemma 4.10 affirms (P1) in the definition of a metric plane in Section 6.

######  Lemma 4.11.

Let $\ell$ be a line and let $P$ be a finite point incident with $\ell$. Then
there exists a unique line $\ell^{\prime}$ through $P$ perpendicular to
$\ell$.

###### Proof.

Suppose we have a line $m$ not intersecting and not parallel to $\ell$. (Such
a line exists by (A3) and (A5)). Let $Q$ be the point of intersection of
$\ell$ and $\bot(\ell,m)$. By Lemma 3.15, there is a ‘midpoint’ $M$ incident
with $\ell$ such that $P=Q^{M}$. (If $P=Q$, then $M=P$). Let $Y_{1}$ and
$Y_{2}$ be the rimpoints of $\bot(\ell,m)$, and take the line $\ell^{\prime}$
joining $Y_{1}^{M}$ and $Y_{2}^{M}$. (If $P=Q$, then
$\ell^{\prime}=\bot(\ell,m)$ and we are done). Now $Q$ is incident with
$Y_{1}Y_{2}$, so by Theorem 3.8, $P$ is incident with $\ell^{\prime}$.
Therefore, by Lemma 4.6, $\ell^{\prime}$ is perpendicular to $\ell$. (In fact,
$\ell^{\prime}=\bot(\ell,m^{M})$ and so $\ell^{\prime}$ is the unique line
through $P$ perpendicular to $\ell$). ∎

######  Lemma 4.12.

If two lines have a common perpendicular, then they are not concurrent in a
finite point.

###### Proof.

Suppose $\ell$ and $m$ have a common perpendicular $n$, and suppose for a
proof by contradiction that $\ell$ and $m$ meet in a finite point $P$. By
Lemma 4.11, $n$ is not incident with $P$. By Lemma 4.10, $n^{P}$ is non-
intersecting and non-parallel to $n$, and by Lemma 4.6, $\ell^{P}$ and $m^{P}$
are perpendicular to $n^{P}$. By Corollary 3.9, $\ell^{P}=\ell$ and $m^{P}=m$.
However, then $\ell$ and $m$ would be two distinct lines perpendicular to both
$n$ and $n^{P}$, contradicting Lemma 4.7. Therefore, $\ell$ and $m$ do not
intersect in a finite point. ∎

######  Lemma 4.13.

Let $\ell$ be a line. Then:

  1. (i)

If $\mathcal{C}$ is a cyclic order associated to $//$, then the reflection
about $\ell$ reverses $\mathcal{C}$.

  2. (ii)

The reflection about $\ell$ preserves the separation relation $//$;

  3. (iii)

If $P$ is a finite point, then $\sigma_{\ell}P\sigma_{\ell}$ is a half-turn
and $P\sigma_{\ell}P$ is a reflection (about $\ell^{P}$).

  4. (iv)

If $\ell$ and $m$ are lines, then $\sigma_{m}\sigma_{\ell}\sigma_{m}$ is a
reflection.

  5. (v)

The reflection about $\ell$ is a perpendularity-preserving automorphism of
$\mathcal{I}$, and it fixes a point $P$ if and only if $P$ is incident with
$\ell$.

  6. (vi)

The group $G$ generated by the half-turns is a normal subgroup of the group
$H$ generated by the reflections.

###### Proof.

(i) and (ii):

Let $\mathcal{C}$ be a cyclic order associated to $//$ and let $\ell$ be a
line. Consider three distinct rimpoints $X$, $Y$, $Z$ such that
$\mathcal{C}(X,Y,Z)$, and suppose first that none of these three points are
incident with $\ell$. Without loss of generality, we may suppose that $X$ and
$Z$ lie on opposite sides of $YY^{\sigma_{\ell}}$. So by Remark 2.5, we have
$\mathcal{C}(Y,Y^{\sigma_{\ell}},X)$ and $\mathcal{C}(Y^{\sigma_{\ell}},Y,Z)$,
or $\mathcal{C}(Y,Y^{\sigma_{\ell}},Z)$ and
$\mathcal{C}(Y^{\sigma_{\ell}},Y,X)$. Then by definition of a reflection,
$XX^{\sigma_{\ell}}$, $YY^{\sigma_{\ell}}$, $ZZ^{\sigma_{\ell}}$ are
perpendicular to $\ell$, so no two of these lines meet in a finite point (by
Lemma 4.12). Then $X^{\sigma_{\ell}}$ and $Z^{\sigma_{\ell}}$ also lie on
opposite sides of $YY^{\sigma_{\ell}}$; that is,
$\mathcal{C}(Y,Y^{\sigma_{\ell}},X^{\sigma_{\ell}})$ and
$\mathcal{C}(Y^{\sigma_{\ell}},Y,Z^{\sigma_{\ell}})$, or
$\mathcal{C}(Y,Y^{\sigma_{\ell}},Z^{\sigma_{\ell}})$ and
$\mathcal{C}(Y^{\sigma_{\ell}},Y,X^{\sigma_{\ell}})$. We can write the first
conjunction as
$\mathcal{C}(Y^{\sigma_{\ell}},X^{\sigma_{\ell}},Y)\wedge\mathcal{C}(Y^{\sigma_{\ell}},Y,Z^{\sigma_{\ell}})$
(by (C3)), and hence
$\mathcal{C}(Y^{\sigma_{\ell}},X^{\sigma_{\ell}},Z^{\sigma_{\ell}})$ by (C4).
Therefore,
$\mathcal{C}(Z^{\sigma_{\ell}},Y^{\sigma_{\ell}},X^{\sigma_{\ell}})$ (by
(C3)).

Now let us suppose at least one of our points, $X$ say, is incident with
$\ell$. Then $X=X^{\sigma_{\ell}}$. Let $X^{\prime}$ be the other rimpoint
incident with $\ell$. If $Y=X^{\prime}$, then $Y=Y^{\sigma_{\ell}}$ and $Z$ is
not incident with $\ell$. Moreover, $Z^{\sigma_{\ell}}$ lies on the opposite
side of $\ell$ to $Z$, and hence $\mathcal{C}(X,Z^{\sigma_{\ell}},Y)$; that
is, $\mathcal{C}(Z^{\sigma_{\ell}},Y^{\sigma_{\ell}},X^{\sigma_{\ell}})$ by
(C3). A similar argument yields the same conclusion for when $Z=X^{\prime}$.
So suppose $Y$ and $Z$ are not incident with $\ell$.

Without loss of generality, we will suppose that
$\mathcal{C}(X,Y,X^{\prime})$. Since $Y^{\sigma_{\ell}}$ lies on the other
side of $\ell$, it follows that $\mathcal{C}(X^{\prime},Y^{\sigma_{\ell}},X)$.
If $Y$ and $Z$ lie on the same side of $\ell$, then $X$ and $Z$ lie on
opposite sides of $YY^{\sigma_{\ell}}$. So the argument above works in this
case, and we have
$\mathcal{C}(Z^{\sigma_{\ell}},Y^{\sigma_{\ell}},X^{\sigma_{\ell}})$. So
suppose $Y$ and $Z$ lie on opposite sides of $\ell$. Then $Y$ and
$Z^{\sigma_{\ell}}$ lie on the same side of $\ell$. Hence $X$ and
$Z^{\sigma_{\ell}}$ lie on opposite sides of $YY^{\sigma_{\ell}}$. Since
$X=X^{\sigma_{\ell}}$, it follows that $X$ and $Z$ lie on opposite sides of
$YY^{\sigma_{\ell}}$. So the same argument as before yields
$\mathcal{C}(Z^{\sigma_{\ell}},Y^{\sigma_{\ell}},X^{\sigma_{\ell}})$.

Therefore, $\sigma_{\ell}$ reverses the cyclic order $\mathcal{C}$. It then
follows immediately that $\sigma_{\ell}$ preserves $//$, since

$AB//CD\iff\left(\mathcal{C}(A,B,C)\wedge\mathcal{C}(A,D,B)\right)\vee\left(\mathcal{C}(A,C,B)\wedge\mathcal{C}(A,B,D)\right).$

for all distinct rimpoints $A$, $B$, $C$ and $D$.

(iii):

Let $P$ be a finite point and consider the map
$g:=\sigma_{\ell}P\sigma_{\ell}$. Now $P$ is a fixed-point-free involution of
the rimpoints such that each pair of transpositions of $P$ are separating. So
since $g$ has the same cycle-structure as $P$, and $\sigma_{\ell}$ preserves
the separation relation (from (ii)), it follows from Corollary 3.13 $g$ is a
half-turn.

Now consider the map $h:=P\sigma_{\ell}P$. Recall that the half-turn by $P$
acts by automorphisms on $\mathcal{I}$ (by Theorem 3.8) and so $\ell^{P}$ is a
line whose rimpoints are the images of the rimpoints of $\ell$ under $P$. Let
$P$ be a rimpoint incident with $\ell^{P}$. Then $X^{P}$ is incident with
$\ell$ and hence $X^{P}=(X^{P})^{\sigma_{\ell}}=X^{h}$. Therefore, $h$ fixes
the rimpoints of $\ell^{P}$. Now take a rimpoint $Y$ not incident with
$\ell^{P}$. So $Y^{P}$ is not incident with $\ell$ and hence the line
$(Y^{P})^{\sigma_{\ell}}(Y^{P})$ is perpendicular to $\ell$ (by definition of
$\sigma_{\ell}$). Therefore, $(Y^{P})^{\sigma_{\ell}P}Y$ is perpendicular to
$\ell^{P}$ (by Lemma 4.6) and so $h$ is the reflection about $\ell^{P}$.

(iv):

Let $L_{1}$ and $L_{2}$ be the rimpoints of $\ell$. Let $n$ be the line
spanned by $L_{1}^{\sigma_{m}}$ and $L_{2}^{\sigma_{m}}$. We will show that
$\sigma_{m}\sigma_{\ell}\sigma_{m}=\sigma_{n}$. Now

$(L_{1}^{\sigma_{m}})^{\sigma_{m}\sigma_{\ell}\sigma_{m}}=L_{1}^{\sigma_{\ell}\sigma_{m}}=L_{1}^{\sigma_{m}}$

and similarly, $L_{2}^{\sigma_{m}}$ is fixed by
$\sigma_{m}\sigma_{\ell}\sigma_{m}$. Therefore, the rimpoints of $n$ are fixed
by $\sigma_{m}\sigma_{\ell}\sigma_{m}$. Now suppose $X$ is a rimpoint not
incident with $n$. We will show that the line
$XX^{\sigma_{m}\sigma_{\ell}\sigma_{m}}$ is perpendicular to $n$. Let
$Y:=X^{\sigma_{m}}$. Since $YY^{\sigma_{\ell}}\perp\ell$, there exists a line
$a$ such that $YY^{\sigma_{\ell}}=\bot(\ell,a)$. Let $\\{A_{1},A_{2}\\}$ be
the rimpoints of $a$ and let $a^{\prime}$ be the line spanned by
$A_{1}^{\sigma_{m}}$ and $A_{2}^{\sigma_{m}}$. So
$YY^{\sigma_{\ell}}=A_{2}^{g}L_{2}^{h}$ where $g$ and $h$ are order-preserving
axial maps such that (i) $g$ has axis $\ell$, (ii) $h$ has axis $a$, (iii)
$A_{2}^{g^{2}}=A_{1}$, (iv) $L_{2}^{h^{2}}=L_{1}$. Without loss, we may assume
that $Y=A_{2}^{g}$ and $Y^{\sigma_{\ell}}=L_{2}^{h}$ (since a line has two
rimpoints).

Now an order-preserving axial map is a product of two half-turns, by Lemma
3.19. So the conjugate of an order-preserving axial map by a reflection is an
order-preserving axial map, by (iii). Hence $\sigma_{m}g\sigma_{m}$ and
$\sigma_{m}h\sigma_{m}$ are order-preserving axial maps. Moreover, it is
straight-forward to see that $\sigma_{m}g\sigma_{m}$ has axis $n$ and
$\sigma_{m}h\sigma_{m}$ has axis $a^{\prime}$. Also,

$(A_{2}^{\sigma_{m}})^{(\sigma_{m}g\sigma_{m})^{2}}=A_{2}^{g^{2}\sigma_{m}}=A_{1}^{\sigma_{m}}$

and similarly,
$(L_{2}^{\sigma_{m}})^{(\sigma_{m}h\sigma_{m})^{2}}=L_{2}^{h^{2}\sigma_{m}}=L_{1}^{\sigma_{m}}$.
Therefore,

$\bot(n,a^{\prime})=(A_{2}^{\sigma_{m}})^{\sigma_{m}g\sigma_{m}}(L_{2}^{\sigma_{m}})^{\sigma_{m}h\sigma_{m}}=A_{2}^{g\sigma_{m}}L_{2}^{h\sigma_{m}}=Y^{\sigma_{m}}Y^{\sigma_{\ell}\sigma_{m}}$

and hence
$XX^{\sigma_{m}\sigma_{\ell}\sigma_{m}}=Y^{\sigma_{m}}Y^{\sigma_{\ell}\sigma_{m}}\perp
n$.

(v):

Let $P$ be a point and $a$ be a line such that $P$ is incident with $a$. Let
$A$ and $A^{\prime}$ be the rimpoints of $a$. By (iii), $P^{\sigma_{\ell}}$ is
the unique point fixed by the half-turn $\sigma_{\ell}P\sigma_{\ell}$. Now

$(A^{\sigma_{\ell}})^{\sigma_{\ell}P\sigma_{\ell}}=A^{P\sigma_{\ell}}=(A^{\prime})^{\sigma_{\ell}}$

and similarly, $(A^{\prime})^{\sigma_{\ell}}$ is mapped to $A^{\sigma_{\ell}}$
under the half-turn $\sigma_{\ell}P\sigma_{\ell}$. So the line
$A^{\sigma_{\ell}}(A^{\prime})^{\sigma_{\ell}}$ is fixed by
$\sigma_{\ell}P\sigma_{\ell}$ and hence it is incident with
$P^{\sigma_{\ell}}$ (by Lemma 3.9). Therefore, $\sigma_{\ell}$ preserves
incidence and we write $a^{\sigma_{\ell}}$ for the line
$A^{\sigma_{\ell}}(A^{\prime})^{\sigma_{\ell}}$.

A similar argument shows that $\sigma_{\ell}$ preserves perpendicularity by
noting that from (iv), $a^{\sigma_{\ell}}$ is the axis of the reflection
$\sigma_{\ell}\sigma_{a}\sigma_{\ell}$. The other lines fixed by this
reflection are perpendicular to $a^{\sigma_{\ell}}$. So if $b$ is
perpendicular to $a$, then

$(b^{\sigma_{\ell}})^{\sigma_{\ell}\sigma_{a}\sigma_{\ell}}=b^{\sigma_{a}\sigma_{\ell}}=b^{\sigma_{\ell}}.$

Hence $b^{\sigma_{\ell}}\perp a^{\sigma_{\ell}}$. Therefore, $\sigma_{\ell}$
is a perpendicularity-preserving automorphism of $\mathcal{I}$.

Now let $P$ be a point not incident with a line $\ell$, and let $L$ be a
rimpoint incident with $\ell$. Then $L^{P}$ and $L^{P\sigma_{\ell}}$ lie on
different sides of $\ell$. Therefore, $P^{\sigma_{\ell}}$ lies on the opposite
of $\ell$ to $P$, because $P^{\sigma_{\ell}}$ is incident with
$LL^{P\sigma_{\ell}}$. Hence $P\neq P^{\sigma}_{\ell}$. Suppose now that $P$
is incident with $\ell$. Then $P^{\sigma_{\ell}}$ is also incident with $\ell$
since the rimpoints of $\ell$ are fixed by $\sigma_{\ell}$. By Lemma 4.11,
there exists a line $m$ incident with $P$ such that $\ell\perp m$. Let $X$ be
a rimpoint on $m$. Then $X^{\sigma_{\ell}}=X^{P}$ and $P=\ell\cap
XX^{\sigma_{\ell}}$. Since $XX^{\sigma_{\ell}}$ is fixed by $\sigma_{\ell}$,
it follows that $P^{\sigma_{\ell}}=P$.

(vi):

Follows straight from (iii).∎

######  Lemma 4.14.

If $m$ is a line fixed by $\sigma_{\ell}$, then $m=\ell$ or $m\perp\ell$.

###### Proof.

Let $X$ and $Y$ be the rimpoints of $m$. Then $X^{\sigma_{\ell}}$ and
$Y^{\sigma_{\ell}}$ are rimpoints incident with $m$ (by Lemma 4.13(v)). If
$X^{\sigma_{\ell}}=X$, then $Y^{\sigma_{\ell}}=Y$ and $m=\ell$, since the only
rimpoints fixed by $\sigma_{\ell}$ are those incident with $\ell$ (Lemma
4.13(v)). If $\sigma_{\ell}$ interchanges $X$ and $Y$, then this means
precisely that $m$ is perpendicular to $\ell$. ∎

######  Lemma 4.15.

Let $P$ be a point and let $\ell$ be a line of $\mathcal{I}$. Then the
following are equivalent:

  1. (a)

$P$ and $\ell$ are incident;

  2. (b)

the half-turn about $P$ and the reflection $\sigma_{\ell}$ about $\ell$
commute;

  3. (c)

$(P\sigma_{\ell})^{2}=1$.

###### Proof.

Since half-turns and reflections are involutions, it is clear that (b) and (c)
are equivalent. By Lemma 4.13(v), $P$ and $\ell$ are incident if and only if
$\sigma_{\ell}$ fixes $P$. Since the half-turn about $P$ fixes $P$, we have
that $P^{\sigma_{\ell}}=P$ is equivalent to $P^{P\sigma_{\ell}P}=P$. Now the
permutation $P\sigma_{\ell}P$ is the reflection about $\ell^{P}$, by Lemma
4.13(iii). So $P$ and $\ell$ are incident if and only if
$P\sigma_{\ell}P=\sigma_{\ell}$, and so (a) and (c) are equivalent. ∎

The following result allows us to more easily compute the image of a finite
point under a reflection.

######  Lemma 4.16.

Let $Q$ be a finite point not incident with a line $m$. Then
$Q^{\sigma_{m}}=Q^{P}$ where $P$ is the point of intersection of $m$ and
$QQ^{\sigma_{m}}$.

###### Proof.

Let $M$ and $N$ be the rimpoints of $m$. Note that $M^{Q}$ and $N^{Q}$ lie on
the same side of $\ell$ and hence $M^{Q}N^{Q}$ does not meet
$M^{Q\sigma_{m}}N^{Q\sigma_{m}}$. By Lemma 4.12, $N^{Q}N^{Q\sigma_{m}}$ does
not meet $M^{Q}M^{Q\sigma_{m}}$ as they are both perpendicular to $m$. So by
(S4), we have $M^{Q}N^{Q\sigma_{m}}//N^{Q}M^{Q\sigma_{m}}$, and hence there
exists a finite point $T$ such that $T:=M^{Q}N^{Q\sigma_{m}}\cap
N^{Q}M^{Q\sigma_{m}}$. Consider the hexagon
$M^{Q}MM^{Q\sigma_{m}}N^{Q}NN^{Q\sigma_{m}}$. The diagonal points are then
$Q$, $T$, and $Q^{\sigma_{m}}=MM^{Q\sigma_{m}}\cap NN^{Q\sigma_{m}}$. So $T$
lies on the line $QQ^{\sigma_{m}}$ by (A7). Now

$T^{\sigma_{m}}=M^{Q\sigma_{m}}N^{Q}\cap N^{Q\sigma_{m}}M^{Q}=T$

and hence $T$ is incident with $m$ (by Lemma 4.13(v)). Therefore,
$T=QQ^{\sigma_{m}}\cap m=P$. So $M^{Q\sigma_{m}}=(M^{Q})^{P}$ and
$N^{Q\sigma_{m}}=(N^{Q})^{P}$, and hence

$Q^{\sigma_{m}}=MM^{Q\sigma_{m}}\cap NN^{Q\sigma_{m}}=MM^{QP}\cap
NN^{QP}=\left(MM^{Q}\cap NN^{Q}\right)^{P}=Q^{P}.\qed$

######  Theorem 4.17.

Let $\ell$ and $m$ be two lines meeting in a finite point $P$. Then $\ell\perp
m$ if and only if $\sigma_{\ell}\sigma_{m}=P$.

###### Proof.

Suppose firstly that $\ell$ and $m$ are perpendicular. Let $A$, $B$, $C$, $D$
be rimpoints such that $\ell=AC$ and $m=BD$. Let $Q$ be a finite point
incident with $\ell$, not equal to $P$. Since $B$ is fixed by $\sigma_{m}$,
$Q^{\sigma_{m}}=Q^{P}$ (by Lemma 4.16) and $\sigma_{m}$ is an automorphism of
$\mathcal{I}$ (by Lemma 4.13(v)), we have

$B^{Q\sigma_{m}P}=B^{\sigma_{m}Q\sigma_{m}P}=B^{Q^{\sigma_{m}}P}=B^{Q^{P}P}=B^{PQ}=D^{Q}.$

Therefore,

$Q^{\sigma_{\ell}\sigma_{m}P}=Q^{\sigma_{m}P}=(BB^{Q}\cap
AC)^{\sigma_{m}P}=DD^{Q}\cap AC=Q.$

So $\sigma_{\ell}\sigma_{m}P$ fixes every point of $\ell$. Now let $X$ be a
rimpoint not incident with $\ell$ and suppose without loss of generality that
$X$ lies on the opposite side of $\ell$ as $B$ does. Then $XB$ meets $\ell$ in
a point $R$ and hence $X=B^{R}$. Since $B$ and $R$ are fixed by
$\sigma_{\ell}\sigma_{m}P$, it follows that $X$ is also fixed by
$\sigma_{\ell}\sigma_{m}P$. So every point not incident with $\ell$ is fixed
by $\sigma_{\ell}\sigma_{m}P$, and the rimpoints $A$ and $C$ are clearly fixed
by $\sigma_{\ell}\sigma_{m}P$. Therefore, $\sigma_{\ell}\sigma_{m}P=1$ as
required.

Conversely, suppose $\sigma_{\ell}\sigma_{m}=P$. In particular, $\ell\neq m$
since the half-turn about $P$ acts nontrivially on the rimpoints. Since $P$ is
an involution, we also have $\sigma_{m}\sigma_{\ell}=P$. Now the half-turn $P$
fixes the line $m$ (by Lemma 4.15), and $\sigma_{m}$ fixes $m$, and so

$m=m^{\sigma_{m}\sigma_{\ell}}=m^{\sigma_{\ell}}.$

Therefore, by Lemma 4.14, $m$ is perpendicular to $\ell$. ∎

The following lemmas will be used in the next section, but they also yield
fundamental results about reflections.

######  Lemma 4.18.

Let $\ell$ and $m$ be distinct lines. Then $\sigma_{\ell}\sigma_{m}$ fixes at
most two rimpoints.

###### Proof.

Suppose $\sigma_{\ell}\sigma_{m}$ fixes a rimpoint $X$. Then
$X^{\sigma_{\ell}}=X^{\sigma_{m}}$ and hence $XX^{\sigma_{\ell}}$ is
perpendicular to $m$. By Lemma 4.14, either $\ell$ is perpendicular to $m$ and
$X$ is incident with $\ell$, or $\ell$ and $m$ are not concurrent in a finite
point or rimpoint and $X$ is incident with the unique common perpendicular of
$\ell$ and $m$. In both cases, there are at most two possibilities for $X$. ∎

######  Lemma 4.19.

Let $\ell$, $m$, $n$ be lines concurrent in a point $P$, and let $L$ be a
rimpoint on $\ell$. Let $o$ be the unique perpendicular line to
$LL^{\sigma_{m}\sigma_{n}}$ through $P$, and let
$g:=\sigma_{\ell}\sigma_{m}\sigma_{n}\sigma_{o}$. Then $g$ fixes at least four
rimpoints and the finite point $P$.

###### Proof.

Let $\ell$, $m$, $n$ be three lines concurrent in a point $P$. Let $L$ be a
rimpoint on $\ell$ and let $o$ be the unique perpendicular line to
$LL^{\sigma_{m}\sigma_{n}}$ through $P$. Now $g$ fixes $L$ because

$L^{g}=L^{\sigma_{\ell}\sigma_{m}\sigma_{n}\sigma_{o}}=(L^{\sigma_{m}\sigma_{n}})^{\sigma_{o}},$

and $g$ fixes $P$ because $P$ is fixed by each of the four reflections
comprising $g$. Therefore, $g$ fixes $L^{P}$ (because $g$ is an automorphism
by Lemma 4.13(v)) and the line $\ell^{\prime}$ perpendicular to $\ell$ through
$P$ (because such a line is unique). Let $\mathcal{C}$ be a cyclic order
compatible with the separation relation $//$. Now $g$ is a product of four
reflections and so preserves $\mathcal{C}$ (Lemma 4.13(i)). Let $L^{\prime}$
be a rimpoint on $\ell^{\prime}$ such that $\mathcal{C}(L,L^{\prime},L^{P})$.
Then $\mathcal{C}(L,(L^{\prime})^{g},L^{P})$ and hence $L^{\prime}$ is fixed
by $g$ (as it lies on $\ell^{\prime}$ and is on the same side of $\ell$ as
$L^{\prime}$). So we have four fixed rimpoints (i.e., $L$, $L^{P}$,
$L^{\prime}$, $(L^{\prime})^{P}$) and a fixed finite point $P$. ∎

##  5\. Glide reflections

A glide reflection is the product $P\sigma_{\ell}$ where $P$ and $\ell$ are
non-incident. It is not difficult to see that a glide reflection fixes
precisely two rimpoints. Let $P$ and $\ell$ be a point and line, and suppose
$P\sigma_{\ell}$ fixes a rimpoint $X$. Then $X^{P}=X^{\sigma_{\ell}}$ and
hence $XX^{P}$ is perpendicular to $\ell$. So it follows that the only
rimpoints fixed by $P\sigma_{\ell}$ are those lying on the unique
perpendicular line to $\ell$, through $P$.

######  Lemma 5.1.

Let $A$ and $E$ be two finite points, and let $d$ be a line. Then there exists
a finite point $X$ and a line $y$ such that $AE\sigma_{d}=X\sigma_{y}$.

###### Proof.

Let $x$ be the line joining $A$ and $E$. Suppose $A$ and $E$ are incident with
$d$, that is, $x=d$. If $e$ is the perpendicular to $d$ at $E$, then by
Theorem 4.17, $AE\sigma_{d}=A\sigma_{e}$ and we are done. So we can assume
that $x\neq d$: we have three cases.

Case 1: $x$ and $d$ meet in a finite point $Y$:

This case includes the possibility that $Y=A$ or $Y=E$. By Theorem 3.7, there
is a point $X$ on $x$ such that $AEY=X$. Let $y$ be the perpendicular to $d$
incident with $Y$; so by Theorem 4.17, $Y\sigma_{d}=\sigma_{y}$. Therefore,
$AE\sigma_{d}=XY\sigma_{d}=X\sigma_{y}$, as required.

Case 2: $x$ and $d$ have a common perpendicular $q$:

Let $X=q\cap x$ and let $Q:=q\cap d$. By Theorem 3.7, there is a point $Y$ on
$x$ such that $AEX=Y$. By Lemma 4.10, there exists a line $y$ perpendicular to
$x$ and incident with $Y$. Then

$AE\sigma_{d}=YX\sigma_{d}=Y\sigma_{x}\sigma_{q}\sigma_{d}=\sigma_{y}Q.$

Finally, we can rewrite $\sigma_{y}Q$ as $Q(Q\sigma_{y}Q)$, which is the
product of a half-turn and reflection (by Lemma 4.13(iii)).

Case 3: $x$ and $d$ meet in a rimpoint:

That is, $x$ and $d$ are parallel. Let $a$ be the line perpendicular to $x$
and passing through $A$, and let $e$ be another line perpendicular to $x$ and
passing through $E$. Let $D$ be a finite point on $d$, and let $x^{\prime}$ be
the perpendicular to $a$ passing through $D$. In particular, $x^{\prime}\neq
x$ and $x^{\prime}$ is also perpendicular to $e$ (because if
$M:=A^{\prime}E\cap AE^{\prime}$, then
$\sigma_{e}\sigma_{x^{\prime}}=\sigma_{a}^{M}\sigma_{x}^{M}=M\sigma_{a}\sigma_{x}M=\sigma_{x}^{M}\sigma_{a}^{M}=\sigma_{x^{\prime}}\sigma_{e}$).
Let $A^{\prime}:=a\cap x^{\prime}$ and let $E^{\prime}:=e\cap x^{\prime}$.
Since $x^{\prime}$ meets $d$ in a finite point, by Case 1 above, there exists
a finite point $X$ and a line $y$ such that
$A^{\prime}E^{\prime}\sigma_{d}=X\sigma_{y}$. Therefore (by Theorem 4.17),

$AE\sigma_{d}=(\sigma_{a}\sigma_{x})(\sigma_{x}\sigma_{e})\sigma_{d}=\sigma_{a}\sigma_{e}\sigma_{d}=(A^{\prime}\sigma_{x^{\prime}})(\sigma_{x^{\prime}}E^{\prime})\sigma_{d}=A^{\prime}E^{\prime}\sigma_{d}=X\sigma_{y}.$

In all cases, we have shown that $AE\sigma_{d}=X\sigma_{y}$ for some point $X$
and line $y$. ∎

The following theorem can be thought of as a weakening of _Pascal’s Theorem on
rimpoints_ (A7) where we have at least two internal diagonal points.

######  Theorem 5.2.

If $A$, $B$, $C$, $D$, $E$, $F$ are rimpoints such that two of the three
diagonal points $P:=AB\cap DE$, $Q:=BC\cap EF$ of the hexagon exist, then
either (i) $CD$ meets $FA$ in a finite point incident with $PQ$, or (ii) $CD$,
$FA$, $PQ$ have a common perpendicular.

###### Proof.

If $CD\cap FA$ is a finite point, then it is incident with $PQ$ by (A7). So we
will suppose that $CD$ does not meet $FA$ in a finite point. Let $X:=PQ\cap
BF$ and $Y:=PQ\cap CE$. (Note that $X$ and $Y$ are finite points since $B$ and
$F$ lie on different sides of $PQ$, because $B$ and $C$ lie on different sides
of $PQ$, and $A$, $C$ and $F$ lie on the same side of $PQ$; otherwise $AF$
would meet $CD$ in a finite point. Similarly, $C$ and $E$ lie on different
sides of $PQ$). Let $U$ be the midpoint of $\overline{PY}$ and let $V$ be the
midpoint of $\overline{PX}$. So by Lemma 3.16, $P^{V}=X$ and $P^{U}=Y$. Now by
Lemma 3.8, half-turns are automorphisms, and so

$(BF)^{Q}=B^{Q}F^{Q}=CE$

and hence $Y=X^{Q}$. By Lemma 3.7, $VQU=UQV$. Therefore,

$(UPVQ)^{2}=UP(VQU)(PVQ)=UP(UQV)(PVQ)=P^{U}QP^{V}Q=YQXQ=YY=1$

and hence $UPVQ$ is an involution. Now $UPV$ is a half-turn (by Lemma 3.7), so
either $UPV=Q$ or $(UPV)Q$ is an order-preserving axial element. No order-
preserving axial element has order $2$ (by Lemma 3.21), and so $UPV=Q$. Hence

$E^{UV}=C^{UPV}=C^{Q}=B$

as $E^{UPU}=D^{(PU)^{2}}=C$.

Let $k=\bot(PQ,CD)$. Let $L$ and $M$ be the rimpoints of $PQ$ such that we
have arranged the points in the order $L$, $P$, $Q$, $M$. By Lemma 4.7, there
are unique order-preserving axial maps $g$ and $h$ such that (i) $g$ has axis
$PQ$ and $C^{g^{2}}=D$, (ii) $h$ has axis $CD$ and $M^{h^{2}}=L$, and thus,
$k=C^{g}M^{h}$. By Lemma 3.22, $g=UP$. Thus

$O:=C^{g}=C^{UP}=E^{U}.$

So $k$ is the unique perpendicular line to $PQ$ incident with the rimpoint
$O$. Similarly, the common perpendicular $k^{\prime}$ of $PQ$ and $AF$ is the
unique perpendicular line to $PQ$ incident with

$O^{\prime}:=F^{VP}=B^{V}$

(n.b., the analogue of $g$ is $VP$ here). We have shown above that
$B^{V}=E^{U}$ and hence $O=O^{\prime}$, and therefore, $k=k^{\prime}$. So, it
follows that $PQ$, $CD$, $AF$ have a common perpendicular. ∎

######  Theorem 5.3.

Let $P$ be a finite point and $\ell$ be a line. Suppose $X$, $Y$, $Z$ are
rimpoints such that $XX^{P\sigma_{\ell}}$, $YY^{P\sigma_{\ell}}$,
$ZZ^{P\sigma_{\ell}}$ have a common perpendicular. Then $P$ is incident with
$\ell$.

###### Proof.

Let $o$ be the common perpendicular to $XX^{P\sigma_{\ell}}$,
$YY^{P\sigma_{\ell}}$, $ZZ^{P\sigma_{\ell}}$. By Lemma 4.8, $o$ is the Pascal
line of the hexagon
$XY^{P\sigma_{\ell}}ZX^{P\sigma_{\ell}}YZ^{P\sigma_{\ell}}$. Since
$XX^{P\sigma_{\ell}}$, $YY^{P\sigma_{\ell}}$ have a common perpendicular,
$XY^{P\sigma_{\ell}}\cap X^{P\sigma_{\ell}}Y$ is a finite point $U$ (by
Theorem 3.5 and Lemma 4.12). Consider the hexagon
$XX^{P}X^{P^{\sigma_{\ell}}}YY^{P}Y^{P^{\sigma_{\ell}}}$. Since $XX^{P}\cap
YY^{P}=P$, $X^{P^{\sigma_{\ell}}}Y\cap XY^{P^{\sigma_{\ell}}}=U$ and
$X^{P}X^{P^{\sigma_{\ell}}},Y^{P}Y^{P^{\sigma_{\ell}}}\perp\ell$, it follows
from Theorem 5.2 that $UP\perp\ell$.

Similarly, $XZ^{P\sigma_{\ell}}\cap X^{P\sigma_{\ell}}Z$ is a finite point $V$
and $VP\perp\ell$. Therefore, (by properties of perpendicularity),
$UP=VP=UV=o$. So in particular, $o$ is fixed by $P\sigma_{\ell}$. Let
$A:=\ell\cap o$. Then $\sigma_{\ell}\sigma_{o}=A$ and hence
$P\sigma_{\ell}\sigma_{o}=PA$. Therefore, $P\sigma_{\ell}\sigma_{o}$ is either
trivial (i.e., $P=A$) or axial. Now $P\sigma_{\ell}\sigma_{o}$ has three fixed
points, and so by Lemma 3.17, we must have the former case:
$P\sigma_{\ell}\sigma_{o}=1$. Hence $\sigma_{\ell}\sigma_{o}=P$ and so
$\sigma_{\ell}\sigma_{o}$ is an involution. Therefore,
$\sigma_{o}=\sigma_{o^{\sigma_{\ell}}}$ and so $o$ is fixed by
$\sigma_{\ell}$. Therefore, $\ell\perp o$ and $P$ is incident with $\ell$ (by
Lemma 4.14). ∎

##  6\. A proof of Theorem 1.2 using a theorem of F. Bachmann

A metric plane is an incidence structure $\mathcal{M}$ of points and lines
equipped with a symmetric binary relation of _perpendicularity_ on the lines
of $\mathcal{M}$, satisfying the following axioms. In this context, a
reflection in a line $a$ is a collineation of $\mathcal{M}$ that preserves the
perpendicularity relation and fixes $a$ pointwise. We call $a$ the _axis_ of
the reflection.

  1. (I1)

Every pair of distinct points is incident with a unique line.

  2. (I2)

Every line is incident with at least three points.

  3. (I3)

There exist three points with the property that no line is incident with all
of them.

  4. (P1)

For any point $P$ and any line $a$, there is a line $b$ incident with $P$ and
perpendicular to $a$. If $P$ is incident with $a$, then $b$ is unique.

  5. (P2)

If $a$ is perpendicular to $b$, then there is a unique point incident with
both $a$ and $b$.

  6. (M1)

Every line is the axis of at least one reflection.

  7. (M2)

Reflections are involutions (on the points).

  8. (M3)

If three lines $a$, $b$, $c$ are concurrent, then a composition of reflections
in the lines $a$, $b$ and $c$ is equal to a reflection in another line $d$.

  9. (M4)

If three lines $a$, $b$, $c$ have a common perpendicular, then a composition
of reflections in the lines $a$, $b$ and $c$ is equal to a reflection in
another line $d$.

These axioms are due to Bachmann [5, §2.3], except for (I3), which is due to
Ewald [10] and simply ensures non-degeneracy. One consequence of these axioms
is that for every line $\ell$, there is at most one reflection with axis
$\ell$ (see [5, p.27, Satz 3]).

The main result of our paper centres around the creation of a perpendicularity
relation $\perp$ for an incidence structure that gives rise to reflections
yielding a metric plane. Now (A1), (A2), (A3) imply (I1), (I2), (I3), and
(M1), (M2) follow from the definition of a reflection. So the important
results will be the affirmations of (M3) and (M4).

######  Lemma 6.1.

An incidence structure satisfying Axioms (A1) – (A7) plus (A10) and (A11),
also satisfies (M4).

###### Proof.

Suppose $a$, $b$, $c$ have a common perpendicular $m$. Let $A$, $B$ and $C$ be
the intersections of $a$, $b$, $c$ with $m$, respectively. Now

$\sigma_{a}\sigma_{b}\sigma_{c}=\sigma_{m}A\sigma_{m}B\sigma_{m}C=ABC\sigma_{m}^{3}=ABC\sigma_{m}$

by Theorem 4.17 and Lemma 4.15. Since $A$, $B$, $C$ are collinear, there
exists a finite point $D$ on $m$ such that $ABC=D$ (by Lemma 3.7). Let $d$ be
the unique line on $D$ that is perpendicular to $m$ (c.f., Lemma 4.10). Then
by Theorem 4.17, $\sigma_{d}\sigma_{m}=D$ and hence
$\sigma_{a}\sigma_{b}\sigma_{c}=D\sigma_{m}=\sigma_{d}$. ∎

Let $\mathcal{I}$ be an incidence structure satisfying Axioms (A1) – (A7) and
(A10), (A11). We have shown that $\mathcal{I}$ satisfies eight of the nine
axioms for a metric plane. The remaining axiom is (M3); the _three reflection
theorem_ for a pencil of concurrent lines.

######  Lemma 6.2.

Let $a$, $b$, $c$ be three distinct lines concurrent in a point $P$. Then
there exists a line $a^{\prime}$ perpendicular to $a$ and having a common
perpendicular $\ell$ with $b$ such that $\ell$ does not intersect $c$ nor is
parallel to $c$.

###### Proof.

Let $A$, $B$, $C$ be rimpoints on $a$, $b$, $c$ respectively, such that
$\mathcal{C}(A,B,C)$. Then $\mathcal{C}(B,C,A^{\sigma_{b}})$ or
$\mathcal{C}(B,A^{\sigma_{b}},C)$ (by (C2)); so we can find a rimpoint $L$
such that $\mathcal{C}(B,L,A^{\sigma_{b}})$ and $\mathcal{C}(B,L,C)$ (in both
cases, $L$ is between $B$ and $C$). Therefore,
$\mathcal{C}(A,L^{\sigma_{b}},B)$ (as $\sigma_{b}$ reverses $\mathcal{C}$, by
Lemma 4.13(i), and fixes $B$) and the line $\ell:=LL^{\sigma_{b}}$ is
perpendicular to $b$ and not meeting $a$ or $c$ in a finite point or rimpoint.
So, by Lemma 4.9, $\ell$ has a common perpendicular $a^{\prime}$ to $a$. ∎

######  Lemma 6.3.

An incidence structure satisfying Axioms (A1) – (A7) plus (A10) and (A11),
also satisfies (M3).

###### Proof.

Let $\ell$, $m$, $n$ be three distinct lines concurrent in a point $P$. By
Lemma 6.2, there is a line $\ell^{\prime}$ perpendicular to $\ell$ and having
a common perpendicular $k$ with $m$ such that $k$ does not intersect $n$ nor
is parallel to $n$. That is, there is a line $n^{\prime}$ perpendicular to $k$
and $n$. Let $A$ be the point of intersection of $\ell$ and $\ell^{\prime}$,
and let $B$ be the point of intersection of $n$ and $n^{\prime}$. Then (by
Theorem 4.17),

$\sigma_{\ell}\sigma_{m}\sigma_{n}=(\sigma_{\ell}\sigma_{\ell^{\prime}})\sigma_{\ell^{\prime}}\sigma_{m}\sigma_{n^{\prime}}(\sigma_{n^{\prime}}\sigma_{n})=A(\sigma_{\ell^{\prime}}\sigma_{m}\sigma_{n^{\prime}})B.$

Now, $k$ is perpendicular to $\ell^{\prime}$, $m$ and $n^{\prime}$, so by
(M4), $\sigma_{\ell^{\prime}}\sigma_{m}\sigma_{n^{\prime}}$ is a reflection
$\sigma_{d}$ in a line $d$ perpendicular to $k$.

Let $L$ be a rimpoint incident with $\ell$, and let $o$ be the line through
$P$ perpendicular to $XX^{\sigma_{m}\sigma_{n}}$. Let
$g=\sigma_{\ell}\sigma_{m}\sigma_{n}\sigma_{o}$. Then
$g=A\sigma_{d}B\sigma_{o}=AB^{\sigma_{d}}\sigma_{d}\sigma_{o}$. Let
$E:=B^{\sigma_{d}}$. By Lemma 5.1, we can write $AE\sigma_{d}$ as
$X\sigma_{y}$ for some $X$ and $y$. Recall that $g$ fixes at least three
rimpoints by Lemma 4.19; call them $U$, $V$, $W$. So $o$ is perpendicular to
each of $UU^{X\sigma_{y}}$, $VV^{X\sigma_{y}}$, $WW^{X\sigma_{y}}$. Hence by
Theorem 5.3, $X$ is incident with $y$. So $X\sigma_{y}=\sigma_{y^{\prime}}$
where $y^{\prime}$ is the line through $X$ perpendicular to $y$, and thus,
$g=\sigma_{y^{\prime}}\sigma_{o}$. However, $g$ fixes more than two rimpoints,
and so by Lemma 4.18, $g=1$ and $y^{\prime}=o$. Therefore,
$\sigma_{\ell}\sigma_{m}\sigma_{n}=\sigma_{o}$. ∎

######  Theorem 6.4.

The incidence structure $\mathcal{I}$, equipped with the perpendicularity
relation $\perp$, is a metric plane.

###### Proof.

The perpendicularity relation $\perp$ is symmetric from Lemma 4.5 and the
reflections are automorphisms of $\mathcal{I}$ that preserve $\perp$, by Lemma
4.13. Recall that (I1), (I2), (I3), (M1), (M2) hold trivially. By Lemmas 6.1
and 6.3, we have that (M3) and (M4) hold. The axiom (P2) follows from Theorem
4.17, and the axiom (P1) follows from Lemma 4.10. ∎

The following theorem is a culmination of the work in Sections 15.1 and 15.2
(in particular, Theorem 9, Theorem 11 and Satz 4) of [5]; see also [23, §6],
[26], [30, §4].

######  Theorem 6.5 (Bachmann’s Theorem).

A metric plane satisfying Axiom ($\neg V^{*}$) and Axiom ($H$) (see below),
and such that every line is an element of an end, is isomorphic to a Cayley-
Klein hyperbolic plane over a Euclidean field with characteristic not equal to
$2$.

Axiom ($\neg V^{*}$):

There are two lines $a$ and $b$ that share no point nor perpendicular in
common. (Alternatively: there exists an end).

Axiom ($H$):

If a point $P$ and line $\ell$ are not incident then there are at most two
lines through $P$ which have neither a common point nor a common perpendicular
with $\ell$.

An _end_ is a certain type of pencil of lines (see [30, p. 335]), which for
our purposes, is synonymous with a rimpoint (see [21]).

### Proof of Theorem 1.2

Let $\mathcal{I}$ be an incidence structure of points and lines satisfying
(A1) through to (A7), plus axioms (A10) and (A11). By Theorem 6.4,
$\mathcal{I}$ is a metric plane, when equipped with the particular
perpendicularity relation that we defined in Section 4. In our setting, a
rimpoint is an ‘end’ and every line of $\mathcal{I}$ is incident with two
rimpoints [1, Theorem 1.7]. So it remains to check that Axiom ($H$) holds. Let
$P$ be a point and let $\ell$ be a line that are not incident, and let $m$ be
a line on $P$ that does not meet $\ell$ in a finite point nor has a common
perpendicular with $\ell$. Recall that two lines are parallel if they share
neither a point nor a perpendicular (Lemma 4.9), hence, $\ell$ and $m$ are
parallel. In the introduction, we mentioned that there are at most two lines
through $P$ parallel to $\ell$, by [9, Corollary 2.4]. Therefore, we can
directly apply Bachmann’s Theorem to establish the truth of Theorem 1.2.

######  Remark 6.6.

Another way to prove Theorem 1.2 is via _$H$ -groups_. An $H$-group is an
$S$-group (as defined in [17, §4]) such that there exist parallels and if $a$
and $b$ are parallel and $c$ and $d$ are parallel then there is a element
$v\in S$ with $abv$ and $cdv$ involutions. It can be shown, after a little bit
of work, that our group $H$ is an $H$-group (pardon the clash of
terminology!). So we could also use a result of Bachmann [5, Theorem 9, p.198]
to establish Theorem 1.2.

##  7\. Abstract ovals

An abstract oval $\mathcal{B}=(O,S)$ is a set $O$ of elements (where
$|O|\geqslant 3$) together with a set of involutory permutations $S$ such that
for every $A_{1},A_{2},B_{1},B_{2}\in O$ with
$\\{A_{1},A_{2}\\}\cap\\{B_{1},B_{2}\\}=\varnothing$ there is exactly one
element $s$ of $S$ such that $A_{1}^{s}=A_{2}$ and $B_{1}^{s}=B_{2}$. Here we
allow the identity permutation to be an involution. An abstract oval gives
rise to an incidence structure $\mathbb{P}(\mathcal{B})$ as follows:

Points:

the elements of $O$ and the involutions $S$;

Lines:

each line $\ell_{A,B}$ is defined by a pair of (not necessarily distinct
elements) $A$, $B$ of $O$, and is incident with $A$, $B$ and all elements of
$S$ that map $A$ to $B$. (Such a line is secant if $A\neq B$ and tangent if
$A=B$.)

A line $\ell$ of $\mathbb{P}(\mathcal{B})$ is said to be Pascalian if, for
every three involutions $I,J,K$ of $S$ incident with $\ell$, the product $IJK$
is also an involution of $S$ incident with $\ell$. An abstract oval is said to
be Pascalian if every line of its associated incidence structure is Pascalian.
A conic $\mathcal{K}$ of a Pappian projective plane $\Pi$ gives rise to a
Pascalian abstract oval as follows. Given a point $P$ of
$\Pi\backslash\mathcal{K}$ and a point $X$ of $\mathcal{K}$, we define $X^{P}$
to be the other point of $\mathcal{K}$ incident with $PX$ when $PX$ is a
secant line, otherwise we set $X^{P}:=X$ if $PX$ is tangent to $\mathcal{K}$.
Each point $P$ of $\Pi\backslash\mathcal{K}$ then gives rise to an involution
$s_{P}:X\mapsto X^{P}$, and we let $S$ be the set of all such involutory
permutations. Then $\mathcal{B}:=(\mathcal{K},S)$ is a special abstract oval
known as an abstract conic. The incidence structure $\mathbb{P}(\mathcal{B})$,
in this case, is isomorphic to the incidence structure we obtain from $\Pi$ by
removing the lines that do not intersect $\mathcal{K}$ (i.e., the _external_
lines). In general, if the incidence structure $\mathbb{P}(\mathcal{B})$ of an
abstract oval $\mathcal{B}$ is isomorphic to the incidence structure arising
from some abstract conic then we say that $\mathcal{B}$ is also an abstract
conic.

######  Theorem 7.1.

Let $\mathcal{I}$ be an incidence structure satisfying (A1) – (A7), plus (A10)
and (A11). Then the set of all reflections and all half-turns forms an
abstract oval with Pascalian secant lines.

###### Proof.

Our underlying set is the set of rimpoints $\mathcal{O}$, and we let $T$ be
the set of all reflections and half-turns. We will show that $(\mathcal{O},T)$
is an abstract oval. Recall that in the definition of an abstract oval, we
consider $A_{1},A_{2},B_{1},B_{2}\in\mathcal{O}$ with
$\\{A_{1},A_{2}\\}\cap\\{B_{1},B_{2}\\}=\varnothing$, but $A_{1}$ and $A_{2}$
could be equal, and $B_{1}$ and $B_{2}$ could be equal. So there are three
cases to consider according to whether $\\{A_{1},A_{2},B_{1},B_{2}\\}$ has
size 2, 3, or 4. Given a pair $X$, $Y$ of distinct rimpoints there is a unique
reflection fixing $X$ and $Y$, namely $\sigma_{XY}$, and no half-turn can fix
a rimpoint. Next, given a triple of distinct rimpoints $X$, $Y$, $Z$, there is
a unique perpendicular $\ell$ to $YZ$ on $X$ and so a unique reflection fixing
$X$ and interchanging $Y$ and $Z$, and again, no half-turn can fix a rimpoint.
Finally, given a quadruple of distinct rimpoints $X$,$Y$,$Z$,$W$, (i) if $XY$
and $ZW$ meet in a finite point $P$, there is a unique half-turn (namely that
around $P$) and no reflection interchanging both $X$ and $Y$ and $Z$ and $W$;
(ii) if $XY$ and $ZW$ are disjoint, there is a unique perpendicular $\ell$ to
$XY$ and $ZW$, and so a unique reflection (namely that in $\ell$) and no half-
turn interchanging both $X$ and $Y$ and $Z$ and $W$. Therefore,
$(\mathcal{O},T)$ is an abstract oval.

Now suppose $X$, $Y$ are two rimpoints, and $s_{1},s_{2},s_{3}$ are
involutions taking $X$ to $Y$. If $s_{1}$, $s_{2}$, $s_{3}$ are reflections
about the line $XY$, then $s_{1}s_{2}s_{3}\in T$ by Theorem 6.1. If $s_{1}$,
$s_{2}$, $s_{3}$ are half-turns about points incident with $XY$, then
$s_{1}s_{2}s_{3}\in T$ by Theorem 3.7. If $s_{1}$, $s_{2}$ are half-turns
about points incident with $XY$, and $s_{3}$ is a reflection about a line
$\ell$ perpendicular to $XY$, then by Theorem 4.17, there exist lines $p$ and
$q$ such that $s_{1}=\sigma_{p}\sigma_{XY}$ and $s_{2}=\sigma_{XY}\sigma_{q}$.
Then, $s_{1}s_{2}s_{3}=\sigma_{p}\sigma_{q}\sigma_{\ell}$, which lies in $T$
by Theorem 6.1. The final case to consider is that $s_{1}$ is a half-turn
about a point incident with $XY$, and $s_{2},s_{3}$ are reflections about
lines $\ell$ and $m$ perpendicular to $XY$. Again, by Theorem 4.17, there
exists a line $p$ such that $s_{1}=\sigma_{XY}\sigma_{p}$. Then,
$s_{1}s_{2}s_{3}=\sigma_{XY}\sigma_{p}\sigma_{\ell}\sigma_{m}$. By Theorem
6.1, there is a line $n$ perpendicular to $XY$ such that
$\sigma_{p}\sigma_{\ell}\sigma_{m}=\sigma_{n}$. Let $N:=n\cap XY$. Then
$\sigma_{n}\sigma_{XY}=N$ and hence $s_{1}s_{2}s_{3}=N\in T$. Hence the
product of three elements of $T$ taking $X$ to $Y$ is another element of $T$
taking $X$ to $Y$. Therefore, $(\mathcal{O},T)$ has Pascalian secant lines. ∎

If $(O,B)$ is an abstract oval, then we say it is _regular_ if $B$ is
invariant under conjugation by elements of $B$.

######  Theorem 7.2 (Bamberg, Harris, Penttila [6]).

An abstract oval $(O,S)$ such that all elements of $S$ are regular and all
secant lines are Pascalian is an abstract conic.

######  Corollary 7.3.

Let $\mathcal{I}$ be an incidence structure satisfying (A1) – (A7), plus (A10)
and (A11). Then the set of all reflections and all half-turns forms an
abstract conic.

###### Proof.

Notice that the set of half-turns and reflections is regular by Remark 3.12
(or Corollary 3.13), Lemma 4.13(iii), and Lemma 4.13(iv). ∎

To see the connection between Corollary 7.3 and Theorem 1.2, we will first
realise the Cayley-Klein model of the hyperbolic plane $\mathcal{H}$ over a
field $F$ with an incidence structure arising from the elements of the group
$\mathsf{PGL}(2,F)$ (see [5, §15.1]). To do this, we use the language of
permutation group theory. The points of $\mathcal{H}$ are the points of
$\mathsf{PG}(2,F)$ that are internal with respect to $\mathcal{O}$. The lines
of $\mathcal{H}$ are the lines of $\mathsf{PG}(2,F)$ that intersect
$\mathcal{O}$ in two points; the _secant lines_. By Frégier’s Theorem (see
[28, Theorem 25, p. 68]), the harmonic homologies of $\mathsf{PGL}(2,F)$ with
centre an internal point and axis the polar external line, are precisely the
fixed-point-free involutions $\hat{T}_{0}$ of $\mathsf{PGL}(2,F)$. Dually, the
harmonic homologies of $\mathsf{PGL}(2,F)$ with axis a secant line and centre
the external point that is the pole, are precisely the involutions
$\hat{T}_{2}$ of $\mathsf{PGL}(2,F)$ that fix two points of $\mathcal{O}$. So
we have permutational isomorphisms where the action of $\mathsf{PGL}(2,F)$ on
$\hat{T}_{0}$ and $\hat{T}_{2}$ is by conjugation:

$\displaystyle\left(\hat{T}_{0},\mathsf{PGL}(2,F)\right)$
$\displaystyle\cong\left(\text{points of
}\mathcal{H},\mathsf{PGL}(2,F)\right),$
$\displaystyle\left(\hat{T}_{2},\mathsf{PGL}(2,F)\right)$
$\displaystyle\cong\left(\text{lines of
}\mathcal{H},\mathsf{PGL}(2,F)\right).$

An involution in $\hat{T}_{0}$ is incident with an involution in $\hat{T}_{2}$
if and only if they commute, and hence we obtain an isomorphism of the
incidence structures $(\hat{T}_{0},\hat{T}_{2})$ and $\mathcal{H}$.

So we have another proof of Theorem 1.2: Theorem 7.3 says that the half-turns
and reflections of $\mathcal{O}$ are those of an abstract conic over some
field $F$. This allows us to identify the points of $\mathcal{I}$ with fixed-
point-free involutions in $\mathsf{PGL}(2,F)$, and identify the lines of
$\mathcal{I}$ with the involutions of $\mathsf{PGL}(2,F)$ that fix two points.
Therefore, $\mathcal{I}$ is isomorphic to the Cayley-Klein hyperbolic plane
over a field $F$, and by [5, Satz 4, p. 237], $F$ is a Euclidean field.

##  8\. Concluding remarks

  1. 1.

A metric plane is said to have _free mobility_ if its group of motions
$\Gamma$ (i.e., automorphisms that preserve perpendicularity) acts
transitively on flags; pairs $(P,\ell)$ where $P$ is a point incident with
$\ell$. It turns out (see [5, Satz 2, p.126]) that free mobility is equivalent
to each of the following three properties:

    1. (1)

$\Gamma$ is transitive on points and lines;

    2. (2)

Every two points have a midpoint, and every two lines have a bisector;

    3. (3)

Every element of $\Gamma$ that has a fixed line or fixed point is a square.

Note that the third condition above resembles the fundamental property of
order-preserving axial maps (Lemma 3.25) that enables us to define
perpendicularity of lines. The bisectors in the second condition are in
analogy with midpoints for half-turns. That is, if $\ell$ and $m$ are lines,
then $n$ is a bisector of $\ell$ and $m$ if
$\sigma_{\ell}\sigma_{n}=\sigma_{n}\sigma_{m}$. We can show directly that the
incidence structure $\mathcal{I}$ studied in this paper is a metric plane with
free mobility:

######  Lemma 8.1 (Bisectors exist).

Every two distinct lines $\ell$ and $m$ have a (unique) bisector.

###### Proof.

Take the rimpoints $X$, $Y$ of $\ell$ and $X^{\prime}$, $Y^{\prime}$ of $m$.
We may suppose that we have labelled our rimpoints so that one of the
following holds: (i) $XX^{\prime}$ and $YY^{\prime}$ do not meet and are not
parallel, (ii) $X=X^{\prime}$ or (iii) $Y=Y^{\prime}$. In all cases, thee is
no half-turn interchanging $X$ and $X^{\prime}$, and $Y$ and $Y^{\prime}$.
Since the half-turns and the reflections form the set of permutations of an
abstract oval (Theorem 7.1), there is a reflection $\sigma_{n}$ interchanging
$X$ and $X^{\prime}$ and $Y$ and $Y^{\prime}$. Note that this is valid in
cases (ii) and (iii). Hence, by Lemma 4.13, $\ell^{\sigma_{n}}=m$ and hence
$\sigma_{n}\sigma_{\ell}\sigma_{n}=\sigma_{m}$ (by Lemma 4.13(iv)). Therefore,
$\sigma_{\ell}\sigma_{n}=\sigma_{n}\sigma_{m}$ and $n$ is a bisector of $\ell$
and $m$. ∎

  2. 2.

We show that we can use Skala’s Theorem for a fourth proof of Theorem 1.2. In
each of the three proofs that we have given, we have an intermediate step that
shows that our incidence structure $\mathcal{I}$ can be embedded into a
Pappian projective plane $\Pi$ such that collinearity of three points is
preserved. Since Pappus’ Theorem and Desargues’ Theorem hold true in $\Pi$, it
follows that Skala’s axioms (A8) and (A9) hold (if two of the diagonal points
exist, then so does the third), and so by Theorem 1.1, our result follows.

  3. 3.

Hyperbolic planes over non-Euclidean fields do not satisfy (A10), however, we
do not know if there are hyperbolic planes that have rimpoints and do not
satisfy (A11). Thus we leave it as an open question whether (A11) is derivable
from the other axioms.

## Acknowledgements

The first author acknowledges the support of the Australian Research Council
Future Fellowship FT120100036, and would like to give special thanks to Sue
Barwick for accommodating the first author’s research visit to the University
of Adelaide where much of this work was done.

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