# Klein’s ten planar dessins of degree $11$,
and beyond

Gareth A. Jones and Alexander K. Zvonkin

###### Abstract

We reinterpret ideas in Klein’s paper on transformations of degree $11$ from
the modern point of view of dessins d’enfants, and extend his results by
considering dessins of type $(3,2,p)$ and degree $p$ or $p+1$, where $p$ is
prime. In many cases we determine the passports and monodromy groups of these
dessins, and in a few small cases we give drawings which are topologically
(or, in certain examples, even geometrically) correct. We use the Bateman–Horn
Conjecture and extensive computer searches to support a conjecture that there
are infinitely many primes of the form $p=(q^{n}-1)/(q-1)$ for some prime
power $q$, in which case infinitely many groups ${\rm PSL}_{n}(q)$ arise as
permutation groups and monodromy groups of degree $p$ (an open problem in
group theory).

2010 Mathematics Subject Classification. Primary 14H57, secondary 05C10,
11G32, 11N13, 11N32, 20B20, 20B25.

Key words and phrases. Plane tree, dessin d’enfant, map, monodromy group,
projective groups of prime degree, groups containing a cycle, dessins on
elliptic curves, Belyĭ functions, Bateman–Horn Conjecture.

##  1 Introduction

In 1878–9 Klein published two papers [50, 51] with almost identical titles.
The first concerned equations of degree $7$, while the second concerned those
of degree $11$, with the properties and actions of the groups ${\rm
PSL}_{2}(p)$, for $p=7$ and $11$ respectively, playing a major role in them.
The paper [50] is deservedly famous, not least for introducing the Hauptfigur,
the iconic diagram of a tessellation of a hyperbolic $14$-gon which, after
identification of sides, yields the quartic curve of genus $3$ now named after
Klein: see [55], a book entirely devoted to this curve. In contrast with the
largely geometric approach taken in [50], Klein’s emphasis in [51] was more
algebraic. However, that paper contained a diagram111Regrettably this diagram,
like that of the Hauptfigur in [50], is omitted from the on-line version of
the paper, both having been provided on separate inserted sheets in the
original journal. However, both diagrams can be found in Klein’s collected
works [53, vol. 3, pp. 126, 143], integrated into updated versions of the two
papers; see also [55, pp. 115, 320] for the Hauptfigur. showing ten plane
trees, which anticipated Grothendieck’s concept of dessins d’enfants [35] by
over a century. Our aim here is to reinterpret this diagram and the
mathematics underlying it from a modern point of view, and to explore some
generalisations which Klein might have been able to achieve if he had had
access to such tools as character theory, computer algebra and the
classification of finite simple groups.

In particular, we study dessins of type $(3,2,p)$ and of degree $p$ or $p+1$
for arbitrary primes $p$, extending Klein’s work in [51] on the case $p=11$.
These dessins are best studied through their monodromy groups, which are
permutation groups of these degrees, and in principle such groups are all
known, as a consequence of the classification of finite simple groups.
However, because of certain open problems in number theory it is unknown
whether two families of these groups, consisting of projective and affine
groups of degrees $p$ and $p+1$, are finite or infinite. Using the
Bateman–Horn Conjecture and extensive computer searches, we present what we
believe is strong evidence that the first of these two families is infinite.
(The second family is based on the Mersenne primes, and we have nothing new to
add about its cardinality.)

In Section 2 we describe Klein’s ten plane trees, firstly as he presented them
in [51], as pictorial representions of the possible 11-sheeted coverings of
the Riemann sphere $\Sigma={\mathbb{P}}^{1}({\mathbb{C}})$ with branching
patterns $3^{3}1^{2}$, $2^{4}1^{3}$ and $11^{1}$ over $0$, $1$ and $\infty$.
We then use more modern terminology and graphic conventions to describe them
as dessins d’enfants, specifically as bipartite maps (Figure 2) and as maps
with free edges (Figure 3). Klein asserted without proof that his list was
complete, so we outline a method he might have used to see this.

An important invariant of any dessin is its monodromy group (equivalently the
monodromy group of the covering of the sphere which it represents), discussed
in Section 3. This is a transitive permutation group on the sheets of the
covering (more precisely, on the fibre over a base-point), generated by the
monodromy permutations describing the branching over the critical values.
Klein showed that just one chiral pair of his trees, corresponding to our maps
${\mathcal{M}}_{1}$ and $\overline{\mathcal{M}}_{1}$, have monodromy groups
isomorphic to ${\rm PSL}_{2}(11)$, acting with degree $11$; we show that the
other eight have the alternating group ${\rm A}_{11}$ as their monodromy
group.

A powerful technique for enumerating dessins with a given monodromy group, but
unavailable to Klein when he wrote his papers, is the character-theoretic
triple-counting formula introduced by Frobenius [25] in 1896. We use this in
Section 4, with the aid of character tables provided by GAP [28] and by the
ATLAS [17], to confirm Klein’s enumeration of the trees and our description of
their monodromy groups; this includes the elimination of the Mathieu group
${\rm M}_{11}$, another possible candidate.

Any dessin $\mathcal{D}$ with monodromy group $G$ has a unique minimal regular
cover, a regular dessin $\mathcal{R}$ with ${\rm Aut}\,{\mathcal{R}}\cong G$.
Much of Klein’s paper [51] is devoted to the common regular cover
${\mathcal{R}}_{1}$ of ${\mathcal{M}}_{1}$ and $\overline{\mathcal{M}}_{1}$, a
map on the modular curve $X(11)$ of genus $26$. We show in Section 5 that the
other eight trees, with $G\cong{\rm A}_{11}$, have mutually non-isomorphic
regular covers of genus $756\,001$.

Each of Klein’s ten trees corresponds to an $11$-sheeted covering
$\Sigma\to\Sigma$ realised by a Shabat polynomial $P$ of degree $11$. In
Section 6 we determine these polynomials for ${\mathcal{M}}_{1}$ and
$\overline{\mathcal{M}}_{1}$: they are defined over
${\mathbb{Q}}(\sqrt{-11})$, and are transposed by the Galois group of that
field, while those for the other eight trees form a single Galois orbit.
Whereas the illustrations of dessins given so far are only topologically
correct, we compute and show in Figure 8 a geometrically correct drawing of
${\mathcal{M}}_{1}$ as the inverse image under $P$ of the unit interval.

In Section 7 we extend Klein’s classification by determining all the dessins
of degree $11$ and type $(3,2,r)$ for $r\neq 11$: there are sixteen of them
(see Figure 9), all of genus $0$ and with the symmetric group ${\rm S}_{11}$
as their monodromy group.

In Sections 8, 9 and 10 we generalise Klein’s work further by considering
dessins $\mathcal{D}$ of degree $p$ and type $(3,2,p)$ for arbitrary primes
$p$. The investigation is quite difficult, because we encounter some important
open problems in group theory and number theory. The monodromy group $G$ of
$\mathcal{D}$ must be a transitive permutation group of degree $p$, and if
$p>3$ (as we shall assume) then as a quotient of the triangle group
$\Delta(3,2,p)$ with mutually coprime periods, $G$ must be perfect. In Section
8 we use the classification of finite simple groups, and hence of permutation
groups of prime degree, to restrict the possibilities for $G$ to the
alternating groups ${\rm A}_{p}$, ${\rm PSL}_{2}(11)$ and the Mathieu groups
${\rm M}_{11}$ and ${\rm M}_{23}$ for $p=11$, $11$ and $23$, and ${\rm
PSL}_{n}(q)$ in cases where its natural degree $(q^{n}-1)/(q-1)$ is a prime
$p$. We show that ${\rm A}_{p}$ arises for $p=5$ and each $p\geq 11$, with
exponentially many dessins (far too many to classify) as $p\to\infty$. The
group ${\rm PSL}_{2}(11)$, acting with degree $11$, was dealt with by Klein,
as we have already seen, while the Frobenius triple-counting formula
eliminates the Mathieu groups as quotients of $\Delta(3,2,p)$.

This leaves the groups ${\rm PSL}_{n}(q)$, acting naturally with degree
$(q^{n}-1)/(q-1)$ for some prime power $q$. In Section 9 we consider the open
problem of whether this degree is prime for finitely or infinitely many pairs
$(n,q)$; such projective primes, as we call them, include the Fermat and
Mersenne primes, for $n=2$ and $q=2$ respectively. For each $n\geq 3$
(necessarily prime) the Bateman–Horn Conjecture (a powerful but apparently
little-known open problem in number theory) provides a heuristic estimate for
the number of primes $t\leq x$ such that $q=t^{e}$ and $(q^{n}-1)/(q-1)$ is
prime; since these estimates agree closely with results obtained from computer
searches revealing large numbers of examples, we conjecture that for each
prime $n\geq 3$ there are infinitely many projective primes $p$. This case is
explored in more detail in [46].

In Section 10 we return to the dessins of type $(3,2,p)$ and degree $p$ for
projective primes $p=(q^{n}-1)/(q-1)$. Using character theory we show that in
the cases $n=2$ and $3$ there are such dessins with monodromy group ${\rm
PSL}_{n}(q)$ for each projective prime $p$; in Theorems 10.1 and 10.3 we
enumerate them and give their genus. Theorem 10.11 is a similar but weaker
result for $q=2$, proving the existence of such dessins for each Mersenne
prime $p=2^{n}-1$, and we conjecture that it extends to all projective primes.

In Sections 11 and 12 we consider dessins of type $(3,2,p)$ and degree $p+1$
for arbitrary primes $p$. The parameters $3$ and $2$ in the types of the
dessins we have considered imply that all the monodromy groups and
automorphism groups associated with them are quotients of the modular group
${\rm PSL}_{2}({\mathbb{Z}})\cong{\rm C}_{3}*{\rm C}_{2}$. In most cases the
corresponding kernels are non-congruence subgroups, but in Section 11 we
consider the modular dessins ${\mathcal{D}}_{0}(p)$ arising from reduction of
coefficients mod $(p)$, corresponding to the action of ${\rm PSL}_{2}(p)$ on
the projective line ${\mathbb{P}}^{1}({\mathbb{F}}_{p})$. We do this firstly
in the planar cases $p\leq 7$ and $p=13$, and then in the more difficult case
$p=11$, where we outline a calculation by John Voight which determines both
the underlying elliptic curve $E$ and the Belyĭ function $E\to\Sigma$. For
$p=11$ and $13$ we compute and display geometrically correct drawings of these
dessins.

We extend this further in Section 12 by considering arbitrary dessins of type
$(3,2,p)$ and degree $p+1$ for primes $p>3$. We use the classification by
Müller [65] of primitive permutation groups containing a cycle with one fixed
point to restrict the possible monodromy groups to ${\rm A}_{p+1}$, ${\rm
PSL}_{2}(p)$, ${\rm AGL}_{n}(2)$ for $p=2^{n}-1$, ${\rm M}_{11}$ and ${\rm
M}_{12}$ for $p=11$, and ${\rm M}_{24}$ for $p=23$. We show that ${\rm
A}_{p+1}$ arises for all $p>7$, again with too many dessins to classify for
all except small $p$, whereas ${\rm PSL}_{2}(p)$ arises only for
${\mathcal{D}}_{0}(p)$. The Mathieu groups ${\rm M}_{12}$ and ${\rm M}_{24}$
each yield two chiral pairs of dessins, whereas as before ${\rm M}_{11}$
yields none. Finally we show that the affine group ${\rm AGL}_{n}(2)$ yields
at least one dessin for each Mersenne prime $p=2^{n}-1>7$; however, this fails
for $p=7$ since ${\rm AGL}_{3}(2)$ is not a Hurwitz group.

In Section 13 we briefly outline another realisation of ${\rm PSL}_{2}(11)$,
this time as the isometry group of the hendecachoron, a tessellation of a non-
orientable $3$-orbifold by $11$ hemi-icosahedra, discovered independently by
Grünbaum [36] and Coxeter [18].

Section 14 contains monodromy permutations and diagrams for some of the
dessins which appear in this paper.

##  2 Klein’s ten plane trees

In [51] Klein drew a diagram showing ten plane trees, each with eleven edges
and twelve vertices. In each tree there is a bipartite partition of the
vertices, with five vertices (coloured white) of valencies dividing $3$, and
seven vertices (indicated by short cross-bars, perpendicular to their incident
edges) of valencies dividing $2$; in each case the unique face is an $11$-gon.
Among these plane trees there are four chiral (mirror-image) pairs, numbered I
to IV, and two others, numbered V and VI, which exhibit bilateral symmetry.
The chiral pair I, which play a major role in Klein’s paper, are shown
(slightly distorted for simplicity, but combinatorially correct) in Figure 1.

Figure 1: The chiral pair I of Klein’s plane trees.

###  2.1 Klein’s trees as dessins d’enfants

If we regard the underlying surface as the complex plane ${\mathbb{C}}$, and
compactify it to give the Riemann sphere or complex projective line
$\Sigma={\mathbb{P}}^{1}({\mathbb{C}})={\mathbb{C}}\cup\\{\infty\\}$, we can
recognise these plane trees as dessins d’enfants in Grothendieck’s sense [35],
namely bipartite graphs embedded in compact Riemann surfaces $X$; as such they
represent finite covers $X\to\Sigma$ of the sphere, unbranched outside
$\\{0,1,\infty\\}$, or equivalently, by Belyĭ’s Theorem [6], projective
algebraic curves $X$ defined over algebraic number fields. (See [30, 44, 54]
for background on dessins d’enfants.) All ten of these dessins have genus $0$,
so that $X=\Sigma$. They have degree $11$, meaning that they represent
$11$-sheeted coverings of $\Sigma$; they have type $(3,2,11)$, meaning that
there are branch-points of orders dividing $3,2$ and $11$ over $0,1$ and
$\infty$, represented in Klein’s diagrams by the white vertices, cross-bars
and the point at infinity. A modern convention is to use black and white
vertices for points over $0$ and $1$, with the edges corresponding to the
points over the unit interval $[0,1]\subset\Sigma$, and to regard those over
$\infty$ as the face-centres. We will denote the dessins corresponding to
Klein’s chiral pairs I to IV by ${\mathcal{M}}_{i}$ and
$\overline{\mathcal{M}}_{i}$ for $i=1,\ldots,4$, with ${\mathcal{M}}_{i}$ on
the left in Klein’s diagrams and ours; similarly ${\mathcal{M}}_{i}\;(i=5,6)$
is the dessin corresponding to the tree V or VI in [51]222The temptation to
follow Klein by using Roman numerals as subscripts here was almost (but not
totally) irresistible.. The dessins ${\mathcal{M}}_{1}$ and
$\overline{\mathcal{M}}_{1}$ corresponding to Klein’s plane trees I are shown
in Figure 2.

Figure 2: The dessins ${\mathcal{M}}_{1}$ and $\overline{\mathcal{M}}_{1}$
corresponding to Klein’s plane trees I.

When the order of branching over $1$ divides $2$, as in these dessins, so that
white vertices all have valencies $1$ or $2$, it is sometimes convenient to
omit these vertices, leaving a map with only black vertices and possibly some
free edges where white vertices of valency $1$ have been removed. Of course,
this operation is reversible, so no information is lost. With this convention,
the dessins corresponding to all ten of Klein’s plane trees are shown in
Figure 3, with the same layout as in his original diagram.

${\mathcal{M}}_{1}$$\overline{\mathcal{M}}_{1}$${\mathcal{M}}_{2}$$\overline{\mathcal{M}}_{2}$${\mathcal{M}}_{3}$$\overline{\mathcal{M}}_{3}$${\mathcal{M}}_{4}$$\overline{\mathcal{M}}_{4}$${\mathcal{M}}_{5}$${\mathcal{M}}_{6}$
Figure 3: The maps ${\mathcal{M}}_{i}$ and $\overline{\mathcal{M}}_{i}$
corresponding to Klein’s plane trees I – VI.

In fact these ten dessins form a complete list (up to isomorphism) of the
dessins of type $(3,2,11)$ and degree $11$, which is the least possible degree
for dessins of this type. Klein presented his diagram as representing a
classification of the $11$-sheeted coverings of the sphere $\Sigma$ with
branching patterns $3^{3}1^{2}$ and $2^{4}1^{3}$ over $0$ and $1$. He
explained this in terms of the modular function $J:{\mathbb{H}}\to\Sigma$,
which parametrises isomorphism classes of elliptic curves, an approach which
leads naturally to the modular group $\Gamma={\rm PSL}_{2}({\mathbb{Z}})$, the
subgroup leaving $J$ invariant in the automorphism group ${\rm
PSL}_{2}({\mathbb{R}})$ of the Riemann surface $\mathbb{H}$. This part of
Klein’s work will not be discussed in our paper.

###  2.2 The classification problem

Klein presented his ten planar trees in [51, §1] in answer to a question he
had posed in [49, §5]: in modern terminology, he had asked for the number of
dessins $\mathcal{D}$ of degree $11$ and passport $(3^{3}1^{2};2^{4}1^{3};-)$,
that is, with three black vertices of valency $3$ and two of valency $1$,
together with four white vertices of valency $2$ and three of valency $1$, and
with no restrictions on the faces.

In his diagram Klein drew ten planar trees, equivalent to these dessins, but
did not explain why this list was complete, writing simply ‘Dass es auch nicht
mehr giebt, ist ebenso evident’333‘It is also clear that there are no
others.’. The argument, which we will now give in the language of dessins, is
indeed straightforward. Since each dessin $\mathcal{D}$ has eleven edges and
twelve vertices the embedded graph, being connected, must be a tree, so the
underlying surface is the sphere and there is a single face, of valency $11$.
Since the white vertices have valency $1$ or $2$ we can omit them and for
simplicity represent $\mathcal{D}$ as a map $\mathcal{M}$, which must have
three vertices of valency $3$ and two of valency $1$, and seven edges, three
of them free. By connectedness, one of the three vertices of valency $3$ must
be adjacent to the other two, so $\mathcal{M}$ must be formed from the basic
map $\mathcal{M}_{0}$ in Figure 4 by choosing two of its five free edges and
adding a vertex to each. This gives us
$\displaystyle\left(\\!\\!\begin{array}[]{c}5\\\ 2\end{array}\\!\\!\right)=10$
non-isomorphic maps $\mathcal{M}$; they are shown in Figure 3, with the same
layout as the corresponding plane trees in Klein’s paper [51].

Figure 4: The planar map $\mathcal{M}_{0}$.

An alternative approach to this and similar problems, based on the character
theory and the Frobenius formula, will be developed in Section 4.

##  3 Monodromy groups

Any dessin $\mathcal{D}$ of degree $n$ can be represented as an ordered pair
$x,y$ of permutations of its $n$ edges, obtained by using the chosen
orientation of the underlying surface to follow the rotation of edges around
their incident black and white vertices. The monodromy group $G=\langle
x,y\rangle$ of $\mathcal{D}$, through its action permuting the $n$ edges, can
be identified with a subgroup of the symmetric group ${\rm S}_{n}$. By the
connectedness of the embedded graph it is transitive. It is convenient to
introduce a third permutation $z:=(xy)^{-1}$, so that $xyz=1$. It is easy to
verify that, while the cycles of $x$ and $y$ correspond respectively to black
and white vertices, the cycles of $z$ correspond to faces.

######  Convention 3.1 (Where to put labels)

We put a label of an edge on its left side while moving from its black end to
its white one. In this way the labels corresponding to a cycle of $z$ or,
equivalently, to a face, will be situated inside this face and will be rotated
by $z$ around the center of the face in the direction corresponding to the
orientation of the surface. This convention is illustrated in Figure 5.

(When a dessin is drawn on the plane, its outer face gives an impression that
its labels are rotated in the direction opposite to the orientation of the
plane. But this is an illusion. In fact, the dessin should be considered not
on the plane but on a sphere, and the center of the outer face is situated on
the “opposite side” of this sphere. Looking from this center, the rotation of
the labels of the outer face does correspond to the orientation of the
sphere.)

When $y^{2}=1$ and the white vertices are omitted, so that $\mathcal{D}$ is a
map, one can regard $G$ as permuting its half-edges, or else directed edges.
Thus, an edge of a map bears two labels while a free edge bears only one. A
free edge may be considered as an outgoing directed edge.

Figure 5: Monodromy permutations: here $\alpha x$, $\alpha y$ and $\alpha z$
are the images of $\alpha$ under the permutations $x$, $y$ and $z=(xy)^{-1}$.

###### Example 3.2

The permutations corresponding to the map in Figure 6 are as follows:

$\displaystyle x$ $\displaystyle=$
$\displaystyle(1,15,16)(2,3,4)(5,9)(6,7,8,10,11,12),$ $\displaystyle y$
$\displaystyle=$ $\displaystyle(1,6)(2,9)(3,13,8,14)(4,7)(5,10)(11,12),$
$\displaystyle z$ $\displaystyle=$
$\displaystyle(1,12,10,9,4,6,16,15)(2,5,8,13)(3,14,7).$

The monodromy group of this map is the alternating group ${\rm A}_{16}$. See
also Figures 17 and 27 for other examples of maps with labelled edges.

Figure 6: A map with labelled edges.

The ten dessins $\mathcal{D}$ considered here have degree $11$, so their
monodromy groups $G$ are transitive subgroups of ${\rm S}_{11}$. In each case
$x^{3}=y^{2}=1$, so $G$ is a quotient of the modular group

$\Gamma={\rm PSL}_{2}({\mathbb{Z}})=\langle X,Y,Z\mid
X^{3}=Y^{2}=XYZ=1\rangle\cong C_{3}*C_{2},$

where the generators

$X=\pm\left(\,\begin{array}[]{cc}0&1\\\ -1&-1\end{array}\right),\quad
Y=\pm\left(\,\begin{array}[]{cc}0&1\\\ -1&0\end{array}\right),\quad
Z=\pm\left(\,\begin{array}[]{cc}1&0\\\ 1&1\\\ \end{array}\right)$

correspond to the Möbius transformations

$t\mapsto\frac{-1}{t-1},\quad t\mapsto\frac{-1}{t}\quad\hbox{and}\quad
t\mapsto t+1.$

(Here we have associated Möbius transformations $\displaystyle
t\mapsto\frac{at+b}{ct+d}$ with pairs of matrices
$\displaystyle\pm\,\left(\begin{array}[]{cc}a&c\\\ b&d\end{array}\right)$, the
transposes of those normally used in this context. This is because we will
later need to regard Möbius transformations, especially those over finite
fields, as permutations; it is algebraically and computationally convenient to
compose these from left to right, instead of using the analytic right to left
convention, so in order to have a homomorphism from matrices to Möbius
transformations we will ignore the deceptively suggestive notation for the
latter and use transposes.)

Klein was interested in which of these monodromy groups $G$ are congruence
quotients, that is, which are quotients by congruence subgroups of $\Gamma$.
The only possible level for such a congruence subgroup is $11$ (since this is
the order of $z:=(xy)^{-1}$), and indeed the resulting quotient group $L:={\rm
PSL}_{2}(11)$, a simple group of order $660$, does act as a transitive group
of degree $11$, namely on the cosets of a subgroup $H\cong{\rm A}_{5}$, as we
will now show. This is one of the three cases, all known to Galois (see [27],
pp. 411–412), in which a simple group ${\rm PSL}_{2}(p)$, where $p$ is prime,
has a proper subgroup of index less than $p+1$; the others are ${\rm
A}_{4}<{\rm PSL}_{2}(5)$ and ${\rm S}_{4}<{\rm PSL}_{2}(7)$.

As generators of $L$ one can take the images

$x=\pm\left(\,\begin{array}[]{cc}0&1\\\ -1&-1\\\ \end{array}\right),\quad
y=\pm\left(\,\begin{array}[]{cc}0&1\\\ -1&0\\\ \end{array}\right),\quad
z=\pm\left(\,\begin{array}[]{cc}1&0\\\ 1&1\\\ \end{array}\right)$

of $X,Y$ and $Z$, satisfying the relations

$x^{3}=y^{2}=z^{11}=xyz=1.$ (1)

The elements

$u=\pm\left(\,\begin{array}[]{cc}4&-1\\\ 2&-3\\\ \end{array}\right),\quad
v=y=\pm\left(\,\begin{array}[]{cc}0&1\\\ -1&0\\\ \end{array}\right),\quad
w=\pm\left(\,\begin{array}[]{cc}2&-4\\\ -3&1\\\ \end{array}\right)$

of $L$ (note that $\det(u)=\det(w)=-10\equiv 1$ mod $(11)$) satisfy

$u^{3}=v^{2}=w^{5}=uvw=1,$

so they generate a quotient $H\leq L$ of the triangle group
$\Delta(3,2,5)\cong{\rm A}_{5}$. By the simplicity of ${\rm A}_{5}$ we have
$H\cong{\rm A}_{5}$, so $|L:H|=11$. Conjugation by the element

$\pm\left(\,\begin{array}[]{cc}-1&0\\\ 0&1\\\ \end{array}\right)\in{\rm
PGL}_{2}(11)\cong{\rm Aut}\,L$

induces an outer automorphism of $L$, transposing $u,v$ and $w$ with elements

$\overline{u}=\pm\left(\,\begin{array}[]{cc}4&1\\\ -2&-3\\\
\end{array}\right),\quad\overline{v}=v=y=\pm\left(\,\begin{array}[]{cc}0&-1\\\
1&0\\\
\end{array}\right),\quad\overline{w}=\pm\left(\,\begin{array}[]{cc}2&4\\\
3&1\\\ \end{array}\right)$

which satisfy the same relations and therefore generate a subgroup
$\overline{H}\cong{\rm A}_{5}$. This subgroup is not conjugate in $L$ to $H$:
if it were, some non-identity element of ${\rm PGL}_{2}(11)$ would commute
with $u,v$ and $w$, whereas it is straightforward to check that only the
identity element does this. It follows that $L$ has two inequivalent
transitive representations of degree $11$, on the cosets of the subgroups $H$
and $\overline{H}$, or equivalently (since these subgroups are equal to their
normalisers in $L$), by conjugation on the two conjugacy classes of subgroups
they represent.

Nowadays, by using a program such as GAP [28] one can easily determine the
monodromy groups of the ten dessins, showing that those with monodromy group
$L={\rm PSL}_{2}(11)$ are the chiral pair $\mathcal{M}_{1}$ and
$\overline{\mathcal{M}}_{1}$. However, one can also distinguish this pair by
hand: for $i=1,\ldots,6$ the commutator $[x,y]=x^{-1}y^{-1}xy$ has cycle
structure
$5^{2}1^{1},7^{1}3^{1}1^{1},8^{1}2^{1}1^{1},5^{1}4^{1}2^{1},7^{1}2^{2}$ and
$4^{2}3^{1}$ in ${\mathcal{M}}_{i}$ (and hence also in
$\overline{\mathcal{M}}_{i}$), so only when $i=1$ does it have one of the
possible orders $1,2,3,5,6$ or $11$ of an element of $L$. Conversely, it is
easy to see that when $L$ is represented on the cosets of $H$ or
$\overline{H}$ the generators $x$ and $y$ have cycle structures $3^{3}1^{2}$
and $2^{4}1^{3}$, so $L$ is the monodromy group of at least one of the ten
dessins, and this must be the pair $\mathcal{M}_{1}$ and
$\overline{\mathcal{M}}_{1}$, with the two dessins corresponding to the two
inequivalent representations of degree $11$.

Klein [51, §2] used a similar argument, based on a word equivalent to our
$yz^{3}$, to show that the plane trees labelled II–VI in his diagram do not
have monodromy group $L$; he gave details only for tree V, corresponding to
the dessin ${\mathcal{M}}_{5}$. His paper is otherwise entirely devoted to the
two trees labelled I (our Figure 1) with monodromy group $L$, together with
their minimal regular cover, which correspond to congruence subgroups of the
modular group; the other eight trees, labelled II – VI and corresponding to
noncongruence subgroups, are ignored, as are their monodromy groups.

In fact, the cycle structures of $[x,y]$ given above show that for $i=2,4,5$
and $6$ the monodromy group $G$ is ${\rm A}_{11}$. Being transitive of prime
degree, $G$ must be primitive; now a theorem of Jordan (see [72, Theorem
13.9], for example) shows that any finite primitive permutation group
containing a cycle of prime length with at least three fixed points must
contain the alternating group; in each of these four cases a suitable power of
$[x,y]$ has this property, so $G\geq{\rm A}_{11}$. Since the generators $x$
and $y$ are even permutations, it follows that $G={\rm A}_{11}$. However, a
separate argument is required for $i=3$, for instance using the fact that in
this case $xz^{3}xz^{5}$ has cycle structure $5^{1}3^{1}1^{3}$. (The theorem
of Jordan used here follows immediately from results in his papers [47, 48]
published in 1871 and 1873, so Klein could have known and used this argument;
for further historical comments on Jordan’s Theorem, and for a modern
extension of it, see [41].)

The fact that these eight dessins have monodromy groups $G\cong{\rm A}_{11}$
shows that, as claimed above, they correspond to noncongruence subgroups of
$\Gamma$: the only nonabelian composition factors of any congruence quotient
of $\Gamma$ are isomorphic to ${\rm PSL}_{2}(p)$ for primes $p\geq 5$ dividing
the level [64], and by comparing orders it is easily seen that the simple
group ${\rm A}_{11}$ does not have this form.

##  4 Counting triples

We have already given a very simple argument showing why there are exactly ten
dessins with the given characteristics. However, this simple example allows us
an opportunity to illustrate another important enumeration technique based on
character theory. Quite often, in more complicated cases where the degree or
the genus is greater, this method is the only one available. It is based on
Frobenius’s formula [25]

$\displaystyle\frac{|{\mathcal{X}}|\cdot|{\mathcal{Y}}|\cdot|{\mathcal{Z}}|}{|G|}\sum_{\chi}\frac{\chi(x)\chi(y)\chi(z)}{\chi(1)}$
(2)

for the number of triples
$(x,y,z)\in\mathcal{X}\times\mathcal{Y}\times\mathcal{Z}$ with $xyz=1$, where
$\mathcal{X}$, $\mathcal{Y}$ and $\mathcal{Z}$ are conjugacy classes in a
finite group $G$, and the sum is over the irreducible complex characters
$\chi$ of $G$. (In specific applications, one often finds that many characters
take the value $0$ on $\mathcal{X}$, $\mathcal{Y}$ or $\mathcal{Z}$, so they
can be omitted from the summation.) Since [25] was published 17 years after
[51], this powerful technique was not available to Klein. Here we will show
how it works in this fairly simple situation, ignoring results explained
earlier; as a bonus we obtain the monodromy groups of all ten dessins.

Before doing this, we will give an alternative form of (2) which is sometimes
more convenient, for instance when using the ATLAS [17] where orders of
centralisers, rather than conjugacy classes, are given. We have
$|\mathcal{X}|=|G|/|C(x)|$ where $C(x)=C_{G}(x)$ denotes the centraliser of
$x$ in $G$, with similar equations for $|\mathcal{Y}|$ and $|\mathcal{Z}|$, so
we can rewrite (2) as

$\displaystyle\frac{|G|^{2}}{|C(x)|\cdot|C(y)|\cdot|C(z)|}\sum_{\chi}\frac{\chi(x)\chi(y)\chi(z)}{\chi(1)}.$
(3)

If a dessin $\mathcal{D}$ has type $(3,2,11)$ and degree $11$, then its
monodromy permutations $x,y,z\in{\rm S}_{11}$ satisfy the relations (1);
having odd order, $x$ and $z$ are even, and hence so is $y$, so the monodromy
group $G=\langle x,y,z\rangle$ is a subgroup of $A:={\rm A}_{11}$. Our aim is
to count the triples $x$, $y$, $z$ in $A$ satisfying (1), and to determine the
corresponding dessins $\mathcal{D}$ and their monodromy groups $G$. In more
general applications of (2) one has to exclude triples generating intransitive
subgroups $G$, since these cannot be monodromy groups of connected covers.
However, here the fact that $z$ must be an $11$-cycle guarantees transitivity,
so this problem does not arise. Since the orders $3,2$ and $11$ of $x,y$ and
$z$ are mutually coprime, $G$ must be perfect (have trivial abelianisation),
and therefore cannot be solvable. If $G\neq A$ then $G$ must be contained in a
nonsolvable maximal subgroup $M$ of $A$, of order divisible by $11$. According
to the list of maximal subgroups of ${\rm A}_{11}$ in [17], the only such
maximal subgroups $M$ are isomorphic to the Mathieu group ${\rm M}_{11}$,
forming two conjugacy classes of subgroups, of index 2520, transposed by ${\rm
Out}\,A$. Similarly the list of maximal subgroups of ${\rm M}_{11}$ in [17]
tells us that if $G\neq M$ then $G$ is isomorphic to $L={\rm PSL}_{2}(11)$,
each subgroup $M\cong{\rm M}_{11}$ containing one conjugacy class of such
subgroups, of index $12$ in $M$. Thus $G$ is equal to $A$, or is isomorphic to
${\rm M}_{11}$ or ${\rm PSL}_{2}(11)$; we will count the triples generating
these three groups in turn, starting with the smallest.

###### Example 4.1

We will show that the group $L={\rm PSL}_{2}(11)$ has, up to automorphisms
(equivalently, up to conjugacy in ${\rm PGL}_{2}(11)\cong{\rm Aut}\,L$), a
single generating triple $x,y,z$ satisfying (1). The character table of $L$,
produced by GAP [28] and given here, has columns corresponding to the
conjugacy classes, with the headings $n$a, $n$b, etc. ($n$A, $n$B, etc in
ATLAS notation [17]) indicating successive classes of elements of order $n$;
the rows, labelled X.j, correspond to irreducible characters $\chi_{j}$ for
$j=1,2,\ldots$. The table includes entries A, ∗A $:=(-1\mp\sqrt{5})/2$ and
${\rm B},/{\rm B}:=(-1\pm i\sqrt{11})/2$, while a full stop denotes an entry
$0$.

>
>            1a 3a 2a 5a 5b 6a 11a 11b
>
>     X.1     1  1  1  1  1  1   1   1
>     X.2     5 -1  1  .  .  1   B  /B
>     X.3     5 -1  1  .  .  1  /B   B
>     X.4    10  1 -2  .  .  1  -1  -1
>     X.5    10  1  2  .  . -1  -1  -1
>     X.6    11 -1 -1  1  1 -1   .   .
>     X.7    12  .  .  A *A  .   1   1
>     X.8    12  .  . *A  A  .   1   1
>
>     A = E(5)^2+E(5)^3
>       = (-1-Sqrt(5))/2 = -1-b5
>     B = E(11)+E(11)^3+E(11)^4+E(11)^5+E(11)^9
>       = (-1+Sqrt(-11))/2 = b11
>

In this example we take $\mathcal{X}$ and $\mathcal{Y}$ to be the classes 3a
and 2a, and $\mathcal{Z}$ to be either of the classes 11a or 11b. For each
choice of $\mathcal{Z}$ the character sum in formula (3) has the value

$\Sigma_{L}=1+\frac{(-1)\cdot 1\cdot{\rm B}}{5}+\frac{(-1)\cdot 1\cdot/{\rm
B}}{5}+\frac{1\cdot(-2)\cdot(-1)}{10}+\frac{1\cdot
2\cdot(-1)}{10}=1-\frac{{\rm B}+/{\rm B}}{5}=\frac{6}{5}.$

Now the centralisers $C_{L}(x)$, $C_{L}(y)$ and $C_{L}(z)$ of typical elements
$x,y$ and $z$ of $\mathcal{X}$, $\mathcal{Y}$ and $\mathcal{Z}$ have orders
$6$, $12$ and $11$, so that

$\frac{|L|^{2}}{|C(x)|\cdot|C(y)|\cdot|C(z)|}=\frac{2^{4}\cdot 3^{2}\cdot
5^{2}\cdot 11^{2}}{6\cdot 12\cdot 11}=2\cdot 5^{2}\cdot 11.$

Multiplying this by $\Sigma_{L}$ gives $2^{2}\cdot 3\cdot 5\cdot 11$ triples
$(x,y,z)\in{\mathcal{X}}\times{\mathcal{Y}}\times{\mathcal{Z}}$ with $xyz=1$
for each choice of $\mathcal{Z}$, and hence $2^{3}\cdot 3\cdot 5\cdot 11$ in
total. Each triple must generate $L$ since this group has no proper subgroups
of order divisible by $66$ (in fact, none of index less that $11$). Since
${\rm Aut}\,L$ acts semi-regularly on generating triples, dividing by its
order $|{\rm PGL}_{2}(11)|=2^{3}\cdot 3\cdot 5\cdot 11$ shows that ${\rm
Aut}\,L$ has a single orbit on such triples, as claimed.

Since the normal subgroups of any group $\Delta$ with quotient isomorphic to a
group $G$ are in bijective correspondence with the orbits of ${\rm Aut}\,G$,
acting by composition, on epimorphisms $\Delta\to G$, it follows that the
triangle group

$\Delta:=\Delta(3,2,11)=\langle X,Y,Z\mid X^{3}=Y^{2}=Z^{11}=XYZ=1\rangle$

has a single normal subgroup with quotient isomorphic to $L$. Now $L$ has two
conjugacy classes of subgroups of index $11$, all isomorphic to ${\rm A}_{5}$,
and these classes are transposed by its automorphism group ${\rm PGL}_{2}(11)$
(see [17] or [22, §259]); they lift to two conjugacy classes of subgroups of
index $11$ in $\Delta$, transposed by conjugation in the extended triangle
group

$\Delta[3,2,11]=\langle R_{0},R_{1},R_{2}\mid
R_{i}^{2}=(R_{1}R_{2})^{3}=(R_{2}R_{0})^{2}=(R_{0}R_{1})^{11}=1\rangle,$

which contains $\Delta$ with index $2$, and they correspond to a chiral pair
of dessins $\mathcal{D}$ of type $(3,2,11)$ degree $11$ with monodromy group
isomorphic to $L$. This calculation shows that they are the only such dessins,
so they are the dessins ${\mathcal{M}}_{1}$ and $\overline{\mathcal{M}}_{1}$
discussed earlier.

###### Example 4.2

Next we count triples satisfying (1) in the simple group $M={\rm M}_{11}$ of
order $7920=2^{4}\cdot 3^{2}\cdot 5\cdot 11$. The character table of $M$,
given by GAP, is shown here:

>
>             1a 11a 11b 2a 4a 8a 8b 3a 6a 5a
>
>     X.1      1   1   1  1  1  1  1  1  1  1
>     X.2     10  -1  -1  2  2  .  .  1 -1  .
>     X.3     10  -1  -1 -2  .  B -B  1  1  .
>     X.4     10  -1  -1 -2  . -B  B  1  1  .
>     X.5     11   .   .  3 -1 -1 -1  2  .  1
>     X.6     16   A  /A  .  .  .  . -2  .  1
>     X.7     16  /A   A  .  .  .  . -2  .  1
>     X.8     44   .   .  4  .  .  . -1  1 -1
>     X.9     45   1   1 -3  1 -1 -1  .  .  .
>     X.10    55   .   . -1 -1  1  1  1 -1  .
>
>     A = E(11)^2+E(11)^6+E(11)^7+E(11)^8+E(11)^10
>       = (-1-Sqrt(-11))/2 = -1-b11
>     B = -E(8)-E(8)^3
>       = -Sqrt(-2) = -i2
>

We must take $\mathcal{X}$ and $\mathcal{Y}$ to be the unique conjugacy
classes 3a and 2a of elements of order $3$ and $2$, and $\mathcal{Z}$ to be
either of the classes 11a or 11b of elements of order $11$. We see that the
character sum for $M$ is

$\Sigma_{M}=1+\frac{1\cdot
2\cdot(-1)}{10}+\frac{1\cdot(-2)\cdot(-1)}{10}+\frac{1\cdot(-2)\cdot(-1)}{10}=\frac{6}{5}.$

By [17] the centralisers in $M$ of $x,y$ and $z$ have orders $18$, $48$ and
$11$, so

$\frac{|M|^{2}}{|C(x)|\cdot|C(y)|\cdot|C(z)|}=\frac{2^{8}\cdot 3^{4}\cdot
5^{2}\cdot 11^{2}}{18\cdot 48\cdot 11}=2^{3}\cdot 3\cdot 5^{2}\cdot 11.$

Multiplying this by $2\Sigma_{M}$ gives $2^{5}\cdot 3^{2}\cdot 5\cdot 11$
triples in $M$. Now $M$ has a single conjugacy class of $12$ subgroups
$L\cong{\rm PSL}_{2}(11)$, and we saw in Example 4.1 that each of these is
generated by $2^{3}\cdot 3\cdot 5\cdot 11$ triples, so $2^{5}\cdot 3^{2}\cdot
5\cdot 11$ triples in $M$ generate proper subgroups. Thus none of these
triples can generate $M$. (See [15, 73] for confirmation that ${\rm M}_{11}$
is not a quotient of $\Delta$.)

The groups ${\rm M}_{11}$ and ${\rm M}_{12}$ were introduced by Mathieu [62]
in 1861, so they would have been known to Klein; unlike ${\rm M}_{12}$ (see
[44, §2.1.6], for example), ${\rm M}_{11}$ is not the monodromy group of any
orientable cubic map, since an extension of the argument used here shows that
it is not a quotient of $\Gamma$.

###### Example 4.3

Now we count triples satisfying (1) in $A={\rm A}_{11}$. In this case there
are three conjugacy classes of elements of order $3$, and two classes of
elements of order $2$. A simple argument shows that if elements $x,y\in A$ of
orders $3$ and $2$ generate a transitive group than they must have cycle
structures $3^{3}1^{2}$ and $2^{4}1^{3}$, so they belong to the conjugacy
classes $\mathcal{X}={\rm 3c}$ and $\mathcal{Y}={\rm 2b}$ of $A$, while $z$
must be in one of the two classes 11a and 11b of $11$-cycles in $A$.

The character table of $A$ is much too large to give here. However, the only
irreducible characters $\chi$ with $\chi(x)\chi(y)\chi(z)\neq 0$ are the
principal character $\chi_{1}$ and the characters $\chi_{2}$, $\chi_{6}$ and
$\chi_{11}$ of degrees $10$, $120$ and $210$, so we can give here the relevant
portion of the character table, concerning these classes and characters:

>
>               1a  2b  3c 11a 11b
>
>     X.1        1   1   1   1   1
>     X.2       10   2   1  -1  -1
>     X.6      120  -8   3  -1  -1
>     X.11     210   2   3   1   1
>

The character sum

$\Sigma_{A}:=\sum_{\chi}\frac{\chi(x)\chi(y)\chi(z)}{\chi(1)}$

has the value

$1+\frac{1\cdot 2\cdot(-1)}{10}+\frac{3\cdot(-8)\cdot(-1)}{120}+\frac{3\cdot
2\cdot
1}{210}=1-\frac{1}{5}+\frac{1}{5}+\frac{1}{35}=\frac{36}{35}=\frac{2^{2}\cdot
3^{2}}{5\cdot 7}.$

Now $|A|=11!/2=2^{7}\cdot 3^{4}\cdot 5^{2}\cdot 7\cdot 11$, and by [17] the
centralisers of $x,y$ and $z$ in $A$ have orders $162=2\cdot 3^{4}$,
$1152=2^{7}\cdot 3^{2}$ and $11$, so

$\frac{|A|^{2}}{|C(x)|\cdot|C(y)|\cdot|C(z)|}=2^{6}\cdot 3^{2}\cdot 5^{4}\cdot
7^{2}\cdot 11.$

Multiplying this by $2\Sigma_{A}$ shows that the number of triples in $A$
satisfying (1) is

$t:=2^{9}\cdot 3^{4}\cdot 5^{3}\cdot 7\cdot 11=10\times 11!.$

From this total we must subtract the number of triples generating proper
subgroups of $A$. As we have seen, the only such subgroups are those
isomorphic to ${\rm PSL}_{2}(11)$. Each subgroup $L\cong{\rm PSL}_{2}(11)$ in
$A$ is contained in a unique subgroup $M\cong{\rm M}_{11}$; there are two
conjugacy classes of 2520 such subgroups $M$ in $A$, each containing one
conjugacy class of $12$ subgroups $L$, so the number of such subgroups $L$ in
$A$ is $2\cdot 2520\cdot 12=60480=2^{6}\cdot 3^{3}\cdot 5\cdot 7$. As shown in
Example 4.1, each subgroup $L$ is generated by $2^{3}\cdot 3\cdot 5\cdot 11$
triples satisfying (1), so the number of triples in $A$ generating subgroups
$L\cong{\rm PSL}_{2}(11)$ is $2^{9}\cdot 3^{4}\cdot 5^{2}\cdot 7\cdot
11=t/5=2\times 11!$.

Since there are $10\times 11!$ triples in $A$, of which $2\times 11!$ generate
proper subgroups, there are $8\times 11!$ triples generating $A$, forming
eight orbits under ${\rm Aut}\,A={\rm S}_{11}$; these correspond to eight
normal subgroups of $\Delta$ with quotient $A$, and hence to eight regular
dessins with (orientation-preserving) automorphism group $A$. Now $A$ has a
single conjugacy class of subgroups of index $11$, namely the natural point-
stabilisers isomorphic to ${\rm A}_{10}$, so these lift to eight more
conjugacy classes of subgroups of index $11$ in $\Delta$, in addition to the
two resulting from Example 4.1. These give eight more dessins $\mathcal{D}$ of
type $(3,2,11)$ and degree $11$, all with monodromy group $G\cong{\rm
A}_{11}$. Thus we have a total of ten dessins of the required degree and type.
This confirms the earlier enumeration, and also confirms that for
${\mathcal{M}}_{1}$ and $\overline{\mathcal{M}}_{1}$ the monodromy group is
${\rm PSL}_{2}(11)$, whereas for the other eight dessins it is ${\rm A}_{11}$.
Generating triples for these dessins are given in Subsection 14.1.

In each of the examples considered here, it was a fairly easy matter to
evaluate the Frobenius formulae (2) or (3) by hand, but in other cases, if
there are many non-zero summands, or if irrational character values arise, it
may be necessary to appeal to GAP for this. For instance, the following GAP
commands repeat the calculation in Example 4.1.

>
>
>     > G:=PSL(2,11);;
>     > T:=CharacterTable(G);;
>     > OrdersClassRepresentatives(T);
>       [ 1, 3, 2, 5, 5, 6, 11, 11 ]
>     > ClassStructureCharTable(T,[2,3,7]);
>       660
>     > ClassStructureCharTable(T,[2,3,8]);
>       660
>
>

Here $2,3,7$ and $2,3,8$ indicate the positions of the chosen classes
$\mathcal{X}$, $\mathcal{Y}$ and $\mathcal{Z}$ in the preceding list. The
double semicolons after a command ask GAP to execute the command but not to
output the result. As in Example 4.1 we obtain $1320$ triples.

##  5 The dessins and their regular covers

Our enumeration of triples in $L$ shows that there is a single regular dessin
$\mathcal{R}_{1}$ of type $(3,2,11)$ with automorphism group $L\cong{\rm
PSL}_{2}(11)$; this is the minimal regular cover of each of the two dessins
$\mathcal{M}_{1}$ and $\overline{\mathcal{M}}_{1}$ with monodromy group $L$,
and these are quotients of $\mathcal{R}_{1}$ by non-conjugate subgroups of $L$
isomorphic to ${\rm A}_{5}$. By the Riemann–Hurwitz formula a regular dessin
of this type with automorphism group $G$ has genus $1+\frac{5}{132}|G|$, in
this case equal to $26$. By its uniqueness, or by an observation of Singerman
[70] concerning triples in ${\rm PSL}_{2}(q)$, $\mathcal{R}_{1}$ is regular as
a map, corresponding to a subgroup of $\Delta$ which is normal in the extended
triangle group $\Delta[3,2,11]$. It is, in fact, the dual of the unique
orientable regular map of genus $26$ and type $\\{3,11\\}$, denoted by R26.2
in Conder’s catalogue of regular maps [16]; its full automorphism group as a
map, containing also the automorphisms reversing the orientation, is
isomorphic to ${\rm PGL}_{2}(11)$. For a drawing of R26.2 (combinatorially
though not geometrically correct), with $60$ $11$-valent vertices, $330$ edges
and $220$ triangular faces, see [40, Fig. 6 and Table 1], where it is shown as
a $198$-gon with side identifications; in this paper, Ivrissimtzis, Singerman
and Strudwick reinterpret the regular maps associated with Klein’s papers [50]
and [51] in terms of Farey fractions.

The eight dessins $\mathcal{D}$ with monodromy group ${\rm A}_{11}$ consist of
the three chiral pairs $\mathcal{M}_{i}$ and $\overline{\mathcal{M}}_{i}$ with
$i=2,3,4$, together with $\mathcal{M}_{5}$ and $\mathcal{M}_{6}$. The minimal
regular covers of these dessins are eight regular dessins $\mathcal{R}$ of
genus $756\,001$, and each dessin $\mathcal{D}$ is the quotient of
$\mathcal{R}$ by a subgroup isomorphic to ${\rm A}_{10}$ in its automorphism
group ${\rm A}_{11}$. These eight dessins $\mathcal{R}$ are mutually non-
isomorphic, since ${\rm A}_{11}$ has a unique conjugacy class of such
subgroups (the point-stabilisers in the natural representation), so that any
isomorphism between the dessins $\mathcal{R}$ would induce an isomorphism
between their quotients $\mathcal{D}$, which is visibly impossible (see Figure
3).

The bilateral symmetry of the dessins $\mathcal{M}_{5}$ and $\mathcal{M}_{6}$
is easily explained: in each case there is a permutation in $S_{11}$
simultaneously inverting the generating permutations $x$ and $y$ for the
monodromy group $A_{11}$, and this induces an isomorphism of the dessin with
its mirror image. (Of course, since $y^{2}=1$, inverting $y$ means the same as
commuting with it.) Equivalently, each of the corresponding map subgroups $M$
in $\Delta(3,2,11)$ is contained with index $2$ in a subgroup of
$\Delta[3,2,11]$ which induces this isomorphism. The same applies to the
regular covers of $\mathcal{M}_{5}$ and $\mathcal{M}_{6}$, each corresponding
to the core (intersection of conjugates) of $M$ in $\Delta(3,2,11)$; the other
six regular covers form three chiral pairs, like their quotients.

For example, Figure 7 shows the map $\mathcal{M}_{5}$ corresponding to
$\mathcal{D}_{5}$; directed edges are labelled $1,2,\ldots,11$, so that

$x=(1,11,6)(2,5,4)(7,10,8)\quad{\rm and}\quad y=(1,5)(2,3)(6,10)(8,9),$

and hence

$z=(1,2,\ldots,11).$

The permutation $t=(1,6)(2,8)(3,9)(4,7)(5,10)$ inverts $x$ and $y$, inducing
the automorphism of $\mathcal{M}_{5}$ given by reflection in the vertical
axis. The situation is similar for $\mathcal{M}_{6}$, with generators

$x=(1,11,6)(2,5,3)(7,10,9)\quad{\rm and}\quad y=(1,5)(3,4)(6,10)(7,8)$

inverted by $t=(1,6)(2,9)(3,7)(4,8)(5,10)$.

$11$$6$$1$$10$$5$$8$$2$$7$$4$$9$$3$ Figure 7: The map $\mathcal{M}_{5}$
corresponding to Klein’s plane tree V.

##  6 Computations and Galois orbits

Each of the ten dessins $\mathcal{D}$ discussed here represents a Belyĭ
function, in this case an $11$-sheeted covering $\beta:\Sigma\to\Sigma$. As
such, $\beta$ is a rational function of degree $11$, and having a single pole
at $\infty$ it must be a polynomial, called a Shabat polynomial.

A computation using MAPLE shows that the Shabat polynomial $P:\Sigma\to\Sigma$
for the dessins $\mathcal{M}_{1}$ and $\overline{\mathcal{M}}_{1}$ in Figure 2
has the form

$P=\frac{1}{2^{12}3^{14}}p_{1}^{3}p_{2}^{3}p_{3}$

where

$p_{1}(x)=2x+(11-3\sqrt{-11}),$
$p_{2}(x)=2x^{2}-(11-3\sqrt{-11})x-(22+6\sqrt{-11}),$
$p_{3}(x)=x^{2}+11x+(55+9\sqrt{-11}),$

so that

$P-1=-\frac{1}{2^{11}3^{14}}q_{1}^{2}q_{2}^{2}q_{3}$

where

$q_{1}(x)=2x+(5+3\sqrt{-11}),$
$q_{2}(x)=2x^{3}+(15-3\sqrt{-11})x^{2}-(12-12\sqrt{-11})x+(56+96\sqrt{-11}),$
$q_{3}(x)=2x^{3}-18x^{2}+(21+45\sqrt{-11})x-(175+279)\sqrt{-11}).$

The two possible choices for $\sqrt{-11}$ give the two dessins. They are
defined over ${\mathbb{Q}}(\sqrt{-11})$ (which is not surprising, since
$\sqrt{-11}$ appears in the character table of ${\rm PSL}_{2}(11)$), and they
are transposed by the Galois group of this field, generated by complex
conjugation.

Figure 8: A geometrically correct drawing of $\mathcal{M}_{1}$.

In each dessin, the black vertices are at the zeros of $P$: the three vertices
of valency $3$ correspond to the zeros of $p_{1}$ and $p_{2}$, each appearing
with multiplicity $3$ as a zero of $P$, and the two vertices of valency $1$
are at the zeros of $p_{3}$. The white vertices are at the zeros of $P-1$: the
four vertices of valency $2$ correspond to the zeros of $q_{1}$ and $q_{2}$,
each appearing with multiplicity $2$ as a zero of $P-1$, and the three
vertices of valency $1$ are the zeros of $q_{3}$. The unique face-centre is at
$\infty$, the pole of $P$ with multiplicity $11$. For example, Figure 8 shows
such a geometrically correct version of $\mathcal{M}_{1}$ in the plane
$\mathbb{C}$. With a few exceptions (Figures 15, 16 and 18) all plane trees,
dessins and maps shown in this paper are only combinatorially and
topologically correct.

The other eight dessins form a single Galois orbit. A computation using
Pari/GP produced several defining polynomials for the field of moduli
corresponding to this orbit, of which the simplest appears to be

$Z^{8}+2Z^{6}-3Z^{5}+10Z^{4}-14Z^{3}+14Z^{2}-8Z+1.$

The Galois group of this field is ${\rm S}_{8}$, inducing all possible
permutations of the eight dessins. The field does not contain
${\mathbb{Q}}(\sqrt{-11})$, nor, indeed, does it contain
${\mathbb{Q}}(\sqrt{21})$, even though $\sqrt{21}$ appears in the character
table of ${\rm A}_{11}$.

##  7 Other related dessins of degree $11$

The ten dessins discussed in this paper are not the only dessins of degree
$11$ and type $(3,2,r)$ for some $r$, the least common multiple of the face
valencies. It is easy to see that in any such dessin the distribution of
valencies of the black vertices must be $3^{3}1^{2}$, while that of the white
vertices must be $2^{4}1^{3}$ or $2^{5}1^{1}$. In the former case we obtain
the ten dessins based on Klein’s diagram, all with $r=11$, but in the latter
case we have planar dessins with two faces, rather than one, not considered by
Klein.

The maps corresponding to these additional dessins can all be formed from the
basic map ${\mathcal{M}}_{0}$ in Figure 4 by adding vertices of valency $1$ to
two of its free edges, as in the construction of the maps
${\mathcal{M}}={\mathcal{M}}_{i}$ and $\overline{\mathcal{M}}_{i}$, and then
joining together two of the remaining three free edges to form a single non-
free edge. Equivalently one can perform the latter operation, in three
possible ways, on each of the ten maps ${\mathcal{M}}$, giving $30$ maps of
the required form. However, one finds that some of these maps are mutually
isomorphic, and that up to isomorphism we obtain just $16$ dessins (confirmed
using GAP), consisting of four with bilateral symmetry (in the first row of
Figure 9) and six chiral pairs (those in the second row and their mirror
images).

(a)(b)(c)(d)(e)(f)(g)(h)(i)(j) Figure 9: Maps of degree $11$ and type
$(3,2,r)$, $r\neq 11$.

There are six where the face valency split is $10+1$, three each where it is
$9+2$ or $7+4$, and two each where it is $8+3$ or $6+5$. These dessins
correspond to sixteen more conjugacy classes of subgroups of index $11$ in the
modular group $\Gamma$, in addition to the ten corresponding to Klein’s plane
trees. (This is confirmed by [31, Theorem 1], which gives a recurrence
relation for the number $N_{n}$ of subgroups of index $n$ in $\Gamma$; here
$N_{11}=286$, corresponding to $10+16=26$ conjugacy classes each containing
$11$ subgroups.) In all cases except $10+1$, one of the face valencies is a
prime $p$, and $z^{11-p}$ is a $p$-cycle with $11-p\geq 3$ fixed points; being
transitive of prime degree the monodromy group $G=\langle x,y\rangle$ is
primitive, so it follows from Jordan’s Theorem and the fact that $y$ is odd
that $G={\rm S}_{11}$; in the case $10+1$ we may apply an extension of
Jordan’s Theorem in [41] to the cycles of length $10$ and $11$ to show that
$G$ acts as ${\rm S}_{11}$ or ${\rm AGL}_{1}(11)$, with the latter excluded
since it has no elements of order $3$. It follows that these dessins
correspond to non-congruence subgroups of $\Gamma$. Their regular covers have
genus

$1+\frac{(r-6)}{12r}11!$

as given in the following table:

face valencies | 10, 1 | 9, 2 | 8, 3 | 7, 4 | 6, 5
---|---|---|---|---|---
$r$ | $10$ | $18$ | $24$ | $28$ | $30$
genus | $1\,330\,561$ | $2\,217\,601$ | $2\,494\,801$ | $2\,613\,601$ | $2\,661\,121$
dessins | e, f, g | a, h | b, c | d, i | j
number of dessins | 6 | 3 | 2 | 3 | 2
Table 1: Regular coverings of dessins of type $(3,2,r)$.

##  8 Monodromy groups of dessins of type $(3,2,p)$ and degree $p$

We can generalise our earlier arguments by considering dessins $\mathcal{D}$
of type $(3,2,p)$ and degree $p$ for all primes $p$, not just for $p=11$. The
monodromy group $G$ of such a dessin $\mathcal{D}$ must be a transitive
permutation group of degree $p$, and must be perfect if $p>3$, as we will
assume, since $3,2$ and $p$ are mutually coprime. The transitive groups of
prime degree $p$ are all known (modulo a very difficult number-theoretic
problem which we will discuss in the next section), and are described in [13,
23, 39], for example. They are as follows:

###### Theorem 8.1

The transitive permutation groups $G$ of prime degree are the following:

  * (a)

subgroups $G$ of ${\rm AGL}_{1}(p)$ containing the translation subgroup;

  * (b)

$G={\rm A}_{p}$ or ${\rm S}_{p}$ for a prime $p\geq 5$;

  * (c)

$G={\rm PSL}_{2}(11)$, ${\rm M}_{11}$ or ${\rm M}_{23}$ with $p=11,11$ or
$23$;

  * (d)

groups $G$ such that ${\rm PSL}_{n}(q)\leq G\leq{\rm P\Gamma L}_{n}(q)$ where
the degree $(q^{n}-1)/(q-1)$ of its natural representation(s) is prime.

By a result of Galois [26], the solvable groups of prime degree $p$ are those
in (a). Burnside (see [11] or [12, §251]) showed that any nonsolvable group
$G$ of prime degree $p$ is doubly transitive; in this case a straightforward
argument shows that $S\leq G\leq{\rm Aut}\,S$ where $S$, the unique minimal
normal subgroup of $G$, is a doubly transitive nonabelian simple group. By the
classification of finite simple groups, the possibilities for $S$ are $S={\rm
A}_{p}$ for a prime $p\geq 5$, giving the groups in (b), the three groups
$S=G$ in (c), and $S={\rm PSL}_{n}(q)$, giving the groups in (d).

In (a) and (b) these are the natural actions of the groups listed. In (c),
${\rm PSL}_{2}(11)$ has two actions of degree $11$, on the cosets of two
conjugacy classes of subgroups isomorphic to ${\rm A}_{5}$, while the Mathieu
groups ${\rm M}_{11}$ and ${\rm M}_{23}$ act on Steiner systems with $11$ and
$23$ points. In (d) the groups act on the points of the projective geometry
${\mathbb{P}}^{n-1}({\mathbb{F}}_{q})$, with a second dual action on its
hyperplanes if $n\geq 3$. In particular, since ${\rm Out}\,S$ is solvable in
all cases, we have:

###### Corollary 8.2

The only perfect groups of prime degree are the simple groups ${\rm A}_{p}$,
${\rm PSL}_{2}(11)$, ${\rm M}_{11}$, ${\rm M}_{23}$ and ${\rm PSL}_{n}(q)$
appearing in parts (b), (c) and (d) of Theorem 8.1.

These simple groups $G$ all have elements $x$, $y$ and $z$ of orders $3$, $2$
and $p$ since they have orders divisible by these primes, so there remain the
questions of whether there exist such triples with $xyz=1$, whether these can
generate $G$, and if so, of determining (or at least enumerating) the
corresponding dessins $\mathcal{D}$. We will address these questions, on a
case by case basis, later in this and the following two sections. First we
calculate the genera of these maps and of their regular covers.

If $x$ and $y$ have cycle structures $3^{a}1^{\alpha}$ and $2^{b}1^{\beta}$,
with $\alpha=p-3a$ and $\beta=p-2b$, then $\mathcal{D}$ has $p-2a$ black
vertices, $p-b$ white vertices, $p$ edges and one face, so it has Euler
characteristic

$\chi=(p-2a)+(p-b)-p+1=p-2a-b+1\,=\frac{4\alpha+3\beta-p}{6}+1$

and hence genus

$1-\frac{\chi}{2}=\frac{2a+b+1-p}{2}\;=\frac{p-4\alpha-3\beta+6}{12}.$

(Note that this implies that $2a+b\geq p-1$, or equivalently
$4\alpha+3\beta\leq p+6$.) If it exists, such a dessin $\mathcal{D}$ has a
regular cover $\mathcal{R}$ of type $(3,2,p)$ with automorphism group $G$.
Since $\mathcal{R}$ has $|G|/3$ black vertices of degree 3, $|G|/2$ white
vertices of degree 2, $|G|$ edges, and $|G|/p$ faces of degree $p$, its Euler
characteristic is

$\chi=\left(\frac{1}{3}+\frac{1}{2}+\frac{1}{p}-1\right)\cdot|G|=\frac{6-p}{6p}|G|,$

so that its genus is

$g=1-\frac{\chi}{2}=\frac{p-6}{12p}|G|+1.$

###  8.1 Case (c): sporadic examples

It is convenient to deal with case (c) first. We have seen in Examples 4.1 and
4.2, that if $p=11$ then ${\rm M}_{11}$ does not arise as the monodromy group
of a dessin of type $(3,2,p)$ and degree $p=11$, whereas ${\rm PSL}_{2}(11)$
does, for the chiral pair ${\mathcal{M}}_{1}$ and
$\overline{\mathcal{M}}_{1}$. The Frobenius formula (2) eliminates ${\rm
M}_{23}$: this group contains no triples $(x,y,z)$ of type $(3,2,23)$ such
that $xyz=1$.

###  8.2 Case (b): alternating groups

In case (b) we have $G={\rm A}_{p}$ for some prime $p$. The following result
shows that almost all such groups arise as monodromy groups in the required
context.

###### Theorem 8.3

For each prime $p\neq 2,3,7$ there is a dessin $\mathcal{D}$ of degree $p$ and
type $(3,2,p)$ with monodromy group ${\rm A}_{p}$.

$\mathcal{D}$${\mathcal{D}}^{\prime}$ Figure 10: Maps $\mathcal{D}$ and
${\mathcal{D}}^{\prime}$ for $p=3k+2$ and $p=3k+1$.

Proof. First suppose that $p\equiv 2$ mod $(3)$, say $p=3k+2$. There is an
obvious map for $p=5$, so we may assume that $p\geq 11$, and hence $k\geq 3$.
Let $\mathcal{D}$ be the map on the left in Figure 10, where there are $k$
vertices of degree $3$, so that $\mathcal{D}$ has passport
$(3^{k}1^{2},2^{k+1}1^{k},p^{1})$. We aim to show that its monodromy group $G$
is ${\rm A}_{p}$. By Corollary 8.2 and the discussion of sporadic examples in
case (c) it is sufficient to show that $G$ is not ${\rm PSL}_{2}(11)$ acting
with degree $11$ or a group ${\rm PSL}_{n}(q)$ acting naturally for some
$n\geq 2$ and prime power $q$. In fact, when $p=11$, so that $k=3$,
$\mathcal{D}$ is isomorphic to $\overline{\mathcal{M}}_{3}$ in Figure 3, and
we have seen that this has monodromy group ${\rm A}_{11}$. We may therefore
suppose, for a contradiction, that $G={\rm PSL}_{n}(q)$ acting naturally, and
that $p\geq 17$, so that $k\geq 5$.

If $\pi$ denotes the permutation character of ${\rm S}_{p}$, counting fixed
points of elements, we see that the involution $y$ corresponding to
$\mathcal{D}$ has fixed point ratio

$\frac{\pi(y)}{\pi(1)}=\frac{k}{p}=\frac{k}{3k+2}\geq\frac{5}{17}.$

We will compare this with the corresponding values for involutions in ${\rm
PSL}_{n}(q)$. Any involution in ${\rm PSL}_{2}(q)$ fixes at most two points,
so $n\neq 2$ since $\pi(y)=k\geq 5$. For $n\geq 3$ the involutions $g\in{\rm
PSL}_{n}(q)$ have fixed point ratios $\pi(g)/\pi(1)$ bounded above by

$b_{e}=\frac{q^{n-2}+q^{n-3}+\cdots+q+1}{q^{n-1}+q^{n-2}+\cdots+q+1}\quad\hbox{or}\quad
b_{o}=\frac{(q^{n-3}+\cdots+q+1)+(q+1)}{q^{n-1}+q^{n-2}+\cdots+q+1}$

as $q$ is even or odd: these bounds are attained when $g$ fixes a hyperplane
or two disjoint subspaces of dimensions $n-3$ and $1$, respectively,
corresponding to matrices in ${\rm SL}_{n}(q)$ with Jordan normal form given
by block matrices

$\Big{(}\,\begin{array}[]{cc}J&O\\\
O&I\end{array}\,\Big{)}\quad\hbox{where}\quad
J=\Big{(}\,\begin{array}[]{cc}1&1\\\
0&1\end{array}\,\Big{)}\quad\hbox{or}\quad\Big{(}\,\begin{array}[]{cc}-1&0\\\
0&-1\end{array}\,\Big{)}\quad\hbox{and}\quad I=I_{n-2}.$

It is straightforward to check that if $q$ is odd then $b_{o}<5/17$ and hence
$G\neq{\rm PSL}_{n}(q)$, unless $n=q=3$, when $b_{o}=5/13$; however, this
exceptional case corresponds to $p=13\not\equiv 2$ mod $(3)$. Thus $q$ is
even, so $q=2$ since $b_{e}<1/q$. This gives $3k+2=p=2^{n}-1$ and hence
$2^{n}\equiv 0$ mod $(3)$, which is impossible.

We can apply a similar argument when $p\equiv 1$ mod $(3)$, say $p=3k+1$. We
may assume that $p\geq 13$, that is, $k\geq 4$. By deleting a vertex of
valency $1$ from $\mathcal{D}$ we obtain the map ${\mathcal{D}}^{\prime}$ of
degree $p$, type $(3,2,p)$ and passport $(3^{k}1^{1},2^{k}1^{k+1},p^{1})$
shown on the right in Figure 10. The corresponding involution $y$ now has
fixed point ratio

$\frac{\pi(y)}{\pi(1)}=\frac{k+1}{p}=\frac{k+1}{3k+1}\geq\frac{5}{13}.$

As before, if $G={\rm PSL}_{n}(q)$ then $\pi(y)>2$ implies that $n\geq 3$. We
have already seen that if $q$ is odd then $b_{o}<5/17$, so that $b_{o}<5/13$
and hence $G\neq{\rm PSL}_{n}(q)$, unless $n=q=3$; however, in this case
$p=13$, and GAP shows that ${\mathcal{D}}^{\prime}$ has monodromy group ${\rm
A}_{13}$ rather than ${\rm PSL}_{3}(3)$, so $q$ must be even. Again, $q=2$
since $b_{e}<1/q$, so if any element of $G={\rm PSL}_{n}(2)$ has fixed points
they must form a projective subspace. Thus $k=\pi(y)$ is odd and hence
$p=3k+1$ is even, a contradiction. $\square$

It is easy to see that there is no corresponding dessin if $p=2$ or $3$. When
$p=7$ there is a chiral pair of dessins of the required degree and type,
represented by ${\mathcal{D}}^{\prime}$ in Figure 10 with $k=2$, but they have
monodromy group ${\rm PSL}_{2}(7)$, not ${\rm A}_{7}$ (which is not a Hurwitz
group). Simple modifications to the dessins in Figure 10 show that in the
proof of Theorem 8.3 the number of choices for $\mathcal{D}$ and
$\mathcal{D}^{\prime}$ increases exponentially with $p$, even though that
construction produces only planar dessins.

###  8.3 Case (d): projective groups

For the rest of this section, we will therefore assume that we are in case
(d), with $G={\rm PSL}_{n}(q)$ acting on points or hyperplanes of
${\mathbb{P}}^{n-1}({\mathbb{F}}_{q})$. (This case, together with associated
number theoretic estimates and computer searches covered in Section 9, is
investigated in more detail in [46].)

Warning: here, in spite of an almost universal convention on notation, the
unnamed prime of which $q$ is a power is not the prime $p$; instead, $p$ will
denote the degree $(q^{n}-1)/(q-1)$ of $G$.

Before proceeding, we need to make a short digression into number theory. Let
us define a prime $p$ to be projective if it has the form

$p=\frac{q^{n}-1}{q-1}=1+q+q^{2}+\cdots+q^{n-1}$

for some prime power $q\geq 2$ and integer $n\geq 2$. These primes include the
Fermat primes $p=2^{2^{k}}+1$ with $q=2^{2^{k}}$ and $n=2$, the Mersenne
primes $p=2^{n}-1$ with $q=2$ and $n$ prime, together with other examples with
$q,n\geq 3$, such as $p=13$ with $q=n=3$, and $p=31$ with $q=5$ and $n=3$
(notice that $p=31$ is also a Mersenne prime). Of course, it is an open
problem whether there are infinitely many Fermat or Mersenne primes; at the
time of writing, only five Fermat primes (with $k=0,\ldots,4$) and $51$
Mersenne primes are known to exist. More generally, the existence of
infinitely many projective primes seems to be an open problem. In [46] we give
heuristic arguments based on the Bateman–Horn Conjecture, together with
associated computational evidence, to support a conjecture that there are
infinitely many such primes, even for each fixed prime $n\geq 3$. These are
summarised in the next section. However, first we need to prove some necessary
conditions for $(q^{n}-1)/(q-1)$ to be prime.

###### Lemma 8.4

Given integers $n\geq 2$ and $e\geq 1$, the polynomial

$f(t)=\frac{t^{ne}-1}{t^{e}-1}=1+t^{e}+t^{2e}+\cdots+t^{(n-1)e}$

is irreducible in ${\mathbb{Z}}[t]$ if and only if $n$ is prime and $e$ is a
power $n^{i}\;(i\geq 0)$ of $n$.

Proof. If $k\in\mathbb{N}$ the cyclotomic polynomial $\Phi_{k}(x)$ is, by
definition, the polynomial with integer coefficients whose roots are the
primitive $k$th roots of unity. It is irreducible and has degree $\varphi(k)$,
where $\varphi$ is the Euler totient function. For any $n\in\mathbb{N}$ we
have $x^{n}-1=\prod_{d|n}\Phi_{d}(x)$ (see [8, §5.2.1] or [66, §4.3, Problem
26]). Putting $x=t^{e}$ gives

$f(t)=\frac{t^{ne}-1}{t^{e}-1}=\prod_{d}\Phi_{d}(t),$ (4)

with the product over all $d$ which divide $ne$ but not $e$. Thus $f$ is
irreducible if and only if there is just one such divisor $d$ (which is $ne$
itself, of course). By considering the prime power decompositions of $e$ and
$ne$ one can see that this happens if and only if $n$ is prime and $e$ is a
power of $n$. $\square$

###### Lemma 8.5

Let $q=r^{e}$ where $r$ is prime, and let $n\geq 2$. If $p=(q^{n}-1)/(q-1)$ is
prime then

  * (a)

$n$ is prime and $e$ is a power $n^{i}\;(i\geq 0)$ of $n$;

  * (b)

$q\not\equiv 1$ mod $(n)$;

  * (c)

${\rm PSL}_{n}(q)={\rm SL}_{n}(q)$.

Proof. (a) This follows from Lemma 8.4, using the fact that
$p=f(r)=\prod_{d}\Phi_{d}(r)$ as in (4), with each $\Phi_{d}(r)\neq\pm 1$. (To
see the latter, note that $\Phi_{d}(r)=\prod_{\zeta}(r-\zeta)$ where $\zeta$
ranges over the primitive $d$-th roots of unity, so that $|r-\zeta|>1$ for
each $\zeta$.)

(b) If $q\equiv 1$ mod $(n)$ then $p=1+q+\cdots+q^{n-1}\equiv 0$ mod $(n)$
with $p>n$.

(c) For any $q$ and $n$ we have ${\rm PSL}_{n}(q)={\rm SL}_{n}(q)/Z$, where
$Z=\\{\lambda I\in{\rm GL}_{n}(q)\mid\lambda^{n}=1\\}\cong{\rm C}_{d}$ is the
centre of ${\rm SL}_{n}(q)$ and $d=\gcd(q-1,n)$. If $p$ is prime then $d=1$ by
(a) and (b). $\square$

##  9 The Bateman–Horn Conjecture

The Bateman–Horn Conjecture [3] is a quantified version of Schinzel’s
Hypothesis H [68]. Both of them concern the question of whether a finite set
of polynomials $f_{1}(t),\ldots,f_{k}(t)\in{\mathbb{Z}}[t]$ simultaneously
take prime values $f_{i}(t)$ at infinitely many $t\in{\mathbb{N}}$.
(Schinzels’s Hypothesis H is in turn a generalisation of the Bunyakovsky
Conjecture (1857, see [9, 10]), which deals with the case $k=1$.) Clearly the
following conditions are necessary for this to happen:

  * •

the leading coefficient of each $f_{i}$ is positive,

  * •

each $f_{i}$ is irreducible,

  * •

$f:=f_{1}\ldots f_{k}$ is not identically zero modulo any prime.

In any specific case, the final condition can be verified as follows: $f$
cannot be identically zero modulo any prime $p>\deg(f)$ unless $p$ divides all
of its coefficients, and for the remaining primes $p$ one can simply evaluate
$f(t)$ at each element $t\in{\mathbb{Z}}_{p}$. Schinzel (following Bunyakovsky
in the case $k=1$) conjectured that these conditions are also sufficient. If
true, this would imply several other conjectures in Number Theory. For
instance, taking $f_{1}(t)=t^{2}+1$ would imply the Euler–Landau Conjecture
that there are infinitely many primes of the form $t^{2}+1$, while taking
$f_{1}(t)=t$ and $f_{2}(t)=t+2$ or $2t+1$ would imply the conjectures that
there are infinitely many twin primes and Sophie Germain primes.

The Bateman–Horn Conjecture goes further by proposing a heuristic estimate,
based on the Prime Number Theorem, for the number $Q(x)$ of $t\leq x$ in
$\mathbb{N}$ such that each $f_{i}(t)$ is prime. This has the form

$Q(x)\sim E(x):=\frac{C}{\prod_{i=1}^{k}\deg f_{i}}\int_{2}^{x}\frac{dt}{(\ln
t)^{k}}\quad\hbox{as}\quad x\to\infty$ (5)

where

$C=C(f_{1},\ldots,f_{k}):=\prod_{r}\left(1-\frac{1}{r}\right)^{-k}\left(1-\frac{\omega_{f}(r)}{r}\right)$
(6)

with the product over all primes $r$, and $\omega_{f}(r)$ is the number of
$t\in{\mathbb{F}}_{r}$ such that $f(t)=0$. If the above conditions on
$f_{1},\ldots,f_{k}$ are satisfied then the infinite product in (6) converges
to a limit $C>0$ (this is far from obvious: see [2], Theorem 5.4.3); since the
integral in (5) diverges as $x\to\infty$, this would imply that
$f_{1}(t),\ldots,f_{k}(t)$ are simultaneously prime for infinitely many
$t\in{\mathbb{N}}$. So far, these three conjectures have been proved only in
the case $k=1$, $\deg f_{1}=1$: in the case of the Bunyakovsky and Schinzel
conjectures, this is Dirichlet’s theorem on primes in an arithmetic
progression $at+b$; in the case of Bateman–Horn it is equivalent to the
stronger form of this theorem, giving an asymptotically equal distribution of
primes among the congruence classes of units $b$ mod $(a)$ (see [8, §5.3.2]
for a proof, and [71] for an error estimate).

Nevertheless, the Bateman–Horn Conjecture has provided estimates remarkably
close to experimental evidence in cases such as the twin primes conjecture. To
give just one example, the number of twin prime pairs up to $10^{18}$ is
$808\,675\,888\,577\,436$ (entry A007508 of [67]), while the Bateman–Horn
estimate gives $808\,675\,900\,456\,220$. The relative error is
$-0.000000147\%$. See also [2, Figure 10]. The constant $C$ is, in this case,
equal to $1.320323630\ldots$, while $C/2=0.660161815\ldots$ is traditionally
called the “twin primes constant”.

###### Remark 9.1

Li [57] has suggested a modified version of the Bateman–Horn Conjecture in
which $\prod_{i}\deg f_{i}$ in the denominator is removed and each factor
$1/\ln(t)$ within the integral is replaced with $1/\ln(f_{i}(t))$; the lower
limit of integration may also be adjusted to avoid a logarithmic singularity
at $f_{i}(t)=1$. The result is asymptotically equivalent to the original
conjecture, but in cases such as Sophie Germain primes, involving a non-monic
polynomial $f_{i}$, the approximation is significantly better.

In our case, if we take $k=2$, with $f_{1}(t)=t$ and
$f_{2}(t)=1+t^{e}+t^{2e}+\cdots+t^{(n-1)e}$ for a fixed prime $n\geq 3$ and a
fixed exponent $e=n^{i}$, then $f_{2}$ is irreducible by Lemma 8.4, so the
conjecture gives an estimate $E(x)$ for the number $Q(x)$ of primes $t\leq x$
such that $1+q+q^{2}+\cdots+q^{n-1}$ is prime, where $q=t^{e}$. Computer
searches confirm the obvious guess that $Q(x)$ grows fastest when $e$ and $n$
take their minimum values $n=3$ and $e=1$ (so $q$ is prime), so that
$f_{2}(t)=1+t+t^{2}$. For example,444The computation was carried out by our
colleague Jean Bétréma [5]. He used the system Julia which, on this occasion,
turned out to be more efficient than Maple. all except $301$ of the
$1\,974\,311$ projective primes $p\leq 10^{18}$ have the form $p=1+q+q^{2}$
for some prime $q$.

Putting $n=3$ and $e=1$, so that $f=t(1+t+t^{2})$, we get

$\omega_{f}(r)\,=\,\left\\{\begin{array}[]{ll}3&\mbox{if }r\equiv 1\mbox{ mod
}(3),\\\ 2&\mbox{for }r=3,\\\ 1&\mbox{otherwise}.\end{array}\right.$

We use Maple to calculate $C\approx 1.52173006$ (taking the product over all
primes $r<10^{9}$) and $\int_{2}^{x}(\ln t)^{-2}dt$ (by numerical quadrature)
for $x=i\cdot 10^{10}\;(i=1,2,\ldots,10)$. In Table 2 the resulting estimates
$E(x)$ are compared with the true values $Q(x)$ found by computer searches.
(Here, Li’s improvement in Remark 9.1 made negligible difference: for
instance, using $1/\ln(1+t+t^{2})$ instead of $1/2\ln(t)$ with $x=10^{11}$
gave the estimate $129\,297\,407.4$, compared with $129\,297\,407.9$ in Table
2.) The proportional errors shown are less that $0.00034$, that is, $0.034\%$.
Similar comparisons for other values of $n$ and $e$ were reassuring but less
convincing, as the numbers of projective primes within our computing range
were much smaller: for example, with $n=5$ and $e=1$ we obtained an estimate
of $246.718\ldots$ for the number of projective primes
$p=1+q+\cdots+q^{4}<10^{18}$, whereas the correct number is 252. (Note that in
order to estimate the number of primes $p=1+q+q^{2}\leq 10^{18}$ we have to
compute the integral in (5) up to $x=10^{18/2}$; however, for
$p=1+q+\cdots+q^{4}<10^{18}$ the integral is up to $10^{18/4}$, over a much
smaller interval.)

$x$ | $Q(x)$ | $E(x)$ | $E(x)/Q(x)$
---|---|---|---
$1\cdot 10^{10}$ | 15 801 827 | $1.579642126\times 10^{7}$ | 0.9996579044
$2\cdot 10^{10}$ | 29 684 763 | $2.968054227\times 10^{7}$ | 0.9998578150
$3\cdot 10^{10}$ | 42 963 858 | $4.296235691\times 10^{7}$ | 0.9999650617
$4\cdot 10^{10}$ | 55 877 571 | $5.587447496\times 10^{7}$ | 0.9999445924
$5\cdot 10^{10}$ | 68 522 804 | $6.852175590\times 10^{7}$ | 0.9999847043
$6\cdot 10^{10}$ | 80 962 422 | $8.096382889\times 10^{7}$ | 1.0000173771
$7\cdot 10^{10}$ | 93 236 613 | $9.323905289\times 10^{7}$ | 1.0000261688
$8\cdot 10^{10}$ | 105 372 725 | $1.053741048\times 10^{8}$ | 1.0000130940
$9\cdot 10^{10}$ | 117 383 505 | $1.173885689\times 10^{8}$ | 1.0000431394
$10^{11}$ | 129 294 308 | $1.292974079\times 10^{8}$ | 1.0000239757
Table 2: Comparing the BHC estimates $E(x)$ with true values $Q(x)$

On the basis of this evidence, we make the following conjecture:

###### Conjecture 9.2

There are infinitely many projective primes of the form $p=1+q+q^{2}$ where
$q$ is prime.

On the basis of similar evidence for $n>3$ or $e>1$, but necessarily involving
smaller numbers of projective primes, we make the following more general
conjecture:

###### Conjecture 9.3

For each fixed prime $n\geq 3$ and each fixed power $e$ of $n$, there are
infinitely many projective primes of the form $p=1+q+\cdots+q^{n-1}$ where $q$
is the $e$-th power of a prime.

Of course, rather weaker statements than these would also imply the existence
of infinitely many projective primes. For example, it would be sufficient to
prove that for infinitely many primes $n$ (resp. primes power $q$) there is at
least one prime power $q$ (resp. prime $n$) such that $(q^{n}-1)/(q-1)$ is
prime. This would give an infinite set of such pairs $(n,q)$, with any
specific prime represented by at most finitely many of them.

##  10 Groups and dessins of projective prime degrees

Returning to the groups, the elements $z$ of order $p=(q^{n}-1)/(q-1)$ in
$G={\rm PSL}_{n}(q)$ are the Singer cycles (see [39, Satz II.7.3], for
example), induced by appropriate non-zero elements of ${\mathbb{F}}_{q^{n}}$
acting linearly by multiplication on that field, regarded as an
$n$-dimensional vector space over ${\mathbb{F}}_{q}$, and permuting its
$1$-dimensional subspaces regularly. It follows from results of Berecky [7]
that when a Singer cycle $z$ has prime order $p$ the only maximal subgroup of
$G$ containing it is the normaliser $N_{G}(\langle z\rangle)\cong{\rm
C}_{p}\rtimes{\rm C}_{n}$, where the complement is induced by the Galois group
${\rm Gal}\,{\mathbb{F}}_{q^{n}}/{\mathbb{F}}_{q}=\langle t\mapsto
t^{q}\rangle\cong{\rm C}_{n}$. Being solvable, $N_{G}(\langle z\rangle)$
cannot contain a perfect subgroup $\langle x,y,z\rangle$, so any triple of
type $(3,2,p)$, if it exists, must generate $G$. Moreover, there are $(p-1)/n$
conjugacy classes $\mathcal{Z}$ of Singer cycles $z$ in $G$, one for each
orbit of ${\rm Gal}\,{\mathbb{F}}_{q^{n}}/{\mathbb{F}}_{q}$ on elements of
multiplicative order $p$ in ${\mathbb{F}}_{q^{n}}$. (Note that since $n$ is
prime we have $p-1=q+q^{2}+\cdots+q^{n-1}\equiv 0$ mod $(n)$: this is obvious
if $q\equiv 0$ or $1$ mod $(n)$, and otherwise since $n$ is prime the summands
represent the non-zero congruence classes mod $(n)$, so they cancel in pairs.

For the small projective primes $p=5$, $7$, $13$ and $17$ the groups ${\rm
PSL}_{2}(4)\;(\cong{\rm PSL}_{2}(5)\cong{\rm A}_{5})$, ${\rm
PSL}_{3}(2)\;(\cong{\rm PSL}_{2}(7))$, ${\rm PSL}_{3}(3)$ and ${\rm
PSL}_{2}(16)$ arise as monodromy groups. We will systematically examine
various possibilities for monodromy groups ${\rm PSL}_{n}(q)$, starting with
small values of $n$.

###  10.1 The case $n=2$: Fermat primes

In the case of the Fermat primes, where $n=2$, we have the following:

###### Theorem 10.1

For each Fermat prime $p=2^{2^{k}}+1\geq 5$ there are $2^{2^{k}-k-1}$ dessins
$\mathcal{D}$ of type $(3,2,p)$ and degree $p$ with monodromy group $G={\rm
PSL}_{2}(2^{2^{k}})$, acting naturally. They have genus $(p-5)/12$, while
their regular covers $\mathcal{R}$, which are mutually non-isomorphic, have
genus $p(p-4)(p-5)/12$. Dessins $\mathcal{D}$ and $\mathcal{R}$ are both
mirror symmetric.

Let $G={\rm PSL}_{2}(q)$ where $q=2^{2^{k}}\geq 4$ and $p=q+1$ is prime. The
conjugacy classes $\mathcal{X}$ and $\mathcal{Y}$ of elements $x$ and $y$ of
orders $3$ and $2$ are unique, while there are $q/2$ classes $\mathcal{Z}$ of
elements $z$ of order $p$. Since $3$ and $p$ divide $q-1$ and $q+1$
respectively, the character table of $G$ (see [44, §5.5], for example) shows
that only the principal character $\chi=\chi_{1}$ has
$\chi(x)\chi(y)\chi(z)\neq 0$, so for each choice of $\mathcal{Z}$ we have a
character sum $\Sigma=1$. Since $|C(x)|=q-1$, $|C(y)|=q$, $|C(z)|=q+1$ and
$|G|=q(q^{2}-1)$, the number of triples of type $(3,2,p)$ in $G$ is

$\frac{q}{2}\cdot\frac{|G|^{2}}{|C(x)|\cdot|C(y)|\cdot|C(z)|}\cdot\Sigma=\frac{q}{2}\cdot|G|.$

As explained earlier, results of Berecky [7] on Singer cycles imply that each
triple generates $G$. We have ${\rm Aut}\,G\cong G\rtimes{\rm
Gal}\,{\mathbb{F}}_{q}\cong G\rtimes{\rm C}_{2^{k}}$, acting semi-regularly on
these triples, so the number of its orbits on them is

$\frac{q}{2}\cdot\frac{1}{2^{k}}=2^{2^{k}-k-1}.$

This is therefore the number of normal subgroups of $\Delta=\Delta(3,2,p)$
with quotient group $G$, and thus the number of regular dessins $\mathcal{R}$
of type $(3,2,p)$ with automorphism group $G$.

Now $G$ has a unique conjugacy class of subgroups of index $p=q+1$, the
stabilisers of points in ${\mathbb{P}}^{1}({\mathbb{F}}_{q})$. These lift back
to $2^{2^{k}-k-1}$ conjugacy classes of subgroups of the same index in
$\Delta$, so they correspond to that number of dessins $\mathcal{D}$ of type
$(3,2,p)$ and degree $p$ with monodromy group $G$ and regular cover one of the
dessins $\mathcal{R}$. Since $x$, $y$ and $z$ have cycle structures
$3^{(q-1)/3}1^{2}$, $2^{q/2}1^{1}$ and $p^{1}$ on
${\mathbb{P}}^{1}({\mathbb{F}}_{q})$, the dessins $\mathcal{D}$ have
characteristic

$\chi=\frac{q-1}{3}+2+\frac{q}{2}+1-(q+1)+1=\frac{16-q}{6}$

and hence have genus

$g=1-\frac{\chi}{2}=\frac{q-4}{12}=\frac{p-5}{12}.$

By the Riemann–Hurwitz formula the regular dessins $\mathcal{R}$ have genus

$\frac{p-6}{12p}|G|+1=\frac{p(p-4)(p-5)}{12}.$

Each of the conjugacy classes $\mathcal{Z}$ consists of mutually inverse pairs
$z,z^{q}=z^{-1}$, so the maps $\mathcal{D}$ and $\mathcal{R}$ are isomorphic
to their mirror images. (In fact, as noted by Singerman [70], it follows from
results of Macbeath [58] that for any prime power $q$, each generating pair
for ${\rm PSL}_{2}(q)$ are simultaneously inverted by an automorphism,
implying this result more generally.)

###### Example 10.2

If $k=1$ and $p=5$ we obtain one planar dessin $\mathcal{D}$, shown on the
left in Figure 11, with monodromy group $G={\rm PSL}_{2}(4)\cong{\rm A}_{5}$;
its regular cover $\mathcal{R}$ is the dodecahedron, and $\mathcal{D}$ is the
quotient of $\mathcal{R}$ by a point-stabiliser in $G$, isomorphic to ${\rm
A}_{4}$. If $k=2$ and $p=17$ we obtain two dessins of genus $1$; they are
shown in Section 14, Figure 25. Their regular covers are of genus $221$. If
$k=3$ and $p=257$ we obtain $16$ dessins of genus $21$, with regular covers of
genus $1\,365\,441$. If $k=4$ and $p=65\,537$ we obtain $2048$ dessins of
genus $5461$, with regular covers of genus $23\,454\,100\,602\,881$. No Fermat
primes are known for $k\geq 5$, and it is conjectured that none exist, but if
any do, they are covered by Theorem 10.1.

Figure 11: Dessins of type $(3,2,p)$ and degree $p$ for $p=5,7$.

Having dealt with the case $n=2$, we will now assume that $n\geq 3$. In this
case ${\rm PSL}_{n}(q)$ has two natural representations, on points and
hyperplanes of ${\mathbb{P}}^{n-1}({\mathbb{F}}_{q})$, transposed by the outer
automorphism induced by the duality of this geometry. Another difference
between this case and the case $n=2$ is that when $n\geq 3$ the conjugacy
classes $\mathcal{Z}$ of Singer cycles are not inverse closed: instead, they
form mutually inverse pairs of classes, transposed by duality. This implies
that the dessins $\mathcal{D}$ occur in chiral pairs, each pair having the
same regular cover $\mathcal{R}$, which is regular as a map, that is,
isomorphic to its mirror image.

###  10.2 The case $n=3$: projective planes

The great majority of projective primes, up to a given bound, have $n=3$, so
that the group $G={\rm PSL}_{3}(q)$ acts on the points and lines of the
projective plane ${\mathbb{P}}^{2}({\mathbb{F}}_{q})$. In this case one can
enumerate and describe the associated dessins by using the generic character
table for $G$ given by Simpson and Frame in [69], in addition to those in the
ATLAS [17] for $q\leq 9$.

###### Theorem 10.3

If $q$ is a prime power such that $p:=1+q+q^{2}$ is prime, there are
$(p-1)/3e$ dessins of type $(3,2,p)$ and degree $p$ with monodromy group ${\rm
PSL}_{3}(q)$ acting naturally, where $q$ is the $e$-th power of a prime. They
are all chiral. If $q$ is odd they have passport
$(3^{q(q+1)/3}1^{1},2^{(q^{2}-1)/2}1^{q+2},p^{1})$ and genus $(q-3)(q+1)/12$,
whereas if $q=2^{e}$ their passport is
$(3^{q(q+1)/3}1^{1},2^{q^{2}/2}1^{q+1},p^{1})$ and their genus is $q(q-2)/12$.

Let $G:={\rm PSL}_{3}(q)$. By Lemma 8.5(c) we have $G={\rm SL}_{3}(q)$. We
will apply the Frobenius formula (2) to $G$, using the generic character table
for ${\rm SL}_{3}(q)$ in [69, Table 1b].

First assume that $q$ is coprime to $6$. There are unique conjugacy classes
$\mathcal{X}$ and $\mathcal{Y}$ of elements of order $3$ and $2$ in $G$, and
there are $(p-1)/3=q(q+1)/3$ classes $\mathcal{Z}$ of elements (Singer cycles)
of order $p$; these classes have types ${\mathcal{C}}_{7}$,
${\mathcal{C}}_{4}$ and ${\mathcal{C}}_{8}$ in the notation of [69, Table 1a],
and their elements have centralisers of orders $q^{2}-1$, $q(q-1)^{2}(q+1)$
and $p$. The only non-principal irreducible character of $G$ which does not
vanish on ${\mathcal{X}}$, ${\mathcal{Y}}$ or ${\mathcal{Z}}$ is the Steinberg
character, the unique irreducible character of degree $q^{3}$, which takes the
values $-1$, $q$ and $1$ on them. Since $|G|=q^{3}(q-1)^{2}(q+1)(q^{2}+q+1)$
we find that the number of triples of type $(3,2,p)$ in $G$ is

$\frac{p-1}{3}\cdot\frac{(q^{3}(q-1)^{2}(q+1)(q^{2}+q+1))^{2}}{(q^{2}-1)\cdot
q(q-1)^{2}(q+1)\cdot p}\left(1+\frac{(-1)\cdot q\cdot
1}{q^{3}}\right)=\frac{p-1}{3}|G|.$

By [7] the only maximal subgroup of $G$ containing $z$ is $N_{G}(\langle
z\rangle)$, of order $3p$, so any such triple generates $G$. If $q$ is the
$e$-th power of a prime then $|{\rm Out}\,G|=2e$, the two factors
corresponding to duality of the projective plane
${\mathbb{P}}^{2}({\mathbb{F}}_{q})$ on which $G$ acts and the Galois group of
${\mathbb{F}}_{q}$, so $\Delta(3,2,p)$ has $(p-1)/6e$ normal subgroups with
quotient $G$. Since $G$ has two conjugacy classes of subgroups of index $p$,
the stabilisers of points and lines in ${\mathbb{P}}^{2}({\mathbb{F}}_{q})$,
we obtain $(p-1)/3e$ dessins of type $(3,2,p)$ and degree $p$ with monodromy
group $G$.

In each of the two actions of degree $p$ of $G$, the elements $x$, $y$ and $z$
fix $1$, $q+2$ and $0$ points, so these dessins all have passport
$(3^{q(q+1)/3}1^{1},2^{(q^{2}-1)/2}1^{q+2},p^{1})$ and hence genus
$(q-3)(q+1)/12$.

If $q=3^{e}$ there are two conjugacy classes of elements of order $3$, of
types ${\mathcal{C}}_{2}$ and ${\mathcal{C}}_{3}$, with centralisers of order
$q^{3}(q-1)$ and $q^{2}$. If $x$ is chosen from the first class, then only the
principal character and the constituent of the permutation character of degree
$p-1=q(q+1)$ contribute to the character sum $\Sigma$, and this sum is equal
to

$\Sigma=1+\frac{q\cdot(q+1)\cdot(-1)}{q(q+1)}=0.$

If $x$ is chosen from the second class, then only the principal character
contributes to $\Sigma$, so the resulting number of triples in $G$ is

$\frac{p-1}{3}\cdot\frac{(q^{3}(q-1)^{2}(q+1)(q^{2}+q+1))^{2}}{q^{2}\cdot
q(q-1)^{2}(q+1)\cdot p}=\frac{p-1}{3}|G|$

as before. The number of dessins is therefore again $(p-1)/3e$, and the
passport and genus are also as before.

If $q=2^{e}$ then the main changes are that the unique class $\mathcal{Y}$ of
involutions has type ${\mathcal{C}}_{2}$, with centralisers of order
$q^{3}(q-1)$, and that every non-principal irreducible character vanishes on
$x$, $y$ or $z$, so $\Sigma=1$. The number of triples of type $(3,2,p)$ in $G$
is therefore

$\frac{p-1}{3}\cdot\frac{(q^{3}(q-1)^{2}(q+1)(q^{2}+q+1))^{2}}{(q^{2}-1)\cdot
q^{3}(q-1)\cdot p}=\frac{p-1}{3}|G|,$

so the number of dessins is again $(p-1)/3e$. The other change is that $y$ now
fixes $q+1$ points, so that the passport is
$(3^{q(q+1)/3}1^{1},2^{q^{2}/2}1^{q+1},p^{1})$ and the genus is $q(q-2)/12$.

In all cases, chirality of the dessins follows from the facts that the two
dual representations of $G$ of degree $p$ correspond to a mutually inverse
pair of conjugacy classes $\mathcal{Z}$, and that ${\rm Out}\,G$ acts
semiregularly on the set of such pairs.

###### Example 10.4

The smallest example for the above theorem is the chiral pair of dessins of
degree $p=7$ with monodromy group ${\rm PSL}_{3}(2)\cong{\rm PSL}_{2}(7)$; one
of this pair is shown on the right in Figure 11.

###### Example 10.5

Let $p=13$ and $G={\rm PSL}_{3}(3)$. The required conjugacy classes are (in
ATLAS [17] notation) ${\mathcal{X}}=3B$, ${\mathcal{Y}}=2A$ and
${\mathcal{Z}}=13A,13B,13C$ or $13D$, giving four dessins $\mathcal{D}$ of
type $(3,2,13)$ and degree $13$; they form two chiral pairs, each pair having
the same regular cover. Each dessin $\mathcal{D}$ (see Figure 12, or [1, Orbit
13.1, Fig. 8.22(A)]) has passport $(3^{4}1^{1},2^{4}1^{5},13^{1})$ and genus
$0$, while the two regular covers have genus $253$. (These are the duals of
the regular maps R253.1 and R253.2 of type $\\{3,13\\}_{8}$ and
$\\{3,13\\}_{12}$ in [16].)

Figure 12: Two dessins with monodromy group ${\rm PSL}_{3}(3)$.

###### Example 10.6

Let $p=31$, with $G={\rm PSL}_{3}(5)$. In this case ${\mathcal{X}}=3A$,
${\mathcal{Y}}=2A$ and there are ten possible classes
${\mathcal{Z}}=31A,31B,\ldots$. These give ten dessins $\mathcal{D}$ of type
$(3,2,31)$ and degree $31$, forming five chiral pairs with five regular
covers. Each dessin $\mathcal{D}$ has passport
$(3^{10}1^{1},2^{12}1^{7},31^{1})$ and genus $1$, while the regular covers
have genus $25\,001$. Generators of the monodromy groups of these ten dessins
$\mathcal{D}$, as well as the diagram of one of them, are given in Subsection
14.4.

###### Example 10.7

Let $p=73$ and $G={\rm PSL}_{3}(8)$. Here ${\mathcal{X}}=3A$ and
${\mathcal{Y}}=2A$, and there are $24$ classes ${\mathcal{Z}}=73A,\ldots$ of
elements of order $73$. Since $e=3$ we obtain $24/3=8$ dessins $\mathcal{D}$.
They have passport $(3^{24}1^{1},2^{32}1^{9},73^{1})$ and genus $4$; their
permutation representations are given in Subsection 14.7. They form four
chiral pairs, each pair having a regular cover of genus $1\,260\,673$.

Of course, we do not consider cases such as $q=4$, $7$, $9$ and $11$ since the
corresponding values of $1+q+q^{2}$ are not prime. Indeed, squares of primes
can be ignored, since for $p$ to be prime we require $e$ to be a power of
$n=3$. For any such $e$, the Bateman–Horn conjecture, applied to the
polynomials $t$ and $1+t^{e}+t^{2e}$, suggests that infinitely many projective
primes will be obtained. However, even for $e=3$ there are so few examples
within our computing range ($10$ primes $p=1+q+q^{2}<10^{18}$, compared with
an estimate of about $12$) that the computational evidence is not convincing.

###  10.3 Fixed $n>3$

In the case of a prime $n>3$ we do not have generic character tables for ${\rm
PSL}_{n}(q)$. Nevertheless, the Bateman–Horn Conjecture still applies, again
suggesting that if we fix $n$ and take $e$ to be a fixed power of $n$ we will
obtain infinitely many projective primes. The only case in which the
computational evidence is persuasive is when $n=5$ and $e=1$, with $252$
projective primes $p<10^{18}$, compared with an estimate of about $246$.

###  10.4 The case $q=2$: Mersenne primes

When $q=2$ we have $G={\rm PSL}_{n}(2)={\rm GL}_{n}(2)$, and $p=2^{n}-1$,
currently known to be prime for $51$ values of $n$. No generic character
tables for ${\rm GL}_{n}(2)$ are available, although there is a general theory
of characters for ${\rm GL}_{n}(q)$, largely based on the work of Green [34];
for a concise and readable summary, see [56, §4] or [59, Ch. IV]. Here are two
small examples where character tables in [17] can be used.

###### Example 10.8

Let $p=2^{3}-1=7$ and $G={\rm PSL}_{3}(2)\;(={\rm GL}_{3}(2)\cong{\rm
PSL}_{2}(7))$. The only conjugacy classes of elements of order $3$ or $2$ in
$G$ are $3A$ and $2A$, with cycle structures $3^{2}1^{1}$ and $2^{2}1^{3}$ in
both of the natural representations of $G$, on points and lines of the Fano
plane ${\mathbb{P}}^{2}({\mathbb{F}}_{2})$; there are two classes $7A,7B$ of
elements of order $p=7$. The only character which does not vanish in the
character sum is $\chi_{1}$, so $\Sigma=1$. Since $|C(x)|=3$,
$|C(y)|=8=2^{3}$, $|C(z)|=7$ and $|G|=168=2^{3}\cdot 3\cdot 7$, the number of
triples of type $(3,2,7)$ in $G$ is

$2\cdot\frac{2^{6}\cdot 3^{2}\cdot 7^{2}}{3\cdot 2^{3}\cdot 7}=2^{4}\cdot
3\cdot 7=2|G|=|{\rm Aut}\,G|.$

(Note that ${\rm Out}\,G\cong{\rm C}_{2}$, induced by the duality of the Fano
plane, or equivalently by conjugation of ${\rm PSL}_{2}(7)$ in ${\rm
PGL}_{2}(7)$.) These triples all generate $G$, so $\Delta=\Delta(3,2,7)$ has a
unique normal subgroup with quotient $G$, corresponding to a regular dessin
$\mathcal{R}$ of type $(3,2,7)$ with automorphism group $G$. There are two
conjugacy classes of subgroups of index $7$ in $G$, the stabilisers of points
and lines in the Fano plane; these lift to two conjugacy classes of subgroups
of index $7$ in $\Delta$, corresponding to two dessins $\mathcal{D}$ of type
$(3,2,7)$ and degree $7$, each having regular cover $\mathcal{R}$. Each dessin
$\mathcal{D}$ has passport $(3^{2}1^{1},2^{2}1^{3},7^{1})$, so it has genus
$0$, while $\mathcal{R}$ has genus $3$. One of the two dessins $\mathcal{D}$
is shown on the right in Figure 11, and the other is its mirror image
$\overline{\mathcal{D}}$. These two maps, corresponding to the two choices for
the conjugacy class $\mathcal{Z}$ or to the two choices for the class of
subgroups of index $7$, are not isomorphic since the outer automorphism of $G$
transposes both pairs. The dessin $\mathcal{R}$ is Klein’s heptagonal map on
his quartic curve, introduced in [50]; it is denoted by $\\{7,3\\}_{8}$ in
[20] and is the dual of R3.1 in [16]. As the common regular cover of a chiral
pair of dessins (or by its uniqueness) this map is isomorphic to its mirror
image.

###### Example 10.9

Let $p=31$ (again), but now in its role as a Mersenne prime $2^{5}-1$, with
$G={\rm PSL}_{5}(2)={\rm GL}_{5}(2)$. There are two classes each of elements
of orders $3$ and $2$, namely $3A$ and $3B$, with cycle structures
$3^{8}1^{7}$ and $3^{10}1^{1}$, and $2A$ and $2B$, with cycle structures
$2^{8}1^{15}$ and $2^{12}1^{7}$, together with six classes $31A,\ldots$ of
elements of order $31$. The inequality $2a+b\geq p-1$ discussed earlier
implies that $\mathcal{X}$ and $\mathcal{Y}$ must be $3B$ and $2B$. For each
of the six choices of $\mathcal{Z}$, only $\chi_{1}$ and $\chi_{6}$ appear in
the character sum, so

$\Sigma=1+\frac{8\cdot(-5)\cdot 1}{280}=1-\frac{1}{7}=\frac{6}{7}.$

We have $|C(x)|=180=2^{2}\cdot 3^{2}\cdot 5$, $|C(y)|=1536=2^{9}\cdot 3$,
$|C(z)|=31$ and $|G|=9\,999\,360=2^{10}\cdot 3^{2}\cdot 5\cdot 7\cdot 31$, so
the total number of triples in $G$ is

$6\cdot 2^{9}\cdot 3\cdot 5\cdot 7^{2}\cdot 31\cdot\frac{6}{7}=2^{11}\cdot
3^{3}\cdot 5\cdot 7\cdot 31=6|G|=3|{\rm Aut}\,G|.$

Each triple generates $G$, so $\Delta$ has three normal subgroups with
quotient $G$. There are two conjugacy classes of subgroups of index $31$ in
$G$, the stabilisers of points and hyperplanes in
${\mathbb{P}}^{4}({\mathbb{F}}_{2})$, giving six dessins $\mathcal{D}$ of
degree $31$, with three chiral pairs each having the same regular cover
$\mathcal{R}$. Each dessin $\mathcal{D}$ has passport
$(3^{10}1^{1},2^{12}1^{7},31^{1})$, so it has genus $1$, while the regular
covers $\mathcal{R}$ have genus $672\,001$. Generators of the monodromy groups
of these six dessins $\mathcal{D}$, as well as the diagram of one of them, are
given in Subsection 14.5.

######  Remark 10.10 (Goormaghtigh conjecture)

Note that $31$ appears twice as a projective prime, as both $1+5+5^{2}$ and
$1+2+2^{2}+2^{3}+2^{4}$. In 1917 Goormaghtigh [32] conjectured that this
example and $1+2+2^{2}+\cdots+2^{12}=8191=1+90+90^{2}$ are the only
repetitions among numbers of the form $(q^{n}-1)/(q-1)$ for integers $q,n\geq
2$. Although $8191$ is prime (and hence a projective prime for
$(q,n)=(2,13)$), $90$ is not a prime power, so only the first example is a
repetition of projective primes. A computer search of prime degrees up to
$10^{18}$ has revealed no further examples. The conjecture is still open. For
additional information see [33].

Before attempting to find triples $(x,y,z)$ of type $(3,2,p)$ in $G={\rm
GL}_{n}(2)$ for general primes $n\geq 3$, it is useful first to consider the
possibilities for elements of orders $3$, $2$ and $p$ in $G$. Any element
$x\in G$ of order $3$ has a rational canonical form with $r$ and $(n-r)/2$
blocks of the form

$\left(\,\begin{array}[]{c}1\\\ \end{array}\,\right)\quad{\rm
or}\quad\left(\,\begin{array}[]{cc}0&1\\\ 1&1\\\ \end{array}\right)$

for some odd $r<n$, so it fixes a subspace of
${\mathbb{P}}^{n-1}({\mathbb{F}}_{2})$ of dimension $r-1$; this contains
$\alpha=2^{r}-1$ points, so $x$ has $a=(p-2^{r}+1)/3$ nontrivial cycles.
Similarly, any element $y\in G$ of order $2$ has a Jordan normal form with $s$
and $(n-s)/2$ blocks of the form

$\left(\,\begin{array}[]{c}1\\\ \end{array}\,\right)\quad{\rm
or}\quad\left(\,\begin{array}[]{cc}1&1\\\ 0&1\\\ \end{array}\right)$

for some odd $s<n$; it therefore fixes a subspace of dimension $(n+s-2)/2$
containing $\beta=2^{(n+s)/2}-1$ points, so it has $b=(p-2^{(n+s)/2}+1)/2$
nontrivial cycles. The element $z$ is a $p$-cycle, so provided a triple
$(x,y,z)$ of type $(3,2,p)$ exists, it will generate $G$ by [7]; the resulting
dessin $\mathcal{D}$ will have passport
$(3^{a}1^{\alpha},2^{b}1^{\beta},p^{1})$ and genus $(2a+b+1-p)/2$, while its
minimal regular cover $\mathcal{R}$ will have genus

$\frac{p-6}{12p}|G|+1=\frac{p-6}{12p}\prod_{i=0}^{n-1}(2^{n}-2^{i})+1=(p-6)2^{(n^{2}-n-4)/2}\prod_{i=3}^{n-1}(2^{i}-1)+1,$

independent of $r$ and $s$. In particular, if we choose the most ‘active’
elements $x$ and $y$ by taking $r=s=1$ (as in the earlier examples with $n=3$
and $5$) we will have $\alpha=1$ and $\beta=2^{(n+1)/2}-1$, so $a=(p-1)/3$ and
$b=(p-2^{(n+1)/2}+1)/2$ and hence $\mathcal{D}$ will have genus

$\frac{1}{3}(2^{n-2}+1)-2^{(n-3)/2}.$ (7)

###### Theorem 10.11

For each Mersenne prime $p=2^{n}-1\geq 7$ there are at least $(p-1)/2n$ chiral
pairs of dessins of type $(3,2,p)$, degree $p$ and genus given by (7), with
monodromy group $G={\rm PSL}_{n}(2)={\rm GL}_{n}(2)$.

Proof. We will use the Frobenius formula (2) to show that for each conjugacy
class $\mathcal{Z}$ of Singer cycles in $G$ there a triple $(x,y,z)$ of type
$(3,2,p)$ where $x$ and $y$ satisfy $r=s=1$ (as above) and
$z\in{\mathcal{Z}}$. The calculation is aided by the fact that most of the
non-principal irreducible characters of $G$ vanish on the Singer cycles $z$
(see [56, §4.2], for example), and hence do not contribute to the character
sum $\Sigma$. Among the other characters, that of degree $p-1$, the non-
principal constituent of the natural permutation character, vanishes on $x$,
while the Steinberg character of degree $2^{n(n-1)/2}$ vanishes on $y$, so
they can also be ignored. The remaining characters have sufficiently large
degrees, compared with their values at $x,y$ and $z$ (see [56, §4.1,4.2] for
full details), that their total contribution to $\Sigma$ has modulus less than
the contribution $1$ of the principal character, so that $\Sigma>0$ and hence
the required triples exist. As before, they all generate $G$ by [7]. There are
$(p-1)/n$ conjugacy classes $\mathcal{Z}$ of Singer cycles $z$ in $G$, forming
$(p-1)/2n$ mutually inverse pairs; each pair are transposed by the outer
automorphism group of $G$, as are the two permutation representations of $G$
of degree $p$; thus we obtain (at least) $(p-1)/2n$ chiral pairs of dessins
$\mathcal{D}$, all with the same passport and genus, as calculated above.
$\square$

On the basis of Theorem 10.11, and of other examples involving small
projective primes, we conjecture the following extension of this theorem:

###### Conjecture 10.12

For each projective prime $p=1+q+\cdots+q^{n-1}>3$ there exists at least one
dessin of type $(3,2,p)$ and degree $p$ with monodromy group ${\rm
PSL}_{n}(q)$.

###  10.5 Summary

We can summarise the results of this rather long section as follows:

###### Theorem 10.13

Suppose that $p$ is a prime such that there is a dessin of type $(3,2,p)$ and
degree $p$ with monodromy group $G$. Then $p\geq 5$, and one of the following
holds:

  * (a)

$G={\rm A}_{p}$, acting naturally, with $p\neq 7$;

  * (b)

$p=11$ and $G={\rm PSL}_{2}(p)$, acting on the cosets of a subgroup isomorphic
to ${\rm A}_{5}$;

  * (c)

$p$ is a projective prime $(q^{n}-1)/(q-1)$ for some prime $n$ and prime-power
$q$, and $G={\rm PSL}_{n}(q)={\rm SL}_{n}(q)$ acting naturally on points or
hyperplanes of ${\mathbb{P}}^{n-1}(q)$.

In case (a), if $p=5$ there is a unique dessin, shown on the left in Figure
11; if $p>7$ there are exponentially many dessins as $p\to\infty$, even if we
restrict to planar dessins. In case (b) there is a single chiral pair of
dessins; these are the dessins ${\mathcal{M}}_{1}$ and
$\overline{\mathcal{M}}_{1}$ studied by Klein in [51] and shown in Figures 2
and 3. In case (c) we conjecture that there are infinitely many projective
primes $p$, even for any fixed prime $n\geq 3$; when $n=2$ (Fermat primes) or
$3$ (projective planes) the dessins are all as described in Theorem 10.1 or
Theorem 10.3. In the case $q=2$ (Mersenne primes) we have an existence result
rather than a complete classification.

##  11 Modular dessins and curves

Just as we generalised our arguments about dessins of type $(3,2,p)$ and
degree $p$ from $p=11$ to all primes $p$, we can do the same for dessins of
this type and of degree $p+1$. In this section we will consider the dessins
${\mathcal{D}}_{0}(p)$ arising naturally from congruence subgroups of level
$p$ in the modular group, and in the next section we will consider more
general dessins of this type and degree.

###  11.1 ${\mathcal{D}}_{0}(p)$ and its passport

As in the case $p=11$ considered earlier, for any prime $p$ the natural action
of ${\rm PSL}_{2}(p)$ on the projective line
${\mathbb{P}}^{1}({\mathbb{F}}_{p})$ yields a dessin ${\mathcal{D}}_{0}(p)$ of
type $(3,2,p)$ and degree $p+1$. One can find its passport by using the fact
that a non-identity Möbius transformation (over any field) has at most two
fixed points. The cases $p=2$ and $3$ are trivial, so let us assume that
$p\geq 5$.

The permutation $x$ representing the black vertices has order $3$, so

  * •

if $p\equiv 1$ mod $(3)$ then $x$ has two fixed points, and its cycle type is
$3^{(p-1)/3}1^{2}$;

  * •

if $p\equiv-1$ mod $(3)$ then $x$ has no fixed points, and its cycle type is
$3^{(p+1)/3}$.

Similarly the permutation $y$ representing the white vertices has order $2$,
and is even, so

  * •

if $p\equiv 1$ mod $(4)$ then $y$ has two fixed points, and its cycle
structure is $2^{(p-1)/2}1^{2}$;

  * •

if $p\equiv-1$ mod $(4)$ then $y$ has no fixed points, and its cycle structure
is $2^{(p+1)/2}$.

Finally the permutation $z$ representing the faces has order $p$, so its cycle
structure is $p^{1}1^{1}$.

From the passport one can compute the Euler characteristic $\chi$ and hence
the genus $g$ of ${\mathcal{D}}_{0}(p)$. There are four cases:

  * •

if $p\equiv 1$ mod $(12)$ the passport is
$(3^{(p-1)/3}1^{2},2^{(p-1)/2}1^{2},p^{1}1^{1})$, giving the genus
$g=(p-13)/12$;

  * •

if $p\equiv 5$ mod $(12)$ the passport is
$(3^{(p+1)/3},2^{(p-1)/2}1^{2},p^{1}1^{1})$, giving the genus $g=(p-5)/12$;

  * •

if $p\equiv 7$ mod $(12)$ the passport is
$(3^{(p-1)/3}1^{2},2^{(p+1)/2},p^{1}1^{1})$, giving the genus $g=(p-7)/12$;

  * •

if $p\equiv 11$ mod $(12)$ the passport is
$(3^{(p+1)/3},2^{(p+1)/2},p^{1}1^{1})$, giving the genus $g=(p+1)/12$;

Thus the genus of ${\mathcal{D}}_{0}(p)$ is $(p-c)/12$ for primes $p\equiv
c=-1,5,7$ or $13$ mod $(12)$, and $0$ for $p=2$ or $3$. Table 3 gives the
values of $g$ for the first few primes. Figure 13 shows the planar dessins
from this table.

$p$ | 2 | 3 | 5 | 7 | 11 | 13 | 17 | 19 | 23 | 29 | 31 | 37 | 41 | 43 | 47 | 53
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$g$ | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 2 | 2 | 2 | 2 | 3 | 3 | 4 | 4
Table 3: The genus of the dessin ${\mathcal{D}}_{0}(p)$. Figure 13:
${\mathcal{D}}_{0}(p)$ for $p=2,3,5,7$ and $13$; white vertices are omitted.

###  11.2 The dessin ${\mathcal{D}}_{0}(13)$

The dessin ${\mathcal{D}}_{0}(p)$ is planar if and only if $p=2,3,5,7$ or
$13$. These five dessins are shown in Figure 13. It is easy to see that the
first four are uniquely determined by their passports, and the same applies to
the torus dessin ${\mathcal{D}}_{0}(11)$ which we will show in Figure 17.
However, for $p=13$ there are 30 dessins with passport
$(3^{4}1^{2},2^{6}1^{2},p^{1}1^{1})$, consisting of two mirror symmetric
dessins, including ${\mathcal{D}}_{0}(13)$, and 14 chiral pairs. In fact, a
simple argument shows that these dessins can all be formed from the three
basic maps in Figure 14, together with the mirror images of the first and
last, by adding black vertices to two of their four free edges. This gives
$\displaystyle 5\times\left(\\!\\!\begin{array}[]{c}4\\\
2\end{array}\\!\\!\right)=30$ dessins.

Figure 14: Three spanning dessins out of which the dessins with the same
passport as that of ${\mathcal{D}}_{0}(13)$, are constructed. White vertices
are once again omitted.

Whereas ${\mathcal{D}}_{0}(13)$ has monodromy group ${\rm PSL}_{2}(13)$, the
other $29$ dessins all have monodromy group ${\rm A}_{14}$; this follows from
the fact that such a group must be a doubly transitive subgroup of ${\rm
A}_{14}$, and apart from ${\rm}A_{14}$ itself the only other possibility is
${\rm PSL}_{2}(13)$, which by the Frobenius formula corresponds to just one
dessin, namely ${\mathcal{D}}_{0}(13)$.

The fact that ${\mathcal{D}}_{0}(13)$ is planar gives us some hope of
computing its Belyĭ function $f:\Sigma\to\Sigma$, which is as follows:

$f=-\frac{1}{1728}\cdot\frac{(t^{4}-7t^{3}+20t^{2}-19t+1)^{3}(t^{2}-5t+13)}{t},$
$f-1=-\frac{1}{1728}\cdot\frac{(t^{6}-10t^{5}+46t^{4}-108t^{3}+122t^{2}-38t-1)^{2}(t^{2}-6t+13)}{t}.$

In this case, of course, $f$ is not a polynomial, since there is a pole at $0$
(the face-centre enclosed by the loop) in addition to the pole at $\infty$
(the centre of the outer face).

Figure 15: A geometrically correct drawing of ${\mathcal{D}}_{0}(13)$.

Figure 15 shows a geometrically correct version of ${\mathcal{D}}_{0}(13)$,
with black and white vertices. In this picture, the loop is so small that it
is obscured by its incident black and white vertices. (We will meet a similar
problem for ${\mathcal{D}}_{0}(11)$ in Figure 18.) Figure 16 shows a magnified
drawing of the loop, together with nearby parts of the real and imaginary
axes.

Figure 16: The loop in ${\mathcal{D}}_{0}(13)$. Its horizontal width is,
approximately, $0.08$, while the length of the horizontal segment attached to
it (which goes beyond the limits of the picture) is, approximately, $0.46$.

###  11.3 The dessin ${\mathcal{D}}_{0}(11)$

Whereas ${\mathcal{D}}_{0}(p)$ is planar for each prime $p\leq 13$ except
$11$, the dessin ${\mathcal{D}}_{0}(11)$ has genus $1$, so we have the
additional problem of determining the corresponding elliptic curve. Two views
of this dessin are shown in Figure 17 as maps on a torus, with opposite sides
of the outer square or hexagon identified in the usual way. Like the dessins
${\mathcal{M}}_{1}$ and $\overline{\mathcal{M}}_{1}$, ${\mathcal{D}}_{0}(11)$
has the dessin $\mathcal{R}_{1}$ as its minimal regular cover. However, in
this case it is the quotient of $\mathcal{R}_{1}$ by the subgroup of $L\,={\rm
PSL}_{2}(11)$ fixing a point in ${\mathbb{P}}^{1}({\mathbb{F}}_{11})$, of the
form ${\rm C}_{11}\rtimes{\rm C}_{5}$ rather than ${\rm A}_{5}$. In addition
to coming from a more natural representation of $L$, ${\mathcal{D}}_{0}(11)$
is also more pleasing because of its greater uniformity: the permutations $x$
and $y$ have no fixed points, so the black vertices all have valency $3$ and
there are no free edges. Moreover, the generators $y:t\mapsto-1/t$ and
$z:t\mapsto t+1$ of $L$ can both be seen from the edges and faces through the
labelling of directed edges with elements of
${\mathbb{P}}^{1}({\mathbb{F}}_{11})$. Since $xyz=1$, the permutation $x$ can
now easily be computed as $x:t\mapsto-1/(t-1)$.

Figure 17: Two views of the torus map ${\mathcal{D}}_{0}(11)$. The labels of
the half-edges are the elements of
$\mathbb{P}^{1}(\mathbb{F}_{11})=\mathbb{F}_{11}\cup\\{\infty\\}$. One may see
that if the label of a half-edge is $t$ than the label of the opposite half-
edge of the same edge is $-1/t$, while the cycle going around the outer face
is $(0,1,2,\ldots,10)$.

Another remarkable feature of this dessin is the fact that it is unique for
its passport, which is $(3^{4},2^{6},11^{1}1^{1})$. One does not have to
consider ten different trees and sort out the corresponding groups. We may
wonder why Klein overlooked this example. Maybe the reason was that the
representation of maps via permutations was not yet invented. The first
publications which introduced this construction, by Dyck [24] (1888) and
Heffter [38] (1891), appeared only a decade after Klein’s paper. In fact, the
pioneer in this subject was Hamilton [37] (1856), but he explained the
construction in a private letter. Even at the time close to ours, when the
permutational model of maps became a part of a common knowledge, the
researchers studied mainly the automorphism groups of maps while their
monodromy groups were largely neglected. Only with the advent of the theory of
dessins d’enfants the monodromy groups rose to the fore since they are Galois
invariants.

From the point of view of dessins d’enfants, the uniqueness of the dessin
${\mathcal{D}}_{0}(11)$ implies that it is defined over $\mathbb{Q}$. However,
a difficulty of a different sort appears. While in the planar case what we
need is a Belyĭ function, which is a rational function $f:\Sigma\to\Sigma$, in
the case of a dessin of genus $g\geq 1$ we need a Belyĭ pair $(X,f)$ where $X$
is a Riemann surface (or an algebraic curve) of genus $g$, and $f$ is a
meromorphic function $f:X\to\Sigma$ with no critical values outside
$\\{0,1,\infty\\}$. There are uncountably many curves of any given genus
$g\geq 1$, and the dessin determines one of them in a unique way. To find such
a curve (and a Belyĭ function defined on it) is an incredibly difficult task.
A relatively simple answer below hides numerous obstacles one has to surmount
in order to find it. The corresponding computation was made for us by John
Voight, to whom we are greatly indebted.

In our case $g=1$, so the curve in question is an elliptic curve $E$. One
possible presentation is

$\displaystyle y^{2}+y=x^{3}-x^{2}-10x-20.$ (8)

The Belyĭ function $f=f(x,y)$ is then constructed as follows:

$\displaystyle a$ $\displaystyle=$
$\displaystyle-x^{5}-23x^{4}+697x^{3}-1031x^{2}-2170x-353;$ $\displaystyle b$
$\displaystyle=$
$\displaystyle-11x^{6}+148x^{5}+643x^{4}-2704x^{3}-6780x^{2}+1781x+3308;$
$\displaystyle f$ $\displaystyle=$ $\displaystyle\frac{ay+b}{1728(x-16)}.$ (9)

(Alternatively, putting $y=z-\frac{1}{2}$, $x=t+\frac{1}{3}$ gives

$z^{2}=t^{3}-\frac{31}{3}t-\frac{2501}{108},$

a Weierstrass form for $E$. The $J$-invariant of the curve is $\displaystyle
J(E)=-\frac{2^{12}31^{3}}{11^{5}}$.)

Even a verification of the above result is not an easy task. We give an
outline of the main steps of such a verification.

Step 0. Since the map ${\mathcal{D}}_{0}(11)$ has 12 directed edges, we expect
its Belyĭ function to have degree 12. However, when one looks at the
expression (9), this property does not jump to the eyes, to put it mildly. We
have an impression that this function is rather of degree 6. The clue is that
it is considered to be defined on the curve $E$. Let us denote
$h(x,y)=y^{2}+y-x^{3}+x^{2}+10x+20$ (see equation (8)). Expressing $y$ from
the equation $f(x,y)=z$ and substituting the result in $h$ we indeed get a
function of degree 12 in $x$ since $h$ contains $y^{2}$.

Step 1. In order to find critical points on the curve we must ensure that the
gradients of the functions $f$ and $h$ are collinear. Hence the following
determinant must be equal to zero:

$\left|\begin{array}[]{cc}{\displaystyle\frac{\partial f}{\partial
x}}&{\displaystyle\frac{\partial f}{\partial y}}\\\ &\\\
{\displaystyle\frac{\partial h}{\partial x}}&{\displaystyle\frac{\partial
h}{\partial y}}\end{array}\right|\,=\,0.$

This gives us an equation in two variables $x$ and $y$. The second equation is
$h(x,y)=0$ since we are looking not for arbitrary points of collineation but
for those which belong to $E$.

Step 2. A good idea is to find a Gröbner basis for the system thus obtained,
and to do it twice. Using the ‘pure lexicographic’ monomial order plex(y,x) we
get as the first element of the basis a polynomial in $x$; using plex(x,y) we
get a polynomial in $y$. Here they are:

$\displaystyle P$ $\displaystyle=$
$\displaystyle(x^{4}-20x^{3}+62x^{2}+116x+97)^{2}(x-16)^{2}\times$
$\displaystyle(x^{6}-30x^{5}+243x^{4}-256x^{3}-1053x^{2}+654x+7793),$
$\displaystyle Q$ $\displaystyle=$
$\displaystyle(y^{4}+68y^{3}+590y^{2}+556y+7489)^{2}(y-60)^{2}\times$
$\displaystyle(y^{6}+102y^{5}+2619y^{4}+11272y^{3}+131211y^{2}+91074y+253312).$

Step 3. The roots of these two polynomials are the $x$-coordinates and the
$y$-coordinates of the critical points of $f$ on $E$. Two factors of degree 4,
being squared, indicate that there are four critical points of multiplicity 3.
We guess that they are the black vertices of our map. But is it true that the
value of $f$ at all these points is equal to zero? We must also ‘couple’ $x$\-
and $y$-coordinates in order to show which $y$ correspond to which $x$.

These questions are easy to answer. We solve the system $f=0$, $h=0$ and see
that the values of $x$ are indeed roots of
$p_{1}=x^{4}-20x^{3}+62x^{2}+116x+97$, while

$y=-\frac{1}{23}(x^{3}-16x^{2}+90x+62).$

It is also easy to verify that these values of $y$ are indeed roots of the
polynomial $q_{1}=y^{4}+68y^{3}+590y^{2}+556y+7489$. We just substitute these
expressions of $y$ into $q_{1}$ and compute the result modulo $p_{1}$: it is
equal to zero.

Beside these four solutions there is one more: $(x,y)=(16,60)$. What does it
mean? There are two points on $E$ over $x=16$, namely $(16,60)$ and
$(16,-61)$. Substituting $y=60$ into the numerator of $f$ we get

$-(x-16)(11x^{5}+88x^{4}+2145x^{3}-4796x^{2}-8096x-1117).$

Thus, $x-16$ in the numerator and in the denominator of $f$ cancel each other:
this is a removable singularity. We conclude that $f$ has a simple pole at the
point $(16,-61)$ on the curve.

Now, the factors of degree 6 of the polynomials $P$ and $Q$ correspond to
simple critical points, which are mid-points of edges. And, indeed, solving
the system $f=1$, $h=0$, we obtain, beside the superfluous solution $(16,60)$,
the roots of $p_{2}=x^{6}-30x^{5}+243x^{4}-256x^{3}-1053x^{2}+654x+7793$,
while

$y=-\frac{1}{101617}(104x^{5}-2811x^{4}+15943x^{3}+64714x^{2}-44258x-217861).$

Once again, it is easy to verify that these values of $y$ are roots of the
polynomial

$q_{2}=y^{6}+102y^{5}+2619y^{4}+11272y^{3}+131211y^{2}+91074y+253312.$

Step 4. Since the function $f$ has only one simple pole on the finite part of
the curve $E$, we may conclude that it also has a pole of multiplicity 11
‘over infinity’. We might stop here. However, we prefer to make the last
statement more explicit.

First of all, let us introduce the projective coordinates
$(X\\!:\\!Y\\!:\\!Z)$, where $x=X/Z$, $y=Y/Z$, and projectivize the equation
(8) of the curve:

$\displaystyle Y^{2}Z+YZ^{2}=X^{3}-X^{2}Z-10XZ^{2}-20Z^{3}.$ (10)

The finite part of the curve corresponds to $Z=1$; the point ‘at infinity’
corresponds to $Z=0$, which implies also $X=0$; thus, we must take $Y=1$. Near
this point, we may introduce new affine coordinates $u=X/Y=x/y$, $v=Z/Y=1/y$.
Substituting $x=u/v$, $y=1/v$ in the initial equation we get the equation of
our curve near infinity:

$\displaystyle v+v^{2}=u^{3}-u^{2}v-10uv^{2}-20v^{3}.$ (11)

Notice that the presence of the term $v$ at the left-hand side of this
equation ensures that the gradient is non-zero and therefore the curve in the
vicinity of the point $(u,v)=(0,0)$ is smooth, and $u$ can be taken as a local
coordinate.

Let us represent $v$ as a series
$v=c_{0}+c_{1}u+c_{2}u^{2}+c_{3}u^{3}+c_{4}u^{4}+\ldots$ and insert it into
the difference

$(v+v^{2})-(u^{3}-u^{2}v-10uv^{2}-20v^{3}).$

We would like to get as the result $o(u^{3})$. Then, equating to zero the
coefficients in front of the degrees up to the 3rd, we get
$c_{0}=c_{1}=c_{2}=0$, $c_{3}=1$. What remains is to make the following two
operations:

  * 1.

Substitute $x=u/v$, $y=1/v$ in the expression (9) for the Belyĭ function.

  * 2.

Substitute $v=u^{3}$ in the resulting expression.

Then the Belyĭ function, up to smaller terms, becomes

$f\,\sim\,-\frac{1}{1728}\frac{1+11u+23u^{2}-148u^{3}-\ldots-3308u^{13}}{u^{11}(1-16u^{2})}\,\sim\,\frac{\rm
Const}{u^{11}}\quad\mbox{when}\quad u\to 0.$

Now a pole of degree 11 at the point $u=0$ is apparent.

Figure 18: A geometrically correct drawing of ${\mathcal{D}}_{0}(11)$ lifted
to the complex plane.

Figure 18 shows a geometrically correct picture of the dessin
${\mathcal{D}}_{0}(11)$, lifted from the torus $E$ to the complex plane to
give a doubly periodic map. In addition to the black vertices of degree $3$,
the white vertices of degree $2$ are also shown, as is a fundamental
parallelogram for the associated lattice. The repeated copies of
${\mathcal{D}}_{0}(11)$ are aligned as in the right-hand map in Figure 17.
However, the loop (the face of valency $1$), which is obvious there, is in
reality too small to be clearly visible here. Changing accuracy parameters to
make this face and its incident vertices visible would create a mess
elsewhere, so in this case we have to accept that nature, technology and human
eyesight are incompatible. We met a similar problem earlier for
${\mathcal{D}}_{0}(13)$, in connection with Figures 15 and 16.

To obtain this diagram we use the substitution

$y=\frac{z-1}{2},\quad x=t+\frac{1}{3}$

to give the elliptic curve

$z^{2}=4t^{3}-\frac{124}{3}t-\frac{2501}{27}$

in Weierstrass normal form. Then Algorithm 7.4.7 in Cohen’s book [14], using
the arithmetic-geometric mean, gives a basis

$\omega_{1}=1.269209304,\quad\omega_{2}=-0.6346046520+1.458816617i$

for the lattice, as shown in Figure 18. The associated modulus is

$\tau=\omega_{2}/\omega_{1}=-0.5000000000+1.149390106i,$

on the left-hand vertical border of the standard fundamental region for
$\Gamma$, which is consistent with the fact that $J(\tau)<0$. To obtain the
embedded graph we lift the unit interval $[0,1]\subset\Sigma$, with black and
white vertices at $0$ and $1$, to the curve $E$ via the Belyĭ function $f$,
and then use the standard parametrisation of $E$ by the Weierstrass functions,
namely $t=\wp(w)$ and $z=\wp^{\prime}(w)$ for $w\in\mathbb{C}$, to lift the
resulting dessin to the plane.

###  11.4 The dessins ${\mathcal{D}}(p)$

The minimal regular cover of ${\mathcal{D}}_{0}(p)$ is a regular dessin
${\mathcal{D}}(p)$ of type $(3,2,p)$ with automorphism group ${\rm
PSL}_{2}(p)$; for $p>2$ it has degree $p(p^{2}-1)/2$ and genus
$(p+2)(p-3)(p-5)/24$, while for $p=2$ it has degree $6$ and genus $0$. The
underlying curves of ${\mathcal{D}}(p)$ and ${\mathcal{D}}_{0}(p)$ are the
modular curves $X(p)=\overline{{\mathbb{H}}/\Gamma(p)}$ and
$X_{0}(p)=\overline{{\mathbb{H}}/\Gamma_{0}(p)}$ (hence the notation for these
two dessins); here $\Gamma(p)$ is the principal congruence subgroup of level
$p$ in $\Gamma$, the kernel of the reduction mod $(p):\Gamma\to{\rm
PSL}_{2}(p)$ (see [42, §6.9–10], for example), while $\Gamma_{0}(p)$ is the
inverse image in $\Gamma$ of the subgroup of ${\rm PSL}_{2}(p)$ fixing
$\infty$.

###### Theorem 11.1

For each prime $p$ the dessin ${\mathcal{D}}(p)$ is the only regular dessin of
type $(3,2,p)$ with automorphism group $G={\rm PSL}_{2}(p)$, and
${\mathcal{D}}_{0}(p)$ is the only dessin of degree $p+1$ and type $(3,2,p)$
with monodromy group $G$.

To prove the uniqueness of ${\mathcal{D}}(p)$ it is sufficient to show that
$G$ has, up to automorphisms, only one generating triple $(x,y,z)$ of type
$(3,2,p)$. The elements $z\in G$ of order $p$ are all equivalent under
automorphisms, while the elements of order $3$ and $2$ are those with traces
$\pm 1$ and $0$. We may therefore assume that

$y=\pm\left(\,\begin{array}[]{cc}a&b\\\ c&-a\\\
\end{array}\right)\quad\hbox{and}\quad z=\pm\left(\,\begin{array}[]{cc}1&0\\\
1&1\\\ \end{array}\right),\quad\hbox{so that}\quad
yz=\pm\left(\,\begin{array}[]{cc}a+b&b\\\ c-a&-a\\\ \end{array}\right).$

Since we require $yz=x^{-1}$ to have order $3$ we must have $b=\pm 1$, and
without loss of generality we can take $b=1$. Since $\det y=1$ we have
$c=-a^{2}-1$, giving a $1$-parameter set of solutions

$y=y_{a}=\pm\left(\,\begin{array}[]{cc}a&1\\\ -a^{2}-1&-a\\\
\end{array}\right)\quad(a\in{\mathbb{F}}_{p}).$

Now conjugating the triple $(x,y,z)$ with $z^{i}$ fixes $z$ and replaces
$y=y_{a}$ with $y_{a+i}$, so all such triples are equivalent under
automorphisms, as required. One could also prove this by using the Frobenius
triple-counting formula (2), as in Example 4.1 for $p=11$ (see [44, §5.5] for
the character table of $G$); this is easy if $p\equiv\pm 5$ mod $(12)$ since
in this case only one nonprincipal irreducible character, of degree $p$,
appears in the character sum, but if $p\equiv\pm 1$ mod $(12)$ then other
characters appear and some work with algebraic numbers is required. Since $G$
has a unique conjugacy class of subgroups of index $p+1$, namely the
stabilisers of points in ${\mathbb{P}}^{1}({\mathbb{F}}_{p})$, it follows that
${\mathcal{D}}_{0}(p)$ is the only dessin of degree $p+1$ and type $(3,2,p)$
with monodromy group $G$.

##  12 Dessins of type $(3,2,p)$ and degree $p+1$

Having considered the dessins ${\mathcal{D}}_{0}(p)$ in the preceding section,
we will now consider arbitrary dessins of their type $(3,2,p)$ and degree
$p+1$. The monodromy group $G$ of such a dessin $\mathcal{D}$ is a transitive
permutation group of degree $p+1$, and since it contains a $p$-cycle $z$ it is
doubly transitive, and hence primitive. As before, if $p>3$, as we shall
assume, $G$ is perfect. The finite primitive groups containing a cycle with
one fixed point were determined by Müller in [65, Theorem 6.2], and as a
corollary we have:

###### Theorem 12.1

The perfect permutation groups of degree $p+1$ ($p$ prime), containing a
$p$-cycle, are the following:

  * (a)

${\rm A}_{p+1}$ for primes $p\geq 5$,

  * (b)

${\rm PSL}_{2}(p)$ for primes $p\geq 5$,

  * (c)

affine groups ${\rm AGL}_{n}(2)$ for Mersenne primes $p=2^{n}-1\geq 5$,

  * (d)

Mathieu groups ${\rm M}_{11}$ and ${\rm M}_{12}$ for $p=11$ and ${\rm M}_{24}$
for $p=23$,

all except ${\rm M}_{11}$ (see below) in their natural representations.

We will consider these groups in turn, together with their analogues for
$p<5$, to see which of them can serve as monodromy groups of dessins of type
$(3,2,p)$. The results will be summarized in Theorem 12.5. It is convenient to
deal with the simplest cases (b) and (d) first, and to treat the alternating
groups last.

The groups ${\rm PSL}_{2}(p)$ in Theorem 12.1(b) arise as monodromy groups in
this context for all primes $p$, each of them associated with a unique dessin,
the modular dessin ${\mathcal{D}}_{0}(p)$ (see Theorem 11.1).

In (d), although ${\rm M}_{11}$ has a permutation representation of degree
$12$, on the cosets of a subgroup ${\rm PSL}_{2}(11)$, we have seen in Example
4.2 that it is not a quotient of $\Delta(3,2,11)$, since all triples of this
type in ${\rm M}_{11}$ generate not ${\rm M}_{11}$ but its subgroup ${\rm
PSL}_{2}(11)$; thus it does not arise as a monodromy group in this context.

The group $G={\rm M}_{12}$ in (d) has two representations of degree $12$, on
the cosets of two conjugacy classes of subgroups isomorphic to ${\rm M}_{11}$,
transposed by ${\rm Out}\,G$. It is the monodromy group of two chiral pairs of
planar dessins of type $(3,2,11)$ and degree $12$. One dessin from each pair
is shown in Figure 19: the dessin on the left has passport
$(3^{3}1^{3},2^{6},11^{1}1^{1})$, with ${\mathcal{X}}=3A$ and
${\mathcal{Y}}=2A$, while that on the right has passport
$(3^{4},2^{4}1^{4},11^{1}1^{1})$, with ${\mathcal{X}}=3B$ and
${\mathcal{Y}}=2B$. Their regular covers are a pair of dessins of type
$(3,2,11)$ and genus $3601$. There are no other dessins of this type and
degree with monodromy group $G={\rm M}_{12}$: taking ${\mathcal{X}}=3B$ and
${\mathcal{Y}}=2A$, or ${\mathcal{X}}=3A$ and ${\mathcal{Y}}=2B$ yields two
more orbits of ${\rm Aut}\,G$ on triples, but these generate transitive or
intransitive subgroups isomorphic to ${\rm PSL}_{2}(11)$. The dessin
corresponding to these transitive subgroups has already been shown in Figures
17 and 18, while the intransitive subgroups, with orbits of length $11$ and
$1$, are the monodromy groups of the dessins ${\mathcal{M}}_{1}$ and
$\overline{\mathcal{M}}_{1}$ of degree $11$ in Figures 1, 2 and 3.

Figure 19: Two dessins with monodromy group ${\rm M}_{12}$. The field of
moduli for both of them is $\mathbb{Q}(\\!\sqrt{-11})$.

For $p=23$, the group ${\rm M}_{24}$ in (d), found by Mathieu [63] in 1873, is
the monodromy group of two chiral pairs of planar dessins, with passports
$(3^{6}1^{6},2^{12},23^{1}1^{1})$ and $(3^{8},2^{8}1^{8},23^{1}1^{1})$: see
orbits 24.2 and 24.1 in [1, pp. 129–130]. For the first pair $\mathcal{X}=3A$,
$\mathcal{Y}=2B$ and $\mathcal{Z}=23A$ or $23B$, while for the second pair
$\mathcal{X}=3B$, $\mathcal{Y}=2A$ and $\mathcal{Z}=23A$ or $23B$. A member of
each pair is shown in Figure 20. Unlike in the case of ${\rm M}_{12}$, there
is a single permutation representation of degree $p+1$ (the natural
representation, on the cosets of a subgroup ${\rm M}_{23}$), and there is no
outer automorphism transposing the two mutually inverse conjugacy classes of
elements of order $p$.

Figure 20: Two dessins with monodromy group ${\rm M}_{24}$. The field of
moduli for both of them is $\mathbb{Q}(\\!\sqrt{-23})$.

We now consider case (c) of Theorem 12.1, the affine groups ${\rm
AGL}_{n}(2)$, where $n$ is prime and $p$ is a Mersenne prime $2^{n}-1$. The
case $p=3$ does not arise since ${\rm AGL}_{2}(2)\cong{\rm S}_{4}$, so this
group is not a quotient of $\Delta(3,2,2)\cong{\rm S}_{3}$. For $p=7$ the only
dessin of type $(3,2,7)$ and degree $8$ is ${\mathcal{D}}_{0}(7)$, with
monodromy group ${\rm PSL}_{2}(7)$, so ${\rm AGL}_{3}(2)$ does not arise.
(Note that the Hurwitz group of genus $17$ is a nonsplit extension of ${\rm
V}_{8}$ by ${\rm GL}_{3}(2)$, not isomorphic to ${\rm AGL}_{3}(2)$, and it has
no transitive permutation representation of degree $8$; see [45] for details.)
On the other hand, for $p=31$ orbit 32.1 in [1, pp. 134–135] consists of six
planar dessins of type $(3,2,31)$ and degree $32$ with monodromy group ${\rm
ASL}_{5}(2)={\rm AGL}_{5}(2)$. These are shown in Section 14, Figure 28.

###### Theorem 12.2

For each Mersenne prime $p=2^{n}-1>7$ there are dessins of type $(3,2,p)$ and
degree $p+1$ with passport $(3^{(p-1)/3}1^{2},2^{(p+1)/2},p^{1}1^{1})$, genus
$(p+5)/12$ and monodromy group ${\rm AGL}_{n}(2)$.

Proof. We will use the fact that $G:={\rm AGL}_{n}(2)$ is a semidirect product
of its translation subgroup $T\cong({\rm C}_{2})^{n}$ by $G_{0}:={\rm
GL}_{n}(2)$, the stabiliser of $0$. Let $(x,y,z)$ be a generating triple for
${\rm GL}_{n}(2)$ of type $(3,2,p)$, chosen as in the proof of Theorem 10.11
by taking $r=s=1$, so that $\alpha=1$ and $\beta=(n+1)/2$. Now let $u=xa$,
$v=yb$ and $w=zc$, where $a,b,c\in T$. We will try to choose $a,b$ and $c$ so
that $(u,v,w)$ is a generating triple for $G$ of type $(3,2,p)$.

Since $x^{3}=1$ we have

$u^{3}=(xa)^{3}=x^{3}a^{x^{2}}a^{x}a=a^{x^{2}+x+1},$

so $u^{3}=1$ if and only if $a\in T_{x}:=\ker(x^{2}+x+1)$. (Here we regard
$x^{2}+x+1$ as a linear transformation of the vector space $T$ over
${\mathbb{F}}_{2}$.) Similarly $v^{2}=1$ if and only if $b\in
T_{y}:=\ker(y+1)$, and $w^{p}=1$ if and only if $c\in
T_{z}:=\ker(z^{p-1}+z^{p-2}+\cdots+z+1)$. Also

$uvw=xa.yb.zc=xya^{y}zb^{z}c=xyza^{yz}b^{z}c=a^{yz}b^{z}c,$

so $uvw=1$ if and only if $a^{yz}b^{z}c=1$.

Since $z$ is a Singer cycle in $G_{0}$, $T_{z}=T$ and hence the condition on
$c$ is vacuous, whereas the Jordan forms for $x$ and $y$ used in the proof of
Theorem 10.11 show that the subspaces $T_{x}$ and $T_{y}$ have dimensions
$n-1$ and $(n+1)/2$. Within these subspaces one can choose $a$ and $b$
arbitrarily, and then define $c=a^{yz}b^{z}$, so that $a^{vyz}b^{z}c=1$ and
hence $(u,v,w)$ is a triple of type $(3,2,p)$.

The cycle structures of $u$, $v$ and $w$ can be found as follows. One can
regard ${\rm AGL}_{n}(2)$ as acting naturally on $T$, with elements of $T$ and
${\rm GL}_{n}(2)$ acting by translations and by linear transformations. The
element $u=xa$ acts on $T$ by $t\mapsto t^{x}a$, so it fixes an element $t\in
T$ if and only if $a=t(1-x)$, using additive notation for the vector space
$T$. Now the Jordan form for $x$ shows that $T(1-x)=T_{x}$, so for every
choice of $a\in T_{x}$ there are $|\ker(1-x)|=2$ fixed points $t\in T$ for
$u$. Since $u$ has order $3$ it has $(p-1)/3$ cycles of length $3$. In the
case of $v$ the Jordan form for $y$ shows that $T(1-y)$ is a subspace of
codimension $1$ in $T_{y}$, so by choosing $b\in T_{y}\setminus T(1-y)$ we can
ensure that $v$ has no fixed points, and therefore consists of $(p+1)/2$
transpositions. Since $w$ has order $p$ it has cycle structure $p^{1}1^{1}$.

Now let us define $H:=\langle u,v,w\rangle\leq G$. We will prove that if $p>7$
then $H=G$. Factoring out $T$ maps $H$ onto $\langle x,y,z\rangle=G_{0}$, so
$HT=G$. If $H$ contains a non-identity translation $t\in T$ then since $H$
contains $w$ it contains all the non-identity translations $t^{w}$, so $H\geq
T$ and hence $H=G$, as required.

We may therefore assume, for a contradiction, that $H\cap T=1$, so that $H$ is
a complement for $T$ in $G$. Since the first cohomology group $H^{1}({\rm
GL}_{n}(2),T)$ for ${\rm GL}_{n}(2)$ on its natural module $T$ is trivial for
all $n>3$ (see [4]) there is a single conjugacy class of such complements, so
$H$ is a point-stabiliser in the natural action of $G$. This contradicts the
fact that $v$ has no fixed points, by our choice of the translation $b$.

Thus $(u,v,w)$ is a generating triple for $G$, so it corresponds to a dessin
of type $(3,2,p)$ and degree $p+1$ with monodromy group $G$. Its genus follows
from its passport, determined above. $\square$

The above proof fails in the case $p=7$ since $|H^{1}({\rm GL}_{3}(2),T)|=2$
(see [4]), so that there are two conjugacy classes of complements for $T$ in
${\rm AGL}_{3}(2)$, consisting of the point-stabilisers and of subgroups
acting transitively on $T$ as ${\rm PSL}_{2}(7)$; in this case every triple of
type $(3,2,7)$ generates a complement. (The fact that all complements are
isomorphic to $G/T$ provides a simple proof that ${\rm GL}_{3}(2)\cong{\rm
PSL}_{2}(7)$.)

###### Example 12.3

Taking $n=5$ we obtain a dessin of type $(3,2,31)$, degree $32$ and genus $3$,
with passport $(3^{10}1^{2},2^{16},31^{1}1^{1})$. This should not be confused
with the Galois orbit 32.1 of six planar dessins with the same type, degree
and monodromy group in [1]: these have passport
$(3^{10}1^{2},2^{12}1^{8},31^{1}1^{1})$, and they correspond to taking $b\in
T(1-y)$, so that $v$ fixes $|\ker(1-y)|=8$ points (see Subsection 14.6).

Finally, we consider the alternating groups ${\rm A}_{p+1}$, in case (a) of
Theorem 12.1, as monodromy groups.

###### Theorem 12.4

The alternating group ${\rm A}_{p+1}$, $p$ prime, is the monodromy group of a
dessin of type $(3,2,p)$ and degree $p+1$ if and only if $p=3$ or $p\geq 11$.

Proof. For $p=2,5$ or $7$ the triangle group $\Delta(3,2,p)$ does not map onto
${\rm A}_{p+1}$: this is obvious if $p=2$ or $5$, and for $p=7$ it is well
known that ${\rm A}_{8}$ is not a Hurwitz group: the triples of type $(3,2,7)$
in this group all generate proper subgroups isomorphic to ${\rm PSL}_{2}(7)$.
On the other hand, ${\rm A}_{4}$ is the monodromy group of the dessin of type
$(3,2,3)$ and degree $4$ on the left in Figure 21, the quotient of the
tetrahedron $\\{3,3\\}$ by ${\rm C}_{3}$. (Note that ${\rm A}_{4}$ is