Construction of chiral 4-polytopes with alternating or symmetric automorphism group

Abstract

In this paper we describe a construction for finite abstract chiral 4-polytopes with Schläfli type $\{3,3,k\}$ (with tetrahedral facets) and with an alternating or symmetric group as automorphism group. We use it to prove that for all but finitely many $n$, both $A_n$ and $S_n$ are the automorphism groups of such a polytope. We also show that the vertex-figures of the polytopes obtained from our construction are chiral.

1 Introduction

Abstract polytopes generalise the classical notion of convex geometric polytopes to more general structures. Highly symmetric examples include not only classical regular polytopes such as the Platonic solids and more exotic structures such as the $120$-cell and $600$-cell, but also non-degenerate regular maps on surfaces (such as Klein’s quartic, of genus $3$).

Roughly speaking, an abstract polytope $\mathcal{P}$ is a partially ordered set endowed with a rank function, satisfying certain conditions that arise naturally from a geometric setting. Such objects were proposed by Grünbaum in the 1970s, and their definition (initially as ‘incidence polytopes’) and theory were developed by Danzer and Schulte.

Every automorphism of an abstract polytope is uniquely determined by its effect on any flag, which is a maximal chain in $\mathcal{P}$ (when this is regarded as a poset). The most symmetric examples are regular, with all flags lying in a single orbit, and a comprehensive description of these is given in a book on the subject by McMullen and Schulte [6]. These objects are also known as ‘thin residually connected geometries with a linear diagram’.

An interesting class of examples which are not quite regular are the chiral polytopes, for which the automorphism group has two orbits on flags, with any two flags that differ in a single element lying in different orbits. The study of chiral abstract polytopes was pioneered by Schulte and Weiss (see [10, 11] for example). Chiral polytopes of rank $3$ are essentially the same as chiral maps on surfaces, with some modest extra geometric conditions.

For quite some time, the only known finite examples of chiral polytopes had ranks $3$ and $4$, but then some finite examples of rank $5$ were constructed by Conder et al. [3], and now quite a few such examples are known. Many small examples of regular or chiral polytopes have been assembled in collections, as in [4, 5], for example.

In early 2009, the first author and Alice Devillers devised a construction for chiral polytopes whose facets are simplices, and used this to construct examples of finite chiral polytopes of ranks $6, 7$ and $8$ (unpublished). At about the same time, the fourth author of this paper devised a quite different method for constructing finite chiral polytopes with given regular facets, and used this construction to prove the existence of finite chiral polytopes of every rank $d \ge 3$; see [7]. The latter polytopes are enormous, however, and not easy to describe. It is still an open problem to find alternative constructions for families of chiral polytopes of relatively small order, or which have more easily described automorphism groups. A large number of other open questions about chiral polytopes are given by the fourth author in [8].

In this paper, we make a contribution towards producing infinite families of chiral polytopes with well-known groups. Specifically, we describe a construction for chiral $4$-polytopes of type $\{3,3,k\}$, with tetrahedral facets, using a way of combining together permutation representations of the tetrahedral group $A_4$ into the automorphism group.

Our main result is the following:

Theorem 1.1

For all but finitely many positive integers $n$, both $A_n$ and $S_n$ are the automorphism groups of chiral $4$-polytopes of type $\{3,3,k\}$ for some $k$.

In fact our construction proves this theorem for all $n \ge 50$, but thanks to an easy computation with Magma [1], we know it is also true for $20 \le n \le 49$ and hence for all $n \ge 20$. In addition, we know that the only smaller values of $n$ for which $A_n$ is the automorphism group of such a chiral $4$-polytope are $9, 13, 14, 15, 17$ and $18$, while the only such values of $n$ for $S_n$ are $12, 16, 17, 18$ and $19$. Examples of generating permutations for $A_n$ and $S_n$ in the cases not covered by our construction are given in [2].

In a planned sequel, we will extend the ideas presented here to the construction of infinite families of chiral polytopes of larger rank $d$, using permutation representations of the alternating group $A_d$ (as the rotation group of the regular ($d$ $-$1) simplex) to build their automorphism groups.

Here, we give some further background on regular and chiral polytopes in Sect. 2, and then in Sect. 3 we set up some of the building blocks and other things needed for our construction. We describe our construction and prove Theorem 1.1 in Sect. 4. Finally, in Sect. 5 we show that the vertex-figures of the chiral $4$-polytopes resulting from our construction are all chiral.

2 Abstract polytopes and chirality

An abstract $d$ -polytope (or abstract polytope of rank $d$) is a partially ordered set $\mathcal {P}$, the elements and maximal totally ordered subsets of which are called faces and flags respectively, such that certain properties are satisfied, which we explain below.

2.1 Definition of abstract polytopes

First, $\mathcal {P}$ contains a minimum face $F_{-1}$ and a maximum face $F_d$, and there is a rank function from $\mathcal {P}$ to the set $\{-1, 0, \dots , d\}$ such that $\mathrm{rank}(F_{-1})= -1$ and $\mathrm{rank}(F_d)=d$. Every flag of $\mathcal {P}$ contains precisely $d+2$ elements, including $F_{-1}$ and $F_d$. The faces of rank $i$ are called $i$ -faces, the $0$-faces are called vertices, the $1$-faces are called edges, and the $(d-1)$-faces are called facets. If $F$ and $G$ are faces of ranks $r$ and $s$ with $F \le G$, then we say that $F$ and $G$ are incident, we define $G/F := \{H \ |\ F\le H\le G\}$, and call this a section of $\mathcal {P}$, of rank $s - r - 1$. When convenient, we identify the section $G/F_{-1}$ with the face $G$ itself in $\mathcal {P}$, and if $v = F_0$ is a vertex, then the rank $d- 1$ section $F_d/F_0 := \{H \,|\, F_0 \le H\}$ is called the vertex-figure of $\mathcal {P}$ at $v$.

Whenever $G/F$ is a rank $1$ section (with $\mathrm{rank}(G) - \mathrm{rank}(F) = 2$), there are precisely two faces $H_1$ and $H_2$ such that $F < H_i < G$. This property is called the diamond condition. It implies that for any flag $\Phi$ and for every $i \in \{0, \dots , d-1\}$, there is a unique flag $\Phi ^i$ differing from $\Phi$ in precisely the $i$-face. We call $\Phi ^i$ the $i$ -adjacent flag for $\Phi$.

Finally, for any two flags $\Phi$ and $\Phi '$ of $\mathcal {P}$, there exists a sequence $\Psi _0, \Psi _1, \dots , \Psi _m$ of flags of $\mathcal {P}$ from $\Psi _0 = \Phi$ to $\Psi _m = \Phi '$ such that $\Psi _{k-1}$ is adjacent to $\Psi _k$, and $\Phi \cap \Phi ' \subseteq \Psi _k$, for $1 \le k \le m$. The last condition is known as strong flag-connectivity and completes the definition of an abstract $d$-polytope.

In this paper, we will deal with finite polytopes (namely those with finite rank and only finitely many faces of each rank).

Every rank $2$ section $G/F$ between an $(i-2)$-face $F$ and an incident $(i+1)$-face $G$ of a finite abstract polytope $\mathcal {P}$ is isomorphic to the face lattice of a polygon, and by convention, we assume that each such polygon is non-degenerate (having at least $3$ sides). If the number of sides of each such polygon depends only on $i$, and not on $F$ or $G$, then we say that $\mathcal {P}$ is equivelar. Regular and chiral polytopes (defined below) are examples of equivelar polytopes. We define the Schläfli type of an equivelar $d$-polytope $\mathcal {P}$ as $\{p_1, \dots , p_{d-1}\}$, when each section between an $(i-2)$-face and an $(i+1)$-face is an abstract $p_i$-gon. By finiteness, $p_i < \infty$ for all $i$, and by our non-degeneracy assumption, $p_i > 2$ for all $i$.

2.2 Automorphisms and regular polytopes

An automorphism of an abstract polytope $\mathcal {P}$ is an order-preserving permutation of its faces. We denote the group of automorphisms of $\mathcal {P}$ by $\Gamma (\mathcal {P})$. By the diamond condition and strong flag-connectivity, every automorphism is uniquely determined by its effect on any flag, and it follows that the number of automorphisms of $\mathcal {P}$ is bounded above by the number of flags of $\mathcal {P}$.

A $d$-polytope $\mathcal {P}$ is said to be regular whenever $\Gamma (\mathcal {P})$ acts transitively (and therefore regularly) on the set of all flags of $\mathcal {P}$. When that happens, the automorphism group $\Gamma (\mathcal {P})$ is generated by involutions $\rho _0, \dots , \rho _{d-1}$, where $\rho _i$ is the unique automorphism mapping a given base flag $\Phi$ to its $i$-adjacent flag $\Phi ^i$. Moreover, the generators $\rho _0, \dots , \rho _{d-1}$ satisfy

$$\begin{aligned} \rho _i^2&= 1 \quad \text{ for } \text{ all } \ i, \end{aligned}$$

(1)

$$\begin{aligned} (\rho _i \rho _j)^2&= 1 \quad \text{ whenever } |i-j| \ge 2. \end{aligned}$$

(2)

These generators also satisfy the following intersection condition:

$$\begin{aligned} \langle \, \rho _i \,|\, i \in I \,\rangle \cap \langle \, \rho _i \,|\, i \in J \,\rangle = \langle \, \rho _k \,|\, k \in I \cap J \,\rangle \quad \text{ for } \text{ all } I, J \subseteq \{0,1, \dots , d-1\}. \end{aligned}$$

(3)

The stabiliser in $\Gamma (\mathcal {P})$ of the $i$-face of the base flag $\Phi$ is generated by $\{\rho _0, \dots , \rho _{d-1}\} \setminus \{\rho _i\}$, for $0 \le i < d$, and the order of the element $\rho _{i-1} \rho _i$ coincides with the $i$-th term $p_i$ of the Schläfli type $\{p_1, \dots , p_{d-1}\}$, for $1 \le i < d$.

These properties of the automorphism group of a regular polytope can be exploited to construct examples from particular groups, called string C-groups. A string C-group of rank $d$ is a finite group $\Gamma$ and an associated set $\{\rho _0, \dots , \rho _{d-1}\}$ of $d$ generators for $\Gamma$ which satisfy (1) and (2), as well as the intersection condition (3). For any such $\Gamma$, we may construct a regular $d$-polytope $\mathcal {P}$ with $\Gamma = \Gamma (\mathcal {P})$, by taking as its $i$-faces the (right) cosets of the subgroup generated by $\{\rho _0, \dots , \rho _{d-1}\} \setminus \{\rho _i\}$, for $0 \le i < d$, and defining incidence by non-empty intersection; see [6, Theorem 2E11].

Hence up to isomorphism, regular $d$-polytopes are in one-to-one correspondence with string C-groups.

Next, we define the rotation group $\Gamma ^+(\mathcal {P})$ of a regular $d$-polytope $\mathcal {P}$ as the subgroup of $\Gamma (\mathcal {P})$ consisting of words of even length in the generators $\rho _0, \dots , \rho _{d-1}$, or equivalently, the subgroup generated by the abstract rotations $\sigma _i = \rho _{i-1} \rho _i$ for $1 \le i < d$. The index of $\Gamma ^+(\mathcal {P})$ in the full automorphism group $\Gamma (\mathcal {P})$ is at most $2$. Motivated by what happens for maps (in rank $3$), we say that $\mathcal {P}$ is orientably regular whenever this index is $2$, and otherwise we say that $\mathcal {P}$ is non-orientably regular.

Note that $\sigma _i = \rho _{i-1} \rho _i$ has order $p_i$ for all $i$. Moreover, these generators satisfy the relations

$$\begin{aligned} (\sigma _i \sigma _{i+1} \cdots \sigma _j)^2 = 1 \ \ \text{ for } \ 1 \le i < j < d. \end{aligned}$$

(4)

The involutory element $\tau _{i, j} = \sigma _i \sigma _{i+1} \cdots \sigma _j$ is called an abstract half-turn, for $1 \le i < j < d$. If we extend this definition of $\tau _{i, j}$ by setting $\tau _{0, i} = \tau _{i, d} = 1$ for $0 \le i \le d$, and $\tau _{i, i} = \sigma _i$ for $0 < i < d$, so that $\tau _{i, j}$ is defined whenever $0 \le i \le j \le d$, and we define the subgroup $H_I = \langle \, \tau _{i+1, j} \ |\ i, j \in I, \ i < j \,\rangle$ for every $I \subseteq \{-1,0, \dots , d\}$, then these subgroups satisfy the intersection condition

$$\begin{aligned} H_I \cap H_J =H_{I \cap J} \ \text{ for } \text{ all } I, J \subseteq \{-1,0, \dots , d\}. \end{aligned}$$

(5)

2.3 Chiral polytopes

The abstract $d$-polytope $\mathcal {P}$ is said to be chiral if its automorphism group $\Gamma (\mathcal {P})$ has two orbits on flags, with every two adjacent flags lying in different orbits. The reason for this terminology is that any such $\mathcal {P}$ has maximum possible ‘rotational’ symmetry (admitting analogues of the abstract rotations $\sigma _i = \rho _{i-1} \rho _i$), without admitting the ‘reflections’ $\rho _i$.

The rank $d$ of a chiral polytope is at least 3, since every abstract $2$-polytope is combinatorially isomorphic to a regular convex polygon with at least $3$ sides (by our non-degeneracy assumption). The facets and vertex-figures of a chiral $d$-polytope $\mathcal {P}$ may be regular or chiral, but the $(d-2)$-faces (and dually the co-edges) are always regular (by a nice argument given in [10, Proposition 9]).

The structure of the automorphism group of a chiral polytope $\mathcal {P}$ closely resembles that of the rotation group of a regular polytope. In particular, $\Gamma (\mathcal {P})$ is generated by elements $\sigma _1, \dots , \sigma _{d-1}$, where $\sigma _i$ maps a given base flag $\Phi$ to the flag $(\Phi ^i)^{i-1}$ which differs from $\Phi$ in its $(i-1)$- and $i$-faces. The rank $2$ section of $\mathcal {P}$ between the $(i-2)$- and $(i+1)$-faces of $\Phi$ is then isomorphic to a regular $p_i$-gon for some $p_i$, and the automorphism $\sigma _i$ permutes the $(i-1)$- and $i$-faces of this section in two cycles of length $p_i$.

Moreover, the generators $\sigma _i$ also satisfy (4), and if we define elements $\tau _{i, j} = \sigma _i \sigma _{i+1} \cdots \sigma _j$ for $1 \le i < j < d$, and exactly as in the previous subsection for other values of $i$ and $j$, then the subgroups $H_I = \langle \, \tau _{i+1, j} \ |\ i, j \in I \,\rangle$ also satisfy the intersection condition (5).

For simplicity and consistency, we still refer to these generators $\sigma _i$ of $\Gamma (\mathcal {P})$ as abstract rotations, and the products $\tau _{i, j}$ for $1 \le i < j < d$ as abstract half-turns. Also we often refer to the automorphism group of the chiral polytope $\mathcal {P}$ as its rotation group and sometimes denote it by $\Gamma ^+(\mathcal {P})$.

Conversely, any finite group $\Gamma$ generated by $d-1$ elements $\sigma _1, \sigma _2, \dots , \sigma _{d-1}$ satisfying (4) and the intersection condition (5) is the rotation subgroup of an abstract $d$-polytope $\mathcal {P}$ that is either (orientably) regular or chiral; see [10, Theorem 1]. Indeed $\mathcal {P}$ is regular if and only if there is a group automorphism $\rho$ of $\Gamma$ of order $2$ such that

$$\begin{aligned} \sigma _i^{\,\rho } = \left\{ \begin{array}{ll} \sigma _i^{-1} &{} \text{ when } i=1,\\ \sigma _1^{\ 2} \sigma _i &{} \text{ when } i=2,\\ \sigma _i &{} \text{ when } 2 < i < d. \end{array}\right. \end{aligned}$$

(6)

Note (for later use) that for rank $3$, the automorphism $\rho$ has to invert $\sigma _1$ and take $\sigma _2$ to $\sigma _1^{\ 2} \sigma _2 = \sigma _1 \sigma _2^{-1} \sigma _1^{-1}$, so the composite of $\rho$ with conjugation by $\sigma _1$ inverts both $\sigma _1$ and $\sigma _2$; the existence of such an automorphism is the more customary test for chirality of maps.

Each chiral $d$-polytope $\mathcal {P}$ occurs in two enantiomorphic forms, which may be understood as $\mathcal {P}$ and its ‘mirror image’ (and hence as a right- and left-handed version of $\mathcal {P}$). The group of the mirror image of $\mathcal {P}$ is also $\Gamma (\mathcal {P})$, but with respect to the generators $\sigma _1^{-1}, \sigma _1^2 \sigma _2, \sigma _3, \dots , \sigma _{d-1}$. Further information can be found in [11].

2.4 Chiral 4-polytopes

In this paper we concentrate on chiral polytopes of rank $d = 4$.

By [10, Lemma 11], the intersection condition for a chiral $4$-polytope $\mathcal {P}$ can be reduced to just three cases, as follows:

$$\begin{aligned} \langle \sigma _1 \rangle \cap \langle \sigma _2 \rangle = \{1\}, \ \ \ \langle \sigma _2 \rangle \cap \langle \sigma _3 \rangle = \{1\} \ \ \text{ and } \ \ \langle \sigma _1, \sigma _2 \rangle \cap \langle \sigma _2, \sigma _3 \rangle = \langle \sigma _2 \rangle . \end{aligned}$$

(7)

We will also make use of an alternative generating set for $\Gamma ^+(\mathcal {P})$, namely $\{\tau _1, \tau _2, \tau _3\}$, where $\tau _i = \tau _{1,i} = \sigma _1 \sigma _2 \cdots \sigma _i$ for $1 \le i \le 3$. In terms of these three generators, the relations $(\sigma _i \sigma _{i+1} \cdots \sigma _j)^2 = 1$ in (4) are equivalent to

$$\begin{aligned} (\tau _1 \tau _3)^2 = \tau _2^2 = \tau _3^2 = 1. \end{aligned}$$

(8)

Furthermore, the test in (6) for regularity of $\mathcal {P}$ simplifies to the existence of a group automorphism $\rho$ of $\Gamma ^+(\mathcal {P})$ such that

$$\begin{aligned} \tau _i ^{\,\rho } = \tau _i^{-1} \ \text{ for } 1 \le i \le 3. \end{aligned}$$

(9)

Finally we note that $\langle \tau _1 \rangle = \langle \sigma _1 \rangle$ and $\langle \tau _1, \tau _2 \rangle = \langle \sigma _1, \sigma _2 \rangle$, but a comparison of orders shows that $\langle \tau _2 \rangle \ne \langle \sigma _2 \rangle$, and similarly it need not be true that $\langle \tau _2, \tau _3 \rangle = \langle \sigma _2, \sigma _3 \rangle$.

3 Actions of $A_4$

In Sect. 4 we will construct families of chiral 4-polytopes whose facets are tetrahedra. The construction involves extending an intransitive action of the rotation group $A_4$ of the tetrahedron on a set with $n$ elements, to the standard action of $A_n$ or $S_n$ on the same set, by adjoining a new permutation that represents a generator of the automorphism group of the $4$-polytope.

In this section we create some building blocks for the construction, via transitive permutation representations of $A_4$. We will be particularly interested in the permutations $\tau _1$ and $\tau _2$ representing the generators of $A_4$ as the rotation group of the tetrahedron. These permutations satisfy the relations $\tau _1^{\,3} = \tau _2^{\,2} = (\tau _1^{-1} \tau _2)^3 = 1$.

3.1 Building blocks

The transitive representations of $A_4$ that we use as building blocks are those on $1, 4, 6$ and $12$ points, as follows:

Representation A: the trivial representation of $A_4$, of degree $1$;

Representation B: the standard representation of $A_4$ on $4$ points, with

$$\begin{aligned} \tau _1 = (1,3,2) (4) \quad \text{ and } \quad \tau _2 = (1,2)(3,4); \end{aligned}$$

Representation C: the transitive representation of $A_4$ on $6$ points, given by

$$\begin{aligned} \tau _1 = (1,2,3)(4,5,6) \quad \text{ and } \quad \tau _2 = (1,4)(2,5); \end{aligned}$$

Representation D: the regular representation of $A_4$ on $12$ points, given by

$$\begin{aligned} \tau _1&= (1,2,3)(4,5,6)(7,8,9)(10,11,12) \quad \text{ and } \quad \\ \tau _2&= (1,4)(2,7)(3,10)(5,12)(6,8)(9,11). \end{aligned}$$

Note that these transitive representations are unique up to re-labelling points, because $A_4$ has a single conjugacy class of subgroups of each of the orders $12, 3, 2$ and $1$.

We will also be interested in the orbits of the subgroup $\langle \tau _1 \rangle$. In Representation B there are two orbits, of lengths $3$ and $1$, respectively, while in Representations C and D there are two of length $3$ and four of length $3$, respectively.

For later use, we illustrate these representations in Fig. 1 by subdivided boxes, with each subdivision giving the length of an orbit of $\langle \tau _1 \rangle$.

Fig. 1

Transitive permutation representations of $A_4= \langle \tau _1, \tau _2 \rangle$ on 1, 4, 6 and 12 points

3.2 Extending the action of $A_4$

Our construction involves extending an intransitive action of $A_4 = \langle \tau _1,\tau _2 \rangle$ to a transitive action of $\langle \tau _1,\tau _2, \tau _3 \rangle$, by a suitable definition of the third generator $\tau _3$.

The first and third of the relations $(\tau _1 \tau _3)^2 = \tau _2^{\,2} = \tau _3^{\,2} = 1$ given in (8) imply that $\tau _3$ must be an involution which conjugates the generator $\tau _1$ to its inverse. For this reason, $\tau _3$ must permute the fixed points of $\langle \tau _1 \rangle$ among themselves, and permute the orbits of length $3$ among themselves. To make the resulting action of $\langle \tau _1,\tau _2, \tau _3 \rangle$ transitive, we must link together the orbits of $A_4 = \langle \tau _1,\tau _2 \rangle$, and this can be achieved by defining $\tau _3$ in such a way as to link together the orbits of $\langle \tau _1 \rangle$, perhaps sometimes linking an orbit to itself.

There is just one way of linking together two orbits of $\langle \tau _1 \rangle$ of length $1$, namely by making $\tau _3$ interchange the single points from the orbits. On the other hand, linking together two different orbits of $\langle \tau _1 \rangle$ of length $3$ can be done in three ways. If $\tau _1$ acts on one orbit as the $3$-cycle $(y_1,y_2,y_3)$, where $y_1 = \min \{y_1,y_2,y_3\}$, and on the other as the $3$-cycle $(z_1,z_2,z_3)$, where $z_1 = \min \{z_1,z_2,z_3\}$, then we have these three possibilities for the effect of $\tau _3$ on the set $\{y_1,y_2,y_3,z_1,z_2,z_3\}$:

$$\begin{aligned}&(y_1, z_1)(y_2, z_3)(y_3, z_2) \qquad \dots \ \hbox {type I} \\&(y_1, z_2)(y_2, z_1)(y_3, z_3) \qquad \dots \ \hbox {type II} \\&(y_1, z_3)(y_2, z_2)(y_3, z_1) \qquad \dots \ \hbox {type III}. \end{aligned}$$

In the special case where these orbits are the same (so that $(y_1,y_2,y_3) = (z_1,z_2,z_3)$), the element $\tau _3$ induces $(y_2,y_3), (y_1,y_2)$ and $(y_1,y_3)$ for types I, II and III, respectively.

Also at this stage, we note that for an orientably regular or chiral 4-polytope $\mathcal {P}$ of type $\{3,3,k\}$, whose facets are tetrahedra, the reduced intersection condition (7) can be simplified even further.

Lemma 3.1

Let $\Gamma$ be a transitive permutation group of degree $n$ generated by three elements $\sigma _1, \sigma _2$ and $\sigma _3$ satisfying

$$\begin{aligned} \sigma _1^{\,3} = \sigma _2^{\,3} = (\sigma _{1}\sigma _{2})^2 = (\sigma _{2}\sigma _{3})^2 = (\sigma _{1}\sigma _{2}\sigma _{3})^2 = 1 \end{aligned}$$

with $\langle \sigma _1, \sigma _2 \rangle \cong A_4$. If $\langle \sigma _2, \sigma _3 \rangle$ is intransitive and $\sigma _2$ is not a power of $\sigma _3$, then the intersection condition (7) holds.

Proof

First $\langle \sigma _1 \rangle \cap \langle \sigma _2 \rangle = \{1\}$, since $\sigma _1$ and $\sigma _2$ are two elements of order $3$ generating $A_4$. Next, observe that $\langle \sigma _1, \sigma _2 \rangle \cap \langle \sigma _2, \sigma _3 \rangle$ is a subgroup of $\langle \sigma _1, \sigma _2 \rangle$, containing $\langle \sigma _2 \rangle$ and that $\langle \sigma _2 \rangle$ is maximal in $\langle \sigma _1, \sigma _2 \rangle$, since every cyclic subgroup of order $3$ in $A_4$ is maximal in $A_4$. It follows that if $\langle \sigma _1, \sigma _2 \rangle \cap \langle \sigma _2, \sigma _3 \rangle \ne \langle \sigma _2 \rangle$, then $\langle \sigma _1, \sigma _2 \rangle \cap \langle \sigma _2, \sigma _3 \rangle = \langle \sigma _1, \sigma _2 \rangle$, and therefore $\sigma _1 \in \langle \sigma _2, \sigma _3 \rangle$, which gives $\Gamma = \langle \sigma _1, \sigma _2, \sigma _3 \rangle = \langle \sigma _2, \sigma _3 \rangle$. But that is clearly impossible, because $\Gamma$ is transitive while $\langle \sigma _2, \sigma _3 \rangle$ is not. Thus $\langle \sigma _1, \sigma _2 \rangle \cap \langle \sigma _2, \sigma _3 \rangle = \langle \sigma _2 \rangle$. Finally, $\langle \sigma _2 \rangle \cap \langle \sigma _3 \rangle = \{1\}$, since the element $\sigma _2$ of order 3 does not lie in $\langle \sigma _3 \rangle$.$\square$

3.3 Other facts needed

To conclude this section we mention some results from group theory that we need for the construction presented in Sect. 4, specifically for recognising when a transitive subgroup of $S_n$ is either $A_n$ or $S_n$, and also about the automorphism groups of $A_n$ and $S_n$.

Theorem 3.2

(Jordan, 1873) Let $G$ be a primitive group of permutations on a set $X$ of degree $n$, and suppose $G$ contains an element that acts as a $p$-cycle, fixing the other $n-p$ points, where $p$ is a prime such that $p \le n-3$. Then $G$ is isomorphic to $A_n$ or $S_n$.

For a proof, see [12, Theorem 13.9]. The next theorem is well known; proofs can be found in [9, Corollary 7.7] for $\mathrm{Aut}(S_n)$, and [13, Theorem 2.3] for $\mathrm{Aut}(A_n)$, for example.

Theorem 3.3

For every $n \ge 7$, every automorphism of $A_n$ and every automorphism of $S_n$ is induced by conjugation by an element of $S_n$. In particular, $\mathrm{Aut}(A_n) \cong \; \mathrm{Aut}(S_n) \cong S_n$ for every $n \ge 7$.

4 Construction of chiral 4-polytopes

In this section we use the building blocks given earlier to construct two families of chiral 4-polytopes, with automorphism groups $S_n$ and $A_n$ respectively, for all $n\ge 46$.

4.1 General approach

We let $X$ be the set $\{1,2, \dots , n\}$, and define permutations $\tau _1, \tau _2$, and $\tau _3\in S_n$ such that $\langle \tau _1, \tau _2\rangle =A_4$ and $\tau _1, \tau _2$, and $\tau _3$ satisfy (8). In order to prove that the construction actually gives a chiral 4-polytope, we need to do three things:

Step (a): Show that $\,\Gamma = \langle \tau _1,\tau _2,\tau _3\rangle$ is $A_n$ or $S_n$.

Our construction ensures that the action of $\Gamma$ is transitive on $X$. We exhibit an element of $\Gamma$ that acts as a cycle of prime length $p$, fixing at least $3$ points, and then use this to prove that $\Gamma$ is primitive on $X$ and apply Theorem 3.2 to give $\Gamma \cong A_n$ or $\Gamma \cong S_n$.

Step (b): Show that $\Gamma$ is the rotation subgroup of an orientably regular polytope or the automorphism group of a chiral polytope.

For this step, all we need to do is prove that the permutations $\sigma _1=\tau _1, \,\sigma _2=\tau _1^{-1}\tau _2$ and $\sigma _3=\tau _2^{-1}\tau _3$ satisfy the reduced form of the intersection condition given in (7). By Lemma 3.1, it is sufficient to show that $\langle \sigma _2,\sigma _3\rangle$ is intransitive on $X$ and that $\sigma _2 \not \in \langle \sigma _3\rangle$.

Step (c): Verify chirality, by ruling out the existence of a permutation $\rho \in S_n$ such that $\tau _i^{\,\rho } = \tau _i^{-1}$ for all $i \in \{1,2,3\}$.

Note the permutations $\tau _1$ and $\tau _2$ are always even, since they come from permutation representations of $A_4$. It follows that once we have completed step (a), we can decide whether $\Gamma$ is $A_n$ or $S_n$ by simply checking whether $\tau _3$ is even or odd. In some cases we will make an adjustment to $\tau _3$ that will still ensure that $\Gamma = \langle \tau _1,\tau _2,\tau _3\rangle$ is the automorphism group of some chiral $4$-polytope of type $\{3,3,k\}$ for some $k$, but has a different parity, in which case we change $\Gamma$ from an alternating group to a symmetric group, or vice versa.

We will consider a number of cases, based on the residue class of $n$ mod $6$. Before that, we give a concrete example (for $n = 46$), which will show how most of the construction works. This can then be adapted in a number of ways for other values of the degree $n$.

4.2 Example: degree $n = 46$

Consider the following three permutations on $46$ points:

$$\begin{aligned} \tau _1&= (1,2,3)(4,5,6)(7,8,9)(10,11,12)(13,14,15)(16,17,18)(19,20,21)\\&\quad (22,23,24)(25,26,27)(28,29,30)(31,32,33)(34,35,36)(37,38,39)\\&\quad (40,41,42)(43,44,45),\\ \tau _2&= (1,4)(2,7)(3,10)(5,12)(6,8)(9,11)(13,16)(14,17)(19,22)(20,23) \\&\quad (25,28)(26,29)(31,34)(32,35)(37,40)(38,41)(43,44)(45,46),\\ \tau _3&= (1,2)(4,7)(5,9)(6,8)(10,13)(11,15)(12,14)(16,19)(17,21)(18,20)(22,25)\\&\quad (23,27)(24,26)(28,31)(29,33)(30,32)(34,39)(35,38)(36,37)(40,43)\\&\quad (41,45)(42,44). \end{aligned}$$

These satisfy the required relations, and generate a transitive subgroup of $S_{46}$. The orbits of $\langle \tau _1,\tau _2 \rangle$ are the sets $\{1,2,\dots ,12\}, \{13,14,\dots ,18\}, \{19,20,\dots ,24\}, \{25,26,\dots ,30\}, \{31,32,\dots ,36\}, \{37,38,\dots ,42\}$ and $\{43,44,45,46\}$, of lengths $12, 6, 6, 6, 6, 6$ and $4$. The way in which the orbits of $\langle \tau _1 \rangle$ are linked together by $\tau _3$ is illustrated in Fig. 2, where the Roman numerals indicate the type of link.

Fig. 2

Orbit links for a permutation representation of degree $46$

In this representation, observe that the elements $\sigma _2 = \tau _1^{-1} \tau _2$ and $\sigma _3 = \tau _2^{-1} \tau _3 = \tau _2 \tau _3$ are as follows:

$$\begin{aligned} \sigma _2&= (1,10,5)(2,4,8)(3,7,11)(6,12,9)(13,15,17)(14,16,18)(19,21,23)\\&\quad (20,22,24)(25,27,29)(26,28,30)(31,33,35)(32,34,36)(37,39,41)\\&\quad (38,40,42)(43,46,45); \\ \sigma _3&= (1,7)(2,4)(3,13,19,25,31,39,34,28,22,16,10)(5,14,21,17,12,9,15,11) \\&\quad (18,20,27,23)(24,26,33,29)(30,32,38,45,46,41,35)\\&\quad (36,37,43,42,44,40). \end{aligned}$$

In particular, the cycle structure of $\sigma _3$ is $1^{2}\,2^{2}\,4^{2}\,6^{1}\,7^{1}\,8^{1}\,11^{1}$, and so its order is $1848$. Also $\sigma _3^{\,168}$ is an $11$-cycle, namely $(3,25,34,16,13,31,28,10,19,39,22).$

We claim that the action of $\Gamma = \langle \tau _1,\tau _2,\tau _3 \rangle$ is primitive on $\{1, \dots , 46\}$. To verify this, we assume the contrary (but we will ignore the fact that $3$ and $11$ do not divide $46$, just to exhibit a more general argument). All the $11$ points moved by $\sigma _3^{\,168}$ would have to belong to the same block of imprimitivity, say $U$, since $11$ is prime and every block containing a fixed point of $\sigma _3^{\,168}$ would be fixed by $\sigma _3^{\,168}$. Next, $\tau _2$ preserves $U$ since it interchanges the points $3$ and $10$ of $U$, and similarly, $\tau _3$ preserves $U$, since it fixes the point $3$. It follows that $\tau _1$ cannot preserve $U$, and so the images of $U$ under $\tau _1$ and its inverse $\tau _1^{\,2}$ must be new blocks $V$ and $W$, containing $\{1, 26, 35, 17, 14, 32, 29, 11, 20, 37, 23\}$ and $\{2, 27, 36, 18, 15, 33, 30, 12, 21, 38, 24\}$, respectively. Now $\tau _2$ preserves both $V$ and $W$, since it interchanges the points $26$ and $29$ and fixes the point $24$, and similarly, $\tau _3$ interchanges $V$ with $W$ since it interchanges the points $1$ and $2$. By transitivity, it follows that there are just three blocks, with $\tau _1, \tau _2$ and $\tau _3$ inducing the permutations $(U,V,W), (U)(V)(W)$ and $(U)(V,W)$ on them. In particular, $\tau _2$ preserves every block, while $\tau _1$ preserves no block. But that is impossible, since $\tau _1$ fixes the point $46$.

By Theorem 3.2, we find that $\Gamma = A_{46}$ or $S_{46}$, and since $\tau _3$ is even, we have $\Gamma = A_{46}$.

Next, we verify the intersection condition. First, $\langle \sigma _2, \sigma _3 \rangle = \langle \tau _1^{-1}\tau _2, \tau _2\tau _3 \rangle$ is intransitive, since it has $\{2,4,8\}$ as an orbit, and second, $\sigma _2 \not \in \langle \sigma _3 \rangle$, since $\sigma _2 = \tau _1^{-1}\tau _2$ induces the $3$-cycle $(1,10,5)$, while $\sigma _3$ interchanges the points $1$ and $7$. Hence by Lemma 3.1, the intersection condition (7) holds.

Thus $A_{46}$ is the rotation group of a regular or chiral $4$-polytope $\mathcal {P}$ (of type $\{3,3,1848\}$).

Next, suppose $\mathcal {P}$ is regular. Then there must exist an involutory group automorphism $\rho$ of $\Gamma$ inverting each of $\tau _1, \tau _2$ and $\tau _3$. By Theorem 3.3, this automorphism $\rho$ can be taken as a permutation in $S_{46}$. In particular, since $\rho$ inverts $\tau _1$, it must permute the orbits of $\langle \tau _1 \rangle$ among themselves, and hence must fix the point $46$. Then since $\rho$ inverts $\tau _2$, it follows that $\rho$ preserves the orbit $\{45, 46\}$ of $\langle \tau _2 \rangle$ and hence fixes the point $45$. In turn, since $\rho$ inverts $\tau _1$, it must interchange the other two points $43$ and $44$ of the $3$-cycle $(43,44,45)$ of $\tau _1$, and then must interchange the orbits $\{40, 43\}$ and $\{42, 44\}$ of $\langle \tau _3 \rangle$, and hence must interchange the points $40$ and $42$. But this is impossible, since $42$ is fixed by $\tau _2$, while $40$ is not. Thus $\mathcal {P}$ is a chiral 4-polytope, of type $\{3,3,1848\}$, with automorphism group $A_{46}$.

To do the same for $S_{46}$, we define $\tau _1$ and $\tau _2$ exactly as above, but now take

$$\begin{aligned} \tau _3&= (1,2)(4,6)(7,8)(10,13)(11,15)(12,14)(16,19)(17,21)(18,20)(22,25)\\&\quad (23,27)(24,26)(28,31)(29,33)(30,32)(34,39)(35,38)(36,37)(40,43)\\&\quad (41,45)(42,44). \end{aligned}$$

This is almost the same as the permutation taken for $\tau _3$ above, but with the three transpositions $(4,7), (5,9)$ and $(6,8)$ replaced by the two transpositions $(4,6)$ and $(7,8)$, and two fixed points $5$ and $9$. With regard to Fig. 2, we have replaced the $\tau _3$-link between the first two orbits of $\langle \tau _1 \rangle$ by self-links for those two orbits, of types III and II respectively. This time we have

$$\begin{aligned} \sigma _3&= (1,6,7)(2,8,4)(3,13,19,25,31,39,34,28,22,16,10)(5, 14, 21, 17, 12)\\&\quad (9, 15, 11)(18,20,27,23)(24,26,33,29)(30,32,38,45,46,41,35)\\&\quad (36,37,43,42,44,40), \end{aligned}$$

which has cycle structure $3^{3}\,4^{2}\,5^{1}\,6^{1}\,7^{1}\,11^{1}$, so its order is $4620$.

Now $\sigma _3^{\,420}$ is an 11-cycle, in fact the 8th power of the one found before, and it can be used to prove that the action of $\Gamma = \langle \tau _1,\tau _2,\tau _3 \rangle$ is primitive, and hence that $\Gamma = A_{46}$ or $S_{46}$. This time $\tau _3$ is odd, so $\Gamma = S_{46}$. Again the intersection condition is satisfied (noting that $\{2,4,8\}$ is still an orbit of $\langle \sigma _2, \sigma _3 \rangle = \langle \tau _1^{-1}\tau _2, \tau _2\tau _3 \rangle$ and that $(1,10,5)$ is still a cycle of $\sigma _2$), and the same argument as before proves chirality. Thus we have another chiral 4-polytope, but now of type $\{3,3,4620\}$, and with automorphism group $S_{46}$.

4.3 Adding extra orbits of $A_4$ of length 6

Now take the above example (for $n = 46$) and insert an additional orbit of length $6$ for $\langle \tau _1,\tau _2 \rangle \cong A_4$ between the last two on the right, with a $\tau _3$-link of type I to the previous final orbit of length $6$ and a $\tau _3$-link of type III to the orbit of length $4$, as in Fig. 3.

Fig. 3

Inserting an extra orbit of $\langle \tau _1,\tau _2 \rangle \cong A_4$ of length $6$

This gives a transitive permutation representation on $46+6 = 52$ points, with the following changes made to the three permutations $\tau _i$ used to generate $A_{46}$:

• $\tau _1$: adjoin the two $3$-cycles $(46,47,48)$ and $(49,50,51)$, and the fixed point $52$,

• $\tau _2$: replace $(43,44)(45,46)$ by $(43,46)(44,47)(49,50)(51,52)$, fixing points $45$ and $48$,

• $\tau _3$: replace the fixed point $46$ by $(46,51)(47,50)(48,49)$, fixing the point $52$.

With these changes, the only effect on the permutation $\sigma _3 = \tau _{2}\tau _{3}$ is to alter cycles containing any of the points numbered greater than $42$, and in fact, it is easy to see that the two cycles $(30, 32, 38, 45, 46, 41, 35)$ and $(36, 37, 43, 42, 44, 40)$ of lengths $7$ and $6$ are replaced by $(30, 32, 38, 45, 41, 35), (36, 37, 43, 51, 52, 46, 40)$ and $(42, 44, 50, 48, 49, 47)$, of lengths $6, 7$ and $6$. In particular, the cycle structure of $\sigma _3$ remains the same except for the addition of one further cycle of length $6$, and $\sigma _3^{\,168}$ is still the same $11$-cycle, namely $(3,25,34,16,13,31,28,10,19,39,22).$

Again this $11$-cycle and the existence of a fixed point of $\tau _1$ can be used to prove that the group $\Gamma = \langle \tau _1,\tau _2,\tau _3 \rangle$ is primitive, and then since the parity of $\tau _3$ has changed from even to odd, we have $\Gamma = S_{52}$. The intersection condition (7) holds for exactly the same reasons as for degree $46$, and the proof of chirality is entirely similar: any involutory group automorphism $\rho$ inverting each of $\tau _1, \tau _2$ and $\tau _3$ would have to fix the points $52$ and $51$, and swap the points $49$ and $50$, and then swap the points $47$ and $48$, which is impossible.

Thus $S_{52}$ is the automorphism group of a chiral $4$-polytope $\mathcal {P}$ (of type $\{3,3,1848\}$).

Furthermore, we can now make the same change to the effect of $\tau _3$ on the first orbit of $\langle \tau _1,\tau _2 \rangle$ (of length $12$) as we did for degree $46$, with a change in parity of $\tau _3$, and the same arguments work again, to prove that $A_{52}$ is the automorphism group of a chiral $4$-polytope of type $\{3,3,4620\}$.

In summary, inserting the extra orbit of $A_4$ of length $6$ increased the degree $n$ by $6$, but retained the properties of the permutations $\tau _1,\tau _2$ and $\tau _3$ needed to prove that $A_n$ and $S_n$ are the automorphism groups of chiral $4$-polytopes of type $\{3,3,k\}$ for some $k$.

But clearly we can do the same kind of thing again. Suppose we insert another new orbit of $A_4$ of length $6$ between the last one and the orbit of length $4$, with a $\tau _3$-link of type III to the previous final orbit of length $6$ and a $\tau _3$-link of type I to the orbit of length $4$. Then the degree $n$ increases by $6$, and we return to a situation similar to that for degree $46$. With the obvious re-numbering of points in the last two orbits of $A_4$, the cycles $(36, 37, 43, 51, 52, 46, 40)$ and $(42, 44, 50, 48, 49, 47)$ of lengths $7$ and $6$ for $\sigma _3$ in the case of degree $52$ are replaced by $(36, 37, 43, 51, 46, 40), (42, 44, 50, 57, 58, 53, 47)$ and $(48, 49, 55, 54, 56, 52)$, of lengths $6, 7$ and $6$. Hence the cycle structure of $\sigma _3$ is again changed only by the addition of another cycle of length $6$. All the previous arguments work in the same way, to prove that $A_{58}$ and $S_{58}$ are the automorphism groups of chiral $4$-polytopes of types $\{3,3,1848\}$ and $\{3,3,4620\}$, respectively.

These insertions can be repeated over and over again, increasing the degree by $6$ through insertion of a new orbit of length $6$ for $A_4$ each time. Provided that the types of the $\tau _3$-links joining successive new orbits of $A_4$ are chosen to alternate between types I and III, the important properties of the the permutations $\tau _1,\tau _2$ and $\tau _3$ will be retained, and all our arguments will go through in the same way as for degrees $46$ and $52$.

Thus we have the following: for every integer $n \ge 46$ such that $n \equiv 4$ mod $6$, both $A_n$ and $S_n$ are the automorphism groups of chiral $4$ -polytopes of type $\{3,3,k\}$ for some $k$.

In fact, $k$ can be taken as $1848$ or $4620$, depending on the residue class of $n$ mod $12$, and in particular, our construction shows there are infinitely many chiral $4$-polytopes of type $\{3,3,k\}$ for each of these two values of $k$.

4.4 Adding an extra point fixed by $A_4$

In all of the cases considered so far in this section, with degree $n \equiv 4$ mod $6$, the subgroup $\langle \tau _1,\tau _2 \rangle \cong A_4$ had single orbits of lengths $12$ and $4$, and $\frac{n-16}{6}$ orbits of length $6$, and the permutation $\tau _1$ had a single fixed point (which we chose to be $n$) and $\frac{n-1}{3}$ cycles of length $3$. We will now consider what happens when we adjoin a single orbit of length $1$.

Necessarily, the permutations $\tau _1$ and $\tau _2$ will fix this point, while $\tau _3$ must interchange it with the only other fixed point of $\tau _1$. The only change to the permutation $\sigma _3$ is to enlarge its unique $7$-cycle (containing the original fixed point of $\tau _1$) to an $8$-cycle. For example, when $n = 46$, the cycle $(30,32,38,45,46,41,35)$ becomes $(30,32,38,45,47,46,41,35)$.

The order of $\sigma _3$ changes from $1848$ to $1848/7 = 264$, or from $4620$ to $2\cdot 4620/7 = 1320$, and in those two cases respectively, the permutations $\sigma _3^{\,24}$ and $\sigma _3^{\,120}$ are $11$-cycles, namely $\xi = (3, 19, 31, 34, 22, 10, 13, 25, 39, 28, 16)$ and $\xi ^5 = (3, 10, 16, 22, 28, 34, 39, 31, 25, 19, 13)$.

In each case, the $11$-cycle and the existence of a fixed point of $\tau _1$ can be used to prove that the resulting permutations $\tau _1,\tau _2$ and $\tau _3$ generate a primitive group, and hence an alternating or symmetric group. Also the intersection condition holds for exactly the same reasons as before. On the other hand, the proof of chirality needs a small variation.

Take $n$ to be the resulting degree, and $n-1$ and $n$ as the fixed points of $\tau _1$, and $n-2$ as the image of $n-1$ under $\tau _2$ in the orbit of $A_4$ of length $4$. Now suppose there exists an involutory group automorphism $\rho$ of $\Gamma = \langle \tau _1,\tau _2,\tau _3 \rangle$ inverting each $\tau _i$. By Theorem 3.3, this automorphism $\rho$ can be taken as an element of $S_n$, and since $\rho$ inverts $\tau _1$, it must fix or interchange the points $n-1$ and $n$. If it fixes both, then the same argument as before gives a contradiction, and so it must interchange them. But that is impossible, since $n-1$ and $n$ lie in cycles of $\tau _2$ of different lengths (namely $2$ and $1$). Hence there is no such automorphism $\rho$, and we have a chiral $4$-polytope.

Thus we have the following: for every integer $n \ge 47$ such that $n \equiv 5$ mod $6$, both $A_n$ and $S_n$ are the automorphism groups of chiral $4$ -polytopes of type $\{3,3,k\}$ for some $k$.

In fact, $k$ can be taken as $264$ or $1320$, depending on the residue class of $n$ mod $12$, and in particular, our construction shows there are infinitely many chiral $4$-polytopes of type $\{3,3,k\}$ for each of these two values of $k$.

4.5 Adding a second orbit of $A_4$ of length 4

Next, we consider what happens when we add a second orbit of length $4$ for $A_4$ to the permutations given earlier for $A_{46}$, but at the ‘first end’, linked to the orbit of length $12$ for $A_4$ by a $\tau _3$ link of type II, as in Fig. 4.

Fig. 4

Adding a second orbit of length $4$ for $A_4$

Specifically (and to avoid altering the numbering too much), we introduce four new points, labelled $v, x, y$ and $z$, with the assumption that $v < x < y < z$ and make the following changes made to the three permutations $\tau _i$ used to generate $A_{46}$:

• $\tau _1$: adjoin the $3$-cycle $(x,y,z)$ and the fixed point $v$,

• $\tau _2$: adjoin the transpositions $(v,z)$ and $(x,y)$,

• $\tau _3$: replace the transposition $(1,2)$ and fixed point $3$ by $(x,2)(y,1)(z,3)$, fixing $v$.

With these changes, the only effect on the permutation $\sigma _3 = \tau _{2}\tau _{3}$ is to alter the cycles containing any of the points numbered $1, 2$ and $3$, namely the transpositions $(1,7)$ and $(2,4)$ and the $11$-cycle $(3,13,19,25,31,39,34,28,22,16,10)$. These cycles are replaced by $(x,1,7), (y,2,4)$ and $(v,3,13,19,25,31,39,34,28,22,16,10,z)$, of lengths $3, 3$ and $13$, respectively.

In particular, the cycle structure of $\sigma _3$ becomes $1^{2}\,3^{2}\,4^{2}\,6^{1}\,7^{1}\,8^{1}\,13^{1}$, and so $\sigma _3$ now has order $2184$. Also $\sigma _3^{\,168}$ is a $13$-cycle, namely $(3,v,z,10,16,22,28,34,39,31,25,$ $19,13).$

We claim that the action of $\Gamma = \langle \tau _1,\tau _2,\tau _3 \rangle$ is primitive on $\{1, \dots , 46\} \cup \{v,x,y,z\}$. If not, then the $13$ points moved by $\sigma _3^{\,168}$ would belong to the same block $U$, and $U$ would be preserved by $\tau _1$ and $\tau _3$, since the point $v$ is fixed by both $\tau _1$ and $\tau _3$, and $U$ would be preserved by $\tau _2$, since $\tau _2$ swaps $v$ with $z$. But then $U$ would be preserved by $\langle \tau _1,\tau _2,\tau _3 \rangle = \Gamma$ and so could not be a block of imprimitivity. Since $\tau _3$ is even, it follows that $\Gamma \cong A_{50}$.

Also the subgroup generated by $\sigma _2$ and $\sigma _3$ is intransitive, because it has $\{y,2,4,8\}$ as an orbit, and $\sigma _2$ does not lie in $\langle \sigma _3 \rangle$, because $\sigma _2$ induces the $3$-cycle $(2,4,8)$, while $\sigma _3$ induces the $3$-cycle $(y,2,4)$ on this orbit. By Lemma 3.1, the intersection condition holds.

We still need to confirm chirality. Suppose there is an involution $\rho$ in $S_{50}$ which conjugates each of $\tau _1,\tau _2$ and $\tau _3$ to its inverse. Then $\rho$ fixes or interchanges the two fixed points of $\tau _1$, namely $v$ and $46$, and if it fixes $46$, then the same argument as before gives a contradiction, so it must interchange them. It follows that $\rho$ swaps $v^{\tau _2} = z$ with $46^{\tau _2} = 45$, and also $z^{\tau _3} = 3$ with $45^{\tau _3} = 41$. But that is impossible, since $3$ and $41$ lie in cycles of $\tau _2$ of different lengths (namely $2$ and $1$).

Thus $A_{50}$ is the automorphism group of a chiral $4$-polytope of type $\{3,3,2184\}$.

Next, if we make the same change to the effect of $\tau _3$ on the orbit $\{1,2,\dots ,12\}$ of $\langle \tau _1,\tau _2 \rangle$ as we did for degree $46$, then we find that the cycles of $\sigma _3$ containing the points of $\{v,x,y,z,1,2,\dots ,12\}$ are $(v, 3, 13, 19, 25, 31, 39, 34, 28, 22, 16, 10, z), (x,1,$ $6, 7), (y, 2, 8, 4), (5, 14, 21, 17, 12)$ and $(9, 15, 11)$. In this case, $\sigma _3$ has cycle structure $3^{1}\,4^{4}\,5^{1}\,6^{1}\,7^{1}\,13^{1}$ and hence order $5460$. Again the existence of the $13$-cycle and the effect of $\tau _1,\tau _2$ and $\tau _3$ on the points $v$ and $z$ imply that $\Gamma = \langle \tau _1,\tau _2,\tau _3 \rangle$ is primitive, and this time the change in parity of $\tau _3$ gives $\Gamma \cong S_{50}$. Also $\{y,2,4,8\}$ is an orbit of $\langle \sigma _2, \sigma _3 \rangle$, on which $\sigma _2$ induces the $3$-cycle $(2,4,8)$ and $\sigma _3$ induces the $4$-cycle $(y,2,8,4)$, and hence the intersection condition holds, again by Lemma 3.1. Chirality follows from the same argument as for $A_{50}$ above.

Thus $S_{50}$ is the automorphism group of a chiral $4$-polytope of type $\{3,3,5460\}$.

Now we can repeat the process begun in Sect. 4.3 and introduce further orbits of length $6$ for $A_4$ near the ‘other end’. As before, this adds extra $6$-cycles to the cycle structure for $\sigma _3$, but does not affect the proof of primitivity, and therefore still gives the group $\Gamma = \langle \tau _1,\tau _2,\tau _3 \rangle$ as either $A_n$ or $S_n$ each time. Also verification of the intersection condition and proof of chirality are entirely analogous to those for the $A_{50}$ case, above.

Thus we have the following: for every integer $n \ge 50$ such that $n \equiv 2$ mod $6$, both $A_n$ and $S_n$ are the automorphism groups of chiral $4$ -polytopes of type $\{3,3,k\}$ for some $k$.

In fact, $k$ can be taken as $2184$ or $5460$, depending on the residue class of $n$ mod $12$, and in particular, our construction shows there are infinitely many chiral $4$-polytopes of type $\{3,3,k\}$ for each of these two values of $k$.

Moreover, we can make the same adjustment as in Sect. 4.4, by adding an extra fixed point of $\langle \tau _1,\tau _2 \rangle \cong A_4$ at the ‘other end’. In this case, the order of $\sigma _3$ changes from $2184$ to $2184/7 = 312$, or from $5460$ to $2\cdot 5460/7 = 1560$, respectively, and the permutations $\sigma _3^{\,24}$ and $\sigma _3^{\,120}$ are $13$-cycles, namely $\zeta = (3, z, 16, 28, 39, 25, 13, v, 10, 22, 34, 31, 19)$ and $\zeta ^5 = (3, 25, 34, 16, v, 19, 39, 22,$ $z, 13, 31, 28, 10)$. Again it is easy to verify primitivity, and deduce that $\Gamma = \langle \tau _1,\tau _2,\tau _3 \rangle$ is isomorphic to $A_n$ or $S_n$. Also the intersection condition holds for exactly the same reasons as before, but again, the proof of chirality needs a small variation. This time there are three fixed points of $\tau _1$, two of which are interchanged by $\tau _3$. If there exists an involutory permutation $\rho$ of the points that inverts each $\tau _i$, then it must fix or interchange those two, and then the argument follows in the same way as in Sect. 4.4.

Thus we have the following: for every integer $n \ge 51$ such that $n \equiv 3$ mod $6$, both $A_n$ and $S_n$ are the automorphism groups of chiral $4$ -polytopes of type $\{3,3,k\}$ for some $k$.

In fact, $k$ can be taken as $312$ or $1560$, depending on the residue class of $n$ mod $12$, and in particular, our construction shows there are infinitely many chiral $4$-polytopes of type $\{3,3,k\}$ for each of these two values of $k$.

4.6 Adding a third orbit of $A_4$ of length 4

We are left with the cases of degree $n \equiv 0$ or $1$ mod $6$. For these, we start with our constructions from the previous subsection for degrees congruent to $2$ or $3$ mod $6$ (beginning with $50$ and $51$), and adjoin a third orbit of length $4$ for $A_4$, at the same end as the second such orbit, with a $\tau _3$-link of type III to itself. This is illustrated in Fig. 5.

Fig. 5

Adding a third orbit of length $4$ for $A_4$

Specifically, we introduce another four new points, labelled $p, q, r$ and $s$, with the assumption that $p < q < r < s$ and make the following changes to the three permutations $\tau _i$ used for generating $A_{n-4}$ or $S_{n-4\,}$:

• $\tau _1$: adjoin the $3$-cycle $(p,q,r)$ and the fixed point $s$,

• $\tau _2$: adjoin the transpositions $(p,q)$ and $(r,s)$,

• $\tau _3$: replace the fixed point $v$ by $(p,r)(s,v)$, fixing $q$.

Obviously this increases the degree by $4$, from $n- 4$ to $n$, and in all cases the only effect on the permutation $\sigma _3$ is to replace the $13$-cycle $(v,3,13,19,25,31,39,34,28,$ $22,16,10,z)$ by the cycle $(v, 3, 13, 19, 25, 31, 39, 34, 28, 22, 16, 10, z, s, p, q, r)$, which has length $17$.

In particular, the order of $\sigma _3$ changes from $2184$ or $5460$ to $2856$ or $7140$ when $n-4 \equiv 4$ mod $6$, and from $312$ or $1560$ to $408$ or $2040$ when $n-4 \equiv 5$ mod $6$.

In all cases, some power of $\sigma _3$ is a single $17$-cycle containing all the points $p, q$ and $r$, and this can be used to prove that $\Gamma = \langle \tau _1,\tau _2,\tau _3 \rangle$ is primitive, since it contains the point $p$ and its images under each of the generators of $\Gamma$. It follows that $\Gamma = A_n$ or $S_n$, again depending on the parity of $\tau _3$.

The intersection condition holds for the same reasons as in the previous subsection, but again a little more care is needed to prove chirality. When $n \equiv 0$ mod $6$, there are three fixed points of $\tau _1$, and two of them (namely $s$ and $v$) are interchanged by $\tau _3$, while the third one (at the ‘other end’) is fixed by $\tau _3$. Hence any permutation $\rho$ in $S_n$ that conjugates each $\tau _i$ to its inverse must fix the third one, and then chirality follows from the same argument as for degree $46$. On the other hand, when $n \equiv 1$ mod $6$, there are four fixed points of $\tau _1$, with two at each end, both interchanged by $\tau _3$. Just one of those, however, is a fixed point of $\tau _2$, and so it is fixed by any such $\rho$, and then chirality follows from the same argument as for degree $47$.

Thus we have the following: for every integer $n \ge 54$ such that $n \equiv 0$ or $1$ mod $6$, both $A_n$ and $S_n$ are the automorphism groups of chiral $4$-polytopes of type $\{3,3,k\}$ for some $k$.

In fact, $k$ can be taken as $2856$ or $7140$ when $n \equiv 0$ mod $6$, and as $408$ or $2040$ when $n \equiv 1$ mod $6$, in both cases depending on the residue class of $n$ mod $12$, and in particular, our construction shows there are infinitely many chiral $4$-polytopes of type $\{3,3,k\}$ for each of these four values of $k$.

5 Vertex-figures

In this section we prove the following:

Theorem 5.1

The vertex-figures of the polytopes constructed in Sect. 4 are all chiral.

Again there is some variation in the argument of different residue classes of $n$ mod $6$, but the approach is much the same in all cases.

Proof

Let $\tau _1, \tau _2$ and $\tau _3$ be the generators of $\Gamma (\mathcal {P})$ as given, and take $\sigma _1=\tau _1, \,\sigma _2=\tau _1^{-1}\tau _2$ and $\sigma _3=\tau _2^{-1}\tau _3=\tau _2\tau _3$ as before. Then the subgroup $\Gamma _0$ generated by $\sigma _2$ and $\sigma _3$ is the rotation group of a vertex-figure of $\mathcal {P}$.

It is easy to verify that the group $\Gamma _0 = \langle \sigma _2,\sigma _3 \rangle$ always has two orbits on the $n$-point set $X$, one of which has length $3$ or $4$, with the other having length $n - 3$ or $n-4$. Indeed if $n \equiv 4$ or $5$ mod $6$, the small orbit $Y$ is $\{2,4,8\}$, while otherwise $Y$ is $\{2,4,8,y\}$, where $y$ is the middle point of the $3$-cycle $(x,y,z)$ of the ‘second’ $A_4$-orbit of length $4$, which is linked by $\tau _3$ to the $A_4$-orbit of length $12$ as in Sects. 4.5 and 4.6.

Also some power $\xi$ of $\sigma _3$ is either the $11$-cycle $(3,25,34,16,13,31,28,10,19,39,22)$, or the $13$-cycle $(v,3,25,34,16,13,31,28,10,19,39,22,z)$, where $v$ and $z$ are another two of the four points of the second $A_4$-orbit of length $4$ introduced in 4.5, or the $17$-cycle $(v,3,25,34,16,13,31,28,10,19,39,22,z,s,p,q,r)$, where $p,q,r$ and $s$ are the four points of the third $A_4$-orbit of length $4$ introduced in 4.6.

We will first show that $\Gamma _0$ acts on the set $Z = X \setminus Y$ as an alternating or symmetric group of degree $|Z| = n -3$ or $n-4$ and then show that $\Gamma _0$ admits no automorphism that inverts both $\sigma _2$ and $\sigma _3$, which is enough to prove chirality of the vertex-figures.

Suppose $\Gamma _0$ is imprimitive on $Z$. Then all the points of the cycle $\xi$ of prime length (obtained as a power of $\sigma _3$) lie in the same block of imprimitivity, say $U$. Now $U$ is preserved by $\sigma _3$ and so cannot be preserved by $\sigma _2$, and furthermore, since $\sigma _2$ has order $3$, the images of $U$ under $\sigma _2$ and its inverse $\sigma _2^{\,2}$ must be new blocks $V$ and $W$. Next, in all cases, $\sigma _2$ takes 10–5, 19–21 and 14–16, while $\sigma _3$ takes 5–14 and 14–21. It follows that $V$ contains $10^{\sigma _2} = 5$ and $19^{\sigma _2} = 21$, while $W$ contains $16^{\sigma _2^{\,2}} = 14$, and therefore $\sigma _3$ interchanges $V$ and $W$. Hence there are just three blocks, cyclically permuted by $\sigma _2$. But also $\sigma _2$ fixes at least one point, namely one of the points of the first $A_4$-orbit of length $4$, and so $\sigma _2$ preserves at least one block, a contradiction.

Thus $\Gamma _0$ is primitive on $Z = X \setminus Y$. Moreover, the existence of the prime cycle $\xi$ shows that $\Gamma _0$ is alternating or symmetric on $Z$ (by Jordan’s theorem).

On the other hand, $\sigma _2$ induces $(2,4,8)$ on $Y$, and $\sigma _3$ induces either $(2,4)$ or $(2,8,4)$ on $Y$ when $|Y| = 3$, or $(2,4,y)$ or $(2,8,4,y)$ on $Y$ when $|Y| = 4$, so $\Gamma _0 = \langle \sigma _2,\sigma _3 \rangle$ acts on $Y$ as $S_3, A_3, A_4$ or $S_4$. It follows that $\Gamma _0$ is isomorphic to a sub-direct product $G_1 \times G_2$ where $G_1 = A_{n-3}, S_{n-3}, A_{n-4}$ or $S_{n-4}$, and $G_2 = S_3, A_3, A_4$ or $S_4$. (Recall that a sub-direct product of groups $G_1$ and $G_2$ is a subgroup $G$ of $G_1 \times G_2$ with the property that the restrictions to $G$ of the projections $\pi _i : G_1 \times G_2 \rightarrow G_i$ are both surjective.)

Now each of $A_{n-3}, S_{n-3}, A_{n-4}$ and $S_{n-4}$ is insoluble, with no non-trivial abelian normal subgroup, while $A_3, S_3, A_4$ and $S_4$ are soluble, and so the kernel $K$ of the action of $\Gamma _0$ on $Z = X \setminus Y$ is the largest soluble normal subgroup of $\Gamma _0$ and is therefore characteristic in $\Gamma _0$ (that is, invariant under all automorphisms of $\Gamma _0$). Thus every automorphism of $\Gamma _0$ induces an automorphism of the group $\Pi _0 \cong \Gamma _0/K$ induced by $\Gamma _0$ on $Z,$ which of course is $A_{n-3}, S_{n-3}, A_{n-4}$ or $S_{n-4}$.

Next, suppose the vertex-figures of $\mathcal {P}$ are regular, so that $\Gamma _0$ has an automorphism that inverts both $\sigma _2$ and $\sigma _3$. Then by the above argument, this automorphism induces an automorphism of $\Pi _0$ which inverts the permutations induced by $\sigma _2$ and $\sigma _3$ on $Z$. Also by Theorem , we know that the latter can be viewed as a permutation on $Z$. We can therefore complete the proof of chirality by showing that there is no permutation $\rho$ in $\mathrm{Sym}(Z)$ that conjugates each of $\sigma _2$ and $\sigma _3$ to its inverse.

In exactly half of the cases we have considered, the permutation $\sigma _3$ has exactly two fixed points, namely $6$ and $8$. These are the cases where $\tau _3$ links the second and third orbits of $\langle \tau _1 \rangle$ in the $A_4$-orbit of length $12$, or equivalently, where $\tau _3$ contains the transpositions $(4,7), (5,9)$ and $(6,8)$. In all these cases, $(1,10,5), (6, 12, 9)$ and $(14,16,18)$ are $3$-cycles of $\sigma _2$, and $(5, 14, 21, 17, 12, 9, 15, 11)$ and $(18,20,27,23)$ are an $8$-cycle and a $4$-cycle of $\sigma _3$, and the point $1$ lies in a $2$-cycle or $3$-cycle of $\sigma _3$.

Now $\rho$ must fix the unique fixed point of $\sigma _3$ on $Z = X \setminus Y$, namely $6$, and therefore $\rho$ interchanges the other two points $9$ and $12$ of the $3$-cycle $(6, 12, 9)$ of $\sigma _2$. It follows that conjugation by $\rho$ inverts the $8$-cycle $(5, 14, 21, 17, 12, 9, 15, 11)$ of $\sigma _3$, and hence interchanges the points $5$ and $14$, and must then conjugate the $3$-cycle $(1,10,5)$ of $\sigma _2$ to the inverse of the $3$-cycle $(14,16,18)$ of $\sigma _2$. Hence $\rho$ interchanges the points $1$ and $18$. But that is impossible, since $18$ lies in a $4$-cycle of $\sigma _3$, while $1$ lies in a $2$-cycle or $3$-cycle of $\sigma _3$.

In the other half of all cases, $\sigma _3$ has no fixed points, but has a unique $5$-cycle, namely $(5, 14, 21, 17, 12)$, and this must be inverted by $\rho$, and the same is true for the prime cycle $\xi$ of length $11, 13$ or $17$. Now each of the four points $5, 14, 21$ and $17$ of the $5$-cycle $(5, 14, 21, 17, 12)$ of $\sigma _3$ lies in a $3$-cycle of $\sigma _2$ that has a point in common with the prime cycle $\xi$, but the fifth point $12$ does not have this property. Hence $\rho$ fixes the point $12$ and therefore must interchange the other two points $6$ and $9$ of the $3$-cycle $(6, 12, 9)$ of $\sigma _2$.

In all these remaining cases, the point $9$ lies in a $3$-cycle of $\sigma _3$, namely $(9,15,11)$, and it follows that the point $6$ must also lie in a $3$-cycle of $\sigma _3$. In the cases where there are two or more $A_4$-orbits of length $4$ (and $\sigma _3$ has no fixed points), the point $6$ lies in the $4$-cycle $(1,6,7,x)$ of $\sigma _3$, and so we can ignore those. This leaves only the cases where there is just one $A_4$-orbit of length $4$, namely those with $n \equiv 4$ or $5$ mod $6$. For these, we consider what happens locally around the single $A_4$-orbit of length $4$.

When $n \equiv 4$ mod $6$ (as in the case $n = 46$ and its extensions considered in Sects. 4.2 and 4.3), we may label the points of $X$ such that the generators $\tau _i$ of $\Gamma$ have the following forms:

$$\begin{aligned} \tau _1&= \dots (n-12,n-11,n-10)(n-9,n-8,n-7)(n-6,n-5,n-4)(n-3,n-2,n-1),\\ \tau _2&= \dots (n-15,n-12)(n-14,n-11)(n-9,n-6)(n-8,n-5)(n-3,n-2)(n-1,n),\\ \tau _3&= \dots (n-12,n-7)(n-11,n-8)(n-10,n-9) (n-6,n-3)(n-5,n-1)(n-4,n-2) \\& \text{ or } \dots (n-12,n-9)(n-11,n-7)(n-10,n-8) (n-6,n-1)(n-5,n-2)(n-4,n-3). \end{aligned}$$

With this labelling, $n-2$ is the only fixed point of $\sigma _2$, and this lies in a $6$-cycle of $\sigma _3$, which is $(n-10,n-9,n-3,n-4,n-2,n-6)$ when $n \equiv 10$ mod $12$ (such as when $n = 46$), or $(n-10,n-8,n-2,n-4,n-3,n-5)$ when $n \equiv 4$ mod $12$ (such as when $n = 52$). Also the unique $7$-cycle of $\sigma _3$ is $(n-16,n-14,n-8,n-1,n,n-5,n-11)$ when $n \equiv 10$ mod $12$, or $(n-16,n-15,n-9,n-1,n,n-6,n-12)$ when $n \equiv 4$ mod $12$.

In both cases $\rho$ must fix the only fixed point of $\sigma _2$, namely $n-2$, and so the $6$-cycle of $\sigma _3$ containing $n-2$ must be inverted by $\rho$. When $n \equiv 10$ mod $12$, this implies that $\rho$ fixes $n-9$ and hence interchanges the two other points $n-7$ and $n-5$ of the $3$-cycle $(n-9,n-7,n-5)$ of $\sigma _2$. But that is impossible, since $n-5$ lies in a $7$-cycle of $\sigma _3$, while $n-7$ does not. Similarly, when $n \equiv 4$ mod $12$, we find that $\rho$ fixes $n-5$, and hence swaps $n-7$ and $n-9$, which is impossible since $n-9$ lies in a $7$-cycle of $\sigma _3$, while $n-7$ does not.

A similar approach works when $n \equiv 5$ mod $6$. In this case $\sigma _2$ has two fixed points, one being the (unique) point fixed by $\langle \tau _1,\tau _2 \rangle \cong A_4$. This lies in an $8$-cycle of $\sigma _3$, while the other lies in a $6$-cycle of $\sigma _3$, and hence both points must be fixed by $\rho$. Then the same argument as in the case $n \equiv 4$ mod $6$ shows that two points from a $3$-cycle of $\sigma _2$ are interchanged by $\rho$, but one of them lies in the $8$-cycle of $\sigma _3$ while the other does not. Hence no such $\rho$ exists, and this completes the proof.