Chiral polytopes of order $2p^m$

Ting-Ting Kong; Yan-Quan Feng; Dong-Dong Hou; Dimitri Leemans; Hai-Peng Qu

arXiv:2508.20654·math.GR·August 29, 2025

Chiral polytopes of order $2p^m$

Ting-Ting Kong, Yan-Quan Feng, Dong-Dong Hou, Dimitri Leemans, Hai-Peng Qu

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TL;DR

This paper classifies chiral polytopes with automorphism groups of order twice a prime power, revealing their structure and conditions for tightness, and constructing examples for various parameters.

Contribution

It provides a detailed structural analysis of chiral polytopes with automorphism groups of order 2p^m, including conditions for tightness and existence of non-tight examples.

Findings

01

G ≅ P ⋉ Z_2 with specific properties

02

Characterization of tight polytopes as metacyclic groups

03

Existence of non-tight polytopes for certain m values

Abstract

Let $P$ be a chiral polytope with type ${k_{1}, k_{2}}$ and $G = A u t (P)$ . Suppose $∣ G ∣ = 2 p^{m}$ , where $k_{1}, k_{2} \geq 3$ and $p$ is an odd prime. Let $P$ be a Sylow $p$ -subgroup of $G$ . We prove that $G ≅ P ⋊ Z_{2}$ , $d (P) = 2$ , $P^{'} \neq = 1$ (so $m \geq 3$ ) and up to duality, ${k_{1}, k_{2}} = {p^{l_{1}}, 2 p^{l_{2}}}$ for some integral $l_{1}, l_{2} \geq 1$ . Moreover, we show that $P$ is tight $(k_{1} k_{2} = 2 p^{m}$ ) if and only if $P$ is metacyclic group. Furthermore, if $m = 3$ or $4$ , then $P$ must be tight, and if $m \geq 5$ , where either $m$ is odd, or $m$ is even and $m \geq p + 3$ , there exists a non-tight chiral polytope $P$ .

Tables1

Table 1. Table 1. The values of u i u_{i} .

[\begin{matrix} \begin{matrix} * & {(s_{1}^{σ})}^{- (\frac{p}{1})} & {(s_{2}^{σ})}^{- (\frac{p}{2})} & {(s_{3}^{σ})}^{- (\frac{p}{3})} & {(s_{4}^{σ})}^{- (\frac{p}{4})} & \dots & \dots & {(s_{p - 1}^{σ})}^{- (\frac{p}{p - 1})} \\ u_{1} & - (\frac{p}{1}) & 0 & 0 & 0 & \dots & \dots & 0 \\ u_{2} & 0 & (\frac{p - 1}{0}) ​ (\frac{p}{2}) & 0 & 0 & \dots & \dots & 0 \\ u_{3} & 0 & (\frac{p - 1}{1}) ​ (\frac{p}{2}) & - (\frac{p - 2}{0}) ​ (\frac{p}{3}) & 0 & \dots & \dots & 0 \\ u_{4} & 0 & (\frac{p - 1}{2}) ​ (\frac{p}{2}) & - (\frac{p - 2}{1}) ​ (\frac{p}{3}) & (\frac{p - 3}{0}) ​ (\frac{p}{4}) & \dots & \dots & 0 \\ u_{5} & 0 & (\frac{p - 1}{3}) ​ (\frac{p}{2}) & - (\frac{p - 2}{2}) ​ (\frac{p}{3}) & (\frac{p - 3}{1}) ​ (\frac{p}{4}) & \dots & \dots & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & \dots & \dots & ⋮ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & \dots & \dots & ⋮ \\ u_{p - 1} & 0 & (\frac{p - 1}{p - 3}) ​ (\frac{p}{2}) & - (\frac{p - 2}{p - 4}) ​ (\frac{p}{3}) & (\frac{p - 3}{p - 5}) ​ (\frac{p}{4}) & \dots & \dots & (\frac{2}{0}) ​ (\frac{p}{p - 1}) \\ u_{p} & 0 & (\frac{p - 1}{p - 2}) ​ (\frac{p}{2}) & - (\frac{p - 2}{p - 3}) ​ (\frac{p}{3}) & (\frac{p - 3}{p - 4}) ​ (\frac{p}{4}) & \dots & \dots & (\frac{2}{1}) ​ (\frac{p}{p - 1}) \\ u_{p + 1} & 0 & (\frac{p - 1}{p - 1}) ​ (\frac{p}{2}) & - (\frac{p - 2}{p - 2}) ​ (\frac{p}{3}) & (\frac{p - 3}{p - 3}) ​ (\frac{p}{4}) & \dots & \dots & (\frac{2}{2}) ​ (\frac{p}{p - 1}) \end{matrix} \end{matrix}]

Equations158

⟨ ρ_{0}, ρ_{1}, \dots, ρ_{n - 1} ∣ ρ_{i}^{2} = 1 for 0 \leq i \leq n - 1, (ρ_{i} ρ_{i + 1})^{k_{i + 1}} = 1 for 0 \leq i \leq n - 2,

⟨ ρ_{0}, ρ_{1}, \dots, ρ_{n - 1} ∣ ρ_{i}^{2} = 1 for 0 \leq i \leq n - 1, (ρ_{i} ρ_{i + 1})^{k_{i + 1}} = 1 for 0 \leq i \leq n - 2,

(ρ_{i} ρ_{j})^{2} = 1 for 0 \leq i < j - 1 < n - 1 ⟩ .

(ρ_{i} ρ_{j})^{2} = 1 for 0 \leq i < j - 1 < n - 1 ⟩ .

M_{p} (1, 1, 1) = ⟨ a, b, c ∣ a^{p}, b^{p}, c^{p}, c = [a, b], [a, c], [b, c]⟩ ≅ (Z_{p} \times Z_{p}) ⋊ Z_{p} .

M_{p} (1, 1, 1) = ⟨ a, b, c ∣ a^{p}, b^{p}, c^{p}, c = [a, b], [a, c], [b, c]⟩ ≅ (Z_{p} \times Z_{p}) ⋊ Z_{p} .

P = ⟨ s_{1}, s_{2}, \dots, s_{r}, s_{r + 1}, \dots, s_{p - 1}, β ⟩ \mbox w i t h \mbox t h e \mbox f o l l o w in g \mbox r e l a t i o n s

P = ⟨ s_{1}, s_{2}, \dots, s_{r}, s_{r + 1}, \dots, s_{p - 1}, β ⟩ \mbox w i t h \mbox t h e \mbox f o l l o w in g \mbox r e l a t i o n s

s_{1}^{p^{e}}, s_{2}^{p^{e}}, \dots, s_{r}^{p^{e}}, s_{r + 1}^{p^{e - 1}}, \dots s_{p - 1}^{p^{e - 1}},

s_{1}^{p^{e}}, s_{2}^{p^{e}}, \dots, s_{r}^{p^{e}}, s_{r + 1}^{p^{e - 1}}, \dots s_{p - 1}^{p^{e - 1}},

β^{p^{2}}, β^{p} = s_{r}^{p^{e - 1}},

β^{p^{2}}, β^{p} = s_{r}^{p^{e - 1}},

[s_{i}, s_{j}], for 1 \leq i, j \leq p - 1, \mbox an d s_{k + 1} = [s_{k}, β], for 1 \leq k \leq p - 2,

[s_{i}, s_{j}], for 1 \leq i, j \leq p - 1, \mbox an d s_{k + 1} = [s_{k}, β], for 1 \leq k \leq p - 2,

[s_{p - 1}, β] = s_{1}^{- (1 p)} s_{2}^{- (2 p)} \dots s_{p - 2}^{- (p - 2 p)} s_{p - 1}^{- (p - 1 p)},

[s_{p - 1}, β] = s_{1}^{- (1 p)} s_{2}^{- (2 p)} \dots s_{p - 2}^{- (p - 2 p)} s_{p - 1}^{- (p - 1 p)},

1 \leq e, 1 \leq r \leq p - 1, m = (p - 1) (e - 1) + r + 1 \geq 3.

1 \leq e, 1 \leq r \leq p - 1, m = (p - 1) (e - 1) + r + 1 \geq 3.

s_{p} = [s_{p - 1}, β] = i = 1 \prod p - 1 s_{i}^{- (i p)},

s_{p} = [s_{p - 1}, β] = i = 1 \prod p - 1 s_{i}^{- (i p)},

s_{p + 1} = [s_{p}, β] = s_{p}^{- 1} s_{p}^{β} = i = 1 \prod p - 1 (s_{i}^{- 1})^{- (i p)} \cdot i = 1 \prod p - 1 (s_{i}^{β})^{- (i p)}

s_{p + 1} = [s_{p}, β] = s_{p}^{- 1} s_{p}^{β} = i = 1 \prod p - 1 (s_{i}^{- 1})^{- (i p)} \cdot i = 1 \prod p - 1 (s_{i}^{β})^{- (i p)}

= i = 1 \prod p - 1 (s_{i}^{- 1} s_{i}^{β})^{- (i p)} = i = 1 \prod p - 1 [s_{i}, β]^{- (i p)} = i = 1 \prod p - 1 s_{i + 1}^{- (i p)} .

= i = 1 \prod p - 1 (s_{i}^{- 1} s_{i}^{β})^{- (i p)} = i = 1 \prod p - 1 [s_{i}, β]^{- (i p)} = i = 1 \prod p - 1 s_{i + 1}^{- (i p)} .

s_{i + 1}^{τ} = [s_{i}, β]^{τ} = [s_{i}^{- 1}, β] = (([s_{i}, β])^{- 1})^{s_{i}^{- 1}} = (s_{i + 1}^{- 1})^{s_{i}^{- 1}} = s_{i + 1}^{- 1} .

s_{i + 1}^{τ} = [s_{i}, β]^{τ} = [s_{i}^{- 1}, β] = (([s_{i}, β])^{- 1})^{s_{i}^{- 1}} = (s_{i + 1}^{- 1})^{s_{i}^{- 1}} = s_{i + 1}^{- 1} .

s_{k}^{β^{1 - k}} = s_{k}^{β^{p + 1 - k}} = (s_{k} s_{k + 1})^{β^{p - k}} = (s_{k} s_{k + 1}^{2} s_{k + 2})^{β^{p - k - 1}} = \dots

s_{k}^{β^{1 - k}} = s_{k}^{β^{p + 1 - k}} = (s_{k} s_{k + 1})^{β^{p - k}} = (s_{k} s_{k + 1}^{2} s_{k + 2})^{β^{p - k - 1}} = \dots

= s_{k}^{(0 p + 1 - k)} s_{k + 1}^{(1 p + 1 - k)} s_{k + 2}^{(2 p + 1 - k)} \dots s_{p}^{(p - k p + 1 - k)} s_{p + 1}^{(p + 1 - k p + 1 - k)} .

= s_{k}^{(0 p + 1 - k)} s_{k + 1}^{(1 p + 1 - k)} s_{k + 2}^{(2 p + 1 - k)} \dots s_{p}^{(p - k p + 1 - k)} s_{p + 1}^{(p + 1 - k p + 1 - k)} .

s_{2}^{σ} = [s_{1}, β]^{σ} = [s_{1}, β^{- 1}] = ([s_{1}, β]^{- 1})^{β^{- 1}} = (s_{2}^{- 1})^{β^{1 - 2}} .

s_{2}^{σ} = [s_{1}, β]^{σ} = [s_{1}, β^{- 1}] = ([s_{1}, β]^{- 1})^{β^{- 1}} = (s_{2}^{- 1})^{β^{1 - 2}} .

s_{k}^{σ} = [s_{k - 1}^{σ}, β^{σ}] = [(s_{k - 1}^{β^{2 - k}})^{- 1}, β^{- 1}] = [s_{k - 1}^{- 1}, β^{- 1}]^{β^{2 - k}} = ([s_{k - 1}, β]^{β^{- 1} s_{k - 1}^{- 1}})^{β^{2 - k}} .

s_{k}^{σ} = [s_{k - 1}^{σ}, β^{σ}] = [(s_{k - 1}^{β^{2 - k}})^{- 1}, β^{- 1}] = [s_{k - 1}^{- 1}, β^{- 1}]^{β^{2 - k}} = ([s_{k - 1}, β]^{β^{- 1} s_{k - 1}^{- 1}})^{β^{2 - k}} .

s_{k}^{σ} = ([s_{k - 1}, β]^{β^{- 1} s_{k - 1}^{- 1}})^{β^{2 - k}} = s_{k}^{(s_{k - 1}^{- 1})^{β} β^{1 - k}} = s_{k}^{β^{1 - k}} .

s_{k}^{σ} = ([s_{k - 1}, β]^{β^{- 1} s_{k - 1}^{- 1}})^{β^{2 - k}} = s_{k}^{(s_{k - 1}^{- 1})^{β} β^{1 - k}} = s_{k}^{β^{1 - k}} .

s_{k}^{σ} = [s_{k - 1}^{σ}, β^{σ}] = [s_{k - 1}^{β^{2 - k}}, β^{- 1}] = [s_{k - 1}, β^{- 1}]^{β^{2 - k}} = (([s_{k - 1}, β]^{- 1})^{β^{- 1}})^{β^{2 - k}} = (s_{k}^{β^{1 - k}})^{- 1} .

s_{k}^{σ} = [s_{k - 1}^{σ}, β^{σ}] = [s_{k - 1}^{β^{2 - k}}, β^{- 1}] = [s_{k - 1}, β^{- 1}]^{β^{2 - k}} = (([s_{k - 1}, β]^{- 1})^{β^{- 1}})^{β^{2 - k}} = (s_{k}^{β^{1 - k}})^{- 1} .

s_{i}^{\sigma}=\left\{\begin{array}[]{ll}s_{i}^{\beta^{1-i}},&\mbox{ for $i$ odd,}\\ (s_{i}^{\beta^{1-i}})^{-1},&\mbox{ for $i$ even.}\end{array}\right.

s_{i}^{\sigma}=\left\{\begin{array}[]{ll}s_{i}^{\beta^{1-i}},&\mbox{ for $i$ odd,}\\ (s_{i}^{\beta^{1-i}})^{-1},&\mbox{ for $i$ even.}\end{array}\right.

[s_{p - 1}^{σ}, β^{σ}] = (s_{1}^{σ})^{- (1 p)} (s_{2}^{σ})^{- (2 p)} (s_{3}^{σ})^{- (3 p)} \dots (s_{p - 1}^{σ})^{- (p - 1 p)},

[s_{p - 1}^{σ}, β^{σ}] = (s_{1}^{σ})^{- (1 p)} (s_{2}^{σ})^{- (2 p)} (s_{3}^{σ})^{- (3 p)} \dots (s_{p - 1}^{σ})^{- (p - 1 p)},

(s_{r}^{p^{e - 1}})^{σ} = (s_{r}^{σ})^{p^{e - 1}} = ((s_{r}^{β^{1 - r}})^{- 1})^{p^{e - 1}} = ((s_{r}^{p^{e - 1}})^{β^{1 - r}})^{- 1} = ((β^{p})^{β^{1 - r}})^{- 1} = β^{- p} = (β^{p})^{σ} .

(s_{r}^{p^{e - 1}})^{σ} = (s_{r}^{σ})^{p^{e - 1}} = ((s_{r}^{β^{1 - r}})^{- 1})^{p^{e - 1}} = ((s_{r}^{p^{e - 1}})^{β^{1 - r}})^{- 1} = ((β^{p})^{β^{1 - r}})^{- 1} = β^{- p} = (β^{p})^{σ} .

s_{1}^{σ} = s_{1}, s_{2}^{σ} = i = 2 \prod p + 1 s_{i}^{- (i - 2 p - 1)}, s_{3}^{σ} = i = 3 \prod p + 1 s_{i}^{(i - 3 p - 2)}, \dots, s_{p - 1}^{σ} = i = p - 1 \prod p + 1 s_{i}^{- (i - ( p - 1 ) 2)} .

s_{1}^{σ} = s_{1}, s_{2}^{σ} = i = 2 \prod p + 1 s_{i}^{- (i - 2 p - 1)}, s_{3}^{σ} = i = 3 \prod p + 1 s_{i}^{(i - 3 p - 2)}, \dots, s_{p - 1}^{σ} = i = p - 1 \prod p + 1 s_{i}^{- (i - ( p - 1 ) 2)} .

u_{p + 1} = j = 2 \sum p - 1 (- 1)^{j} (j p) = (p - 1 p) + j = 2 \sum \frac{p - 3}{2} [(- 1)^{j} (j p) + (- 1)^{p - j} (p - j p)] = (p - 1 p) = p .

u_{p + 1} = j = 2 \sum p - 1 (- 1)^{j} (j p) = (p - 1 p) + j = 2 \sum \frac{p - 3}{2} [(- 1)^{j} (j p) + (- 1)^{p - j} (p - j p)] = (p - 1 p) = p .

u_{p} - (0 1) (p p) = u_{p} - 1 = j = 2 \sum p (- 1)^{j} (p - j p + 1 - j) (j p) .

u_{p} - (0 1) (p p) = u_{p} - 1 = j = 2 \sum p (- 1)^{j} (p - j p + 1 - j) (j p) .

(i - j p + 1 - j) (j p) = [(i - j p - j) + (i - j - 1 p - j)] (j p) = (i - j p - j) (j p) + (i - j - 1 p - j) (j p) = (i p) (j i) + (i - 1 p) (j i - 1) .

(i - j p + 1 - j) (j p) = [(i - j p - j) + (i - j - 1 p - j)] (j p) = (i - j p - j) (j p) + (i - j - 1 p - j) (j p) = (i p) (j i) + (i - 1 p) (j i - 1) .

j = 2 \sum i (- 1)^{j} [(i p) (j i) + (i - 1 p) (j i - 1)]

j = 2 \sum i (- 1)^{j} [(i p) (j i) + (i - 1 p) (j i - 1)]

= (i p) [i - 1 + j = 0 \sum i (- 1)^{j} (j i)] + (i - 1 p) [i - 2 + j = 0 \sum i (- 1)^{j} (j i - 1)]

= (i p) [i - 1 + j = 0 \sum i (- 1)^{j} (j i)] + (i - 1 p) [i - 2 + j = 0 \sum i (- 1)^{j} (j i - 1)]

= (i p) [i - 1 + (- 1)^{i} (i i - 1)] + (i - 1 p) [i - 2 + (- 1)^{i} (i i - 2)]

= (i p) [i - 1 + (- 1)^{i} (i i - 1)] + (i - 1 p) [i - 2 + (- 1)^{i} (i i - 2)]

= (i p) (i - 1) + (i - 1 p) (i - 2)

= (i p) (i - 1) + (i - 1 p) (i - 2)

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TopicsAdvanced Combinatorial Mathematics · Advanced Mathematical Identities · Point processes and geometric inequalities

Full text

Chiral polytopes of order $2p^{m}$

Ting-Ting Kong

Ting-Ting Kong, Department of Mathematics, Shanxi Key Laboratory of Cryptography and Date Security, Shanxi Normal University, TaiYuan, 030031, P.R. China

[email protected]

,

Yan-Quan Feng

Yan-Quan Feng, Department of Mathematics, Beijing Jiaotong University, Beijing, 100044, P.R. China

[email protected]

,

Dong-Dong Hou

Dong-Dong Hou, Department of Mathematics, Shanxi Normal University, TaiYuan, 030031, P.R. China

[email protected]

,

Dimitri Leemans

Dimitri Leemans, Département de Mathématique, Université Libre de Bruxelles, C.P.216, Boulevard du Triomphe, 1050 Bruxelles Belgium

[email protected]

and

Hai-Peng Qu

Hai-Peng Qu, Department of Mathematics, Shanxi Normal University, TaiYuan, 030031, P.R. China

[email protected]

Abstract.

Let $\mathcal{P}$ be a chiral polytope with type $\{k_{1},k_{2}\}$ and $G=\hbox{\rm Aut}(\mathcal{P})$ . Suppose $|G|=2p^{m}$ , where $k_{1},k_{2}\geq 3$ and $p$ is an odd prime. Let $P$ be a Sylow $p$ -subgroup of $G$ . We prove that $G\cong P\rtimes\mathbb{Z}_{2}$ , $d(P)=2$ , $P^{\prime}\neq 1$ (so $m\geq 3$ ) and up to duality, $\{k_{1},k_{2}\}=\{p^{l_{1}},2p^{l_{2}}\}$ for some integral $l_{1},l_{2}\geq 1$ . Moreover, we show that $\mathcal{P}$ is tight $(k_{1}k_{2}=2p^{m}$ ) if and only if $P$ is metacyclic group. Furthermore, if $m=3$ or $4$ , then $\mathcal{P}$ must be tight, and if $m\geq 5$ , where either $m$ is odd, or $m$ is even and $m\geq p+3$ , there exists a non-tight chiral polytope $\mathcal{P}$ .

Key words and phrases:

Chiral polytope, $p$ -group, automorphism group, soluble group

1991 Mathematics Subject Classification:

20B25, 20D15, 52B10, 52B15

1. Introduction

Abstract polytopes are combinatorial structures with properties that generalise those of classical polytopes. In many ways they are more fascinating than convex polytopes and tessellations. Highly symmetric examples of abstract polytopes include not only classical regular polytopes such as the well known Platonic solids, and more exotic structures such as the $120$ -cell and $600$ -cell, but also regular maps on surfaces (such as Klein’s quartic); see [14, Chapter 8] for example.

Roughly speaking, an abstract polytope is a partially ordered set $(\mathcal{P},\leq)$ (or $\mathcal{P}$ for short) endowed with a rank function, satisfying certain conditions that arise naturally from a geometric setting (see Section 2.1 for definitions). Such objects were proposed by Branko Grünbaum in the seventies; their definition (initially as ‘incidence polytopes’) and theory were developed by Ludwig Danzer and Egon Schulte. They are closely related to ‘polyhedral geometries’ that Jacques Tits defined in 1961 [35].

An automorphism of an abstract polytope $\mathcal{P}$ is an order-preserving permutation of the elements of $\mathcal{P}$ , and every automorphism of $\mathcal{P}$ is uniquely determined by its action on any maximal chain in $\mathcal{P}$ (known as a ‘flag’ of $\mathcal{P}$ ). The most symmetric examples of abstract polytopes are the regular ones, where the automorphism group acts transitively on the set of flags. The comprehensive book written by Peter McMullen and Egon Schulte [28] is considered the main reference on this subject.

Another interesting class of examples is the one of chiral polytopes. For these polytopes, their automorphism group has two orbits on the set of flags, with the extra property that any two flags that differ in a single element lie in different orbits. The study of chiral abstract polytopes was pioneered by Schulte and Asia Ivic Weiss (see [32, 33] for example). 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 Marston Conder, Isabel Hubard and Tomaz Pisanski [12]. Daniel Pellicer was the first to show that chiral polytopes of arbitrarily large rank exist [29].

It is a natural question to try to describe all pairs $(\mathcal{P},G)$ , where $\mathcal{P}$ is a chiral polytope and $G$ is the automorphism group of $\mathcal{P}$ . Conder computed all chiral polytopes with automorphism group of order at most 1000 in 2012***These computations have recently been pushed to cover all polytopes with up to 4000 flags [8].. Roughly at the same time, Michael Hartley, Hubard and Dimitri Leemans built in [17] an atlas of all chiral polytopes whose automorphism group is an almost simple group $G$ such that $S\leq G\leq\hbox{\rm Aut}(S)$ and $S$ is a simple group of order less than 1 million appearing in the Atlas of Finite Groups by Conway et al. Recently, Primoz Potočnik, Pablo Spiga and Gabriel Verret computed all chiral 3-polytopes with automorphism group of order at most 12000 [30].

An interesting case consists of pairs $(\mathcal{P},G)$ with $G$ simple or almost simple. In 2015, Conder, Hubard, Pellicer and Eugenia O’Reilly-Regueiro [11] provided a construction of chiral 4-polytopes for $A_{n}$ and $S_{n}$ . After ten years, Conder, Hubard and O’Reilly-Regueiro extended their initial idea and proved that for every $d\geq 5$ , there exists a chiral $d$ -polytope with automorphism group isomorphic to $A_{n}$ , and a chiral $d$ -polytope with automorphism group isomorphic to $S_{n}$ for all but finitely many $n$ , see [10]. Also, Wei-Juan Zhang gave some new construction for chiral $4$ -polytopes with symmetric automorphism groups [37]. Hubard and Leemans dealt with the Suzuki simple groups in [20]. In a series of papers, the case where the automorphism group of a chiral polyhedra is an almost simple group with socle $\hbox{\rm PSL}(2,q)$ has been studied [13, 16, 24, 25]. Moreover, in 2022, Leemans and Adrien Vandenschrick constructed chiral 4-polytopes with automorphism group $\hbox{\rm PSL}(3,q)$ for every prime power $q\geq 5$ [26]. For chiral $3$ -polytopes, Conder, Potočnik and Jozef Širáň proved that the groups $\hbox{\rm PSL}(2,q)$ cannot be automorphism groups of such polyhedra [13]. Leemans and Martin Liebeck proved that every finite simple group acts as the automorphism group of a chiral $3$ -polytope, apart from the groups $\hbox{\rm PSL}(2,q)$ , $A_{7}$ , and maybe also $\hbox{\rm PSL}(3,q),\hbox{\rm PSU}(3,q)$ and [23]. Antonio Breda and Domenico Catanalo then showed that $\hbox{\rm PSL}(3,q)$ and $\hbox{\rm PSU}(3,q)$ cannot be automorphism groups of chiral polyhedra (with the help of Leemans and Liebeck for the groups $\hbox{\rm PSU}(3,q)$ ) [6].

Another interesting case consists of the pairs $(\mathcal{P},G)$ with $G$ a soluble group. When it comes to soluble groups, except for abelian groups, the first families that come to mind are groups of orders $2^{n}$ or $2^{n}p^{m}$ , where $p$ is an odd prime. Schulte and Weiss mentioned in [34] that the order $2^{n}$ and $2^{n}p$ prove to be more difficult than others and suggested to try to classify chiral polytopes with these orders [34, Problem 30]. Zhang constructed in her PhD thesis [39] a number of chiral polytopes of types $\{4,4,4\}$ , $\{4,4,4,4\}$ and $\{4,4,4,4,4\}$ , with automorphism groups of orders $2^{10},2^{11},2^{12}$ , and $2^{15},2^{16},\cdots,2^{22}$ , and $2^{18},2^{19}$ , respectively. Recently, Zhang presented a general construction for chiral polytopes of type $\{4,4,\cdots,4\}$ with rank $4$ , $5$ and $6$ , which are obtained as boolean covers of the unique tight regular polytope of the same type in [38]. Conder, Yanquan Feng and Dong Dong Hou gave two infinite families of chiral 4-polytopes of type $\{4,4,4\}$ with solvable automorphism groups [9]. In 2024, Hou, T. T. Zheng and R. R. Guo proved that for $n\geq 9$ , there exists a chiral 3-polytope of type $\{4,8\}$ with automorphism group of order $2^{n}$ [18]. Gabe Cunningham gave some infinite families of tight chiral $3$ -polytopes of type $\{k_{1},k_{2}\}$ with automorphism group of order $2k_{1}k_{2}$ [15].

Note that if a chiral polytope $\mathcal{P}$ has rank at least three, the order of its automorphism group $G$ must be even. In this paper, we focus on groups $G$ acting on chiral polytopes of rank three, with $|G|=2p^{m}$ , $p$ an odd prime, and $m$ a strictly positive integer. We first prove the following theorem (where $d(G)$ is the size of a smallest generating set of $G$ ).

Theorem 1.1.

Let $p$ be an odd prime and $m$ be a positive integer. Let $G$ be a group of order $2p^{m}$ . Let $(G,\{\sigma_{1},\sigma_{2}\})$ be the automorphism group of a chiral polyhedron $\mathcal{P}$ of type $\{k_{1},k_{2}\}$ . Then

(1)

$G=P\rtimes\langle\sigma_{1}\sigma_{2}\rangle$ * with $P\in\hbox{\rm Syl}_{p}(G)$ ;*

(2)

Up to duality, $\{k_{1},k_{2}\}=\{p^{l_{1}},2p^{l_{2}}\}$ for some positive integers $l_{1},l_{2}$ ;

(3)

$P=\langle\sigma_{1},\sigma_{2}^{2}\rangle,d(P)=2$ * and $P$ is not abelian (so $m\geq 3$ );*

(4)

$\mathcal{P}$ * is tight, that is $k_{1}k_{2}=2p^{m}$ , if and only if $P$ is a metacyclic group.*

We then prove the following theorem, showing that if $m=3$ or $4$ , then $\mathcal{P}$ is a tight polyhedron.

Theorem 1.2.

Let $p$ be an odd prime and $m=3$ or $4$ . Let $G$ be a group of order $2p^{m}$ . let $(G,\{\sigma_{1},\sigma_{2}\})$ be the automorphism group of a chiral polyhedron $\mathcal{P}$ of type $\{k_{1},k_{2}\}$ . Then $\mathcal{P}$ is tight.

Remark 1.3.

As a by-product of the proof of Theorem 1.2, we obtain the following families of non-tight regular polyhedra:

•

regular polyhedra of type $\{p,2p\}$ of order $4p^{3}$ ;

•

regular polyhedra of type $\{p,2p\}$ of order $4p^{4}$ ;

•

regular polyhedra of type $\{p^{2},2p\}$ of order $4p^{4}$ ;

•

regular polyhedra of type $\{p,2p^{2}\}$ of order $4p^{4}$ .

In particular, the first class and part of the second class of these polyhedra are identical to those Feng, Hou, Leemans and Qu constructed in [19].

Cunningham proved in [15] that there exists a tight chiral polyhedron $\mathcal{P}$ of type $\{p^{l_{1}},2p^{l_{2}}\}$ for $l_{1}\geq 2,l_{2}\geq 1$ and $l_{1}\geq l_{2}$ . In the case where $\mathcal{P}$ is not tight, we prove the following theorem. The proof of that theorem is constructive, meaning that we give an explicit way to construct such a polyhedron.

Theorem 1.4.

Let p be an odd prime. Let $m\geq 5$ be an integer such that either $m$ is odd, or $m$ is even and $m\geq p+3$ . Then there exists a non-tight chiral poyhedron $\mathcal{P}$ with $|\hbox{\rm Aut}(\mathcal{P})|=2p^{m}$ .

Let us point out that, as in the regular case (see [19, Section 1]) chiral polytopes constructed from soluble groups are extremely rare. For example, there are 49,910,526,325 groups of order $\leq 2000$ [3]. These groups are readily accessible in Magma [5] except for those of order 1024. Even though those of order 1024 are not in Magma, we know they are all soluble. One can check that out of all groups of order at most 2000, there are 49,910,525,301 that are soluble groups and 1024 that are non-soluble. According to the data collected by Conder [8], the number of chiral polyhedra having an automorphism group of order at most 2000 that is soluble (resp. non-soluble) is 2665 (resp. 165). Hence the ratio (number of chiral group representations/number of groups) is 0.000000053% for soluble groups and 16.11% for non-soluble groups. This shows that it is really not easy to find soluble groups that are the automorphism group of some chiral polytopes. More importantly, if we refine by rank, we get, in the soluble case, 2389 polytopes of rank 3, 276 of rank 4 and none of rank 5. In the non-soluble case, we get 39 polytopes of rank 3, 123 of rank 4 and three of rank 5.

The paper is organised as follows. In Section 2, we give the necessary background to understand this paper. In Section 3, we give several lemmas about finite $p$ -groups of maximal class that are useful in this paper. In Section 4, we prove Theorem 1.1. In Section 5, we prove Theorem 1.2. Finally, in Section 6, we prove Theorem 1.4.

2. Preliminaries

In this section we give some further background that may be helpful for the rest of the paper (see [2, 22, 28, 32, 36] for more details).

2.1. Abstract polytopes: definition, structure and properties

An abstract polytope of rank $n$ is a partially ordered set $(\mathcal{P},\leq)$ endowed with a strictly monotone rank function with range $\{-1,0,\cdots,n\}$ , which satisfies four properties (P1) to (P4) described below.

The elements of $\mathcal{P}$ are called faces of $\mathcal{P}$ . More specifically, the elements of $\mathcal{P}$ of rank $j$ are called $j$ -faces, and a typical $j$ -face is denoted by $F_{j}$ . Two faces $F$ and $G$ of $\mathcal{P}$ are said to be incident with each other if $F\leq G$ or $F\geq G$ in $\mathcal{P}$ . A chain of $\mathcal{P}$ is a totally ordered subset of $\mathcal{P}$ , and is said to have length $i$ if it contains exactly $i+1$ faces. The maximal chains in $\mathcal{P}$ are called the flags of $\mathcal{P}$ . Two flags are said to be $j$ -adjacent if they differ in just one face of rank $j$ , or simply adjacent (to each other) if they are $j$ -adjacent for some $j$ . Also if $F$ and $G$ are faces of $\mathcal{P}$ with $F\leq G$ , then the set $\{\,H\in\mathcal{P}\ |\ F\leq H\leq G\,\}$ is called a section of $\mathcal{P}$ , and is denoted by $G/F$ . Such a section has rank $m-k-1$ , where $m$ and $k$ are the ranks of $G$ and $F$ respectively. A section of rank $d$ is called a $d$ -section.

We can now give the four conditions that are required of $\mathcal{P}$ to make it an abstract polytope.

(P1)

$\mathcal{P}$ contains a least face and a greatest face, denoted by $F_{-1}$ and $F_{n}$ , respectively.

(P2)

Each flag of $\mathcal{P}$ has length $n+1$ (so has exactly $n+2$ faces, including $F_{-1}$ and $F_{n}$ ).

(P3)

$\mathcal{P}$ is strongly flag-connected, which means that any two flags $\Phi$ and $\Psi$ of $\mathcal{P}$ can be joined by a sequence of successively adjacent flags $\Phi=\Phi_{0},\Phi_{1},\cdots,\Phi_{k}=\Psi$ , each of which contains $\Phi\cap\Psi$ .

(P4)

The rank $1$ sections of $\mathcal{P}$ have a certain homogeneity property known as the diamond condition: if $F$ and $G$ are incident faces of $\mathcal{P}$ , of ranks $i-1$ and $i+1$ , respectively, where $0\leq i\leq n-1$ , then there exist precisely two $i$ -faces $H$ in $\mathcal{P}$ such that $F<H<G$ .

An easy case of the diamond condition occurs for polytopes of rank 3 (or polyhedra). In that case, 0-faces are called vertices, 1-faces are called edges and 2-faces are called faces. If $v$ is a vertex of some face $f$ , then there are two edges that are incident with both $v$ and $f$ .

If $F_{n-1}$ is a face of rank $n-1$ , then the section $F_{n-1}/F_{-1}$ is also called a facet of $\mathcal{P}$ , while if $F_{0}$ is a face of rank 0, then the section $F_{n}/F_{0}=\{G\in\mathcal{P}\ |\ F_{0}\leq G\}$ is called a vertex-figure of $\mathcal{P}$ at $F_{0}$ . Every $2$ -section $G/F$ of $\mathcal{P}$ is isomorphic to the face lattice of a polygon. If it happens that the number of sides of every such polygon depends only on the rank of $G$ , and not on $F$ or $G$ itself, then we say that the polytope $\mathcal{P}$ is equivelar. In this case, if $k_{i}$ is the number of edges of a $2$ -section between an $(i-2)$ -face and an $(i+1)$ -face of $\mathcal{P}$ , for $1\leq i\leq n$ , then the ordered set $\{k_{1},k_{2},\cdots,k_{n-1}\}$ is called the Schläfli type of $\mathcal{P}$ . (For example, if $\mathcal{P}$ has rank 3, then $k_{1}$ and $k_{2}$ are respectively the size of each face and the valency of each vertex.)

2.2. Automorphisms of polytopes

An automorphism of an abstract polytope $\mathcal{P}$ is an order-preserving permutation of its elements. In particular, every automorphism preserves the set of faces of any given rank. Under permutation composition, the set of all automorphisms of $\mathcal{P}$ forms a group, called the automorphism group of $\mathcal{P}$ , and denoted by $\hbox{\rm Aut}(\mathcal{P})$ or sometimes more simply as $G$ . Also it is not difficult to use the diamond condition and strong flag-connectedness to prove that if an automorphism preserves one flag of $\mathcal{P}$ , then it fixes every flag of $\mathcal{P}$ and hence every element of $\mathcal{P}$ . It follows that $G$ acts semi-regularly on flags of $\mathcal{P}$ .

A polytope $\mathcal{P}$ is said to be regular if its automorphism group $G$ acts transitively (and hence regularly) on the set of flags of $\mathcal{P}$ . In this case, the number of automorphisms of $\mathcal{P}$ is as large as possible, and equal to the number of flags of $\mathcal{P}$ . In particular, $\mathcal{P}$ is equivelar, and the stabiliser in $G$ of every 2-section of $\mathcal{P}$ induces the full dihedral group on the corresponding polygon. Moreover, for a given flag $\Phi$ and for every $i\in\{0,1,\dots,n-1\}$ , the polytope $\mathcal{P}$ has a unique automorphism $\rho_{i}$ that takes $\Phi$ to its unique $i$ -adjacent flag $\Phi^{i}$ . The automorphisms $\rho_{0},\rho_{1},\dots,\rho_{n-1}$ generate $G$ and satisfy the defining relations for the string Coxeter group $[k_{1},k_{2},\cdots,k_{n-1}]$ , where the $k_{i}$ are as given in the previous subsection for the Schläfli type of $\mathcal{P}$ . Here, the string Coxeter group $[k_{1},k_{2},\cdots,k_{n-1}]$ , is defined as the group with presentation

[TABLE]

They also satisfy an intersection condition, which follows from the diamond and strong flag-connectedness conditions. These and many more properties of regular polytopes may be found in [28].

We now turn to chiral polytopes, for which two good references are [32, 33].

A polytope $\mathcal{P}$ said to be chiral if its automorphism group $G$ has two orbits on its set of flags, with every two adjacent flags lying in different orbits. (Another way of viewing this definition is to consider $\mathcal{P}$ as admitting no ‘reflecting’ automorphism that interchanges a flag with an adjacent flag.) Here the number of flags of $\mathcal{P}$ is $2|G|$ , and $G$ acts regularly on each of the two orbits. Again $\mathcal{P}$ is equivelar, with the stabiliser in $G$ of every 2-section of $\mathcal{P}$ inducing the full cyclic group on the corresponding polygon if $\mathcal{P}$ is a polyhedron and the full dihedral group otherwise.

For a given flag $\Phi$ , denote by $F_{i}$ the $i$ -face in $\Phi$ for each $0\leq i\leq n$ . For every $j\in\{1,2,\dots,n-1\}$ , the chiral polytope $\mathcal{P}$ admits an automorphism $\sigma_{j}$ that fixes each $F_{i}$ with $i\not=j-1,j$ and cyclically permutes consecutive $j$ - and $(j-1)$ -faces in the $2$ -section $F_{j+1}/F_{j-2}$ , that is, $\sigma_{j}$ takes $\Phi$ to the flag $(\Phi)^{j,j-1}$ which differs from $\Phi$ in precisely its $(j-1)$ -and $j$ -faces. This automorphism $\sigma_{j}$ is the analogue of the abstract rotation $\rho_{j-1}\rho_{j}$ in the regular case, for each $j$ . These automorphisms $\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$ generate $G$ , and if $\mathcal{P}$ has Schläfli type $\{k_{1},k_{2},\dots,k_{n-1}\}$ , then the $\sigma_{i}$ ’s satisfy the defining relations for the orientation-preserving subgroup of (index $2$ in) the string Coxeter group $[k_{1},k_{2},\cdots,k_{n-1}]$ . Also they satisfy a chiral form of the intersection condition, which is a variant of the one mentioned earlier for regular polytopes.

Chiral polytopes occur in pairs (or enantiomorphic forms), such that each member of the pair is the ‘mirror image’ of the other. Suppose one of them is $\mathcal{P}$ , and has Schläfli type $\{k_{1},k_{2},\cdots,k_{n-1}\}$ . Then $G$ is isomorphic to the quotient of the orientation-preserving subgroup $\Lambda^{\rm o}$ of the string Coxeter group $\Lambda=[k_{1},k_{2},\cdots,k_{n-1}]$ via some normal subgroup $K$ . By chirality, $K$ is not normal in the full Coxeter group $\Lambda$ , but is conjugated by any orientation-reversing element $c\in\Lambda$ to another normal subgroup $K^{c}$ which is the kernel of an epimorphism from $\Lambda^{\rm o}$ to the automorphism group $G^{c}$ of the mirror image $\mathcal{P}^{c}$ of $\mathcal{P}$ .

The automorphism groups of $\mathcal{P}$ and $\mathcal{P}^{c}$ are isomorphic to each other, but their canonical generating sets satisfy different defining relations. In fact, replacing the elements $\sigma_{1}$ and $\sigma_{2}$ in the canonical generating tuple $(\sigma_{1},\sigma_{2},\sigma_{3},\dots,\sigma_{n-1})$ by $\sigma_{1}^{-1}$ and $\sigma_{1}^{\,2}\sigma_{2}$ gives a set of generators for $G$ that satisfy the same defining relations as a canonical generating tuple for $G^{c}$ , but chirality ensures that there is no automorphism of $G$ that takes $(\sigma_{1},\sigma_{2})$ to $(\sigma_{1}^{-1},\sigma_{1}^{\,2}\sigma_{2})$ and fixes all the other $\sigma_{j}$ ’s.

Conversely, any finite group $G$ that is generated by $n-1$ elements $\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$ which satisfy both the defining relations for $\Lambda^{\rm o}$ and the chiral form of the intersection condition is the ‘rotation subgroup’ of an abstract $n$ -polytope $\mathcal{P}$ that is either regular or chiral. Indeed, $\mathcal{P}$ is regular if and only if $G$ admits a group automorphism $\alpha$ of order $2$ that takes $(\sigma_{1},\sigma_{2},\sigma_{3},\dots,\sigma_{n-1})$ to $(\sigma_{1}^{-1},\sigma_{1}^{\,2}\sigma_{2},\sigma_{3},\dots,\sigma_{n-1})$ .

We now focus our attention on the rank $3$ case. Here the generators $\sigma_{1},\sigma_{2}$ for $\hbox{\rm Aut}(\mathcal{P})$ satisfy the canonical relations $\,\sigma_{1}^{k_{1}}=\sigma_{2}^{k_{2}}=(\sigma_{1}\sigma_{2})^{2}=1$ , and the chiral form of the intersection condition can be abbreviated to $\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle=\{1\}$ . Moreover, $\mathcal{P}$ is chiral if and only if there is no $\rho\in\hbox{\rm Aut}(\mathcal{P})$ with $\alpha(\sigma_{1},\sigma_{2})=(\sigma_{1}^{-1},\sigma_{2}^{-1})$ .

The following proposition is useful for the groups we will deal with in the proof of our main theorem. It is called the quotient criterion for chiral $3$ -polytopes.

Proposition 2.1.

[7, Lemma 3.2]* Let $G$ be a group generated by elements $\sigma_{1},\sigma_{2}$ such that $(\sigma_{1}\sigma_{2})^{2}=1$ , and let $\theta\!:G\to H$ be a group homomorphism taking $(\sigma_{1},\sigma_{2})$ to $(\lambda_{1},\lambda_{2})$ , such that the restriction of $\theta$ to either $\langle\sigma_{1}\rangle$ or $\langle\sigma_{2}\rangle$ is injective. If $(\lambda_{1},\lambda_{2})$ is a canonical generating pair for $H$ as the automorphism group of some chiral $3$ -polytope, then the pair $(\sigma_{1},\sigma_{2})$ satisfies the chiral form of the intersection condition for $G$ .*

2.3. Group theory

In this section we briefly describe some of the knowledge of group theory we need. We use standard notation for group theory, as in [2, 22, 36] for example.

Let $G$ be a group. We define the commutator $[x,y]$ of elements $x$ and $y$ of $G$ by $[x,y]=x^{-1}y^{-1}xy$ . The group $G^{\prime}:=\langle[x,y]|x,y\in G\rangle$ is the commutator subgroup of $G$ . We let $G_{3}:=[G^{\prime},G]=\langle[x,y]|x\in G^{\prime},y\in G\rangle$ and $G_{4}:=[G_{3},G]=\langle[x,y]|x\in G_{3},y\in G\rangle$ .

The following results are elementary and so we give them without proof.

Proposition 2.2.

Let $G$ be a group. Then, for any $x,y,z\in G$ ,

(1)

$[xy,z]=[x,z]^{y}[y,z]$ , $[x,yz]=[x,z][x,y]^{z}$ ;

(2)

$[x,y^{-1}]^{y}=[x,y]^{-1}$ , $[x^{-1},y]^{x}=[x,y]^{-1}$ ;

(3)

$[x^{-1},y^{-1}]^{xy}=[x,y]$ ;

(4)

$[x^{i_{1}},y^{i_{2}},z^{i_{3}}]=[x,y,z]^{i_{1}i_{2}i_{3}}~\quad(\hbox{\rm mod }G_{4})$ .

A finite group $G$ is called metacyclic if it has a cyclic normal subgroup $N$ such that $G/N$ is also cyclic. Huppert proved the following theorem.

Theorem 2.3.

[22, Theorem 11.5]** Let $G=\langle x\rangle\langle y\rangle$ be a $p$ -group with $p>2$ . Then $G$ is metacyclic.

A finite $p$ -group $G$ is a modular $p$ -group if and only if $HK=KH$ , $\forall H,K\leq G$ . Iwaswa proved the following theorem.

Theorem 2.4.

[31, Theorem 2.3.1]** Let $G$ be a metacyclic $p$ -group with $p>2$ . Then $G$ is a modular $p$ -group.

Let $G$ be a group. The Frattini subgroup, denoted by $\Phi(G)$ , is the intersection of all maximal subgroups of $G$ . Obviously, $\Phi(G)$ is a characteristic subgroup†††Recall that a characteristic subgroup $H$ of a group $G$ is a subgroup that is fixed by the automorphism group of $G$ . of $G$ . The following theorem is the well-known Burnside Basis Theorem.

Theorem 2.5.

[2, Theorem 1.12]* Let $G$ be a $p$ -group and $|G:\Phi(G)|=p^{d}$ .*

(1)

$G/\Phi(G)\cong{\mathbb{Z}}_{p}^{d}$ . Moreover, if $N\lhd G$ and $G/N$ is elementary abelian, then $\Phi(G)\leq N$ .

(2)

Every minimal generating set of $G$ contains exactly $d$ elements.

The unique cardinality of all minimal generating sets of a $p$ -group $P$ is called the rank of $P$ , and denoted by $d(P)$ .

We will also need Sylow’s third theorem so we state it here for clarity.

Theorem 2.6 (3rd Sylow’s theorem).

Let $G$ be a group of order $p^{n}m$ with $(p,m)=1$ . Let $n_{p}(G)$ be the number of Sylow $p$ -subgroup of $G$ . Then

(1)

$n_{p}(G)\;|\;m$ ; 2. (2)

$n_{p}(G)\equiv 1\;\hbox{\rm mod }p$ ; 3. (3)

$n_{p}(G)=[G:N_{G}(P)]$ * where $P$ is a Sylow $p$ -subgroup of $G$ .*

Let $c(P)$ denote the nilpotency class of a group $P$ . From the classification of finite groups of order $p^{3}$ and $p^{4}$ (see [21, Chapter 4] and [22, Theorem 12.6]) we have the following propositions.

Proposition 2.7.

Let $P$ be a non-abelian group of order $p^{3}$ for odd prime $p>2$ . Suppose $P$ is non-metacyclic. Then $P\cong M_{p}(1,1,1)$ , and $c(P)=2,P^{\prime}=Z(P)\cong\mathbb{Z}_{p}$ , where

[TABLE]

Proposition 2.8.

Let $P$ be a non-abelian group of order $p^{4}$ for odd prime $p>3$ . Suppose $P$ is non-metacyclic and $d(P)=2$ . Then $c(P)=2$ or $3$ . Furthermore, the following hold:

(1)

if $c(P)=2$ , then $P^{\prime}\cong\mathbb{Z}_{p}$ , and $\Omega_{1}(P)=\langle x\in P\ |\ x^{p}=1\rangle<P$ ;

(2)

if $c(P)=3$ , then $P^{\prime}\cong\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ , and one of the following holds:

(a)

$\exp(P)=p$ , $Z(P)=P_{3}\cong\mathbb{Z}_{p}$ ;

(b)

$\exp(P)=p^{2}$ , $Z(P)=P_{3}=\mho_{1}(P)\cong\mathbb{Z}_{p}$ . Moreover, $\mho_{1}(P)=\langle x^{p}|x\in P,o(x)=p^{2}\rangle$ .

Finally, we state a proposition about binomial coefficients that will be used in the proof of one of our theorems.

Proposition 2.9.

[1, Chapter 1]** Let $n,m,k\in\mathbb{N}$ and let $\tbinom{n}{m}$ denote the binomial coefficient indexed by $n$ and $m$ . Then

(1)

$\tbinom{n}{m}=\tbinom{n}{n-m}$ ;

(2)

$\tbinom{n+2}{m+1}=\tbinom{n+1}{m+1}+\tbinom{n+1}{m}$ ;

(3)

$\tbinom{n}{m}\tbinom{m}{k}=\tbinom{n}{k}\tbinom{n-k}{m-k}$ ;

(4)

$\sum\limits_{k=0}^{m}(-1)^{k}\tbinom{n}{k}=(-1)^{m}\tbinom{n-1}{m}$ ;

Furthermore, if $m<0$ , then $\tbinom{n}{m}=0$ .

3. A few technical lemmas

In this section, we give lemmas about finite $p$ -groups of maximal class that are useful in this paper. We assume in what follows that $p$ is an odd prime and $m$ is a strictly positive integer. A group $P$ of order $p^{m}$ and nilpotency class $m-1$ is said to be of maximal class if $m\geq 3$ . The basic material about these groups can be found in Blackburn [4, pages 83–84] or Huppert [22, Chapter 3].

Let $P$ be a group of maximal class and order $p^{m}$ . If $P$ has an abelian maximal subgroup $A$ , and any $x\in P\backslash A$ has order $p^{2}$ , then, by [36, Chapter 8, Example 8.3.4]),

[TABLE]

where

[TABLE]

Noting that $A=\langle s_{1},s_{2},\cdots,s_{p-1}\rangle$ , the group $A$ is abelian and $A\unlhd P$ . Moreover, we also have $(\beta\sigma_{1})^{p}=\beta^{p}=s_{r}^{p^{e-1}}$ and $o(\beta\sigma_{1})=p^{2}$ . In fact, any $x\in P\backslash A$ has order $p^{2}$ . With the relations given above, it is clear that $P=\langle s_{1},\beta\rangle$ .

Define

[TABLE]

For convenience, let $s_{k+1}:=[s_{k},\beta]$ for $k=p-1$ and $p$ . Then

[TABLE]

We prove the following lemma.

Lemma 3.1.

$\tau\notin\hbox{\rm Aut}(P)$ .

Proof. Suppose $\tau\in\hbox{\rm Aut}(P)$ . Since $A$ is abelian and $s_{i+1}=[s_{i},\beta]$ for $1\leq i\leq p-2$ , by Proposition 2.2, we have

[TABLE]

In particular, $(s_{r}^{p^{e-1}})^{\tau}=(s_{r}^{\tau})^{p^{e-1}}=(s_{r}^{p^{e-1}})^{-1}$ . On the other hand, since $\beta^{p}=s_{r}^{p^{e-1}}$ , $\beta^{p}=(\beta^{p})^{\tau}=(s_{r}^{p^{e-1}})^{\tau}=\beta^{-p}$ . This implies that $\beta^{2p}=1$ . But this is impossible because $o(\beta)=p^{2}$ . ∎

Lemma 3.2.

For any $1\leq k\leq p$ , we have

(1)

$s_{k}^{\beta^{1-k}}=\prod\limits_{i=k}^{p+1}s_{i}^{\tbinom{p+1-k}{i-k}}$ ;

(2)

$s_{k}^{\sigma}=\left\{\begin{array}[]{ll}s_{k}^{\beta^{1-k}},&k\mbox{ is odd,}\\ (s_{k}^{\beta^{1-k}})^{-1},&k\mbox{ is even.}\end{array}\right.$ **

Proof. (1) Since $s_{k+1}=[s_{k},\beta]$ , we have $s_{k}^{\beta}=s_{k}s_{k+1}$ . Since $o(\beta)=p^{2},\beta^{p}\in A$ and $k-1\leq p$ , we have $p-k+1\geq 0$ . It follows that

[TABLE]

(2) The proof uses an induction on $k$ . For $k=1$ , we have $s_{1}^{\sigma}=s_{1}$ . For $k=2$ , by Proposition 2.2, we have

[TABLE]

Hence this lemma is true for $k=1$ and $2$ . We now use induction on $k$ and Proposition 2.2, splitting the analysis in two cases, namely the case where $k$ is odd and the case where $k$ is even.

If $k$ is odd, then

[TABLE]

Recall that $A=\langle s_{1},s_{2},\cdots,s_{p-1}\rangle$ is an abelian maximal subgroup of $G$ and $A\unlhd G$ . It follows that $s_{k}\in A$ and $s_{k}^{\beta}\in A$ for any $k\geq 1$ . Thus,

[TABLE]

If $k$ is even, then

[TABLE]

∎

Lemma 3.3.

If $m$ is odd , then $r$ is even and $\sigma\in\hbox{\rm Aut}(P)$ . Moreover, $o(\sigma)=2$ .

Proof. Since $m=(p-1)(e-1)+r+1$ and $p$ and $m$ are odd, $r$ must be even. The fact that $o(\sigma)=2$ follows from the definition of $\sigma$ . The group $P=\langle s_{1}^{\sigma},\beta^{\sigma}\rangle=\langle s_{1},\beta^{-1}\rangle$ . Thus, we only need to show that the generating set $\{s_{1}^{\sigma},s_{2}^{\sigma},\cdots,s_{p-1}^{\sigma},\beta^{\sigma}\}$ of $P$ satisfies the same relations as $\{s_{1},s_{2},\cdots,s_{p-1},\beta\}$ in order to conclude that $\sigma\in\hbox{\rm Aut}(P)$ . Recall that $A=\langle s_{1},s_{2},\cdots,s_{p-1}\rangle$ is abelian and $A\unlhd P$ . By Lemma 3.2,

[TABLE]

Now $o(s_{i})=o(s_{i}^{\sigma})$ , $[s_{i}^{\sigma},s_{j}^{\sigma}]=1$ for $1\leq i,j\leq p-1$ . Moreover, it is easy to see that $s_{k+1}^{\sigma}=[s_{k}^{\sigma},\beta^{\sigma}]$ for $1\leq k\leq p-2$ . Thus we only need to show that $(\beta^{p})^{\sigma}=(s_{r}^{p^{e-1}})^{\sigma}$ , and

[TABLE]

that is, $s_{p}^{\sigma}=\prod\limits_{i=1}^{p-1}(s_{i}^{\sigma})^{-\tbinom{p}{i}}$ . Since $r$ is even, we have

[TABLE]

Next, by Lemma 3.2, we have

[TABLE]

Then $\prod\limits_{i=1}^{p-1}(s_{i}^{\sigma})^{-\tbinom{p}{i}}=s_{1}^{u_{1}}s_{2}^{u_{2}}\cdots s_{p+1}^{u_{p+1}}$ , where $u_{i}$ comes from Table 1.

This gives $u_{1}=-\tbinom{p}{1}$ . By Proposition 2.9(1), we have

[TABLE]

For $2\leq i\leq p-1$ , we have $u_{i}=\sum\limits_{j=2}^{i}(-1)^{j}\tbinom{p+1-j}{i-j}\tbinom{p}{j}$ . For $i=p$ , we have

[TABLE]

By Proposition 2.9 $(2)$$(3)$ , we have

[TABLE]

Together with Proposition 2.9(4), for $2\leq i\leq p$ , we have

[TABLE]

and hence

[TABLE]

By Lemma 3.2 and the fact that $s_{p+1}=[s_{p},\beta]$ , we have $s_{p}^{\sigma}=s_{p}^{\beta^{1-p}}=s_{p}^{\beta}=s_{p}s_{p+1}.$ The result is now immediate since

[TABLE]

∎

4. Proof of Theorem 1.1

(1) Let $n_{p}(G)$ be the number of Sylow $p$ -subgrougs of $G$ and let $P$ be a Sylow $p$ -subgroup of $G$ . By Theorem 2.6 (1), $n_{p}(G)\mid 2$ and $n_{p}(G)\equiv 1\ \hbox{\rm mod }p$ . It follows that $n_{p}(G)=1$ . Then $P\unlhd G$ .

Since $|\langle\sigma_{1}\sigma_{2}\rangle|=2$ and $|P|=p^{m}$ , $P\cap\langle\sigma_{1}\sigma_{2}\rangle=\{1\}$ . It follows that $|P\langle\sigma_{1}\sigma_{2}\rangle|=2p^{m}=|G|$ and thus $G=P\langle\sigma_{1}\sigma_{2}\rangle$ . In face, since $P\unlhd G$ , $G=P\rtimes\langle\sigma_{1}\sigma_{2}\rangle$ , finishing the proof of (1).

(2) Up to duality, the Schläfli type $\{k_{1},k_{2}\}$ must be one of $\{p^{l_{1}},p^{l_{2}}\},\{p^{l_{1}},2p^{l_{2}}\}$ or $\{2p^{l_{1}},2p^{l_{2}}\}$ . Consider the quotient group $\overline{G}=G/P=\langle\overline{\sigma_{1}},\overline{\sigma_{2}}\rangle$ , then $|\overline{G}|=2$ . If $\{k_{1},k_{2}\}=\{p^{l_{1}},p^{l_{2}}\}$ , then $\sigma_{1},\sigma_{2}\in P$ , and hence $\overline{\sigma_{1}}=\overline{\sigma_{2}}=\overline{1}$ in $\overline{G}$ , which is impossible because $|\overline{G}|=2$ . If $\{k_{1},k_{2}\}=\{2p^{l_{1}},2p^{l_{2}}\}$ , then $\overline{\sigma_{1}}\neq\overline{1}$ , $\overline{\sigma_{2}}\neq\overline{1}$ and $(\overline{\sigma_{1}})^{2}=(\overline{\sigma_{2}})^{2}=\overline{1}$ . Since $(\sigma_{1}\sigma_{2})^{2}=1$ in $G$ , $\overline{\sigma_{1}}\overline{\sigma_{2}}\neq\overline{1}$ and $(\overline{\sigma_{1}}\overline{\sigma_{2}})^{2}=\overline{1}$ in $\overline{G}$ . Thus $\overline{G}=\langle\overline{\sigma_{1}},\overline{\sigma_{2}}\rangle\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ , which is impossible because $|\overline{G}|=2$ . Therefore, $\{k_{1},k_{2}\}=\{p^{l_{1}},2p^{l_{2}}\}$ , finishing the proof of (2).

(3) By Theorem 1.1 $(2)$ , we have $\{k_{1},k_{2}\}=\{p^{l_{1}},2p^{l_{2}}\}$ . Let $H=\langle\sigma_{1},\sigma_{2}^{2}\rangle$ . Then $H\leq P$ . Obviously, $\sigma_{1}\in H\leq N_{G}(H)$ . Moreover, since $(\sigma_{1}\sigma_{2})^{2}=1$ ,

[TABLE]

Then $\sigma_{2}\in N_{G}(H)$ . It follows that $H\unlhd G$ and $H\langle\sigma_{1}\sigma_{2}\rangle=\langle\sigma_{1},\sigma_{2}^{2},\sigma_{1}\sigma_{2}\rangle=G$ . On the other hand, it is easy to see that $|G/H|=2$ . Thus, $P=H$ and $d(P)=d(H)\leq 2$ .

If $d(P)=1$ , then $P$ is cyclic. Since $p\mid k_{1}$ and $p\mid k_{2}$ , we have $\sigma_{1},\sigma_{2}^{2}\in P$ , and so $\langle\sigma_{1}\rangle\leq\langle\sigma_{2}^{2}\rangle$ or $\langle\sigma_{1}\rangle\geq\langle\sigma_{2}^{2}\rangle$ . It follows that $|\langle\sigma_{1}\rangle\cap\langle\sigma_{2}^{2}\rangle|>1$ , which is impossible because $\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle=\{1\}$ . Thus, $d(P)=2$ .

Now, suppose that $P$ is abelian. Then $[\sigma_{1},\sigma_{2}^{2}]=1$ in $P$ and hence $P=\langle\sigma_{1},\sigma_{2}^{2}\rangle=\langle\sigma_{1}\rangle\langle\sigma_{2}^{2}\rangle$ . Since $\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle=\{1\}$ , we have

[TABLE]

It means that $\mathcal{P}$ is tight, that is $|G|=2p^{l_{1}+l_{2}}$ . Moreover, we claim that $G=\mathcal{G}$ , where

[TABLE]

Obviously, $(G,\{\sigma_{1},\sigma_{2}\})$ satisfies all of the defining relations of $(\mathcal{G},\{\sigma_{1},\sigma_{2}\})$ , and so $G$ is the quotient group of $\mathcal{G}$ . On the other hand, since $[\sigma_{1},\sigma_{2}^{2}]=1$ , $\langle\sigma_{2}^{2}\rangle\unlhd\mathcal{G}$ . Consider the quotient group

[TABLE]

It is easy to see that $\overline{\mathcal{G}}\leq D_{2p^{l_{1}}}$ where $D_{2p^{l_{1}}}$ is the dihedral group of order $2p^{l_{1}}$ . Then $|\mathcal{G}|=|\overline{\mathcal{G}}|\cdot|\langle\sigma_{2}^{2}\rangle|\leq 2{p^{l_{1}}}p^{l_{2}}=2p^{l_{1}+l_{2}}$ . So $|\mathcal{P}|=2p^{l_{1}+l_{2}}=|G|$ , and hence $G=\mathcal{G}$ . However, consider the map $\alpha:\sigma_{1}\mapsto\sigma_{1}^{-1},\sigma_{2}\mapsto\sigma_{2}^{-1}$ . It is easy to check that

[TABLE]

Then $\alpha\in\hbox{\rm Aut}(G)$ , which is impossible because $\mathcal{P}$ is chiral. Thus, $P$ is not abelian and $m\geq 3$ . This finishes the proof of (3).

Suppose first that $\mathcal{P}$ is tight, that is $|G|=k_{1}k_{2}=2p^{m}$ . Then $|P|=o(\sigma_{1})o(\sigma_{2}^{2})$ and since $\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle=\{1\}$ , we have $P=\langle\sigma_{1}\rangle\langle\sigma_{2}\rangle$ . Theorem 2.3 implies that $P$ is metacyclic group.

Now suppose that $P=\langle\sigma_{1},\sigma_{2}^{2}\rangle$ is metacyclic. By Theorem 2.4, we have $P=\langle\sigma_{1}\rangle\langle\sigma_{2}^{2}\rangle$ . Note that $\langle\sigma_{1}\rangle\cap\langle\sigma_{2}^{2}\rangle=\{1\}$ , so $|P|=|\langle\sigma_{1}\rangle||\langle\sigma_{2}^{2}\rangle|$ . It follows that $|G|=2\cdot o(\sigma_{1})\cdot o(\sigma_{2}^{2})=o(\sigma_{1})\cdot o(\sigma_{2})=k_{1}k_{2}$ and hence $\mathcal{P}$ is tight.

5. Proof of Theorem 1.2

For $p=3$ or $5$ , we may refer to the lists of all chiral polytopes with up to $4000$ flags computed by Conder [8]. So, in the following discussion, we always assume $p\geq 7$ , and $P\in\hbox{\rm Syl}_{p}(G)$ .

(1) Suppose $|G|=2p^{3}$ and $\mathcal{P}$ is a non-tight chiral polyhedron. By Theorem 1.1(2), we know that $\{k_{1},k_{2}\}=\{p,2p\}$ . Then $o(\sigma_{1})=o(\sigma_{2}^{2})=p$ , which means that $P=\langle\sigma_{1},\sigma_{2}^{2}\rangle$ is generated by two elements of order $p$ . By Proposition 2.7, $\langle[\sigma_{1},\sigma_{2}^{2}]\rangle=P^{\prime}=Z(P)\cong\mathbb{Z}_{p}$ and hence $[\sigma_{1},\sigma_{2}^{2}]^{p}=[[\sigma_{1},\sigma_{2}^{2}],\sigma_{1}]=[[\sigma_{1},\sigma_{2}^{2}],\sigma_{2}^{2}]=1$ in $P$ . Knowing how to represent these two elements of $P$ is enough to define the entire group $G$ . We claim that $G=\mathcal{G}_{1}$ , where

[TABLE]

Obviously, $(G,\{\sigma_{1},\sigma_{2}\})$ satisfies all of the defining relations of $(\mathcal{G}_{1},\{\sigma_{1},\sigma_{2}\})$ , and hence $G$ is a quotient group of $\mathcal{G}_{1}$ . On the other hand, since $(\sigma_{1}\sigma_{2})^{2}=1$ in $\mathcal{G}_{1}$ , we have $\sigma_{1}^{\sigma_{2}}=\sigma_{2}^{-2}\sigma_{1}^{-1}\in\langle\sigma_{1},\sigma_{2}^{2}\rangle$ and hence $\langle\sigma_{1},\sigma_{2}^{2}\rangle\unlhd\mathcal{G}_{1}$ . Consider the quotient group $\overline{\mathcal{G}_{1}}=\mathcal{G}_{1}/\langle\sigma_{1},\sigma_{2}^{2}\rangle=\langle\overline{\sigma_{2}}\ |\ \overline{\sigma_{2}}^{2}\rangle$ . Then $|\overline{\mathcal{G}_{1}}|\leq 2$ . Note that the subgroup $\langle\sigma_{1},\sigma_{2}^{2}\rangle$ of $\mathcal{G}_{1}$ satisfies all of the defining relations of the subgroup $P$ of $G$ , so the subgroup $\langle\sigma_{1},\sigma_{2}^{2}\rangle$ of $\mathcal{G}_{1}$ has order at most $|P|=p^{3}$ . Hence $|\mathcal{G}_{1}|\leq 2p^{3}=|G|$ . Thus, $G=\mathcal{G}_{1}$ .

Consider the map $\alpha:\sigma_{1}\mapsto\sigma_{1}^{-1},\sigma_{2}\mapsto\sigma_{2}^{-1}$ . It is easy to check that

[TABLE]

meaning $\alpha\in\hbox{\rm Aut}(G)$ , which contradicts the hypothesis that $\mathcal{P}$ is chiral. Thus, $\mathcal{P}$ must be tight.

(2) Suppose $|G|=2p^{4}$ and $\mathcal{P}$ is a non-tight chiral polyhedron. By Theorem 1.1(4), $P$ is non-metacyclic. Moreover, by Theorem 1.1(2) and Proposition 2.7, we have $\exp(P^{\prime})=p$ , $c(P)=2$ or $3$ and $\{k_{1},k_{2}\}\in\{\{p,2p\},\{p,2p^{2}\},\{p^{2},2p\}\}$ , meaning we have six cases to consider, that is three for each possible value of $c(P)$ . We show for each of these six cases that they cannot happen.

Case 1: $c(P)=2$ .

Subcase 1.1: $\{k_{1},k_{2}\}=\{p,2p\}$ . Then $o(\sigma_{1})=o(\sigma_{2}^{2})=p$ , and so

[TABLE]

But this is impossible by Proposition 2.8(1).

Subcase 1.2: $\{k_{1},k_{2}\}=\{p,2p^{2}\}$ . Then $o(\sigma_{1})=p$ and $o(\sigma_{2}^{2})=p^{2}$ . Moreover,

[TABLE]

This is enough to define the entire group $G$ . We claim that $G=\mathcal{G}_{2}$ , where

[TABLE]

Obviously, $(G,\{\sigma_{1},\sigma_{2}\})$ satisfies all of the defining relations of $(\mathcal{G}_{2},\{\sigma_{1},\sigma_{2}\})$ , and hence $G$ is a quotient group of $\mathcal{G}_{2}$ . Note that $\langle\sigma_{1},\sigma_{2}^{2}\rangle\unlhd\mathcal{G}_{2}$ . Consider the quotient group $\overline{\mathcal{G}_{2}}=\mathcal{G}_{2}/\langle\sigma_{1},\sigma_{2}^{2}\rangle=\langle\overline{\sigma_{2}}\ |\ (\overline{\sigma_{2}})^{2}\rangle$ . Then $|\overline{\mathcal{G}_{2}}|\leq 2$ . For the subgroup group $\langle\sigma_{1},\sigma_{2}^{2}\rangle$ of $\mathcal{G}_{2}$ , since $[\sigma_{1},\sigma_{2}^{2}]^{p}=[\sigma_{1},\sigma_{2}^{2},\sigma_{1}]=[\sigma_{1},\sigma_{2}^{2},\sigma_{2}^{2}]=1$ , $\langle\sigma_{1},\sigma_{2}^{2}\rangle^{\prime}=\langle[\sigma_{1},\sigma_{2}^{2}]\rangle$ . Then

[TABLE]

It follows that $|\mathcal{G}_{2}|=|\overline{\mathcal{G}_{2}}|\cdot|\langle\sigma_{1},\sigma_{2}^{2}\rangle|\leq 2p^{4}$ , and hence $G=\mathcal{G}_{2}$ .

Consider the map $\alpha:\sigma_{1}\mapsto\sigma_{1}^{-1},\sigma_{2}\mapsto\sigma_{2}^{-1}$ . It is easy to check that

[TABLE]

meaning $\alpha\in\hbox{\rm Aut}(G)$ , which contradicts the hypothesis that $\mathcal{P}$ is chiral.

Subcase 1.3: $\{k_{1},k_{2}\}=\{p^{2},2p\}$ . Then $o(\sigma_{1})=p^{2}$ and $o(\sigma_{2}^{2})=p$ . In a similar way, we obtain $G=\mathcal{G}_{3}$ , where

[TABLE]

But this is impossible because $\alpha\in\hbox{\rm Aut}(G)$ , where $\alpha$ taking $(\sigma_{1},\sigma_{2})$ to $(\sigma_{1}^{-1},\sigma_{2}^{-1})$ , contradicting the chirality of $\mathcal{P}$ .

Case 2: $c(P)=3$ . Then since ${\rm exp}(P^{\prime})=p$ ,

[TABLE]

and as $c(P)=3$ ,

[TABLE]

Subcase 2.1: $\{k_{1},k_{2}\}=\{p,2p\}$ . Then $o(\sigma_{1})=o(\sigma_{2}^{2})=p$ . Since $P^{\prime}\cong\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ , we have

[TABLE]

for some $i,j$ . This is also enough to define the entire group $G$ . We claim that $G=\mathcal{G}_{4}$ , where

[TABLE]

Obviously, $(G,\{\sigma_{1},\sigma_{2}\})$ satisfies all of the defining relations of $(\mathcal{G}_{4},\{\sigma_{1},\sigma_{2}\})$ , and hence $G$ is a quotient group of $\mathcal{G}_{4}$ . On the other hand, consider the quotient group $\overline{\mathcal{G}_{4}}=\mathcal{G}_{4}/\langle\sigma_{1},\sigma_{2}^{2}\rangle=\langle\overline{\sigma_{2}}\ |\ (\overline{\sigma_{2}})^{2}\rangle$ . Then $|\overline{\mathcal{G}_{4}}|\leq 2$ . For the subgroup $\langle\sigma_{1},\sigma_{2}^{2}\rangle$ of $\mathcal{G}_{4}$ , since $[\sigma_{1},\sigma_{2}^{2},\sigma_{1},\sigma_{1}]=[\sigma_{1},\sigma_{2}^{2},\sigma_{1},\sigma_{2}^{2}]=1$ , we have $[\sigma_{1},\sigma_{2}^{2},\sigma_{1}]\in Z(\langle\sigma_{1},\sigma_{2}^{2}\rangle)$ . Let $\widetilde{P}=\langle\sigma_{1},\sigma_{2}^{2}\rangle/\langle[\sigma_{1},\sigma_{2}^{2},\sigma_{1}]\rangle=\langle\widetilde{\sigma_{1}},\widetilde{\sigma_{2}^{2}}\rangle$ . Then $|\langle\sigma_{1},\sigma_{2}^{2}\rangle|\leq|\widetilde{P}|\cdot p$ . Furthermore, since $\widetilde{P}^{\prime}=\langle[\widetilde{\sigma_{1}},\widetilde{\sigma_{2}^{2}}]\rangle$ , we have

[TABLE]

It follows that $|\mathcal{G}_{4}|\leq 2p^{4}$ , and hence $G=\mathcal{G}_{4}$ .

Consider the map $\alpha:\sigma_{1}\mapsto\sigma_{1}^{-1},\sigma_{2}\mapsto\sigma_{2}^{-1}$ . It is easy to check that

[TABLE]

Furthermore, by Proposition 2.2,

[TABLE]

Hence $\alpha\in\hbox{\rm Aut}(G)$ , which contradicts the fact that $\mathcal{P}$ is chiral.

Subcase 2.2: $\{k_{1},k_{2}\}=\{p,2p^{2}\}$ . Then $o(\sigma_{1})=p,o(\sigma_{2}^{2})=p^{2}$ . Moreover,

[TABLE]

Since $P_{3}=\mho_{1}(P)$ , we have $[\sigma_{1},\sigma_{2}^{2},\sigma_{1}]=(\sigma_{2}^{2p})^{i}$ and $[\sigma_{1},\sigma_{2}^{2},\sigma_{2}^{2}]=(\sigma_{2}^{2p})^{j}$ for some $i,j$ . This is enough to define the entire group $G$ . We claim that $G=\mathcal{G}_{5}$ , where

[TABLE]

Obviously, $(G,\{\sigma_{1},\sigma_{2}\})$ satisfies all of the defining relations of $(\mathcal{G}_{5},\{\sigma_{1},\sigma_{2}\})$ , and hence $G$ is the quotient group of $\mathcal{G}_{5}$ . Moreover, the same argument as used in Case 2.1 shows that $|\mathcal{G}_{5}|\leq 2p^{4}$ . Thus, $G=\mathcal{G}_{5}$ .

Consider the map $\alpha:\sigma_{1}\mapsto\sigma_{1}^{-1},\sigma_{2}\mapsto\sigma_{2}^{-1}$ . It is easy to check that

[TABLE]

Furthermore, by Proposition 2.2,

[TABLE]

Hence $\alpha\in\hbox{\rm Aut}(G)$ , which contradicts the fact that $\mathcal{P}$ is chiral.

Subcase 2.3: $\{k_{1},k_{2}\}=\{p^{2},2p\}$ . Then $o(\sigma_{1})=p^{2},o(\sigma_{2}^{2})=p$ . Moreover,

[TABLE]

Since $P_{3}=\mho_{1}(P)$ , we have $[\sigma_{1},\sigma_{2}^{2},\sigma_{1}]=\sigma_{1}^{pi}$ and $[\sigma_{1},\sigma_{2}^{2},\sigma_{2}^{2}]=\sigma_{1}^{pi}$ for some $i,j$ . In a similar way, we obtain $G=\mathcal{G}_{6}$ , where

[TABLE]

But this will again give an element $\alpha\in\hbox{\rm Aut}(G)$ , with $\alpha(\sigma_{1},\sigma_{2})=(\sigma_{1}^{-1},\sigma_{2}^{-1})$ , contradicting the chirality of $\mathcal{P}$ . ∎

6. Proof of Theorem 1.4

The main purpose of this section is to prove Theorem 1.4. In order to do so, we will give a constructive proof. Let $P,\sigma,\tau$ be the group and maps defined in Section 3. We have two cases to consider, namely the case where $m$ is odd and at least 5 and the case where $m$ is even and at least $p+3$ .

Case 1: $m$ is odd and $m\geq 5$ .

Let $G=P\rtimes\langle\sigma\rangle$ . Then $|G|=2p^{m}$ and $G=\langle s_{1},\beta,\sigma\rangle$ . Let $\sigma_{1}=\beta s_{1},\sigma_{2}=s_{1}^{-1}\sigma$ . We will show that $(G,\{\sigma_{1},\sigma_{2}\})$ is the automorphism group of a chiral polytope of type $\{p^{2},2p^{e}\}$ .

Claim 1: $G=\langle\sigma_{1},\sigma_{2}\rangle$ , and $[o(\sigma_{1}),o(\sigma_{2}),o(\sigma_{1}\sigma_{2})]=[p^{2},2p^{e},2]$ .

Obviously, $\sigma_{2}^{2}=(s_{1}^{-1}\sigma)^{2}=s_{1}^{-2}$ . Since $o(s_{1})=p^{e}$ , $s_{1}\in\langle s_{1}^{-2}\rangle=\langle\sigma_{2}^{2}\rangle\leq\langle\sigma_{1},\sigma_{2}\rangle$ . It follows that $\sigma_{1}s_{1}^{-1}=\beta\in\langle\sigma_{1},\sigma_{2}\rangle$ , $s_{1}\sigma_{2}=\sigma\in\langle\sigma_{1},\sigma_{2}\rangle$ and hence $\langle\sigma_{1},\sigma_{2}\rangle=\langle s_{1},\beta,\sigma\rangle=G$ .

Next, since $o(\sigma)=2$ and $\beta^{\sigma}=\beta^{-1}$ , we have

[TABLE]

and hence $o(\sigma_{1}\sigma_{2})=2$ . Moreover, since $\sigma_{1}\in P\backslash A$ and $\sigma_{2}^{2}=s_{1}^{-2}$ , we have $o(\sigma_{1})=p^{2}$ and $o(\sigma_{2})=2p^{e}$ , as claimed.

Claim 2: $\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle=\{1\}$ .

Since $o(\sigma_{1})=p^{2}$ and $\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle\leq\langle\sigma_{1}\rangle$ , $|\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle|=1,p$ or $p^{2}$ . If $|\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle|=p^{2}$ , then $\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle=\langle\sigma_{1}\rangle$ . It follows that $\sigma_{1}\in\langle\sigma_{2}\rangle$ and hence $G=\langle\sigma_{1},\sigma_{2}\rangle=\langle\sigma_{2}\rangle$ , which is impossible because $G$ is non-abelian.

If $|\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle|=p$ , then $\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle=\langle\sigma_{1}^{p}\rangle=\langle\sigma_{2}^{2p^{e-1}}\rangle$ . Observe that $\sigma_{1}^{p}=(\beta s_{1})^{p}=\beta^{p}=s_{r}^{p^{e-1}}$ and $\langle\sigma_{2}^{2p^{e-1}}\rangle=\langle(s_{1}^{-2})^{p^{e-1}}\rangle=\langle s_{1}^{p^{e-1}}\rangle$ . However, by Lemma 3.3, $r$ is even, thus $s_{r}^{p^{e-1}}\notin\langle s_{1}^{p^{e-1}}\rangle$ . It means that $\langle\sigma_{1}^{p}\rangle\neq\langle\sigma_{2}^{2p^{e-1}}\rangle$ , and hence $|\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle|\neq p$ . Thus, $\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle=\{1\}$ , as claimed.

Claim 3: There is no $\alpha\in\hbox{\rm Aut}(G)$ such that $\alpha(\sigma_{1},\sigma_{2})=(\sigma_{1}^{-1},\sigma_{2}^{-1})$ .

Suppose that there exists such an $\alpha$ . Since $P$ is a characteristic subgroup of $G$ , the automorphism $\alpha$ of $G$ induces an automorphism of $P$ , say $\alpha|_{P}$ . Note that $\sigma_{2}^{2}=s_{1}^{-2}$ , so $(s_{1}^{-2})^{\alpha}=((\sigma_{2})^{2})^{\alpha}=((\sigma_{2})^{\alpha})^{2}=\sigma_{2}^{-2}=s_{1}^{2}$ . It follows that $s_{1}^{\alpha}=s_{1}^{-1}$ . Moreover, since $s_{1}^{-1}\beta^{-1}=(\beta s_{1})^{-1}=\sigma_{1}^{-1}=(\sigma_{1})^{\alpha}=(\beta\sigma_{1})^{\alpha}=\beta^{\alpha}s_{1}^{\alpha}=\beta^{\alpha}s_{1}^{-1}$ , we have $\beta^{\alpha}=s_{1}^{-1}\beta^{-1}s_{1}$ . Thus, $\alpha|_{P}$ takes $(s_{1},\beta)$ to $({s_{1}^{-1},(\beta^{s_{1}})^{-1}})$ . Let $\sigma(s_{1})$ be the inner automorphism of $P$ induced by $s_{1}$ . Then we have the following sequence:

[TABLE]

However, by Lemma 3.1, $\tau\notin\hbox{\rm Aut}(P)$ , a contradiction, and hence $\alpha\notin\hbox{\rm Aut}(G)$ , as claimed.

Claims 1, 2 and 3 show that $(G,\{\sigma_{1},\sigma_{2}\})$ is the automorphism group of a chiral polytope of type $\{p^{2},2p^{e}\}$ .

Case 2: $m$ is even, and $m\geq p+3$ .

We define the group

[TABLE]

with the following relations:

[TABLE]

where $1\leq e,1\leq r\leq p-1$ , and $r$ is even.

Obviously, $G^{\ast}=\langle s_{1}^{\ast},\beta^{\ast},\sigma^{\ast}\rangle$ . Let $\sigma_{1}^{\ast}=\beta^{\ast}s_{1}^{\ast},\sigma_{2}^{\ast}=s_{1}^{\ast}{}^{-1}\sigma^{\ast}$ , and $e\geq 2$ .

Claim 4: $o(x)=p$ and $|G^{\ast}|=2p^{(p-1)(e-1)+r+2}\geq 2p^{p+3}$ .

Since $[s_{1}^{\ast},x]=[\beta^{\ast},x]=[\sigma^{\ast},x]=1$ , we have $x\in Z(G^{\ast})$ . Consider the quotient group $G^{\ast}/\langle x\rangle$ . Then $G^{\ast}/\langle x\rangle\cong G$ (which is the group defined in Case 1) and hence $|G^{\ast}/\langle x\rangle|=2p^{(p-1)(e-1)+r+1}$ . We only need to prove that $o(x)=p$ . Then $G^{\ast}$ is a central extension of $G$ by $\langle x\rangle$ . Unfortunately, $x$ not always has order $p$ , which is why we need $e\geq 2$ .

Let $H=\langle a,b\ |\ a^{p^{2}},b^{p},[a,b]\rangle\cong\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$ . It is easy to check that $(H,\{a,b,1\})$ satisfies all of the relations of $(G^{\ast},\{s_{1}^{\ast},\beta^{\ast},\sigma^{\ast}\})$ . Hence the map: $(s_{1}^{\ast},\beta^{\ast},\sigma^{\ast})\mapsto(a,b,1)$ induces an epimorphism $\pi$ from $G^{\ast}$ to $H$ . Moreover, $\pi:x\mapsto a^{-p}$ . Note that $o(a)=p^{2}$ , so $o(\pi(x))=p$ . It follows that $o(x)=p$ and hence $|G^{\ast}|=p|G|=2p^{(p-1)(e-1)+r+2}$ .

From now on, let $|G^{\ast}|=2p^{m}$ . Since $r$ is even and $e\geq 2$ , we get $|G^{\ast}|=2p^{(p-1)(e-1)+r+2}\geq 2p^{p+3}$ which is fine as $m$ is even and $m\geq p+3$ by hypothesis.

Claim 5: $G^{\ast}=\langle\sigma_{1}^{\ast},\sigma_{2}^{\ast}\rangle$ , and $[o(\sigma_{1}^{\ast}),o(\sigma_{2}^{\ast}),o(\sigma_{1}^{\ast}\sigma_{2}^{\ast})]=[p^{2},2p^{e},2]$ .

The same argument as used in Case 1 shows that $G^{\ast}=\langle\sigma_{1}^{\ast},\sigma_{2}^{\ast}\rangle$ and $o(\sigma_{2}^{\ast})=2p^{e},o(\sigma_{1}^{\ast}\sigma_{2}^{\ast})=2$ . Consider the quotient group $G^{\ast}/\langle x\rangle\cong G$ . Note that $\sigma_{1}^{p}=\beta^{p}$ in $G$ implies that $\sigma_{1}^{\ast p}=\beta^{\ast p}x^{i}$ in $G^{\ast}$ for some $i$ . It follows that $\sigma_{1}^{\ast p^{2}}=(\beta^{\ast p}x^{i})^{p}=\beta^{\ast p^{2}}(x^{i})^{p}=1$ in $G^{\ast}$ . Then $o(\sigma_{1}^{\ast})=p^{2}$ .

Claim 6: $\langle\sigma_{1}^{\ast}\rangle\cap\langle\sigma_{2}^{\ast}\rangle=\{1\}$ .

Let $\gamma$ be the epimorphism from $(G^{\ast},\{\sigma_{1}^{\ast},\sigma_{2}^{\ast}\})$ onto $(G,\{\sigma_{1},\sigma_{2}\})$ . Note that $o(\sigma_{1}^{\ast})=o(\sigma_{1})=p^{2}$ and $\langle\sigma_{1}\rangle\cap\langle\sigma_{2}\rangle=\{1\}$ . Then Proposition 2.1 implies that $\langle\sigma_{1}^{\ast}\rangle\cap\langle\sigma_{2}^{\ast}\rangle=\{1\}$ .

Claim 7: There is no $\alpha^{\ast}\in\hbox{\rm Aut}(G^{\ast})$ such that $\alpha^{\ast}(\sigma_{1}^{\ast},\sigma_{2}^{\ast})=(\sigma_{1}^{\ast-1},\sigma_{2}^{\ast-1})$ .

Suppose that such an $\alpha^{\ast}$ exists. Then the same argument as used in Case 1 shows that $\alpha^{\ast}|_{P^{\ast}}$ takes $(s_{1}^{\ast},\beta^{\ast})$ to $({s_{1}^{\ast-1},(\beta^{\ast s_{1}^{\ast}})^{-1}})$ . On the other hand, since $|\langle\sigma^{\ast}\rangle|=2$ and $|P^{\ast}|=p^{m}$ , we get $P^{\ast}\cap\langle\sigma^{\ast}\rangle=\{1\}$ . It follows that $G^{\ast}=P^{\ast}\rtimes\langle\sigma^{\ast}\rangle$ , and hence $\sigma^{\ast}\in\hbox{\rm Aut}(P^{\ast})$ . Let $\tau^{\ast}=\alpha^{\ast}|_{P^{\ast}}\circ\sigma(s_{1}^{\ast})\circ\sigma^{\ast}$ . Then $(s_{1}^{\ast},\beta^{\ast})^{\tau^{\ast}}=(s_{1}^{\ast-1},\beta^{\ast})$ and $\tau^{\ast}\in\hbox{\rm Aut}(P^{\ast})$ . However, the same argument as used in Lemma 3.1 shows that $\tau^{\ast}\notin\hbox{\rm Aut}(P^{\ast})$ , a contradiction.

Claims 4 to 7 above imply that $(G^{\ast},\{\sigma_{1}^{\ast},\sigma_{2}^{\ast}\})$ is the automorphism group of a chiral polytope of type $\{p^{2},2p^{e}\}$ . ∎

7. Acknowledgements

This work was supported by the National Natural Science Foundation of China (12201371, 12271318,12331013,12311530692,12271024,12161141005), the 111 Project of China (B16002), an Action de Recherche Concertée grant of the Communauté Française Wallonie Bruxelles and a PINT-BILAT-M grant from the Fonds National de la Recherche Scientifique de Belgique (FRS-FNRS).

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3[3] H. U. Besche, B. Eick and E. A. O’Brien. The groups of order at most 2000. Electron. Res. Announc. Amer. Math. Soc. 7:1–4, 2001.
4[4] N. Blackburn. On a special class of p p -groups. Acta Math. 100:45–92, 1958.
5[5] W. Bosma, J. Cannon and C. Playoust. The Magma Algebra System. I: the user language. J. Symbolic Comput. 24:235–265, 1997.
6[6] A. Breda d’Azevedo and D. A. Catalano. Strong map-symmetry of SL(3, K K ) and PSL(3, K K ) for every finite field K K . J. Algebra Appl. 20, no. 4, Paper No. 2150048, 2021.
7[7] A. Breda D’Azevedo, G.A. Jones, E. Schulte, Constructions of chiral polytopes of small rank, Canad. J. Math. 63:1254–1283, 2011.
8[8] M.D.E. Conder, Chiral polytopes with up to 4000 flags, https://www.math.auckland.ac.nz/~conder/Chiral Polytopes With Up To 4000 Flags-By Order.txt .

TL;DR

Contribution

Findings

Abstract

Peer Reviews

Videos

Taxonomy

Chiral polytopes of order 2pm2p^{m}2pm

Abstract.

Key words and phrases:

1991 Mathematics Subject Classification:

1. Introduction

Theorem 1.1**.**

Theorem 1.2**.**

Remark 1.3**.**

Theorem 1.4**.**

2. Preliminaries

2.1. Abstract polytopes: definition, structure and properties

2.2. Automorphisms of polytopes

Proposition 2.1**.**

2.3. Group theory

Proposition 2.2**.**

Theorem 2.3**.**

Theorem 2.4**.**

Theorem 2.5**.**

Theorem 2.6** (3rd Sylow’s theorem).**

Proposition 2.7**.**

Proposition 2.8**.**

Proposition 2.9**.**

3. A few technical lemmas

Lemma 3.1**.**

Lemma 3.2**.**

Lemma 3.3**.**

4. Proof of Theorem 1.1

5. Proof of Theorem 1.2

6. Proof of Theorem 1.4

7. Acknowledgements

Chiral polytopes of order $2p^{m}$

Theorem 1.1.

Theorem 1.2.

Remark 1.3.

Theorem 1.4.

Proposition 2.1.

Proposition 2.2.

Theorem 2.3.

Theorem 2.4.

Theorem 2.5.

Theorem 2.6 (3rd Sylow’s theorem).

Proposition 2.7.

Proposition 2.8.

Proposition 2.9.

Lemma 3.1.

Lemma 3.2.

Lemma 3.3.