The structure of $3$-pyramidal groups

TL;DR

This paper classifies groups acting regularly on the points of 3-pyramidal combinatorial block designs, especially Kirkman triple systems, focusing on their involution structure.

Contribution

It provides a classification of groups with specific involution properties acting on 3-pyramidal designs, extending understanding of their automorphism groups.

Findings

Groups with this property have exactly 3 involutions, all conjugate.

Classification results for groups acting on Kirkman triple systems.

Insights into the automorphism structure of 3-pyramidal designs.

Abstract

A combinatorial block design $D$ is called $3$-pyramidal if there exists a subgroup $G$ of $\mbox{Aut}(D)$ fixing $3$ points and acting regularly on the other points. If this happens, we say that the design is $3$-pyramidal under $G$. In case $D$ is a Kirkman triple system, it is known that such a group $G$ has precisely $3$ involutions, all conjugate to each other. In this paper, we obtain a classification of the groups with this property.

Equations

$$X/X\cap N\cong XN/N⩽\mathrm{Aut}(S)/N\cong \text{mbox}Im(\pi )⩽\mathrm{Sym}(c).$$

$$\{1\}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$V^h=((U\cap S{)}^{t_1}\times \cdots \times (U\cap S{)}^{t_m}{)}^{\gamma }=V^{\gamma }=V.$$

$$|N_G(V)|=|G:\mathrm{soc}(G)|\cdot |N_G(V)\cap \mathrm{soc}(G)|=3^t\cdot |U\cap S{|}^m=3^{t-sm}\cdot 2^{dm}\cdot r^m.$$

$$\Delta =\{(s,s,\cdots ,s):s\in S\}⩽S^m,H:=N_G(\Delta ).$$

$$\{1\}=P_0⩽N_0<P_1<N_1<P_2<\dots <P_l⩽N_l=G$$

Full text

The first author acknowledges the support of the CAPES PhD fellowship and the NSF of China - Grant number 12161035. The second author acknowledges the support of Conselho Nacional de Desenvolvimento Científico e Tecnológico (CNPq), Universal

• Grant number 402934/2021-0.

1. Introduction

A Steiner triple system of order $v$, briefly $\operatorname{STS}(v)$, is a pair $(V,B)$ where $V$ is a set of $v$ points and $B$ is a set of $3$-subsets (blocks) of $V$ with the property that any two distinct points are contained in exactly one block. A Kirkman triple system of order $v$, briefly $\operatorname{KTS}(v)$, is an $\operatorname{STS}(v)$ together with a resolution $R$ of its block-set $B$, that is a partition of $B$ into classes (parallel classes) each of which is, in its turn, a partition of the point-set $V$. It is known that a $\operatorname{KTS}(v)$ exists if and only if $v\equiv 3\mod 6$. These structures fall into the broader category of combinatorial designs. A combinatorial design $D$ is said to be $3$-pyramidal if there exists a subgroup of $\operatorname{Aut}(D)$ which fixes $3$ points and acts sharply transitively on all the other points. Bonvicini, Buratti, Garonzi, Rinaldi and Traetta [1] gave a necessary condition for the existence of a $3$-pyramidal $\operatorname{KTS}(v)$, provided some difference methods to construct $3$-pyramidal Kirkman Triple Systems and observed that each group having a $3$-pyramidal action on a Kirkman Triple System must have exactly three involutions, and these involutions are pairwise conjugate (see [1, Theorem 3.8]). In this paper, we are concerned with studying groups with this property.

All groups considered in this paper will be finite. An involution is an element of order $2$.

Definition 1.1** (3-pyramidal group).**

We say that a finite group $G$ is $3$-pyramidal if it has exactly $3$ involutions, which are all conjugate to each other.

In [1], $3$-pyramidal groups are called pertinent. Easy examples of $3$-pyramidal groups are the symmetric group $S_{3}$ and the alternating group $A_{4}$. Observe that a $3$-pyramidal group is necessarily nonabelian of order divisible by $6$, since it contains involutions and it acts transitively on a set of size $3$. Moreover, if $G$ is $3$-pyramidal and $H$ is any group of odd order, then $G\times H$ is also $3$-pyramidal.

There is literature about finite solvable groups in which all the involutions are conjugate, see for example [2, Section 8 of Chapter IX]. The Sylow $2$-subgroups of such groups were classified by Shaw [3], they are cyclic, generalized quaternion (in case there is only one involution), homocyclic (direct products of isomorphic cyclic groups) or Suzuki $2$-groups, i.e. $2$-groups $P$ that admit a solvable subgroup of $\operatorname{Aut}(P)$ whose action on the involutions of $P$ is transitive. We refer to [2, Section 7 of Chapter VIII] for the structure of Suzuki $2$-groups.

Our main theorem is the following.

Theorem 1.2**.**

Let $G$ be a finite group and $O(G)$ the largest normal subgroup of $G$ of odd order. Let $K$ be the subgroup generated by the involutions of $G$. Then $G$ is $3$-pyramidal if and only if one of the following holds.

1. (1)

$G$* is isomorphic to $S_{3}\times H$ where $H$ is a group of odd order.* 2. (2)

$O(G)\leqslant C_{G}(K)$* and $G/O(G)$ is isomorphic to $N\rtimes A$ where $N$ is the Suzuki $2$-group of order $64$ and $A$ is a subgroup of $\operatorname{Aut}(N)$ of order $3$ or $15$.* 3. (3)

$O(G)\leqslant C_{G}(K)$* and $G/O(G)$ is isomorphic to $(C_{2^{n}}\times C_{2^{n}})\rtimes A$ where $A$ is the cyclic group of order $3$ generated by the automorphism $(a,b)\mapsto(b,(ab)^{-1})$.*

In item (1) $K\cong S_{3}$ while in items (2), (3) $K\cong C_{2}\times C_{2}$.

By the Feit-Thompson theorem, all finite groups of odd order are solvable. Therefore, the above theorem implies that all $3$-pyramidal groups are solvable. The quotients $G/O(G)$ in item (2) are SmallGroup(192,1025), SmallGroup(960, 5748).

We can construct infinite families of $3$-pyramidal groups as follows: let $Y$ be a group of odd order with a normal subgroup $X$ of index $3$ and let $N:=C_{2^{n}}\times C_{2^{n}}$. The group $Y$ acts on $N$ as an automorphism of order $3$ as in item (3) of the statement of Theorem 1.2, by composition $Y\to Y/X\cong C_{3}\to\operatorname{Aut}(N)$. The semidirect product $G:=N\rtimes Y$ is $3$-pyramidal and $O(G)=X$. Also, note that there exist $3$-pyramidal groups whose Sylow $2$-subgroups are not normal. An example is SmallGroup(1296,2705).

Our paper is organized as follows. In Section 2 we give some definitions and lemmas. In Section 3 we prove that $3$-pyramidal groups are solvable using the structure of primitive groups and the classification of finite simple groups. In Section 4 we finish the proof of Theorem 1.2 with the help of a theorem of Thompson and the structure of Suzuki $2$-groups.

2. Preliminary results

The following lemma summarizes some properties of $3$-pyramidal groups.

Lemma 2.1**.**

Let $G$ be a $3$-pyramidal group and write $|G|=2^{n}\cdot d$ with $d$ odd. Let $H$ be a subgroup of $G$, let $K$ be the subgroup of $G$ generated by the $3$ involutions and $C:=C_{G}(K)$.

1. (i)

*If $H$ has even order and $HC=G$, then $H$ is $3$-pyramidal. * 2. (ii)

Suppose the involutions of $G$ commute pairwise. If $H$ is normal in $G$ and $|H|$ is odd then $G/H$ is a $3$-pyramidal group. 3. (iii)

If $n=1$ then $K\cong S_{3}$ and $G\cong C\times K$. 4. (iv)

If $n\geqslant 2$ then $n$ is even, $K\cong C_{2}\times C_{2}$ and $|G:C|=3$. 5. (v)

If $H$ is normal in $G$ and $|H|=2^{m}$ then $m$ is even.

Proof.

Almost everything is part of [1, Section 4] (where $3$-pyramidal groups are called pertinent) including the fact that $G/C\cong S_{3}$ if $n=1$. In particular, the fact that if $n\geqslant 2$ then $n$ is even is the content of [1, Theorem 4.6]. The only thing we need to explain here is the fact that, if $n=1$, then $K\cong S_{3}$ and $G\cong C\times K$. Let $i,j,k$ be the three involutions of $G$. Note that, if $n=1$, then $ij\neq ji$, since two commuting involutions generate a subgroup of order $4$. So $i^{j}\neq i$ and of course $i^{j}\neq j$ being $i\neq j$, therefore $i^{j}=k$. The same argument shows that $j^{i}=k$, therefore $ij$ has order $3$ and $K\cong S_{3}$. Since $C\cap K=Z(K)=\{1\}$ and $G/C\cong S_{3}$, we have $|CK|=|C|\cdot|K|=|G|$ hence $CK=G$ and, since $C,K$ are normal in $G$, it follows that $G\cong C\times K$. ∎

The above lemma implies in particular that a $3$-pyramidal group generated by involutions must be isomorphic to $S_{3}$.

We include a result about almost-simple groups that will be useful in the proof. An almost-simple group with socle $S$ is a group $X$ which admits a unique minimal normal subgroup, call it $S$, which is nonabelian and simple. If this is the case, then $C_{X}(S)=\{1\}$ hence we may identify $X$ with a subgroup of $\operatorname{Aut}(S)$. The maximal subgroup $H$ of $S$ is $X$-ordinary if its $X$-class equals its $S$-class, in other words for every $x\in X$ there exists $s\in S$ such that $H^{x}=H^{s}$. Note that, if $S\leqslant X\leqslant Y\leqslant\operatorname{Aut}(S)$ and $H$ is a $Y$-ordinary maximal subgroup of $S$, then $H$ is also $X$-ordinary. We say that $H$ is ordinary if it is $\operatorname{Aut}(S)$-ordinary, so that it is $X$-ordinary for every almost-simple group $X$ with socle $S$.

The $\operatorname{Aut}(S)$-class of $H$ is a disjoint union of $S$-classes: there exist $\varphi_{1},\ldots,\varphi_{c}\in\operatorname{Aut}(S)$ such that the subgroups $H^{\varphi_{i}}\leqslant S$ are pairwise not conjugate in $S$ and for every $\varphi\in\operatorname{Aut}(S)$ there exist $s\in S$, $i\in\{1,\cdots,c\}$ such that $H^{\varphi}=H^{\varphi_{i}s}$. Let $C_{i}=\{H^{\varphi_{i}s}:s\in S\}$ be the $S$-class of $H^{\varphi_{i}}$, for $i=1,\ldots,c$. Then $\operatorname{Aut}(S)$ acts transitively on the set $\{C_{i}\ :\ i=1,\ldots,c\}$ by sending $(C_{i},\varphi)$ to the $S$-class of $H^{\varphi_{i}\varphi}$. This corresponds to a homomorphism $\pi:\operatorname{Aut}(S)\to\operatorname{Sym}(c)$. Note that $H$ is $\operatorname{Aut}(S)$-ordinary if and only if $c=1$.

Lemma 2.2**.**

Let $X$ be an almost-simple group with socle $S$, $H$ a maximal subgroup of $S$ and $\pi$ the map defined above.

1. (1)

If $X\leqslant\ker(\pi)$ then $H$ is $X$-ordinary. 2. (2)

If $X$ is normal in $\operatorname{Aut}(S)$ and $H$ is $X$-ordinary, then $X\leqslant\ker(\pi)$. 3. (3)

If $c=2$ and $X$ does not have $C_{2}$ as a quotient, then $H$ is $X$-ordinary. 4. (4)

If $H$ is $X$-ordinary, then $N_{X}(H)$ is a maximal subgroup of $X$ and the intersection $N_{X}(H)\cap S$ is equal to $H$.

Proof.

Item 1. Assume $X\leqslant\ker(\pi)$ and let $x\in X$. Then $H^{x}$ belongs to the $S$-class of $H$, so there is some $s\in S$ with $H^{x}=H^{s}$. This shows that $H$ is $X$-ordinary.

Item 2. Assume that $X$ is normal in $\operatorname{Aut}(S)$, that $H$ is $X$-ordinary and let $x\in X$. Then there is some $s\in S$ with $H^{x}=H^{s}$. If $i\in\{1,\ldots,c\}$ then, since $\varphi_{i}x{\varphi_{i}}^{-1}\in X$, being $X\unlhd\operatorname{Aut}(S)$, there is some $s\in S$ with $H^{\varphi_{i}x{\varphi_{i}}^{-1}}=H^{s}$, therefore $H^{\varphi_{i}x}=H^{s\varphi_{i}}=H^{\varphi_{i}{\varphi_{i}}^{-1}s\varphi_{i}}$ belongs to the $S$-class of $H^{\varphi_{i}}$ since $S\unlhd\operatorname{Aut}(S)$.

Item 3. Let $N:=\ker(\pi)$. Then

[TABLE]

In particular, if $c=2$ then the index $|X:X\cap N|$ is $1$ or $2$. So if $c=2$ and $X$ does not have $C_{2}$ as a quotient then $|X:X\cap N|=1$, i.e. $X\leqslant N=\ker(\pi)$. This implies that $H$ is $X$-ordinary by item (1).

Item 4. Since $H$ is $X$-ordinary, for every $x\in X$, there is $s\in S$ such that $H^{x}=H^{s}$, hence $H^{xs^{-1}}=H$, and hence $xs^{-1}\in N_{X}(H)$. Therefore $N_{X}(H)S=X$. Since $H$ is not normal in $X$, $X$ has a maximal subgroup $M$ with $N_{X}(H)\leqslant M<X$, so that $MS=X$. Since $M\cap S\unlhd M$, the maximality of $M$ in $X$ implies that $N_{X}(M\cap S)$ is equal to one of $M,X$. Since $\{1\}\neq H\leqslant M\cap S<S$ and $S$ is a simple group, $M\cap S$ cannot be normal in $X$, so $N_{X}(M\cap S)=M$. Since $H\leqslant M\cap S<S$, the maximality of $H$ in $S$ implies that $M\cap S=H$. ∎

Lemma 2.3**.**

Let $X$ be any almost-simple group with socle $S$. Assume that $X/S$ is a (possibly trivial) cyclic $3$-group and that every maximal subgroup of $X$ not containing $S$ is solvable. Let $R$ be a maximal subgroup of $X$ such that $RS=X$. Then $R\cap S$ is a maximal subgroup of $S$.

Proof.

Let $Y$ be a normal subgroup of $X$ containing $S$, assume that $R\cap Y$ is a maximal subgroup of $Y$ and that $Y$ is of minimal order with these properties. This makes sense since $R$ is maximal in $X$. We aim to prove that $Y=S$. Both $Y/S$ and $X/S$ are (possibly trivial) cyclic $3$-groups. The pair $(Y,R\cap Y)$ is listed in [4, Tables 14-20] with the notation $(G_{0},H_{0})$ and we will refer to this table (and its notation) in our discussion. If $S$ is an alternating or sporadic group, then $X/S$ is a $2$-group hence $S=Y=X$. In the case of linear groups, $S={\operatorname{PSL}}(n,q)$ with $q=p^{f}$, $p$ a prime, the inverse-transpose automorphism $\tau$ has order $2$ and, if $\sigma$ is a field automorphism and $\delta$ is a diagonal automorphism (induced by conjugation via an element of ${\operatorname{PGL}}(n,q)$), then $\sigma\tau\neq 1$ has even order dividing $2f$ and $(\delta\tau)^{2}$ is inner. In the symplectic or exceptional case, the graph automorphism $\mu$ defined in [4, Remark 10.4] is a $2$-element and we have $|^{2}F_{4}(2):{{}^{2}}F_{4}(2)^{\prime}|=2$. This implies that, if $S$ is exceptional, then $Y/S$ is a $2$-group hence $Y=S$. We are left to examine the following list of groups. When dealing with small groups, we will use the Atlas [5]. In the case of orthogonal groups, $\sigma$ is a field automorphism and $\tau$ is the triality of ${\operatorname{P\Omega}}_{8}^{+}(q)$.

• •

$Y\cong{\operatorname{PGL}}_{3}(4)$, $S\cong{\operatorname{PSL}}_{3}(4)$, $\operatorname{Out}(S)\cong D_{12}$. Since $X/S$ is a cyclic $3$-group, $X=Y$ has a nonsolvable maximal subgroup $2^{4}:(3\times A_{5})$ supplementing the socle, contradicting the assumption.

• •

$Y\cong{\operatorname{PGU}}_{3}(5)$, $S\cong{\operatorname{PSU}}_{3}(5)$, $\operatorname{Out}(S)\cong S_{3}$. Since $X/S$ is a cyclic $3$-group, $X=Y$ has a nonsolvable maximal subgroup $2S_{5}\times 3$ supplementing the socle, contradicting the assumption.

• •

$Y\cong{\operatorname{P\Omega}}_{8}^{+}(2).\langle\tau\rangle$, $S\cong{\operatorname{P\Omega}}_{8}^{+}(2)$, $\operatorname{Out}(S)\cong S_{3}$. Since $X/S$ is a cyclic $3$-group, $X=Y$ has a nonsolvable maximal subgroup ${\operatorname{PSU}}_{3}(3):2\times 3$ supplementing the socle, contradicting the assumption.

• •

$Y\cong{\operatorname{P\Omega}}_{8}^{+}(3).\langle\tau\rangle$, $S\cong{\operatorname{P\Omega}}_{8}^{+}(3)$, $\operatorname{Out}(S)\cong S_{4}$. Since $X/S$ is a cyclic $3$-group, $X=Y$ has a nonsolvable maximal subgroup $2^{3+6}({\operatorname{PSL}}_{3}(2)\times 3)$ supplementing the socle, contradicting the assumption.

• •

$Y={\operatorname{P\Omega}}_{8}^{+}(q).\langle\tau\sigma\rangle$, $S\cong{\operatorname{P\Omega}}_{8}^{+}(q)$ where $q\neq 3$ and $o(\sigma)=3^{k}$, $k\geqslant 0$. If $q=2$, then $\operatorname{Out}(S)\cong S_{3}$ and the field automorphism $\sigma$ is trivial, this implies that $Y={\operatorname{P\Omega}}_{8}^{+}(2).\langle\tau\rangle$, which was considered above. If $q>3$ then, by [6, Table 8.50] (see also [7, Table I]), $S$ has a nonsolvable ordinary maximal subgroup $M$ in the Aschbacher class $\mathcal{C}_{1}$ of type $[q^{9}](\frac{1}{d}{\operatorname{GL}}_{2}(q)\times\Omega_{4}^{+}(q)).d$, where $d=(q-1,2)$. By Lemma 2.2, we have $N_{X}(M)S=X$ and $N_{X}(M)$ is a nonsolvable maximal subgroup of $X$. This contradicts the assumption.

This concludes the proof. ∎

3. Solvability of $3$-pyramidal groups

In this section, we will discuss the solvability of $3$-pyramidal groups and use the classification theorem for finite simple groups to prove that $3$-pyramidal groups are solvable. We will also use, without mentioning it in all cases, the Feit-Thompson theorem, which says that all finite groups of odd order are solvable.

We prove the result by contradiction. Assume $W$ is a nonsolvable $3$-pyramidal group of minimal order. Let $K$ be the subgroup of $W$ generated by the three involutions and let $C:=C_{W}(K)$. Note that both $K$ and $C$ are normal in $W$. An important observation, which we will use many times without special mention, is that, if $H$ is a proper subgroup of $W$ such that $HC=W$, then $H$ solvable. Indeed, either $|H|$ is odd, in which case $H$ is solvable by the Feit-Thompson theorem, or $|H|$ is even, in which case $H$ is $3$-pyramidal by Lemma 2.1(i), hence it is solvable by minimality of $W$. Write $|W|=2^{n}\cdot d$, where $d$ is odd. By [8, Exercise 1.6.19] we have $n\geqslant 2$. In the light of Lemma 2.1(iv), $n$ is even, $K\cong C_{2}\times C_{2}$ and $W/C\cong A_{3}$, so $C$ is a maximal normal subgroup of $W$. If $H$ is a nontrivial normal subgroup of $W$ with odd order, Lemma 2.1(ii) implies that $W/H$ is $3$-pyramidal. Since $|W/H|<|W|$ and $|H|$ is odd, both $H$ and $W/H$ are solvable, this contradicts the nonsolvability of $W$, so $O(W)=\{1\}$. In particular, the minimal normal subgroups of $W$ have even order, so they contain all the involutions. This implies that the unique minimal normal subgroup of $W$ is $K$.

We divide the proof into several steps. Recall that a chief factor of $W$ is a minimal normal subgroup of a quotient of $W$. Since a minimal normal subgroup is always characteristically simple, it is isomorphic to a direct power of a simple group. As usual, $\Phi(W)$ denotes the Frattini subgroup of $W$.

3.1. Step 1

We will prove that there exists a normal subgroup $N$ of $W$ with $\Phi(W)<N\leqslant C$ such that $W/N$ is a cyclic $3$-group, $N/\Phi(W)$ is a nonabelian chief factor of $W$, isomorphic to $S^{m}$ for some nonabelian simple group $S$, and $\Phi(W)$ is a $2$-group containing $K$. Writing $|\Phi(W)|=2^{a}$, the integer $a$ is even.

[TABLE]

Proof.

Let $N$ be a normal subgroup of $W$ such that $W/N$ is cyclic and $N$ is of minimal order with this property. If $W/N$ is not a $3$-group, then there exists $L/N\unlhd W/N$ such that $|W:L|$ is a prime distinct from $3$, so that $LC=W$ (being $|W:C|=3$) hence $L$ is solvable. Since $W/L$ is solvable, this contradicts the fact that $W$ is nonsolvable. So $W/N$ is a cyclic $3$-group. Moreover $N$ is contained in $C$ because otherwise $NC=W$ and then $N$ would be solvable, so $W$ would be solvable as well. Consider a normal subgroup $R$ of $W$ contained in $N$ with the property that $N/R$ is a minimal normal subgroup of $W/R$. We claim that $R=\Phi(W)$. First, note that $\Phi(W)$ is a $2$-group because otherwise there would exist a nontrivial Sylow subgroup $P$ of $\Phi(W)$ of odd order, and since $P\unlhd_{c}\Phi(W)\unlhd W$, being $\Phi(W)$ nilpotent, $P$ would be a nontrivial normal subgroup of $W$ of odd order, contradicting $O(W)=\{1\}$. Since $\Phi(W)/\Phi(W)\cap N\cong\Phi(W)N/N\leqslant\Phi(W/N)$ is a $3$-group, we deduce that $\Phi(W)\leqslant N$. By minimality of $N$, $W/R$ is not a cyclic $3$-group, so if $x\in W$ is a $3$-element that generates $W/C$, then $R\langle x\rangle$ is a proper subgroup of $W$, moreover $R\langle x\rangle C=W$, so $R$ is solvable. Since $R\leqslant N\leqslant C$, to prove that $R\leqslant\Phi(W)$ it is enough to prove that if $M$ is a maximal subgroups of $W$ distinct from $C$, then $R\leqslant M$. We have $MC=W$, so $M$ is solvable hence $MR\neq W$ being $W$ nonsolvable and $M,R$ solvable. Therefore $R\leqslant M$, implying that $R\leqslant\Phi(W)$. Since $W/N$ is a $3$-group and $W$ is nonsolvable, we deduce that $\Phi(W)\neq N$, in other words $\Phi(W)$ is properly contained in $N$. Since $R\leqslant\Phi(W)<N$ and $N/R$ is a minimal normal subgroup of $W/R$, we deduce that $R=\Phi(W)$. Since $\Phi(W)$ is nilpotent, the nonsolvability of $W$ implies that $N/\Phi(W)$ is a nonabelian chief factor of $W$, isomorphic to a direct power $S^{m}$ for some nonabelian simple group $S$. Observe that $S$ has more than $3$ involutions: indeed, no involution of $S$ is central, being $S$ simple, and if $S$ had three involutions then the kernel of the conjugation action of $S$ on them would be a normal subgroup of $S$ of index at most $3!=6$, a contradiction. We deduce that $\Phi(W)\neq\{1\}$, so $\Phi(W)$, being a $2$-group, contains at least one involution, hence it contains all $3$ of them being $\Phi(W)\unlhd W$. Writing $|\Phi(W)|=2^{a}$, the integer $a$ is even by Lemma 2.1(v). ∎

3.2. Step 2

Recall that a finite group $H$ is said to be primitive if it admits a maximal subgroup $M$ with trivial normal core, in other words the intersection of the conjugates of $M$ in $H$, usually denoted $M_{H}$, is trivial. Note that, if $M$ is any maximal subgroup of a finite group $H$, then $H/M_{H}$ is primitive since $M/M_{H}$ is a maximal subgroup of $H/M_{H}$ with trivial normal core. Let $M$ be a maximal subgroup of $W$ with $M\neq C$. We will prove that the normal core $M_{W}$ of $M$ in $W$ equals $\Phi(W)$ and consequently $W/\Phi(W)$ is a primitive group.

Proof.

By Lemma 2.1(i), the maximal subgroups of $W$ distinct from $C$ are solvable by minimality of $W$, since they supplement $C$. If $M_{1}$ is a maximal subgroup of $W$ such that $M_{1}\neq C$ then $M_{W}\leqslant M_{1}$, since otherwise we would have $M_{1}M_{W}=W$, contradicting the fact that $W$ is nonsolvable, being $M_{1}$ and $M_{W}$ solvable. So every maximal subgroup of $W$ different from $C$ contains $M_{W}$, implying that $\Phi(W)=M_{W}\cap C$. To conclude the proof, we need to prove that $M_{W}\leqslant C$. If this is not the case, then $CM_{W}=W$. Note that $W/M_{W}=CM_{W}/M_{W}\cong C/\Phi(W)$ has a subgroup $N/\Phi(W)\cong S^{m}$. Since every nonabelian simple group has order divisible by $4$, there exist two subgroups $A,B\leqslant W$ containing $M_{W}$ such that $A\leqslant B$ and $|B:A|=|A:M_{W}|=2$. Since $A$ and $B$ contain $M_{W}$, $AC=BC=W$, hence $A$ and $B$ are both $3$-pyramidal by Lemma 2.1(i). Since $K\leqslant\Phi(W)\leqslant M_{W}\leqslant A$ and $|K|=4$, this contradicts Lemma 2.1(iv). ∎

3.3. Step 3

Let $G:=W/\Phi(W)$, it is a primitive group whose socle is $N/\Phi(W)\cong S^{m}$, it is the unique minimal normal subgroup of $G$. Let $T_{1}$ be the first direct factor of $N/\Phi(W)\cong S^{m}$ and let $G_{1}:=N_{G}(T_{1})/C_{G}(T_{1})$. Then $G_{1}$ is an almost-simple group with socle $T_{1}C_{G}(T_{1})/C_{G}(T_{1})\cong S$ and we will identify $G_{1}$ with a subgroup of $\operatorname{Aut}(S)$ containing $S$. We say that a finite group is of even type if, in the prime factorization of its order, the prime $2$ appears with even multiplicity. Lemma 2.1(iv-v) implies that $G$ is of even type. We will prove that, if $U$ is a maximal subgroup of $G_{1}$ such that $US=G_{1}$, then $U$ is a solvable group of even type.

Proof.

We refer to [9, Chapter 1] for the general properties of primitive groups. Since $N/\Phi(W)$ is a nonabelian minimal normal subgroup of $G$ and $W/N$ is cyclic, $G$ is a primitive group of type $2$, meaning that it admits a unique minimal normal subgroup which is nonabelian. Since $W/N$ is a cyclic $3$-group, we have an embedding of $G$ in the standard wreath product $G_{1}\wr P$ where $P$ is a cyclic $3$-subgroup of $\operatorname{Sym}(m)$, the image of the permutation representation of $G$ acting transitively by conjugation on the $m$ minimal normal subgroups of its socle (see [9, Remark 1.1.40.13]). In particular, $m$ is a power of $3$.

Let $V:=(U\cap S)^{m}\leqslant S^{m}=\operatorname{soc}(G)$. We claim that $N_{G}(V)\operatorname{soc}(G)=G$. Let $g=(x_{1},\ldots,x_{m})\gamma\in G$, where $x_{i}\in G_{1}$ for all $i$ and $\gamma\in P$. Since $US=G_{1}$, there exist $s_{i}\in S$, $t_{i}\in U$ such that $x_{i}=s_{i}t_{i}$ for all $i=1,\ldots,m$, and setting $n:=(s_{1},\ldots,s_{m})\in\operatorname{soc}(G)$ we have $h:=(t_{1},\ldots,t_{m})\gamma=n^{-1}g\in G$, therefore

[TABLE]

We deduce that $h=n^{-1}g\in N_{G}(V)$ and since $n\in\operatorname{soc}(G)$ the claim follows.

We claim that $N_{G}(V)\neq G$. If this was not the case, then $V$ would be normal in $G$ and being $N$ a minimal normal subgroup of $G$, and $U\cap S\neq S$ being $US=G_{1}$, since $S$ is simple we deduce that $U\cap S=\{1\}$ and $G_{1}/S\cong U$. It follows that $U$ is isomorphic to a section of $G/\operatorname{soc}(G)\cong W/N$, which is cyclic, therefore $U$ is a cyclic maximal subgroup of the almost-simple group $G_{1}$, contradicting Herstein’s theorem, which says that a nonsolvable finite group cannot have abelian maximal subgroups (see [10, Theorem 5.53]). So the claim follows.

Since $\Phi(W)$ is a nontrivial $2$-group, the preimage of $N_{G}(V)$ in $W$ is a proper subgroup of $W$ of even order and it is not contained in $C$, so it is $3$-pyramidal by Lemma 2.1(i) hence it is solvable by minimality of $W$. So $U\cap S$ is solvable. Since $U/U\cap S\cong US/S=G_{1}/S$ is a cyclic $3$-group, this implies that $U$ is solvable.

Summarizing, we have $N_{G}(V)\operatorname{soc}(G)=G$, $N_{G}(V)\neq G$, $G_{1}/S$ is a cyclic $3$-group and $G_{1}$ has only one nonsolvable maximal subgroup, which contains $S$. By Lemma 2.3, $U\cap S$ is a maximal subgroup of $S$. We claim that $N_{G}(V)\cap\operatorname{soc}(G)=V$. To show this, let $L$ be a maximal subgroup of $G$ containing $N_{G}(V)$. Since $V\leqslant N_{G}(V)\leqslant L$ and $N_{G}(V)\operatorname{soc}(G)=G$, the maximal subgroup $L$ is either of product type or of diagonal type (the two types described in [9, Remark 1.1.40.19]). But since $L\operatorname{soc}(G)=G$, $L$ is solvable, and hence $L$ must be of product type. Thus we can write $L\cap\operatorname{soc}(G)=H_{1}\times\cdots\times H_{m}$ with $H_{i}<S$ for all $i$ and the $H_{i}$’s are pairwise isomorphic. We have $L\cap\operatorname{soc}(G)\neq\{1\}$ since otherwise $L$ would be isomorphic to $G/\operatorname{soc}(G)$, so it would be a cyclic maximal subgroup of $G$. Therefore $H_{i}\neq\{1\}$ for all $i$. Since $U\cap S$ is a maximal subgroup of $S$ and $V\leqslant N_{G}(V)\cap\operatorname{soc}(G)\leqslant L\cap\operatorname{soc}(G)$, we deduce that $H_{i}=U\cap S$ for all $i$, hence $L\cap\operatorname{soc}(G)=V$. On the other hand $L\leqslant N_{G}(L\cap\operatorname{soc}(G))=N_{G}(V)$ hence $L=N_{G}(V)$ by maximality of $L$, therefore $N_{G}(V)\cap\operatorname{soc}(G)=V$.

Since $N_{G}(V)\operatorname{soc}(G)=G$ and $U/U\cap S\cong US/S=G_{1}/S\cong C_{3^{s}}$ is a cyclic $3$-group, writing $|U|=2^{d}\cdot r$ with $r$ odd and $G/\operatorname{soc}(G)\cong C_{3^{t}}$, we have

[TABLE]

Since the preimage of $N_{G}(V)$ in $W$ is $3$-pyramidal by Lemma 2.1(i), and $\Phi(W)$ is of even type by Lemma 2.1(v), $N_{G}(V)$ is of even type by Lemma 2.1(iv), in other words $dm$ is even. Since $m$ is a power of $3$, we conclude that $d$ is even. ∎

3.4. Step 4

$G_{1}=S$.

Proof.

Let $M$ be any maximal subgroup of $W$ different from $C$, so that $M$ is solvable, $\Phi(W)=M_{W}$ and $H:=M/\Phi(W)$ is a point stabilizer of the primitive group $G$. Since $H$ is solvable, it is of product type, so $H$ is conjugate in $G$ to $N_{G}((H_{1}\cap S)^{m})$ for a suitable maximal subgroup $H_{1}$ of $G_{1}$ such that $H_{1}S=G_{1}$. By [4, Theorem 1.1], $G_{1}$ has a normal subgroup $G_{0}$ which is minimal such that $H_{0}:=H_{1}\cap G_{0}$ is maximal in $G_{0}$ and the pair $(G_{0},H_{0})$ is explicitly listed in [4, Tables 14-20] up to conjugacy. Lemma 2.3 implies that $G_{0}=S$.

We have $S\leqslant G_{1}\leqslant\operatorname{Aut}(S)$ and $G_{1}/S$ is a cyclic $3$-group. In our discussion, we will consider suitable nonsolvable maximal subgroups $R$ of $S$ and we will apply Lemma 2.2 with $R$ in place of $H$ and $G_{1}$ in place of $X$. Since $G_{1}/S$ is a cyclic $3$-group, if $c\leqslant 2$ then $R$ is $G_{1}$-ordinary and hence $N_{G_{1}}(R)$ is a nonsolvable maximal subgroup of $G_{1}$ supplementing $S$, this contradicts Step 3. Note that our “$c$” is the same as in [11, Paragraph 3.2] and [6, Chapter 8].

Assume $G_{1}\neq S$ by contradiction. In the following discussion we will use [4, Tables 14-20] recalling that $G_{0}=S$ in our situation. Since $G_{1}/S$ is a $3$-group, if $\operatorname{Out}(S)$ is a $2$-group we immediately have a contradiction. This is the case if $S$ is an alternating or sporadic group. If $S$ is a symplectic group, then by [4, Table 17] again $\operatorname{Out}(S)$ is a $2$-group, a contradiction. In the following discussion, the existence of a nonsolvable $G_{1}$-ordinary maximal subgroup of $S$ gives a contradiction by Lemma 2.2, as explained in the previous paragraph.

If $S$ is a unitary group, $S\cong{\operatorname{PSU}}_{n}(q)$ where $q=p^{f}$ and $n\geqslant 4$, then by [6, Tables 8.10, 8.20, 8.26, 8.37, 8.46, 8.56, 8.62, 8.72, 8.78] and [11, Table 3.5.B], $S$ has an ordinary parabolic maximal subgroup of type $[q^{2n-3}]:{\operatorname{SU}}_{n-2}(q):(q^{2}-1)/d$, with $d=(n,q+1)$, which is nonsolvable unless $(n,q)\in\{(4,2),(4,3),(5,2)\}$, in which case $\operatorname{Out}(S)$ is a $2$-group. Assume $n=3$, so that $q\geqslant 4$ since ${\operatorname{PSU}}_{3}(2)$ is not a simple group and $\operatorname{Out}({\operatorname{PSU}}_{3}(3))\cong C_{2}$. By [6, Table 8.5] $S$ has a nonsolvable ordinary maximal subgroup of type ${\operatorname{GU}}_{2}(q)/d$, where $d=(q+1,3)$. If $n=2$ then $S\cong{\operatorname{PSL}}_{2}(q)$ will be considered below.

If $S$ is an orthogonal group, by [4, Table 19] $S$ is isomorphic to ${\operatorname{P\Omega}}_{7}(3)$, ${\operatorname{P\Omega}}_{8}^{+}(q)$, ${\operatorname{P\Omega}}_{12}^{+}(3)$ or ${\operatorname{P\Omega}}_{16}^{+}(3)$. Since $\operatorname{Out}(S)$ is a $2$-group unless $S\cong{\operatorname{P\Omega}}_{8}^{+}(q)$, we assume this is the case. If $q=2$ then $\operatorname{Out}(S)\cong S_{3}$ and if $q=3$ then $\operatorname{Out}(S)\cong S_{4}$, hence $G_{1}=S.\langle\tau\rangle$ where $\tau$ is the triality automorphism in these two cases, and $G_{1}$ has a nonsolvable maximal subgroup of type ${\operatorname{PSU}}_{3}(3):2\times 3$ if $q=2$ and $G_{2}(3)\times 3$ if $q=3$, not containing $S$ (see [5]), contradicting Step 3. If $q\geqslant 4$, then $S$ has a nonsolvable ordinary maximal subgroup of type $[q^{9}](\frac{1}{d}{\operatorname{GL}}_{2}(q)\times\Omega_{4}^{+}(q)).d$ where $d=(q-1,2)$ by [6, Table 8.50] and [7, Table I].

Assume $S$ is a linear group, $S\cong{\operatorname{PSL}}_{n}(q)$, where $n\geqslant 2$ and $q=p^{f}$. If $n\geqslant 3$ then $S$ has a maximal parabolic subgroup $P_{1}$ of type $[q^{n-1}].{\operatorname{SL}}_{n-1}(q).(q-1)/d$, $d=(q-1,n)$. By [11, Proposition 4.1.17] and Lemma 2.2, $P_{1}$ is $G_{1}$-ordinary. It is nonsolvable unless $n=3$ and $q\leqslant 3$, in which case $\operatorname{Out}(S)\cong C_{2}$ and $G_{1}=S$. Assume $n=2$, so that $S\cong{\operatorname{PSL}}_{2}(q)$. Since $\operatorname{Out}(S)\cong C_{(2,q-1)}\times C_{f}$ and $G_{1}/S$ is a cyclic $3$-group, $3$ divides $f$. By [6, Table 8.1], $S$ has an ordinary maximal subgroup of type ${\operatorname{PSL}}_{2}(q_{0})$, where $q_{0}=p^{f/3}=\sqrt[3]{q}$, if $q_{0}\neq 2$, and this maximal subgroup is nonsolvable if $q_{0}\geqslant 4$. If this is not the case, then $q_{0}\leqslant 3$ i.e. $q\in\{8,27\}$. Thus $G_{1}/S\cong C_{3}$ and $G_{1}$ has a maximal subgroup of type $7:6$ or $13:6$ supplementing $S$ (see [5]), contradicting Step 3, since such subgroups are not of even type.

We now consider exceptional groups. The group $S$ appears in the lines of [4, Table 20] in which $G_{0}$ is a simple group. We first suppose that $S$ is isomorphic to a Suzuki group $\operatorname{Sz}(2^{m})={{}^{2}}B_{2}(2^{m})$, with $m$ odd. If $m$ is not prime, then we can write $m=ab$ with $b$ prime and $a>2$, so by [6, Table 8.16], $S$ has a nonsolvable ordinary maximal subgroup isomorphic to $\operatorname{Sz}(2^{a})$. If $m$ is a prime then, since $3$ divides $|\operatorname{Out}(S)|=m$, we have $m=3$, $S\cong\operatorname{Sz}(8)$ and $G_{1}=S.C_{3}$ has a maximal subgroup of type $7:6$ supplementing $S$ (see [5]). This contradicts Step 3, since such subgroup is not of even type. We use the notation $\operatorname{Ree}(q)={{}^{2}}G_{2}(q)$, as in [4]. Observe that $\operatorname{Ree}(3)^{\prime}\cong{\operatorname{PSL}}_{2}(8)$ was considered in the discussion of linear groups and $\operatorname{Out}(S)\cong C_{2}$ if $S\cong G_{2}(2)^{\prime}$, $G_{2}(3)$, ${}^{2}F_{4}(2)^{\prime}$ or $F_{4}(2)$, so we obtain a contradiction in these cases. By [12, Main Theorem, Tables 5.1, 5.2], taking $L=S$ and $X=G_{1}$, if $S$ is one of the exceptional groups $\operatorname{Ree}({q})$ for $q\geqslant 27$, ${{}^{2}}F_{4}(q)$ for $q\geqslant 8$, ${{}^{3}}D_{4}(q)$, ${{}^{2}}E_{6}(q)$, $E_{6}(q)$ and $E_{8}(q)$, then $G_{1}$ has a maximal subgroup $M=N_{X}(D)$ which is nonsolvable because $N_{G_{\sigma}}(D)$ is nonsolvable and its derived subgroup is contained in $M$, indeed $(G_{\sigma})^{\prime}=L\leqslant X$ and hence $(N_{G_{\sigma}}(D))^{\prime}\leqslant N_{G_{\sigma}}(D)\cap X\leqslant N_{X}(D)=M$. Moreover, $M$ does not have $S$ as a composition factor, therefore $MS=G_{1}$. This gives a contradiction. ∎

3.5. Step 5

We conclude that $G_{1}=S$. Since $G$ is a subgroup of $S\wr P$ containing $S^{m}=\operatorname{soc}(G)$, with $G$ projecting surjectively onto the transitive cyclic group $P\leqslant\operatorname{Sym}(m)$, we deduce that $G$ is isomorphic to the standard wreath product $S\wr P$, where $P$ acts on $S^{m}$ by permuting the coordinates. Consider

[TABLE]

It is clear that $P\leqslant H$, hence $\operatorname{soc}(G)H=G$. Since $G$ has a normal subgroup of index $3$, we have $m\geqslant 2$, therefore $\Delta$ is not normal in $G$, being $\operatorname{soc}(G)$ a minimal normal subgroup of $G$, therefore $H\neq G$. Since $\Delta\cong S$ is nonsolvable, we obtain that the preimage of $H$ in $W$ is a nonsolvable proper subgroup of $W$, of even order and supplementing $C$, so it is $3$-pyramidal by Lemma 2.1(i). This contradicts the minimality of $W$.

This concludes the proof of the fact that $3$-pyramidal groups are solvable. $\square$

4. Proof of the Main Theorem

In this section, we will study the structure of $3$-pyramidal groups and prove our Main Theorem. Recall that, if $G$ is a $p$-solvable group, meaning that it admits a series of normal subgroups $\{1\}=V_{0}<\ldots<V_{n}=G$ such that each factor $V_{i+1}/V_{i}$ is either a $p$-group or a $p^{\prime}$-group, the $p$-length of $G$ is the length $l$ of the upper $p$-series

[TABLE]

where $N_{k}/P_{k}$ is the largest normal $p^{\prime}$-subgroup of $G/P_{k}$ and $P_{k+1}/N_{k}$ the largest normal $p$-subgroup of $G/N_{k}$, for $k=0,\ldots,l-1$.

Recall that a group $G$ is called homocyclic if it is a direct product of isomorphic cyclic groups, and $G$ is called a Suzuki $2$-group if $G$ is a nonabelian $2$-group with more than one involution and there exists a solvable subgroup of $\operatorname{Aut}(G)$ which permutes the set of involutions of $G$ transitively. Since $3$-pyramidal groups are solvable, Thompson’s result [2, Theorem 8.6 of Chapter IX] is now very useful for us. It states the following.

Theorem 4.1** (Thompson).**

Suppose that $G$ is a solvable group of even order and that the Sylow $2$-subgroup of $G$ contains more than one involution. Suppose that all the involutions in $G$ are conjugate. Then the $2$-length of $G$ is $1$ and the Sylow $2$-subgroups of $G$ are either homocyclic or Suzuki $2$-groups.

Let $G$ be a $3$-pyramidal group and write $|G|=2^{n}\cdot d$, where $d$ is odd. Let $K$ be the subgroup of $G$ generated by the three involutions $i,j,k$ and let $C:=C_{G}(K)$. If $n=1$ then the result follows from Lemma 2.1(iii). Now assume that $n\geqslant 2$. Lemma 2.1(iv) implies that $n$ is even, $K\cong C_{2}\times C_{2}$ and $G/C\cong A_{3}$. Since $K$ and $O(G)$ are normal subgroups of $G$ of coprime orders, $K\cap O(G)=\{1\}$ hence $O(G)\leqslant C$. By Lemma 2.1(ii) we may assume that $O(G)=\{1\}$. By Section 3, $G$ is solvable and Theorem 4.1 applies, so the $2$-length of $G$ is $1$. Since $O(G)=\{1\}$, this means that the Sylow $2$-subgroup $N$ of $G$ is normal hence it has a complement $A$ in $G$ by the Schur-Zassenhaus Theorem, i.e., $G=N\rtimes A$. Since $C_{A}(N)$ is a normal subgroup of $G$ of odd order, it is trivial, hence $C_{G}(N)=Z(N)$ is a $2$-group and $A$ is isomorphic to a subgroup of $\operatorname{Aut}(N)$. In the light of Theorem 4.1, $N$ is either homocyclic or a Suzuki $2$-group.

Assume that $N$ is a homocyclic group, i.e. a direct product of pairwise isomorphic cyclic groups. Since $N$ has three involutions, $N\cong C_{2^{m}}\times C_{2^{m}}$ for some positive integer $m$. It is easy to see that $\operatorname{Aut}(N)$ is isomorphic to the group of $2\times 2$ matrices with coefficients in $\mathbb{Z}/2^{m}\mathbb{Z}$ and invertible determinant, therefore $|\operatorname{Aut}(N)|=3\cdot 2^{4m-3}$. By Sylow’s theorem, $A$ is conjugate in $\operatorname{Aut}(N)$ to $\langle\gamma\rangle$ where $\gamma:N\to N$ is defined by $(a,b)\mapsto(b,(ab)^{-1})$. Therefore $G\cong N\rtimes\langle\gamma\rangle$.

We next assume that $N$ is a Suzuki $2$-group. Since $N$ has three involutions, we have $\Phi(N)=N^{\prime}=Z(N)\cong C_{2}^{2}$ by [2, Lemma 7.8, Theorem 7.9 of Chapter VIII]. A simple inspection using [13] shows that no group of order $16$ has these properties, therefore [14] implies that $|N|=64$ and $N$ is the unique Suzuki $2$-group of order $64$, SmallGroup(64, 245), $|\operatorname{Aut}(N)|=3\cdot 5\cdot 2^{10}$ and $G$ is isomorphic to $N\rtimes A$ where $A$ is a subgroup of $\operatorname{Aut}(N)$ of order $3$ or $15$. Since the subgroups of $\operatorname{Aut}(N)$ of order $3$ or $15$ are Hall subgroups and $\operatorname{Aut}(N)$ is solvable, they form a unique conjugacy class of subgroups, so $G$ is determined completely up to isomorphism.

Conversely, it is not difficult to see that $G$ is $3$-pyramidal if $G\cong S_{3}\times H$ where $H$ is a group of odd order. Assume now that $4$ divides $|G|$, let $O:=O(G)$ and assume that $O$ centralizes all involutions of $G$ and that $G/O$ is one of the groups in cases (2) and (3). Since $4$ divides $|G/O|$ and $G/O$ is $3$-pyramidal, it has exactly $3$ involutions $iO$, $jO$ and $kO$. The union $iO\cup jO\cup kO$ contains all involutions of $G$. Since $O$ is a group of odd order centralizing all involutions of $G$, each of $iO$, $jO$, $kO$ contains a unique involution, therefore $G$ has precisely three involutions, which we may assume to be $i$, $j$, $k$. Since $iO$, $jO$ and $kO$ are conjugate in $G/O$, there exists $g\in G$ such that $i^{g}O=(iO)^{gO}=jO$. Since $j$ is the unique involution in $jO$, we obtain that $i^{g}=j$. The same argument shows that $j$ and $k$ are conjugate in $G$, so $G$ is $3$-pyramidal. $\square$