A Generalization of Szep’s Conjecture for Almost Simple Groups

Abstract

We prove a natural generalization of Szep’s conjecture. Given an almost simple group G with socle not isomorphic to an orthogonal group having Witt defect zero, we classify all possible group elements $x,y\in G\setminus \{1\}$ with $G=\textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$, where we are denoting by $\textbf{N}_G(\langle x\rangle )$ and by $\textbf{N}_G(\langle y\rangle )$ the normalizers of the cyclic subgroups $\langle x\rangle$ and $\langle y\rangle$. As a consequence of this result, we classify all possible group elements $x,y\in G\setminus \{1\}$ with $G=\textbf{C}_G(x)\textbf{C}_G(y)$.

1 Introduction

Given a finite group G and $x\in G$, we denote by $\textbf{C}_G(x)$ the centralizer of x in G and by $\textbf{N}_G(\langle x\rangle )$ the normalizer of the cyclic subgroup $\langle x\rangle$ in G. It was conjectured by J. Szep [19], that if $G=\textbf{C}_G(x)\textbf{C}_G(y)$ with $x,y\in G\setminus \{1\}$, then G is not a non-abelian simple group. Over a long period, many authors investigated this conjecture and in 1987, using the Classification of the Finite Simple Groups, E. Fisman and Z. Arad [19, Theorem 1] gave a positive answer to this problem.

More recently, R. Guralnick, G. Malle and P. Tiep have obtained another proof of Szep’s conjecture [11] as a direct application of some results on the product of conjugacy classes in algebraic groups. This new proof, for Lie type groups, actually proves more than the original statement of Szep’s conjecture. Namely, it is shown that, if L is a non-abelian simple group of Lie type and $L\unlhd G\le \textrm{Inndiag}(L)$, then $\textbf{C}_G(x)\textbf{C}_G(y) \ne G$, for every $x,y\in G\setminus \{1\}$. Here $\textrm{Inndiag}(L)$ denotes the group of inner-diagonal automorphisms of L, as defined in [9, Chapter 2].

Moreover, recently Szep’s conjecture has played a crucial role in the investigation of the finite primitive groups having two coprime subdegrees [6]. Indeed, the positive solution of Szep’s conjecture is used in Theorems 1.5 and 1.6 of [6]. In order to simplify some of the arguments in the proofs of these theorems, it would have been necessary to have Szep’s conjecture available for the whole class of the finite almost simple groups.

The following is the main theorem of this paper.

Theorem 1.1

Let G be an almost simple group and let x, y be in $G\setminus \{1\}$. Suppose that the socle of G is not isomorphic to an orthogonal group $\textrm{P}\Omega _{n}^+(q)$, with $n\ge 8$. If $G=\textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$, then (replacing x by y if necessary) (G, x, y) is one of the triples in Table 1. See Sections 1.1 and 1.2, for the notation in Table 1.

Moreover, $G=\textbf{C}_G(x)\textbf{C}_G(y)$ if and only if in the 6th column of Table 1 appears the symbol $\surd$.

In the course of the proof of Theorem 1.1, we show that every triple (G, x, y) in Table 1 gives rise to a genuine example of a factorization $G=\textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$.

Table 1 Triples in Theorem 1.1, $L\ne \textrm{P}\Omega _n^+(q)$. See Sections 1.1 and 1.2 for notation

An immediate application of Theorem 1.1 gives the following corollary.

Corollary 1.2

Let G be an almost simple transitive permutation group on $\Omega$ and let $\omega$ be in $\Omega$. Suppose that the socle of G is not isomorphic to an orthogonal group $\textrm{P}\Omega _{n}^+(q)$, with $n\ge 8$. If the point stabilizer $G_\omega$ normalizes a non-identity cyclic subgroup $\langle x\rangle$ and if G contains an element $y\ne 1$ with $\textbf{N}_G(\langle y\rangle )$ transitive on $\Omega$, then (replacing x by y if necessary) the triple (G, x, y) is in Table 1.

A similar investigation for almost simple groups having socle an orthogonal group $\textrm{P}\Omega _{n}^+(q)$, with $n\ge 8$, seems difficult and, at the moment, we do have 12 different families of factorizations using normalizers. We intend to come back to this question in the future.

It is worth mentioning that our strategy for proving Theorem 1.1 is considerably different from the original proof of Szep’s conjecture [19]. Our main tool uses the classification of the maximal factorizations of the almost simple groups obtained by M. Liebeck, C. Praeger and J. Saxl in [16, 17].

Let G be an almost simple group with socle L. A factorization $G=AB$ is said to be maximal if A and B are both maximal subgroups of G, and is said to be core-free if A and B are core-free in G (that is, $L\nleq A,B$). All the core-free maximal factorizations of the almost simple groups are classified in Tables 1–6 and Theorem D of [16]. In particular, if $G=\textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$ (for some $x,y\in G\setminus \{1\}$) and $\textbf{N}_G(\langle x\rangle )\le A$, $\textbf{N}_G(\langle y\rangle )\le B$ for some core-free maximal subgroups A and B of G, then (G, A, B) is one of the triples classified in [16]. In particular, this reduces the proof of Theorem 1.1 to a case-by-case analysis on Tables 1–6 and on Theorem D of [16]. Moreover, for each of these triples (G, A, B), we have $\textbf{N}_G(\langle x\rangle ) = \textbf{N}_A(\langle x\rangle )$ and $\textbf{N}_G(\langle y\rangle )=\textbf{N}_B(\langle y\rangle )$ and so it suffices to investigate the order and the structure of the normalizers of the non-trivial elements of A and B, respectively.

There is only one more case to consider in our analysis: every maximal subgroup of G containing $\textbf{N}_G(\langle x\rangle )$ (or $\textbf{N}_G(\langle y\rangle )$) contains the socle L of G. The almost simple groups admitting such factorizations are classified in [17, Table 1] and there is only a handful of such examples.

In the process of proving Theorem 1.1 using the work in [16, 17], we have realized that there is one configuration omitted in the proof of Liebeck, Praeger and Saxl [16, 17] classifying the maximal factorizations of the almost simple groups with socle $\textrm{P}\Omega _8^+(2^f)$. Although this missing configuration is of no concern to us here because we are excluding almost simple groups having socle $\textrm{P}\Omega _8^+(2^f)$ in our main results, we discuss this configuration in Section 2 and we show that this configuration does give rise to two extra maximal factorizations omitted in the work of Liebeck, Praeger and Saxl. In Section 2, we comment how these extra factorizations influence other work relying on the classification in [16, 17]. The factorizations in [16] have been extensively used. For instance, recently, this was used in [14] to give a characterization of the factorizations of almost simple groups with a solvable factor, which was then applied to study s-arc-transitive Cayley graphs of solvable groups, leading to a striking corollary that, except for cycles, a non-bipartite connected 3-arc-transitive Cayley graph of a finite solvable group is necessarily a normal cover of the Petersen graph or the Hoffman-Singleton graph. However, Zhou [20] improved this and obtained a remarkable refinement, that is, every non-bipartite connected Cayley graph of a finite solvable group is at most 2-arc-transitive.

1.1 Notation

We use the notation from [9, Chapter 4] and [4] for conjugacy classes of elements in groups of Lie type and, in general, we use the notation from [12] for the subgroups of the classical groups.

Given an almost simple group G, we denote by L the socle of G. Suppose that L is a simple classical group defined over the finite field of size q. For twisted groups our notation for q is such that $\textrm{PSU}_n(q)$ and $\textrm{P}\Omega _{n}^-(q)$ are the twisted groups contained in $\textrm{PSL}_n(q^2)$ and $\textrm{P}\Omega _{n}^+(q^2)$, respectively. We write $q=r^f$, for some prime r and some $f\ge 1$, and we define

$$\begin{aligned} q_0:= \left\{ \begin{array}{ll} q^{2} &{}\quad \text {if { G} is unitary,}\\ q &{}\quad \text {otherwise.} \end{array}\right. \end{aligned}$$

We let V be the natural module defined over the field $\mathbb {F}_{q_0}$ of size $q_0$ for the covering group of L, and we let n be the dimension of V over $\mathbb {F}_{q_0}$.

We consider the following classical groups $\tilde{L}$:

• $\textrm{SL}_n(q)$ with $n\ge 1$,

• $\textrm{SU}_n(q)$ with $n\ge 1$,

• $\textrm{Sp}_{n}(q)$ with n even and $n\ge 2$,

• $\Omega _{n}(q)$ with qn odd and $n\ge 1$, and

• $\Omega _{n}^{\pm }(q)$ with n even and $n\ge 2$.

For some of our proofs, we need to deal with arbitrary classical groups as defined above and hence with no restrictions on n. However, for proving our main results, we take into account the various isomorphisms among classical groups, see [12, Section 2.9]. For instance, $\textrm{SL}_2(q)\cong \textrm{SU}_2(q)\cong \textrm{Sp}_2(q)\cong \Omega _3(q)$ and $\Omega _5(q)\cong \textrm{Sp}_4(q)$. In particular, in Table 1 and in proving Theorem 1.1, we may suppose $n\ge 2$ for linear groups, $n\ge 3$ for unitary groups, $n\ge 4$ for symplectic groups, $n\ge 7$ for odd dimensional orthogonal groups and $n\ge 8$ for even dimensional orthogonal groups.

The corresponding simple classical groups $L:=\tilde{L}/ \textbf{Z} (\tilde{L})$ are

$$\begin{aligned} \textrm{PSL}_n(q),\quad \textrm{PSU}_n(q),\quad \textrm{PSp}_{n}(q),\quad \textrm{P}\Omega _{n}(q),\quad \textrm{ and }\quad \textrm{P}\Omega _{n}^{\pm }(q). \end{aligned}$$

With the restrictions on n as above, these are indeed non-abelian simple groups, except for $\textrm{PSL}_2(2)$, $\textrm{PSL}_2(3)$, $\textrm{PSU}_3(2)$ and $\textrm{PSp}_4(2)$.

We denote by $\pi :\tilde{L}\rightarrow L$ the natural projection of $\tilde{L}$ onto L. By abuse of notation, we refer to the action of $\tilde{L}$ on V simply as the action L on V. We adopt a similar convention for every G with $L\unlhd G\le \textrm{Aut}(L)\cap \textrm{PGL}(V)$. For example, for a subgroup H of L, we say that H acts irreducibly on V when this is true of $\pi ^{-1}(H)$.

Given an integer $\kappa$ and a prime number p, we write $\kappa _p$ for the largest power of p dividing $\kappa$. Given two integers $\kappa$ and $\kappa '$, we denote by $\gcd (\kappa ,\kappa ')$ the greatest common divisor of $\kappa$ and $\kappa '$.

Given a prime power q and an integer $n\ge 2$, a prime t is called a primitive prime divisor of $q^n-1$ if t divides $q^n-1$ and t does not divide $q^i-1$, for each $i\in \{1,\ldots ,t-1\}$. From a celebrated theorem of Zsigmondy [21], the following hold

• for $n\ge 3$, primitive prime divisors exist with the only exception of $(n,q)=(6,2)$,

• for $n=2$, primitive prime divisors exist with the only exception of q being a Mersenne prime, that is, q is prime and $q=2^\ell -1$ for some $\ell \in \mathbb {N}$.

Note that, if t is a primitive prime divisor of $q^n-1$, then q has order n modulo t and thus n divides $t-1$.

1.2 Notation for Table 1

In reading Table 1, we take into account the notation we have established in Section 1.1 and some isomorphisms among classical groups.

When $L=\textrm{PSL}_n(q)$, we suppose $n\ge 2$ and, when

$$(n,q)\in \{(2,4),(2,5),(2,9),(4,2)\},$$

we refer to Lines 1 and 2 of Table 1, because $\textrm{PSL}_2(4)\cong \textrm{PSL}_2(5)\cong \textrm{Alt}(5)$, $\textrm{PSL}_2(9)\cong \textrm{Alt}(6)$ and $\textrm{PSL}_4(2)\cong \textrm{Alt}(8)$. Moreover, when $(n,q)=(3,2)$, we refer to Lines 3 and 4, because $\textrm{PSL}_3(2)\cong \textrm{PSL}_2(7)$.

When $L=\textrm{PSU}_n(q)$, we suppose $n\ge 3$; when $L=\textrm{PSp}_n(q)$, we suppose $n\ge 4$; when $L=\textrm{P}\Omega _n(q)=\Omega _n(q)$ with n odd, we suppose $n\ge 7$; when $L=\textrm{P}\Omega _n^\pm (q)$ with n even, we suppose $n\ge 8$.

1.3 Structure of the Paper

In Section 3, we collect some basic results which we use throughout the whole paper, sometimes without mention.

In Section 4, we prove Theorem 1.1 for the almost simple groups having socle a sporadic, or an exceptional, or an alternating group.

In the rest of the paper we deal with the classical groups. In Section 5, we consider the linear groups $\textrm{PSL}_n(q)$. In Section 6, we consider the unitary groups $\textrm{PSU}_n(q)$. In Section 7, we consider the symplectic groups $\textrm{PSp}_n(q)$. In Section 8, we consider the odd dimensional orthogonal groups $\textrm{P}\Omega _n(q)=\Omega _n(q)$. In Section 9, we consider the even dimensional orthogonal groups $\textrm{P}\Omega _n^-(q)$ having Witt defect 1.

2 An Additional Maximal Factorization of an Almost Simple Group

A computation with the computer algebra system Magma [2] yields that there are factorizations

$$\begin{aligned} \Omega _8^+(4).\langle \phi \rangle =N_2^-\cdot \textrm{SO}_8^-(2)=\textrm{SO}_8^-(2)\cdot N_2^- \end{aligned}$$

(2.1)

and

$$\begin{aligned} \Omega _8^+(16).\langle \phi \rangle =N_2^-\cdot (\textrm{SO}_8^-(4).2)=(\textrm{SO}_8^-(4).2)\cdot N_2^-, \end{aligned}$$

(2.2)

where the subgroup $\textrm{SO}_8^-(q^{1/2}) \le \Omega _8^+(q)$ is the image of a $\mathcal {C}_5$ subgroup under a triality automorphism, and $\phi$ is a non-identity field automorphism of order f, where $q=2^f$. Recall that we are using the notation in [12] and hence, in particular, $N_2^-$ is the stabilizer of a 2-dimensional anisotropic subspace of $V=\mathbb {F}_q^8$. Moreover, these factorizations do not lead to a factorization of the simple group $L=\Omega _8^+(q)$; actually, these factorization exists only in the almost simple groups $\textrm{Aut}(\Omega _8^+(q))$-conjugate to $\Omega _8^+(q).\langle \phi \rangle$ and in no other almost simple groups with socle $\Omega _8^+(q)$. Observe that, using triality, we have exactly three possibilities for $\Omega _8^+(q).\langle \phi \rangle$. Both factors in each of these factorisations are core-free maximal subgroups and so these factorisations are max+ factorisations in the terminology of [17]. Thus [16, Table 4] should have the rows of Table 2 added.

Table 2 Missing maximal factorizations

For $q=4$, the factorisation can be verified by the following steps:

• 1. construct the action of $\Omega ^+_8(4)$ on 2-dimensional subspaces of “minus type”;

• 2. find the normaliser of the induced permutation group in $\textrm{Sym}(6580224)$ to obtain the permutation group G for $\Omega ^+_8(4).\langle \phi \rangle$;

• 3. use the ClassicalMaximals command to construct an appropriate $H=\Omega _8^-(2)$ in $\Omega _8^+(4)$;

• 4. find the image of H in G and then determine its normaliser in G to find the appropriate $\textrm{SO}_8^-(2)$ subgroup;

• 5. check that the $\textrm{SO}_8^-(2)$ is transitive in this action.

For $q=16$ the groups are too large to do many of these steps. It is possible though to use the ClassicalMaximals command to construct an appropriate $\Omega _8^-(4)$ in $\Omega ^+_8(16)$ and then show that it has an orbit on the set of 2-dimensional subspaces of “minus type” whose length is one quarter of the total number of such subspaces. We then explicitly construct $\textrm{SO}_8^-(4).2$ in $\Omega _8^+(16).\langle \phi \rangle$, and the Sylow 2-subgroups of the two potential factors. We can then exhibit that the intersection of the Sylow 2-subgroups has order 8 and is contained in $\Omega _8^-(4)$ and so we do indeed have a factorisation.

2.1 Dealing with the Missing Factorization

Using the notation in [16], the factorizations in (2.1) and (2.2) were erroneously ruled out in [16, pp. 106–107] when considering the possibility

$$\begin{aligned} A\cap L=((q+1)/d\times \Omega _6^-(q)).2^d/Z\quad \text { and }\quad B\cap L=\Omega _8^-(q^{1/2}), \end{aligned}$$

where $d:=\gcd (2,q-1)$ and Z is the group of scalars in $\Omega _8^+(q)$. The argument there for q even only rules out a factorization of the simple group L and misses a subtlety due to triality. We now provide a complete analysis of this case following that in [16].

Let G be an almost simple group having socle $L:=\textrm{P}\Omega _8^+(q)$ and let A and B be maximal subgroups of G. Suppose that $G=AB$ with $A\cap L=((q+1)/d\times \Omega _6^-(q)).2^d/Z$ and $B\cap L=\Omega _8^-(q^{1/2})$. As in Section 1.1, we let V be an 8-dimensional vector space over the finite field $\mathbb {F}_q$ equipped with a non-degenerate quadratic form of plus type with $q=r^f$ for some prime r and some even positive integer f. By applying a suitable triality automorphism if necessary we may assume that $A=N_2^-$, the stabilizer in G of an anisotropic 2-dimensional subspace. Moreover, by [13] the maximality of A in G implies that $G\le \textrm{P}\Gamma \textrm{O}_8^+(q)$. Let $X\cong \Omega _8^-(q^{1/2})$ be a subgroup of L in $\mathcal {C}_5$, that is, a subfield subgroup. By [13], we may take $B\cap L=X^{\tau ^a}$ for some $a\in \{0,1,2\}$, where $\tau$ is a triality automorphism of L. Write $k=\tau ^a$. Now X has a subgroup

$$\begin{aligned} Y=(\textrm{SO}_4^+(q^{1/2})\times \textrm{SO}_4^-(q^{1/2}))\cap X \end{aligned}$$

fixing an orthogonal sum decomposition $V=W_1\perp W_2$, where $W_1$ and $W_2$ are both non-degenerate 4-dimensional subspaces of V of plus type. By [12, Lemma 4.1.1], we have that

$$\begin{aligned} Y=(\Omega _4^+(q^{1/2})\times \Omega _4^-(q^{1/2})).2 \end{aligned}$$

and Y induces $\textrm{SO}_4^+(q^{1/2})$ and $\textrm{SO}_4^-(q^{1/2})$ on $W_1$ and $W_2$ respectively. Therefore, the kernel of the action of Y on $W_1$ is $\Omega _4^-(q^{1/2})$ and the kernel of the action of Y on $W_2$ is $\Omega _4^+(q^{1/2})$. Let M be the stabilizer in L of this decomposition. Now $Y^k\le B\cap L$ and $Y^k\le M^k$. By [13], $\tau$ fixes setwise the L-conjugacy class of M in L and so $W^k$ fixes an orthogonal decomposition $V=W_1'\perp W_2'$, with $W_1',W_2'$ both 4-dimensional non-degenerate subspaces of V of plus type. Note that Y has a subgroup of index 2 and so potentially $Y^k$ interchanges $W_1'$ and $W_2'$. Indeed a Magma [2] calculation shows that this does indeed happen when $q=4$.

This seems to have been overlooked by [16] as their argument seems to assume that $Y^k$ fixes both $W_1'$ and $W_2'$.

We now continue the analysis of this potential factorization, obtaining the missing factorizations in [16].

For $i=1,2$, let $K_i$ be the kernel of the action of the stabilizer in $Y^k$ of $W_i'$ on $W_i'$. Note that $(Y^k)_{W_i'}$ induces a subgroup of $\textrm{GO}^+_4(q)$ on $W_i'$. Now $\Omega _4^+(q^{1/2})\cong \textrm{SL}_2(q^{1/2})\circ \textrm{SL}_2(q^{1/2})$ and $\Omega _4^-(q^{1/2})=\textrm{PSL}_2(q)$. Since $\textrm{GO}_4^+(q)$ does not contain a subgroup isomorphic to an index two subgroup of Y, each $K_i$ is nontrivial.

Suppose first that $q=r^f$ is odd. For $q\ne 9$, by considering the normal subgroups of $Y'$, the unique index 2 subgroup of Y, we see that either $\textrm{SL}_2(q^{1/2})$ or $\Omega _4^-(q^{1/2})$ lies in $K_1$. On the other hand, for $q=9$, we see that the centralizer in $\textrm{GO}^+_4(9)$ of $\Omega ^-_4(3)$ does not contain an element of order 3 and so we may draw the same conclusion about $K_1$. Then, taking a suitable 2-dimensional subspace $U\le W_1'$ and $A:=G_U=\{g\in G~|~U^g=U\}$, we have that $K_1\le A\cap B$ and so $q^{1/2}$ divides $|A\cap B|$. As q is odd, we have that

$$\begin{aligned} \begin{aligned} |A|_r=&|\Omega _6^-(q)|_r|G:L|_r=|G:L|_rq^6,\\ |B|_r=&|\Omega _8^-(q^{1/2})|_r|G:L|_r=q^6|G:L|_r,\\ |A\cap B|_r\ge&q^{1/2},\\ |G|_r=&q^{12}|G:L|_r. \end{aligned} \end{aligned}$$

As G does not contain a triality automorphism of L, we have that $|G:L|_r$ divides $f_r$ and hence we see that $|A|_r|B|_r < |G|_r|A\cap B|_r$, contradicting $G=AB$. Therefore, when q is odd, there are no factorizations, as predicted by [16].

Now suppose that q is even. Then $|A|_2= 2|\Omega _6^-(q)|_2|G:L|_2=2q^6|G:L|_2$ and $|B|_2= q^6|G:L|_2$, while $|G|_2=q^{12}|G:L|_2$. Since $G=AB$, this implies that $|A\cap B|_2= 2|G:L|_2$. Note also that

$$\begin{aligned} |G:L|_2~\text { divides }~2f_2. \end{aligned}$$

(2.3)

Suppose first that $B\cap L=X$. Then $W_1'=W_1$ and $W_2'=W_2$. Choose $U\le W_1$ to be a 2-dimensional subspace of minus type and take $A:=G_U$. Since $Y^{W_1}=\textrm{SO}^+_4(q^{1/2})$ and the order of the stabilizer in $\textrm{SO}_4^+(q^{1/2})$ of a non-degenerate 2-dimensional subspace is divisible by 4, we see that $|A\cap B|_2\ge |Y_U|_2=4q$. However, this contradicts (2.3), because $|A\cap B|_2= 2|G:L|_2\le 4f_2$. Thus

$$\begin{aligned} B\cap L=X^{\tau ^a}, \end{aligned}$$

where $a\in \{1,2\}$. When $q=4$ and 16, we have verified with Magma that we do obtain the maximal factorizations in Table 2. Suppose then $q>16$. Thus $q\ge 64$, because f is even.

Then, by [13], the maximality of B in G implies that |G : L| divides f. As $q\ge 64$, the group Y has a unique index 2 subgroup, namely $Y'$. Then the kernel of the action of $Y'$ on $W_1'$ is non-trivial and it is not hard to show that $K_1\ge \textrm{SL}_2(q^{1/2})$. Therefore,

$$\begin{aligned} ((Y')^{\tau ^a})^{W_1}=\textrm{SL}_2(q^{1/2})\times \textrm{SL}_2(q). \end{aligned}$$

We claim that the stabilizer in this group of a 2-dimensional subspace of $W_1$ of minus type has even order. We argue by contradiction, and we suppose that there exists a 2-dimensional subspace U of minus type of $W_1$ with the property that the stabilizer in $\textrm{SL}_2(q^{1/2})\times \textrm{SL}_2(q)$ has odd order. Let us denote by $S_1$ and $S_2$ the two simple direct factors of $\Omega _4^+(q)$. Let $N:=\textrm{SL}_2(q^{1/2})\times \textrm{SL}_2(q)$ and let $\Omega$ be the collection of all 2-dimensional subspaces of $W_1$ of minus type. Routine computations yield

$$\begin{aligned} |\Omega |=\frac{q^2(q-1)^2}{2}. \end{aligned}$$

(2.4)

Without loss of generality we may suppose that $S_2\subseteq M$. Observe that

$$\begin{aligned} |(\Omega _2^-(q)\times \Omega _2^-(q)).2|=2(q+1)^2\quad \text { and }\quad |N|=q^{3/2}(q-1)(q^2-1). \end{aligned}$$

Since $\gcd (q+1,q-1)=1$ and since we are assuming that $M_U$ has odd order, we deduce that $|N_U|$ divides $q+1$ and $N_U\le S_2$. From one hand we deduce that the N-orbit containing U has cardinality divisible by $q^{3/2}$ and, on the other hand, we deduce that $S_1$ centralizes $N_U$. As $\Omega _4^+(q)=S_1N$, we deduce that each orbit of M on the 2-dimensional subspaces of $W_1$ of minus type has order divisible by $q^{3/2}$. The number of N-orbits is thus $|S_1:S_1\cap N|$ and hence $q^{1/2}\cdot q^{3/2}=q^2$ divides $|\Omega |$; however, this contradicts (2.4).

2.2 Impact of the Missing Factorizations in other Work

The factorizations in [16] have been extensively used. Now, we comment how this extra factorization influences other work relying in the classification in [16].

• 1. Since the Y column for the new factorizations are empty, these new factorizations do not give an example of a primitive almost simple group of degree n as a proper subgroup of a primitive almost simple group of the same degree but different socle. Thus no new examples arise for [15, Table VI].

• 2. The classification of maximal factorizations of almost simple groups was used in [18] to determine all regular subgroups of the primitive almost simple groups. These new factorizations provide four new primitive almost simple groups with a core-free transitive subgroup (namely the action of G on the set of cosets of A and the action of G on the set of cosets of B). We have checked with the help of a computer that none of these new primitive actions admits regular subgroups and hence no exception arises in [18].

• 3. For these new factorizations |G : A| and |G : B| are even and so these are not coprime factorizations and so no new factorization needs to be added to [6, Table 1].

• 4. The maximal factorisations in [16] are used in [14] to determine the factorisations of almost simple group with one of the two factors solvable. Therefore, in principle, the missing factorizations arising when the socle is $\textrm{P}\Omega _8^+(4)$ and $\textrm{P}\Omega _8^+(16)$ could in principle yield factorizations missed by [14] when using [16]. We have checked with Magma and we confirm that no new factorization arises when one of the two factors is solvable.

3 Preliminary Remarks

Lemma 3.1

Let G be a group, let M and N be subgroups of G with $G=MN$ and let T be a subgroup of N. Then $G=MT$ if and only if $N=(M\cap N)T$.

Proof

If $G=MT$, then $N=MT\cap N=(M\cap N)T$, as required. Conversely, if $N=(M\cap N)T$, then $G=MN =M(M\cap N)T = MT$.$\square$

The following elementary lemma is one of the ingredients for our proof of Theorem 1.1 when the socle of G is an alternating group. A permutation $g\in \textrm{Sym}(n)$ is said to be textitsemiregular if all the orbits of $\langle g\rangle$ on $\{1,\ldots ,n\}$ have the same length.

Lemma 3.2

Let g be in $\textrm{Sym}(n)\setminus \{1\}$. If $\textbf{N}_{\textrm{Sym}(n)}(\langle g\rangle )$ is transitive on $\{1,\ldots ,n\}$, then g is semiregular.

Proof

Write $N:=\textbf{N}_{\textrm{Sym}(n)}(\langle g\rangle )$. Clearly, N permutes the orbits of $\langle g\rangle$ having the same length. Since N is transitive on $\{1,\ldots ,n\}$, we obtain that all $\langle g\rangle$-orbits have the same length.$\square$

For the rest of this paper, G denotes an almost simple group with socle L having two group elements $x,y\in G\setminus \{1\}$ with

$$\begin{aligned} G=\textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle ). \end{aligned}$$

Write

$$\begin{aligned} X:=\textbf{N}_G(\langle x\rangle )\quad \hbox { and }\quad Y:=\textbf{N}_G(\langle y\rangle ), \end{aligned}$$

for short.

Lemma 3.3

We have

$$\begin{aligned} G = \textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle ^g) = \textbf{N}_G (\langle x\rangle ^g)\textbf{N}_G(\langle y \rangle ), \end{aligned}$$

for every $g \in G$. Let $\langle g_1\rangle ,\ldots ,\langle g_\ell \rangle$ be a set of representatives for the conjugacy classes of non-identity cyclic subgroups of G. Then there exist $i,j\in \{1,\ldots ,\ell \}$ with $G=\textbf{N}_G(\langle g_i\rangle )\textbf{N}_G(\langle g_j\rangle )$.

Proof

Let g be in G. We have $g=uv$, for some $u\in X=\textbf{N}_G(\langle x\rangle )$ and $v\in Y=\textbf{N}_G(\langle y\rangle )$. Now,

$$\begin{aligned} \begin{aligned} \textbf{N}_G(\langle x\rangle ^g)\textbf{N}_G(\langle y\rangle )=&\textbf{N}_G(\langle x\rangle ^v)\textbf{N}_G(\langle y\rangle )=\textbf{N}_G(\langle x\rangle ^v)\textbf{N}_G(\langle y\rangle ^v)\\ =&\textbf{N}_G(\langle x\rangle )^v\textbf{N}_G(\langle y\rangle )^v=(XY)^v=G^v=G. \end{aligned} \end{aligned}$$

The other case is similar. Now, the rest of the proof follows from the fact that $\langle x\rangle =\langle g_i\rangle ^{h_x}$ and $\langle y\rangle =\langle g_j\rangle ^{h_y}$, for some $i,j\in \{1,\ldots ,\ell \}$ and $h_x,h_y\in G$.$\square$

Lemma 3.3 gives a very efficient test to check whether $G=\textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$. For example, for $M_{11}$ it is immediate to see with [5] that $|\textbf{N}_{M_{11}}(\langle g \rangle )| \le 55$, for every $g\in M_{11}\setminus \{1\}$. As $|M_{11}|>55^2$, we see that Theorem 1.1 holds true for $M_{11}$, that is, $M_{11}$ has no factorization of the form $M_{11}=\textbf{N}_{M_{11}}(\langle x\rangle )\textbf{N}_{M_{11}}(\langle y\rangle )$ with $x,y\in M_{11}\setminus \{1\}$.

We also have the following useful lemma.

Lemma 3.4

Suppose that $G=\textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$ for some $x,y\in G$. Then $G=\textbf{N}_G(\langle x'\rangle )\textbf{N}_G(\langle y'\rangle )$ for some $x',y'\in G$ of prime order.

Proof

Let $p_x$ divide |x| and $p_y$ divide |y| be primes. Then $x':=x^{|x|/p_x}$ and $y':=y^{|y|/p_y}$ have prime order. Moreover, $\textbf{N}_G(\langle x\rangle ) \le \textbf{N}_G(\langle x'\rangle )$ and $\textbf{N}_G(\langle y\rangle )\leqslant \textbf{N}_G(\langle y'\rangle )$. Thus the result follows.$\square$

4 Proof of Theorem 1.1 for Sporadic, Exceptional and Alternating Simple Groups

We are now ready to prove Theorem 1.1 when L is a sporadic simple group, an exceptional group of Lie type or an alternating group.

Proposition 4.1

L is not a sporadic simple group.

Proof

All factorizations of almost simple groups having socle a sporadic simple group are determined in [8]. The result follows by inspection.$\square$

Proposition 4.2

L is not an exceptional group of Lie type.

Proof

Suppose that $G=XY$ and by Lemma 3.4 we may assume that both x and y have prime order. From [16, Table 5], we see that L is one of the following groups:

• $G_2(q)$ with $q=3^f$,

• $F_4(q)$ with $q=2^f$,

• $G_2(4)$.

Note that [16, Table 5] gives all factorisations of G, not just the maximal ones, and so lists the possibilities for X and Y.

Suppose first that $L=G_2(4)$. Then interchanging X and Y if necessary we have that $X\cap L=\textrm{SU}_3(4).4\cap L$. Since $X\cap L$ has trivial centre, it follows that $x\notin L$ and so is a field automorphism of order 2. However, this implies that $C_L(x)\cong G_2(2)$, which does not contain $\textrm{SU}_3(4)$, a contradiction.

Next suppose that $L=G_q(q)$ with $q=3^f$. Then interchanging X and Y if necessary we have that $X\cap L=\textrm{SL}_3(q)$ or $\textrm{SL}_3(q).2$. In both cases $X\cap L$ has trivial centre and so $\textrm{SL}_3(q)\le \textbf{C}_A(x)$. Moreover, $x\notin L$. Hence by [9, Proposition 4.9.1], x is either a field or graph-field automorphism of L and hence $C_L(x)=G_2(q_0)$ with $q=q_0^e$, or ${}^2G_2(q)$, respectively. Neither of these contain $\textrm{SL}_3(q)$ as a subgroup, a contradiction.

Similarly, if $L=F_4(q)$ then $X\cap L=\textrm{Sp}_8(q)$. Hence $\textrm{Sp}_8(q)\le \textbf{C}_A(x)$ and $x\notin L$. Again, x is either a field or graph-field automorphism of L and hence $C_L(x)=F_4(q_0)$ with $q=q_0^e$, or ${}^2F_4(q)$, respectively. Neither of these contain $\textrm{Sp}_8(q)$ as a subgroup, and so we obtain another contradiction.$\square$

Proposition 4.3

If $L=\textrm{Alt}(n)$, then the triple (G, x, y) is in Line 1 or 2 of Table 1.

Proof

For $5\le n\le 6$, the result follows by a computation using Lemma 3.3. We obtain the examples in Line 1 or 2 of Table 1. Assume that $n\ge 7$. In particular, $G=\textrm{Alt}(n)$ or $G=\textrm{Sym}(n)$. Then by [16, Theorem D], interchanging X and Y if necessary we have that either $\textrm{Alt}(n-k)\unlhd X\le \textrm{Sym}(k)\times \textrm{Sym}(n-k)$ for some $k\le 5$ and Y is k-homogeneous on n points, or one of the following holds:

• 1. $n=8$ and $X=\textrm{AGL}_3(2)$;

• 2. $n=10$ and $X=\textrm{PSL}_2(8)$ or $\textrm{PSL}_2(8).3$.

In these two exceptional cases, X is clearly not the normaliser of a nontrivial cyclic subgroup. Hence $\textrm{Alt}(n-k)\unlhd X\le \textrm{Sym}(k)\times \textrm{Sym}(n-k)$ and Y is k-homogeneous. Note that since X has a nontrivial cyclic normal subgroup we must have that $k\ge 2$. Hence by [1, Theorem 6], Y is primitive on n points. As $\langle y\rangle \unlhd Y$, we get that the socle of the primitive group Y is $\langle y\rangle$. Thus n is prime and y is a cycle of prime order. Moreover, $Y\cong \textrm{AGL}(1,n)\cap G$ is 2-homogeneous but not 3-homogeneous. Hence either $k=2$, or $n=7$ and $k=5$. In the first case we deduce that x is a transposition, $G=\textrm{Sym}(n)$ and $X=\textrm{Sym}(2)\times \textrm{Sym}(n-2)$. Thus we do obtain a factorisation. It remains to consider the case where $k=5$ and $n=7$, and we may assume that x is not a transposition. Since $|G|=|X||Y|/|X\cap Y|$, we deduce that X contains a 5-cycle. As x is not a transposition we deduce that x is a 5-cycle. However, we then have $|X|\le 40$ and $|Y|\le 42$, which contradicts $|X||Y|\ge |G|\ge (7!)/2$.

It is easy to verify that in Lines 1 and 2 we have $\textbf{C}_G(x)\textbf{C}_G(y)<G$ and hence there is no symbol $\surd$ in the 6th column.$\square$

5 Classical Groups: Linear Groups

In this section, we assume that G is an almost simple group with socle $L=\textrm{PSL}_n(q)$ with $q=r^f$ for some prime r.

We start our analysis with two technical lemmas, which help to locate the elements x and y with $G=\textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$. Our main reference for these lemmas is [9, Chapter 4] and [4, Chapter 3.1, Tables B.1, B.2, B.3].

Lemma 5.1

Let $n\ge 2$. Suppose $r^{fn}-1$ admits a primitive prime divisor $t_1$. Let $g\in$ $\textrm{Aut}(\textrm{PSL}_n(q))$ with $t_1$ dividing $|\textbf{N}_{\textrm{Aut}(\textrm{PSL}_n(q))}{(\langle g\rangle )}|$ and let $T_1$ be a cyclic subgroup of order $t_1$ in $\textbf{N}_{\textrm{Aut}(\textrm{PSL}_n(q))}{(\langle g\rangle )}$. Then

$$\begin{aligned} g\in \textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}(T_1)= \left\{ \begin{array}{ll} \langle T,\iota \rangle &{}\quad \iota \text { graph automorphism}, \text { if }n \text { is even},\\ \langle T, \iota \rangle &{}\quad \iota \text { graph-field automorphism}, \text { if }n \text { is odd and } f\text { is even},\\ T&{}\quad \text {if }nf\text { is odd}, \end{array}\right. \end{aligned}$$

where T is a maximal torus of $\textrm{PGL}_n(q)$ having order $(q^n-1)/(q-1)$, that is, T is a Singer cycle. In particular, $|\textbf{N}_{\textrm{Aut}(\textrm{PSL}_n(q))}(\langle g\rangle ):\textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}(g)|$ is relatively prime to $t_1$.

Proof

Suppose first that $T_1\le \textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}{( g)}$. Then $g\in \textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}{(T_1)}$. Let T be a Singer cycle of $\textrm{PGL}_n(q)$ containing $T_1$. Using [9], we obtain the structure of $\textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}(T_1)$.

It remains to consider the case that $t_1$ does not divide the order of $\textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}{( g)}$. We aim to prove that this case cannot arise. As $t_1$ divides $|\textbf{N}_{\textrm{Aut}(\textrm{PSL}_n(q))}{(\langle g\rangle )}|$ and as $\textbf{N}_{\textrm{Aut}(\textrm{PSL}_n(q))}{(\langle g\rangle )}/\textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}{(g)}$ acts faithfully as a group of automorphisms on the cyclic group $\langle g\rangle$, we deduce that $t_1$ divides $\varphi (|g|)$, where $\varphi$ is the Euler totient function. In particular, |g| is divisible by a prime p with $t_1\,|\, p-1$. Without loss of generality, replacing g by $g^{|g|/p}$ if necessary, we may suppose that $|g|=p$. As fn divides $t_1-1$, we have $fn<t_1$. As $t_1\,|\, p-1$ and $fn<t_1$, we deduce $g\in \textrm{PSL}_n(q)$. Since $t_1$ is a primitive prime divisor for $r^{fn}-1$, we have $p\ne r$ and hence g is semisimple. Now, as $T_1$ acts non-trivially on $\langle g\rangle$, we deduce that $T_1$ permutes non-trivially the eigenspaces of g. However, this is a contradiction because $t_1>n$.$\square$

Lemma 5.2

Let $n\ge 3$. Suppose $r^{f(n-1)}-1$ admits a primitive prime divisor $t_2$. Let $g\in \textrm{Aut}(\textrm{PSL}_n(q))$ with $t_2$ dividing $|\textbf{N}_{\textrm{Aut}(\textrm{PSL}_n(q))}{(\langle g\rangle )}|$ and let $T_2$ be a cyclic subgroup of order $t_2$ in $\textbf{N}_{\textrm{Aut}(\textrm{PSL}_n(q))}{(\langle g\rangle )}$. Then one of the following holds:

• 1.
  $$\begin{aligned} g\in \textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}(T_2)= \left\{ \begin{array}{ll} \langle T,\iota \rangle &{}\quad \iota \text { graph automorphism}, \text { if }n \text { is odd},\\ \langle T, \iota \rangle &{}\quad \iota \text { graph-field automorphism}, \text { if }n \text { and } f\text { are even},\\ T&{}\quad \text {if }(n-1)f \text { is odd}, \end{array}\right. \end{aligned}$$

  where T is a maximal torus of $\textrm{PGL}_n(q)$ having order $q^{n-1}-1$. In particular, $|\textbf{N}_{\textrm{Aut}(\textrm{PSL}_n(q))}(\langle g\rangle ):\textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}(g)|$ is relatively prime to $t_2$;

• 2. $n=t_2$ is prime, $f=1$, g lies in a Singer cycle of $\textrm{PGL}_n(q)$ and has order divisible by a primitive prime divisor p of $r^{fn}-1$ with $n\,|\, p-1$.

Proof

If $T_2\le \textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}{( g)}$, then the proof follows verbatim the argument in Lemma 5.1 and we obtain that Part 1 holds.

It remains to consider the case that $t_2$ does not divide the order of $\textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}{( g)}$. We aim to prove that Part 2 holds. As $t_2$ divides $|\textbf{N}_{\textrm{Aut}(\textrm{PSL}_n(q))}{(\langle g\rangle )}|$ and as $\textbf{N}_{\textrm{Aut}(\textrm{PSL}_n(q))}{(\langle g\rangle )}/\textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}{(g)}$ acts faithfully as a group of automorphisms on the cyclic group $\langle g\rangle$, we deduce that $t_2$ divides $\varphi (|g|)$. In particular, |g| is divisible by a prime p with $t_2\,|\, p-1$. Without loss of generality, we may suppose that $|g|=p$. As $f(n-1)$ divides $t_2-1$, we have $f(n-1)<t_2$ and hence $g\in \textrm{PSL}_n(q)$. Now, as $T_2$ acts non-trivially on $\langle g\rangle$, we deduce that $T_2$ permutes non-trivially the eigenspaces of g. As $t_2>n-1$, this is possible only when $\langle g\rangle$ has n distinct eigenvalues and $t_2=n$. As $t_2=n$ and $f(n-1)<t_2$, we have $f=1$. Moreover, as g has n distinct eigenvalues in a suitable extension of $\mathbb {F}_q$, we deduce that g is contained in a Singer cycle of $\textrm{PSL}_n(q)$ and, via the embedding of $\mathbb {F}_{q^n}^*$ in $\mathbb {F}_q^n\setminus \{0\}$, g is a field generator.$\square$

In our proofs, we exclude those groups that are isomorphic to alternating groups, as these have already been covered in Proposition 4.3. Thus $n\ge 2$ and

$$\begin{aligned} (n,q)\not \in \{ (2,4), (2,5), (2,9),(4,2)\}. \end{aligned}$$

See Section 1.1, for our notation.

We first deal with 2-dimensional linear groups, because we have little room in this case for using the primitive prime divisors $t_1$ and $t_2$ in Lemmas 5.1 and 5.2.

Lemma 5.3

If $L=\textrm{PSL}_2(q)$ and $q\notin \{4,5,9\}$, then (G, x, y) is in Lines 3, 4 or 5 of Table 1.

Proof

Assume that $f=1$. Here $G=\textrm{PSL}_2(r)$ or $G=\textrm{PGL}_2(r)$. Now, replacing X by Y if necessary, we may assume that r divides |X|. The only elements x of G having normalizer of order divisible by r are the r-elements. Thus $|x|=r$, X is a Borel subgroup of G and

$$\begin{aligned} |X|= \left\{ \begin{array}{ll} \frac{(r-1)r}{2}&{}\quad \text {if }G=\textrm{PSL}_2(r),\\ (r-1)r&{}\quad \text {if } G=\textrm{PGL}_2(r). \end{array}\right. \end{aligned}$$

As $G=XY$ and X fixes a 1-dimensional subspace of V, we see from the Frattini argument that Y acts transitively on the 1-dimensional subspaces of V. With a quick look at the structure of the conjugacy classes of G, we obtain that $y\in T\setminus \{1\}$, with T a maximal torus of $\textrm{PGL}_2(r)$ of order $r+1$. In particular,

$$\begin{aligned} |Y|=|\textbf{N}_G(\langle y\rangle )|= \left\{ \begin{array}{ll} r+1&{}\quad \text {if }G=\textrm{PSL}_2(r),\\ 2(r+1)&{}\quad \text {if }G=\textrm{PGL}_2(r). \end{array}\right. \end{aligned}$$

If $G=\textrm{PGL}_2(r)$, we have

$$\begin{aligned} \frac{|X||Y|}{|X\cap Y|}=\frac{((r-1)r)(2(r+1))}{2}=|G| \end{aligned}$$

and hence $G=XY$: these examples are in Line 3 of Table 1. If $G=\textrm{PSL}_2(r)$, then Y is a dihedral group with $|X\cap Y|=1$ if $r\equiv 3\pmod 4$ and with $|X\cap Y|=2$ if $r\equiv 1\mod 4$. In particular, $G=XY$ only when $r\equiv 3\pmod 4$: these examples are in Line 4 of Table 1.

Assume $f>1$. If $q=8$, then an inspection in [5] shows that there are no non-identity group elements x and y with $G=\textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$. Suppose $q\ne 8$. Let s be a primitive prime divisor of $r^{2f}-1$ (such a prime exists by Zsigmondy’s theorem [21] because we are excluding the case $q=8$). Since $G=XY$, we see that either X or Y has order divisible by s. Replacing X by Y if necessary, we may assume that s divides |X|. Using the subgroup structure of $\textrm{P}\Gamma \textrm{L}_2(q)$ (see [3]), we see that the only elements x having normalizer of order divisible by s are the elements lying in a maximal torus T of $\textrm{PGL}_2(q)$ of order $q+1$. Now, for every $x\in T\setminus \{1\}$, we have $|\textbf{N}_{\textrm{P}\Gamma \textrm{L}_2(q)}(\langle x\rangle )|=2(q+1)f$ and

$$\begin{aligned} \textrm{P}\Gamma \textrm{L}_2(q)=\textbf{N}_{\textrm{P}\Gamma \textrm{L}_2(q)}(\langle x\rangle )\textrm{PSL}_2(q). \end{aligned}$$

Since $G=XY$, we get

$$\begin{aligned} \textrm{P}\Gamma \textrm{L}_2(q)=\textbf{N}_{\textrm{P}\Gamma \textrm{L}_2(q)}(\langle x\rangle )G= \textbf{N}_{\textrm{P}\Gamma \textrm{L}_2(q)}(\langle x\rangle )Y \end{aligned}$$

and so we may assume that $G=\textrm{P}\Gamma \textrm{L}_2(q)$. Thus $|X|=2(q+1)f$. As $|G||X\cap Y|=|X||Y|$, we obtain that $|G|/|X|=(q-1)q/2$ divides |Y|. Another inspection on the maximal subgroups of $\textrm{P}\Gamma \textrm{L}_2(q)$ (for $q\ne 4,9$) shows that $\textbf{N}_{\textrm{P}\Gamma \textrm{L}_2(q)}(\langle y\rangle )$ is divisible by $(q-1)q/2$ only when $q\in \{2^4,2^8\}$ and y is a field automorphism of order 2. If $L=\textrm{PSL}_2(2^8)$ and y is a field automorphism of order 2, then $Y\cong \textrm{PGL}_2(2^4).8$ and $|X\cap Y|=2$. However,

$$\begin{aligned} \frac{|X||Y|}{|X\cap Y|}=\frac{2(2^8+1)8\cdot (2^8-1)2^7}{2}=\frac{|G|}{2} \end{aligned}$$

and hence we have no examples when $q=2^8$. A computation with Magma shows that the only factorization arising with $L=\textrm{PSL}_2(16)$ is in Line 5 of Table 1.

It is easy to verify that in Lines 3, 4 and 5 we have $\textbf{C}_G(x)\textbf{C}_G(y)<G$ and hence there is no symbol $\surd$ in the 6th column (Line 5 can also be verified with an easy computer computation).$\square$

Next we deal with linear groups, where the primitive prime divisor $t_2$ in Lemma 5.2 does not exist.

Lemma 5.4

If $L=\textrm{PSL}_3(q)$ and $q=r=2^\ell -1$, for some $\ell \in \mathbb {N}$, then $G\ne \textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$.

Proof

When $(n,q)=(3,3)$, the proof follows with a computer computation with the computer algebra system Magma. Lemma 3.3 makes the search of factorizations $G=\textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$ very efficient. In particular, for the rest of the argument, we suppose $q\ne 3$.

Observe that $\ell >2$, because we are excluding the case $(n,q)=(3,3)$. Let $t_1$ be a primitive prime divisor of $r^3-1$ and let $T_1$ be a cyclic subgroup of G of order $t_1$. As $t_1$ divides |L|, we have that $t_1$ divides |X| or |Y|. Without loss of generality, we suppose that $t_1$ divides |X| and hence $t_1$ divides $|\textbf{N}_G(\langle x\rangle )|$. From Lemma 5.1, replacing $T_1$ with a suitable conjugate, we have $T_1\le \textbf{C}_G(x)$. Therefore $x\in \textbf{C}_G(T_1)$. Let $T_x$ be the maximal torus of $\textrm{PGL}_3(r)$ containing $T_1$. In particular, $T_x$ is a torus of order $(q^3-1)/(q-1)$. From Lemma 5.1, we deduce

$$\begin{aligned} \textbf{C}_{\textrm{Aut}(L)}(T_1)=T_x. \end{aligned}$$

Thus $x\in T_x$ and hence x is a semisimple element of order dividing $r^2+r+1$. Using this fact, we deduce that $\textbf{N}_G(\langle x\rangle )$ has order a divisor of $6(r^2+r+1)$. Since this number is relatively prime to r, we deduce that Y contains a Sylow r-subgroup R of G. Thus $R\le Y=\textbf{N}_G(\langle y\rangle )$ and $R=\textbf{N}_R(\langle y\rangle )$. A moment’s thought gives that $y\in \textbf{Z}(R)$ and hence y is a transvection of K. Thus $|\textbf{N}_{\textrm{PGL}_3(r)}(\langle y\rangle )|=(r-1)^2r^3$ and hence |Y| divides $2(r-1)^2r^3$. Therefore |X||Y| divides

$$\begin{aligned} 6(r^2+r+1) \cdot 2(r-1)^2r^3=12r^3(r^3-1)(r-1) \end{aligned}$$

and so does |G|, because $G=XY$. As

$$\begin{aligned} |G|=|G:L||L|\ge \frac{r^3(r^3-1)(r^2-1)}{\gcd (3,r-1)}, \end{aligned}$$

we deduce $r+1=2^\ell$ divides 4 and hence $\ell \le 2$, contradicting the fact that $\ell >2$.$\square$

Next, we deal with the exceptional case arising in Part 2 of Lemma 5.2.

Lemma 5.5

If $L=\textrm{PSL}_n(q)$, $q=r$, n is a primitive prime divisor of $q^{n-1}-1$, $n\ge 3$ and $\textbf{C}_{\textrm{PSL}_n(q)}(x)$ or $\textbf{C}_{\textrm{PSL}_n(q)}(y)$ is a maximal torus of order $(q^n-1)/(q-1)$, then $G\ne \textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$.

Proof

Assume by contradiction that $G=XY$. Here, $|\textrm{Aut}(L):L|=2$. Without loss of generality, we may suppose that $\textbf{C}_{\textrm{PSL}_n(q)}(x)$ is a maximal torus of order $(q^n-1)/(q-1)$. Thus $|X|=n(q^n-1)/(q-1)$ when $G=L$, and $|X|=2n(q^n-1)/(q-1)$ when $G>L$. In both cases, $|G:X|=|L|/(n(q^n-1)/(q-1))$.

By consulting the maximal factorizations of almost simple groups with socle $\textrm{PSL}_n(q)$ in [16], we deduce that $Y\le P$, where $P\in \{P_1,P_{n-1}\}$ and $P_1,P_{n-1}$ are maximal parabolic subgroups. As $G=XY$, we deduce that |Y| is divisible by $|G:X|=|L|/(n(q^n-1)/(q-1))$. As $|P|=|L|/((q^n-1)/(q-1))$ we have that |P : Y| divides n. As P has no subgroups having prime index n, we get $Y=P$. However, when $n\ge 3$, P normalizes no cyclic non-identity subgroup.$\square$

Finally, we deal with the general case.

Lemma 5.6

If $L=\textrm{PSL}_n(q)$, $n\ge 3$ and $(n,q)\ne (3,2)$, then (G, x, y) is in Line 6 or 7 of Table 1.

Proof

When $n=3$ and $q=r=2^\ell -1$, for some $\ell \in \mathbb {N}$, the proof follows from Lemma 5.4. Similarly, when $q=r$, n is a primitive prime divisor of $q^{n-1}-1$ and $\textbf{C}_L(x)$ or $\textbf{C}_L(y)$ is a maximal torus of order $(q^n-1)/(q-1)$, the proof follows from Lemma 5.5. Therefore, we exclude these cases from the rest of the proof of this lemma.

When $(n,q)\in \{(3,4),(3,8),(4,4),(6,2),(7,2)\}$, the proof follows with a computer computation with the computer algebra system Magma. Lemma 3.3 makes the search of factorizations $G=\textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$ very efficient. In particular, for the rest of the argument, we suppose

$$\begin{aligned} (n,q)\notin \{(3,4),(3,8),(4,4),(6,2),(7,2)\}. \end{aligned}$$

For the rest of our argument, the primitive prime divisors $t_1$ and $t_2$ in Lemmas 5.1 and 5.2 exist and moreover, Part 2 in Lemma 5.2 does not arise.

Without loss of generality, we may suppose that $t_1$ divides |X|. Let $T_1$ be a cyclic subgroup of X having order $t_1$ and set

$$\begin{aligned} C_1:=\textbf{C}_{\textrm{Aut}(L)}(T_1). \end{aligned}$$

From Lemma 5.1, we have

$$\begin{aligned} x\in C_1= \left\{ \begin{array}{ll} \langle T_x,\iota \rangle &{}\quad \iota \text { graph automorphism of }L, \text { if }n \text { is even},\\ \langle T_x, \iota \rangle &{}\quad \iota \text { graph-field automorphism of }L, \text { if }n \text { is odd and } f\text { is even},\\ T_x&{}\quad \text {if }nf\text { is odd}, \end{array}\right. \end{aligned}$$

(5.1)

where $T_x$ is a maximal torus of $\textrm{PGL}_n(q)$ having order $(q^n-1)/(q-1)$. We now divide the rest of the argument in three cases, depending on whether $x\in T_x$ and on whether n is even.$\square$

5.1 Assume $x\in T_x$

Thus x is a semisimple element and X is a field extension subgroup of G. Thus, using the information in [4, Tables B1, B.2, B.3], we deduce that the order of X divides

$$\begin{aligned} \frac{|G:G\cap \textrm{PGL}_n(q)|\ell }{q-1}|\textrm{GL}_{n/\ell }(q^\ell )|, \end{aligned}$$

(5.2)

for some divisor $\ell$ of n with $\ell >1$.

We now turn our attention to $t_2$. From (5.2), we see that $t_2$ is relatively prime to |X| and hence $t_2$ divides |Y|. Let $T_2$ be a cyclic subgroup of Y having order $t_2$ and set

$$\begin{aligned} C_2:=\textbf{C}_{\textrm{Aut}(L)}(T_2). \end{aligned}$$

From Lemma 5.2, we have

$$\begin{aligned} y\in C_2= \left\{ \begin{array}{lll} \langle T_y,\iota \rangle &{}\quad \iota \text { graph automorphism of }L, \text { if }n \text { is odd},\\ \langle T_y, \iota \rangle &{}\quad \iota \text { graph-field automorphism of }L, \text { if }n \text { and } f\text { are even},\\ T_y&{}\quad \text {if }(n-1)f \text { is odd}, \end{array}\right. \end{aligned}$$

(5.3)

where $T_y$ is a maximal torus of $\textrm{PGL}_n(q)$ having order $q^{n-1}-1$.

Assume $y\in T_y$. Thus the order of Y divides

$$\begin{aligned} |G:G\cap \textrm{PGL}_n(q)|\kappa |\textrm{GL}_{(n-1)/\kappa }(q^\kappa )|, \end{aligned}$$

(5.4)

for some divisor $\kappa$ of $n-1$. Assume first $\kappa >1$. Then, using (5.2) and (5.4), we have

$$\begin{aligned} q^{\frac{n(n-1)}{2}}|G:G\cap \textrm{PGL}_n(q)|_r= & {} |G|_r\le |X|_r|Y|_r\\\le & {} |G:G\cap \textrm{PGL}_n(q)|_r^2(\ell \kappa )_rq^{\frac{n}{2}\left( \frac{n}{\ell }-1\right) }q^{\frac{(n-1)}{2}\left( \frac{(n-1)}{\kappa }-1\right) }. \end{aligned}$$

Therefore

$$\begin{aligned} q^{\frac{n(n-1)}{2}}\le |G:G\cap \textrm{PGL}_n(q)|_r(\ell \kappa )_rq^{\frac{n}{2}\left( \frac{n}{\ell }-1\right) }q^{\frac{(n-1)}{2}\left( \frac{(n-1)}{\kappa }-1\right) }. \end{aligned}$$

A computation yields that this inequality is satisfied only when $(n,q)=(3,2)$. The case $(n,q)=(3,2)$ is of no concern to us here, because we are excluding this case from the statement. Assume next $\kappa =1$. When $\kappa =1$, using the explicit structure of $\textbf{C}_L(x)$ and $\textbf{C}_L(y)$, we deduce $|X\cap Y|_r\ge |\textrm{GL}_{n/\ell -1}(q^\ell )|_r$. Therefore, using (5.2) and (5.4), we have

$$\begin{aligned} \begin{aligned} q^{\frac{n(n-1)}{2}}|G:G\cap \textrm{PGL}_n(q)|_r=&|G|_r= \frac{|X|_r|Y|_r}{|X\cap Y|_r}\\ \le&|G:G\cap \textrm{PGL}_n(q)|_r^2\ell _rq^{\frac{n}{2}\left( \frac{n}{\ell }-1\right) }q^{\frac{(n-1)(n-2)}{2}}q^{-\frac{(n-\ell )}{2}\left( \frac{n}{\ell }-2\right) }. \end{aligned} \end{aligned}$$

Therefore

$$\begin{aligned} q^{\frac{n(n-1)}{2}}\le |G:G\cap \textrm{PGL}_n(q)|_r\ell _rq^{\frac{n}{2}\left( \frac{n}{\ell }-1\right) }q^{\frac{(n-1)(n-2)}{2}}q^{-\frac{(n-\ell )}{2}\left( \frac{n}{\ell }-2\right) }. \end{aligned}$$

A computation yields that this inequality is satisfied only when $q\in \{2,4\}$ and $\ell =2$. The case $q=2$ is impossible because $\textrm{GL}_{n-1}(2)$ is centerless, but $y\in \textbf{Z}(\textrm{GL}_{n-1}(q))$. When $q=4$, we deduce that y is a semisimple element having an eigenspace of dimension $n-1$ and that x is a semisimple element of order 5 with no 1-dimensional eigenspaces on V. In the case $(q,\ell )=(4,2)$, by refining the computation above comparing $|G|_2$ with $|X|_2|Y|_2/|X\cap Y|_2$, we deduce $G\nleq \langle \textrm{PGL}_n(4),\tau \rangle$, where $\tau$ is the inverse-transpose graph automorphism. Thus we obtain the examples in Line 7 of Table 1, with the extra remark concerning G in the fifth column.

Assume $y\notin T_y$. Suppose also that, for the time being, y is an involution. Using [4, Tables B.1, B.2, B.3] (or [9, Chapter 4]), we see that all involutions in $C_2\setminus T_y$ are $\textrm{Aut}(L)$-conjugate to $\iota$. Thus y is $\textrm{Aut}(L)$-conjugate to a graph automorphism when n is odd and $\iota$ is $\textrm{Aut}(K)$-conjugate to a graph-field automorphism when n and f are even. Thus the order of Y divides

$$\begin{aligned} \begin{aligned} 2f(q-1)|\textrm{Sp}_{n-1}(q)|,&\quad \text {when }n \text { is odd},\\ 2f(q-1)|\textrm{GU}_{n}(q^{1/2})|,&\quad \text {when }n, f\text { are even}. \end{aligned} \end{aligned}$$

(5.5)

Using (5.2) and (5.5), we see with a computation similar (but simpler) to the one above that $|X|_r|Y|_r<|G|_r$ and hence this case does not arise. We give details only in the case that Y is of type $\textrm{Sp}_{n-1}(q)$. We have

$$\begin{aligned} |G|_r=|XY|_r\le |X|_r|Y|_r=(\ell |G:G\cap \textrm{PGL}_n(q)|)_rq^{\frac{n}{2}\left( \frac{n}{\ell }-1\right) }\cdot (2f)_rq^{\frac{(n-1)^2}{4}}. \end{aligned}$$

Using $|G|_r=|G:G\cap \textrm{PGL}_n(q)|_rq^{\frac{n(n-1)}{2}}$ and simplifying the expression above, we deduce

$$\begin{aligned} q^{\frac{n^2}{4}+\frac{n}{2}-\frac{1}{4}-\frac{n^2}{2\ell }}\le (2\ell f)_r. \end{aligned}$$

Since $\ell >1$ and n is odd, we have $n^2/2\ell \ge n^2/6$ and hence

$$\begin{aligned} q^{\frac{n^2}{12}+\frac{n}{2}-\frac{1}{4}}\le (2\ell f)_r, \end{aligned}$$

which is never satisfied. Suppose now that y is not an involution. As $y\in C_2\setminus T_y=\langle T_y,\iota \rangle \setminus T_y$, we have $y^2\in T_y$, $y^2\ne 1$ and $G=X\textbf{N}_G(\langle y^2\rangle )$. Therefore, we may apply the argument in the first part of the proof to the triple $(G,x,y^2)$ and we deduce that $(G,x,y^2)$ is in Line 7 of Table 1. Thus n is even, $q=4$, $|x|=5$, $|y^2|=3$, $y^2$ is a semisimple element with an $(n-1)$-dimensional eigenspace and x is a semisimple element with no eigenvalues in $\mathbb {F}_q$. Now, $|y|=6$ and $\textbf{C}_L(y)\cong \textrm{SU}_{n-1}(2)$. Then an easy computation, comparing $|G|_2$ with $|XY|_2$, yields $G\ne XY$; therefore, there are no further examples in this case.

5.2 Assume $x\notin T_x$ and that n is Odd

Then, from (5.1), f is even. Suppose also that, for the time being, x has order 2. Using again the information in [4, Section 3.1] (or in [9, Chapter 4]), we see that x is $\textrm{Aut}(L)$-conjugate to $\iota$. Therefore, x is $\textrm{Aut}(L)$-conjugate to a graph-field automorphism. Hence

$$\begin{aligned} \mathbf{}O^{r'}(X)\cong \textrm{PSU}_n(q^{1/2}). \end{aligned}$$

An inspection on the maximal factorizations in [16, 17] reveals that there are no factorizations $G=XY$ with $\mathbf{}O^{r'}(X)\cong \textrm{PSU}_n(q^{1/2})$. (This analysis could be omitted by comparing the size of a Sylow r-subgroup of X, Y and G in a fashion similar to the computations above.) Suppose now that x is not an involution. Then $x^2\in T_x$ and $x^2\ne 1$. Therefore, we may apply Section 5.1 to the triple $(G,x^2,y)$ and we deduce that $G\ne \textbf{N}_G(\langle x^2\rangle )\textbf{N}_G(y)$, because except for $(n,q)=(3,2)$ there are no factorizations when n is odd and the case $(n,q)=(3,2)$ does not occur here as we require f even.

5.3 Assume $x\notin T_x$, $|x|=2$ and n is Even

Using [4, Section 3.1] (or [9, Chapter 4]), we infer that, when q is even, $C_1\setminus T_x$ contains a unique $\textrm{Aut}(L)$-conjugacy class of involutions and, when q is odd, $C_1\setminus T_x$ contains two $\textrm{Aut}(L)$-conjugacy classes of involutions. Then, using the information in [4, Section 3.1 and Table B.3], we obtain

$$\begin{aligned} \textbf{O}^{2'}(\textbf{C}_{\textrm{PGL}_n(q)}x)\cong \left\{ \begin{array}{ll} \textrm{PSp}_n(q),&{}\quad \text {or},\\ \textrm{P}\Omega _{n}^-(q),&{}\quad q \text { odd}. \end{array}\right. \end{aligned}$$

(5.6)

Then $t_2$ does not divide $|\textbf{N}_G(\langle x\rangle )|$ and so $t_2$ divides |Y|. Thus y is as in (5.3).

Assume $y\in T_y$. We claim that, using (5.4), the second possibility for $\textbf{C}_G(x)$ in (5.6) does not give rise to any factorization $G=XY$. Indeed, we have

$$\begin{aligned} |G|_r\le |X|_r|Y|_r\le (|G:\textrm{PGL}_n(q)|)_rq^{\frac{n(n-2)}{4}}\cdot |G:G\cap \textrm{PGL}_n(q)|_r\kappa _rq^{\frac{(n-1)}{2}\left( \frac{n-1}{\kappa }-1\right) }. \end{aligned}$$

When $\kappa >1$, rearranging the terms and using the fact that $(n-1)/\kappa \le (n-1)/3$, we deduce

$$\begin{aligned} q^{\frac{n^2}{12}+\frac{5n}{6}-\frac{2}{3}}\le (|G:G\cap \textrm{PGL}_n(q)|\kappa )_r, \end{aligned}$$

which is impossible. When $\kappa =1$, a similar computation taking in account that $|X\cap Y|_r\ge |X\cap Y\cap L|_r\ge |\Omega _{n-1}(q)|_r=q^{(n-1)^2/4}$ yields another contradiction. Thus $\textbf{O}^{2'}(\textbf{C}_{\textrm{PGL}_n(q)}(x))\cong \textrm{PSp}_n(q)$. Using the formula for $|\textrm{PSp}_n(q)|$, we obtain

$$\begin{aligned} q^{\frac{n(n-1)}{2}}\le |G|_r\le |X|_r|Y|_r\le (2f)_r^2q^{\frac{n^2}{4}}q^{\frac{n-1}{2\kappa }\left( \frac{n}{\kappa }-1\right) }. \end{aligned}$$

This inequality is satisfied only when $\kappa =1$. Thus y is a semisimple element having an eigenspace of dimension $n-1$. The examples arising in this case are in Line 6 of Table 1.

Assume $y\notin T_y$ and y is an involution. In particular, from (5.3), as n is even, f is also even and y is $\textrm{Aut}(L)$-conjugate to a graph-field automorphism. Therefore $\textbf{O}^{r'}(\textbf{C}_L(y))\cong \textrm{PSU}_{n-1}(q^{1/2})$. Using the formulae for $|\textrm{PSU}_{n-1}(q^{1/2})|$ and for $|\textrm{PSp}_n(q)|$ (or $\textrm{P}\Omega _{n}^-(q)$, depending on the $\textrm{Aut}(L)$-conjugacy class of x) and taking in account whether r is odd or $r=2$, we see with a computation that $|X|_r|Y|_r<|G|_r$. Therefore this case does not arise.

Assume $y\notin T_y$ and y is not an involution. As $y\in C_2\setminus T_y=\langle T_y,\iota \rangle \setminus T_y$, we have $y^2\in T_y$, $y^2\ne 1$ and $G=X\textbf{N}_G(\langle y^2\rangle )$. Therefore, we may apply the argument in the first part to the triple $(G,x,y^2)$ and we deduce that $(G,x,y^2)$ is in Line 6 of Table 1. Thus $|y^2|$ divides $q-1$ and $y^2$ is a semisimple element with an $(n-1)$-dimensional eigenspace. Assume that |y| is divisible by some odd prime p. Replacing y by $y^{|y|/2p}$ if necessary, we may suppose that $|y|=2p$. Now, $y^p\in C_2\setminus T_y=\langle T_y,\iota \rangle \setminus T_y$, $y^p$ is an involution and $G=X\textbf{N}_G(\langle y^p\rangle )$; however, we have shown in the previous paragraph that this is impossible. Therefore, y has order a power of 2 and hence, replacing y by $y^{|y|/4}$ if necessary, we may suppose that $|y|=4$. In particular, q is odd, because when q is even $T_y$ has odd order. To conclude the rest of our analysis we identify $T_y$ with the multiplicative group of $\mathbb {F}_{q^{n-1}}$. (In particular, under this identification, we refer to an element of $\mathbb {F}_{q^{n-1}}^*$ as an element of $\textrm{PGL}_n(q)$.) Let $\lambda$ be a generator of $\mathbb {F}_{q^{n-1}}^*$. Then $y=\lambda ^\ell \iota$, where $\ell$ is a divisor of $q^{n-1}-1$ and $\iota$ is a graph-field automorphism. Now,

$$\begin{aligned} y^2=(\lambda ^\ell \iota )^2=\lambda ^\ell (\lambda ^{\ell })^\iota =\lambda ^\ell \lambda ^{-\ell q^{1/2}}=\lambda ^{\ell (1-q^{1/2})}. \end{aligned}$$

As $y^2$ has order 2, we deduce $\ell (1-q^{1/2})=\kappa (q^{n-1}-1)/2$, for some $\kappa \in \mathbb {N}$. Thus

$$\begin{aligned} \ell =\kappa \frac{q^{n-1}-1}{2(1-q^{1/2})}. \end{aligned}$$

This shows that $\lambda ^\ell$ has order a divisor of $2(q^{1/2}-1)$ and hence $\lambda ^\ell \in \mathbb {F}_q^*$. Since all elements of $\textrm{PGL}_n(q)$ in $\mathbb {F}_q^*\subseteq \mathbb {F}_{q^{n-1}}^*$ have the same centralizer, we have

$$\begin{aligned} \textbf{C}_{\textrm{PGL}_n(q)}(y^2) = \textbf{C}_{\textrm{PGL}_n(q)}(\lambda ^\ell ) \end{aligned}$$

and hence

$$\begin{aligned} \textbf{C}_{\textrm{PGL}_n(q)}(y) = \textbf{C}_{\textrm{PGL}_n(q)}(\langle \lambda ^\ell ,\iota \rangle ). \end{aligned}$$

Now, $\textbf{C}_{\textrm{PGL}_n(q)}(\lambda ^\ell )\cong \textrm{GL}_{n-1}(q)$, from which we deduce that

$$\begin{aligned} \textbf{C}_{\textrm{PGL}_n(q)}(y)\cong \textbf{C}_{\textrm{GL}_{n-1}(q)}(\iota )\cong \textrm{GU}_{n-1}(q^{1/2}). \end{aligned}$$

Using this explicit description of $\textbf{C}_{\textrm{PGL}_n(q)}(y)$ it is not hard to verify that $|X|_r|Y|_r<|G|_r$.

5.4 Assume $x\notin T_x$, $|x|>2$ and n is Even

Here $x^2\in T_x$ and $x^2\ne 1$. Applying Section 5.1 to the triple $(G,x^2,y)$, we deduce that $(G,x^2,y)$ is in Line 7 of Table 1. Thus n is even, $q=4$, $|x^2|=5$, $|y|=3$, $x^5$ has no eigenvalue in $\mathbb {F}_q$ and y has an $(n-1)$-dimensional eigenspace on V. Thus $|x|=10$ and $\textbf{C}_L(x)\cong \textrm{GU}_{n/2}(4)$. However, it is not hard to see that $|G|_2\ne |X|_2|Y|_2/|X\cap Y|_2$. Therefore, no further example arises.

Using the explicit description of $\textbf{N}_G(\langle x\rangle )$, $\textbf{N}_G(\langle y\rangle )$ in Lines 6 and 7, it is readily seen that $G=\textbf{C}_G(x)\textbf{C}_G(y)$ when (G, x, y) is in Line 6 and $\textbf{C}_G(x)\textbf{C}_G(y)<G$ when (G, x, y) is in Line 7. Thus we have the $\surd$ symbol in Line 6, whereas $\surd$ is omitted in Line 7.

6 Classical Groups: Unitary Groups

In this section, we assume that G is an almost simple group with socle $L=\textrm{PSU}_n(q)$. Exactly as in Section 5, we start with three technical lemmas, which help to locate the elements x and y with $G=\textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$.

Lemma 6.1

Let n be even. Suppose $r^{2f(n-1)}-1$ admits a primitive prime divisor $t_2$. Let $g\in \textrm{Aut}(\textrm{PSU}_n(q))$ with $t_2$ dividing $|\textbf{N}_{\textrm{Aut}(\textrm{PSU}_n(q))}{(\langle g\rangle )}|$ and let $T_2$ be a cyclic subgroup of order $t_2$ in $\textbf{N}_{\textrm{Aut}(\textrm{PSU}_n(q))}{(\langle g\rangle )}$. Then

$$\begin{aligned} g\in \textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}(T_2)= T, \end{aligned}$$

where T is a maximal torus of $\textrm{PGU}_n(q)$ having order $q^{n-1}+1$. In particular, $|\textrm{Aut}(\textrm{PSU}_n(q)):\textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}(g)|$ is relatively prime to $t_2$.

Proof

Let T be a maximal torus of $\textrm{PGU}_n(q)$ containing $T_2$. Observe that $T_2$ in its action on $V=\mathbb {F}_{q^2}^n$ fixes a 1-dimensional non-degenerate subspace and acts irreducibly on its complement. Thus $|T|=q^{n-1}+1$. Using the information in [9, Chapter 4] and [4, Chapter 3.2, Tables B.1, B.2, B.3] together with the fact that n is even, we obtain

$$\begin{aligned} \textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}(T_2)=T. \end{aligned}$$

If $T_2\le \textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}{( g)}$, then $g\in \textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}{(T_2)}=T$ and hence $|\textrm{Aut}(\textrm{PSU}_n(q)):$ $\textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}(g)|$ is relatively prime to $t_2$, because so is $|\textrm{Aut}(\textrm{PSU}_n(q)):T|$.

It remains to consider the case that $t_2$ does not divide the order of $\textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}{( g)}$. We aim to prove that this case cannot arise. As $t_2$ divides $\textbf{N}_{\textrm{Aut}(\textrm{PSU}_n(q))}{(\langle g\rangle )}$ and as $\textbf{N}_{\textrm{Aut}(\textrm{PSU}_n(q))}{(\langle g\rangle )}/\textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}{(g)}$ acts faithfully as a group of automorphisms on the cyclic group $\langle g\rangle$, we deduce that $t_2$ divides $\varphi (|g|)$, where $\varphi$ is the Euler’s totient function. In particular, |g| is divisible by a prime p with $t_2\mid p-1$. Without loss of generality, we may suppose that $|g|=p$. As $2f(n-1)$ divides $t_2-1$, we have $2f(n-1)<t_2$ and hence $g\in \textrm{PSU}_n(q)$. Now, as $T_2$ acts non-trivially on $\langle g\rangle$, we deduce that $T_2$ permutes non-trivially the eigenspaces of g. However, this is a contradiction because $t_2>2f(n-1)\ge n$.$\square$

Lemma 6.2

Let n be a positive integer with n/2 even, let $t_1$ be a primitive prime divisor of $r^{fn}-1$, let $g\in \textrm{Aut}(\textrm{PSU}_n(q))$ with $t_1$ dividing $|\textbf{N}_{\textrm{Aut}(\textrm{PSU}_n(q))}{(\langle g\rangle )}|$ and let $T_1$ be a cyclic subgroup of order $t_1$ in $\textbf{N}_{\textrm{Aut}(\textrm{PSU}_n(q))}{(\langle g\rangle )}$. Then

$$\begin{aligned} g\in \textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}(T_1)=\langle T,\iota \rangle , \end{aligned}$$

where T is a maximal torus of $\textrm{PGU}_n(q)$ having order $(q^{n}-1)/(q+1)$, $\iota \in \textrm{Aut}(\textrm{PSU}_n(q))$, $|\iota |=2$ and $\textbf{C}_{\textrm{PGU}_n(q)}(\iota )\cong \textrm{PGSp}_n(q)$. In particular, $|\textbf{N}_{\textrm{Aut}(\textrm{PSU}_n(q))}{(\langle g\rangle )}:\textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}(g)|$ is relatively prime to $t_1$.

Proof

The hypothesis n/2 even is used to guarantee that $t_1$ divides $q^i-(-1)^i$, only when $i=n$. The rest of the proof follows verbatim the proofs of Lemmas 5.1, 5.2 and 6.1. The structure of $\textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}(T_1)$ can be inferred from [4, Section 3.2, Table B.4] or [9, Chapter 4].$\square$

Lemma 6.3

Let $n\ge 4$ be even with n/2 odd. Suppose $r^{fn/2}-1$ admits a primitive prime divisor $t_1'$. Let $g\in \textrm{Aut}(\textrm{PSU}_n(q))$ with $t_1'$ dividing $|\textbf{N}_{\textrm{Aut}(\textrm{PSU}_n(q))}{(\langle g\rangle )}|$ and let $T_1'$ be a cyclic subgroup of order $t_1'$ in $\textbf{N}_{\textrm{Aut}(\textrm{PSU}_n(q))}{(\langle g\rangle )}$. Then

$$\begin{aligned} g\in \textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}(T_1')= \langle T,\iota \rangle , \end{aligned}$$

where T is a maximal torus of $\textrm{PGU}_n(q)$ having order $(q^{n}-1)/(q+1)$, $\iota \in \textrm{Aut}(\textrm{PSU}_n(q))$, $|\iota |=2$ and $\textbf{C}_{\textrm{PGU}_n(q)}(\iota )\cong \textrm{PGSp}_n(q)$. In particular, $|\textbf{N}_{\textrm{Aut}(\textrm{PSU}_n(q))}{(\langle g\rangle )}:\textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}(g)|$ is relatively prime to $t_1'$.

Proof

The hypothesis n/2 odd is used to guarantee that $t_1'$ divides $q^i-(-1)^i$, only when $i=n$. The rest of the proof follows verbatim the other analogous proofs. For this lemma, the only part that is not trivial is showing that, $t_1'$ is relatively prime to $|\textbf{N}_{\textrm{Aut}(\textrm{PSU}_n(q))}(\langle g\rangle ):\textbf{C}_{\textrm{Aut}(\textrm{PSU}_n(q))}(g)|$. Arguing as usual, we have that $t_1'\mid p-1$, $|g|=p$ and g is semisimple because $fn/2<t_1'$. As $t_1'$ is a primitive prime divisor of $r^{fn/2}-1$ we have that fn/2 divides $t_1'-1$ and hence

$$\begin{aligned} \alpha f\frac{n}{2}+1=t_1', \end{aligned}$$

for some $\alpha \in \mathbb {N}$. If $\alpha f>1$, then $t_1'>n$. If $\alpha =f=1$, then $t_1'=n/2+1$ is even because n/2 is odd, which is a contradiction because $n\ge 4$. Therefore, in all cases $t_1'>n$. Now, $T_1'$ permutes the eigenspaces of g, but it must permute the eigenspaces trivially because $t_1'>n$.$\square$

From here onward we make use more intensively of the work of Liebeck, Praeger and Saxl on maximal factorizations [16, 17].

Lemma 6.4

If $L=\textrm{PSU}_n(q)$ with $n\ge 3$, then (G, x, y) is in Line 8 or 9 of Table 1.

Proof

When $(n,q)\in \{(3,3),(3,5),(3,8),(4,2),(4,3),(6,2),(6,4),(9,2),(12,2)\}$, the proof follows with a computer computation with the computer algebra system Magma.

From [16, 17], we see that, when n is odd, G admits no proper factorization, except when $(n,q)\in \{(3,3),(3,5),(3,8),(9,2)\}$. In particular, since we have already dealt with these cases, for the rest of the proof, we may assume that n is even. Set $m:=n/2$. The other pairs that we have excluded with the computer computation allow us to conclude that the only maximal factorizations of G are listed in [16, Table 1] and, via Zsigmondy’s theorem, to guarantee the existence of a primitive prime divisor of $r^{2f(n-1)}-1$ and $r^{fn}-1$ when $n/2=m$ is even, and $r^{fn/2}-1$ when $n/2=m$ is odd.

Let $t_2$ be a primitive prime divisor of $r^{2f(n-1)}-1$. As $t_2$ divides |L|, without loss of generality, we may suppose that $t_2$ divides |Y|. Let $T_2$ be a cyclic subgroup of order $t_2$ in Y and let $T_y$ be a maximal torus of $\textrm{PGU}_n(q)$ containing $T_2$. From Lemma 6.1, we obtain

$$\begin{aligned} y\in T_y, \end{aligned}$$

where $T_y$ is a torus having order $q^{n-1}+1$. As n is even, from [4, Section 3.2], the order of Y divides

$$\begin{aligned} |G:G\cap \textrm{PGU}_n(q)|\kappa |\textrm{GU}_{(n-1)/\kappa }(q^\kappa )|, \end{aligned}$$

(6.1)

for some divisor $\kappa$ of $n-1$. Moreover, from the structure of $T_y$, we also deduce that Y is contained in the stabilizer of a 1-dimensional non-degenerate subspace of V. We claim that

$$\begin{aligned} \kappa =1. \end{aligned}$$

(6.2)

Since we have excluded above the pairs $(n,q)\in \{(4,2),(4,3),(6,2),(12,2)\}$ and since Y fixes a non-degenerate 1-dimensional subspace, using the classification of the factorizations of almost simple groups with socle $L=\textrm{PSU}_n(q)$ in [16, 17], we see that X must be contained in a maximal subgroup A of G with the property that one of the following holds:

• (i) $L\cap A=P_m$,

• (ii) $L\cap A=\textrm{PSp}_{2m}(q)=\textrm{PSp}_n(q)$,

• (iii) $L\cap A={\hat{}\,}\textrm{SL}_{m}(4).2$, $q=2$, $m\ge 3$,

• (iv) $L\cap A={\hat{}\,}\textrm{SL}_m(16).3.2$, $q=4$ and $G\ge L.4$.

In Cases (ii), (iii) and (iv), by comparing the size of a Sylow r-subgroup of Y with a Sylow r-subgroup of A, we deduce that $\kappa =1$, that is, (6.2) holds true. All of these computations are straightforward and we only give details to the Case (ii). Here,

$$\begin{aligned} \begin{aligned} |G|_r=&|G:G\cap \textrm{PSU}_n(q)|_rq^{\frac{n(n-1)}{2}},\\ |X|_r\le&(2f)_rq^{(n/2)^2},\\ |Y|_r=&|G:G\cap \textrm{PSU}_n(q)|_r\kappa _rq^{\frac{(n-\kappa )\left( \frac{n}{\kappa }-1\right) }{2}}. \end{aligned} \end{aligned}$$

Now, a tedious computation using this information yields that $|G|_r\le |X|_r|Y|_r$ is satisfied only when $\kappa =1$.

In Case (i), using the structure of $P_m$ (which can be deduced by fixing a hyperbolic basis for V), we have

$$\begin{aligned} P_m\cap L\cong E_q^{m^2}:\textrm{SL}_m(q^2).(q-1). \end{aligned}$$

We claim that also in this case $\kappa =1$. To prove this claim we use the factorization of the order of $|\textrm{SL}_m(q^2)|$ into cyclotomic polynomials, (6.1) and Zsigmondy’s theorem. Assume first that $n\ge 8$ and let $t'$ be a primitive prime divisor of $q^{2(n-3)}-1$. In particular, $t'$ divides $q^{n-3}+1$ and hence $t'$ divides |L|, because n is even. Now, as $2(n-3)>n$, $t'$ is relatively prime to $|P_m|$ because

$$\begin{aligned} (q^{2m}-1)(q^{2m-2}-1)\cdots (q^4-1)(q^2-1) \end{aligned}$$

cannot be divisible by $t'$. Therefore, $t'$ divides |Y|; but this is only possible when $\kappa =1$. When $n=4$, we have $n-1=3$ and hence $\kappa \in \{1,3\}$. However, if $\kappa =3$, then $|Y|\le 3(q^3+1)|G:G\cap \textrm{PGU}_4(q)|$ and hence

$$\begin{aligned} |G:X|\le 3(q^3+1). \end{aligned}$$

The minimal degree of a faithful permutation representation of $\textrm{PSU}_4(q)$ is $(q+1)(q^3+1)$ (see [10, Table 4]). As $3(q^3+1)<(q+1)(q^3+1)$, we have $L\le X$, which is a contradiction. Finally, when $n=6$, we have $n-1=5$ and hence $\kappa \in \{1,5\}$. However, if $\kappa =5$, then $|Y|\le 5(q^5+1)|G:G\cap \textrm{PGU}_6(q)|$ and hence

$$\begin{aligned} |G:X|\le 5(q^5+1). \end{aligned}$$

The minimal degree of a faithful permutation representation of $\textrm{PSU}_6(q)$ is at least $q^{5}(q^4+q^2+1)$ (see [10, Table 4]). As $5(q^5+1)<q^5(q^4+q^2+1)$, we have $L\le X$, which is a contradiction. This concludes the proof of our claim and hence we have proved (6.2).

From (6.2), $Y\cap L$ is of type ${\hat{}\,}\textrm{GU}_{n-1}(q)$ and $y\in \textbf{Z}({\,{\hat{}\,}\textrm{GU}_{n-1}(q)})$. Thus y is a semisimple element of order a divisor of $q+1$ and y has an $(n-1)$-dimensional eigenspace. In particular, the order of Y divides

$$\begin{aligned} |G:G\cap \textrm{PGU}_n(q)||\textrm{GU}_{n-1}(q)|. \end{aligned}$$

(6.3)

Using the description of Y, we have that $|G:Y|=|X:X\cap Y|$ is divisible by

$$\begin{aligned} q^{n-1}\cdot \frac{q^{n}-1}{\gcd (n,q+1)(q+1)}. \end{aligned}$$

Let $t_1$ be a primitive prime divisor of $r^{fn}-1=q^n-1$ when n/2 is even and let $t_1$ be a primitive prime divisor of $r^{fn/2}-1=q^{n/2}-1$ when n/2 is odd. Observe that in either case, $t_1$ divides |X|. Let $T_1$ be a cyclic subgroup of X of order $t_1$, let $C_1:=\textbf{C}_{\textrm{Aut}(L)}(T_1)$ and let $T_x$ be a maximal torus of $\textrm{PGU}_n(q)$ containing $T_1$. From Lemmas 6.2 and 6.3, we obtain

$$\begin{aligned} C_1=\langle T_x,\iota \rangle ,\quad |\iota |=2,~\iota \in \textrm{Aut}(\textrm{PSU}_n(q))\setminus \textrm{PSU}_n(q)\text { and }\textbf{C}_{\textrm{PGU}_n(q)}(\iota )\cong \textrm{PGSp}_n(q). \end{aligned}$$

6.1 Assume $x\in T_x$

Thus x is a semisimple element and X is a field extension subgroup of G. Thus, using the information in [4, Section 3.2, Tables B.1, B.2, B.3], we deduce that the order of X divides

$$\begin{aligned} \frac{|G:G\cap \textrm{PGU}_n(q)|}{q+1}\ell |\textrm{GU}_{n/\ell }(q^\ell )| \quad \text { or }\quad \frac{|G:G\cap \textrm{PGU}_n(q)|}{q+1}\ell |\textrm{GL}_{n/\ell }(q^\ell )|, \end{aligned}$$

(6.4)

for some divisor $\ell$ of n with $\ell \ge 2$; where the case on the right occurs when $\ell$ is even and the case on the left occurs when $\ell$ is odd.

From the structure of Y, we deduce $|X\cap Y|_r\ge |\textrm{GU}_{n/\ell -1}(q^\ell )|_r$ if $\ell$ is odd, and $|X\cap Y|_r\ge |\textrm{GL}_{n/\ell -1}(q^\ell )|_r$ if $\ell$ is even. Suppose first that $\ell$ is odd. Using (6.3) and (6.4), we have

$$\begin{aligned} \begin{aligned} |G:G\cap \textrm{PGU}_n(q)|_rq^{\frac{n(n-1)}{2}}=&|G|_r\le \frac{|X|_r|Y|_r}{|X\cap Y|_r}\\ \le&|G:G\cap \textrm{PGU}_n(q)|_r^2\ell _rq^{\frac{n}{2}\left( \frac{n}{\ell }-1\right) }q^{\frac{(n-1)(n-2)}{2}}q^{-\frac{(n-\ell )}{2}\left( \frac{n}{\ell }-2\right) }. \end{aligned} \end{aligned}$$

A computation yields that this inequality is never satisfied since $\ell \ge 3$. This shows that $\ell$ is even and X is of type $\textrm{GL}_{n/\ell }(q^\ell ).\ell$. Then, using (6.3) and (6.4), we have again

$$\begin{aligned} q^{\frac{n(n-1)}{2}}\le (|G:G\cap \textrm{PGU}_n(q)|\ell )_r q^{\frac{n}{2}\left( \frac{n}{\ell }-1\right) }q^{\frac{(n-1)(n-2)}{2}}q^{-\frac{(n-\ell )}{2}\left( \frac{n}{\ell }-2\right) }. \end{aligned}$$

(6.5)

Another computation in the same spirit as the one above shows that $\ell =2$ and $q\in \{2,4,16\}$. By refining the computation above, we see that the case $q=16$ does not actually arise. This can be seen by observing that (6.5) is satisfied only when G contains the whole field automorphism of $\mathbb {F}_{q^2}$, but now we can refine the bound $|X\cap Y|_2$ with $2|\textrm{GL}_{n/2-1}(q^2)|_2$. Now, another computation with this refined value for $|X\cap Y|_2$ excludes the case $q=16$. (Observe that this is in line with [16, Table 1], where we see that maximal factorizations using $\mathcal {C}_3$-subgroups arise only when $q\in \{2,4\}$: see Cases (iii) and (iv) above.) When, $q=2$, X is not a local subgroup (see for instance [12, Proposition 4.2.4]) and hence this case does not arise. Thus $q=4$, x is an element of order $(q^2-1)/(q+1)=15/5=3$ having no eigenvalue in $\mathbb {F}_{q^2}$ and y is an element of order 5 having an eigenspace of dimension $n-1$. Moreover, consulting [16, Table 1] or by a careful computation as above, we see that 4 divides $|G:L|_2$. This example is in Line 8 of Table 1.

6.2 Assume $x\notin T_x$ and $|x|=2$

Using the results in [4, Table B.4] (or in [9, Chapter 4]), we see that $\textbf{O}^{r'}(\textbf{C}_L(x))$ is of type

$$\begin{aligned} \textrm{PSp}_n(q),\quad \hbox { or }\textrm{P}\Omega _n^\varepsilon (q) \end{aligned}$$

and the second case only occurs when q is odd. By distinguishing these two possibilities for X, it is not hard to prove that $|G|_r|X\cap Y|_r=|X|_r|Y|_r$ yields that X is of type $\textrm{PSU}_n(q^{1/2})$ or $\textrm{PSp}_n(q)$ (to exclude the case $\textrm{P}\Omega _n^\varepsilon (q)$ we require the fact that q is odd). Incidentally, the case that X is of type $\textrm{P}\Omega _n^\varepsilon (q)$ can also be excluded by checking [16, Table 1].

When X is of type $\textrm{PSp}_n(q)$, we obtain that x is an involution of order 2 with $\textbf{C}_L(x)\cong \textrm{PSp}_n(q)$ and that y is an element of order dividing $q+1$ and having an eigenspace of dimension $n-1$. This example is in Line 9 of Table 1.

6.3 Assume $x\notin T_x$ and $|x|>2$

As $x^2\in T_x\setminus \{1\}$ and $x^2\ne 1$, applying Section 6.1 to the factorization $G=\textbf{N}_G(\langle x^2\rangle )Y$, we obtain that the triple $(G,x^2,y)$ is in Line 8 of Table 1. Therefore, $q=4$, $\textbf{N}_L(\langle x^2\rangle )$ is of type $\textrm{GL}_{n/2}(q^2).2$ and $x^2$ has order 3. Now, $x^3\notin T_x$ and $x^3$ has order 2. Therefore, applying Section 6.2 to the factorization $G=\textbf{N}_G(\langle x^3\rangle )Y$, we obtain that the triple $(G,x^3,y)$ is in Line 9 of Table 1. Thus $x^2\in \textbf{N}_L(\langle x^3\rangle )=\textbf{C}_L(x^3)\cong \textrm{Sp}_n(4)$. Thus

$$\begin{aligned} \textbf{N}_L(\langle x\rangle )=\textbf{N}_L(\langle x^2\rangle )\cap \textbf{C}_L(x^3)\cong \textbf{N}_{\textrm{Sp}_n(4)}(\langle x^2\rangle )\cong 3.\textrm{SL}_{n/2}(4).2. \end{aligned}$$

(6.6)

Now, let p be a primitive prime divisor of $2^{fn}-1=q^{n}-1$. Observe that, from (6.6), p is relatively prime to |X|. When n/2 is even, p is also relatively prime to $|\textrm{GU}_{n-1}(q)|$ and hence p is relatively prime to |Y|. Therefore, in order to have $G=XY$, n/2 must be odd. Assume n/2 is odd. Given $i\in \{1,\ldots ,n\}$, p divides $q^i-(-1)^i$ if and only if $i\in \{n/2,n\}$. Therefore, $|G|_p>|Y|_p$ and hence also in this case we have $G\ne XY$. Summing up, we have shown that this case does not give rise to a factorization of G.

Using the explicit description of $\textbf{N}_G(\langle x\rangle )$, $\textbf{N}_G(\langle y\rangle )$ in Lines 8 and 9, it is readily seen that $G=\textbf{C}_G(x)\textbf{C}_G(y)$ when (G, x, y) is in Line 9 and $\textbf{C}_G(x)\textbf{C}_G(y)<G$ when (G, x, y) is in Line 8. Thus we have the $\surd$ symbol in Line 9, whereas $\surd$ is omitted in Line 8. $\square$

7 Classical Groups: Symplectic Groups

Lemma 7.1

Let $n\ge 4$ be even. Suppose $r^{fn}-1$ admits a primitive prime divisor $t_1$. Let $g\in \textrm{Aut}(\textrm{PSp}_n(q))$ with $t_1$ dividing $|\textbf{N}_{\textrm{Aut}(\textrm{PSp}_n(q))}{(\langle g\rangle )}|$ and let $T_1$ be a cyclic subgroup of order $t_1$ in $\textbf{N}_{\textrm{Aut}(\textrm{PSp}_n(q))}{(\langle g\rangle )}$. Then

$$\begin{aligned} g\in \textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(T_1)= \left\{ \begin{array}{ll} \langle T,\iota \rangle ,&{}\quad \text {when }n=4, r=2,f \text { is odd},\\ &{}\quad \iota \text { graph-field automorphism,}\\ T,&{}\quad \text {otherwise}, \end{array}\right. \end{aligned}$$

where T is a maximal torus of $\textrm{PGSp}_n(q)$ having order $q^{n/2}+1$, that is, T is a Singer cycle. In particular, $|\textrm{Aut}(\textrm{PSp}_n(q)):\textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(g)|$ is relatively prime to $t_1$.

Proof

Let T be a maximal torus of $\textrm{PGSp}_n(q)$ containing $T_1$. Then T is a cyclic group of order $q^{n/2}+1$. The structure of $\textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(T_1)$ follows consulting [4, Section 3.4] and [9, Chapter 4].

Suppose that $T_1\le \textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(g)$. Then $g\in \textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(T_1)$. If $g\in T$, then $|\textrm{Aut}(\textrm{PSp}_n(q)):\textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}(g)|$ is relatively prime to $t_1$, because $T\le \textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(g)$ and $|\textrm{Aut}(\textrm{PSp}_n(q)):T|$ is relatively prime to $t_1$. If $g\notin T$, then $n=4$, $r=2$ and f is odd. Moreover, g is conjugate to $\iota$ because T has odd order. Therefore, g is an involution and hence $\textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(g)=\textbf{N}_{\textrm{Aut}(\textrm{PSp}_n(q))}(\langle g\rangle )$. Moreover, $\textbf{C}_{\textrm{PSp}_n(q)}(g)\cong {}^2B_2(q)$. Since $|{}^2B_2(q)|=(q^2+1)q^2(q-1)$ we have that $|\textrm{Aut}(\textrm{PSp}_n(q)):\textbf{C}_{\textrm{PSp}_n(q)}(g)|$ is relatively prime to $t_1$.

Suppose that $T_1\nleq \textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(g)$. The usual argument using the faithful action of $\textbf{N}_{\textrm{Aut}(\textrm{PSp}_n(q))}(\langle g\rangle )/\textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(g)$ on $\langle g\rangle$ gives that $g\in \textrm{PSp}_n(q)$ and that g has order divisible by a prime p with $t_1\mid p-1$. However, as $t_1>fn\ge n$, we see that $T_1$ cannot permute non-trivially the eigenspaces of g.$\square$

Lemma 7.2

Let $n\ge 6$ be even. Suppose $r^{f(n-2)}-1$ admits a primitive prime divisor $t_2$. Let $g\in \textrm{Aut}(\textrm{PSp}_n(q))$ with $t_2$ dividing $|\textbf{N}_{\textrm{Aut}(\textrm{PSp}_n(q))}{(\langle g\rangle )}|$ and let $T_2$ be a cyclic subgroup of order $t_2$ in $\textbf{N}_{\textrm{Aut}(\textrm{PSp}_n(q))}{(\langle g\rangle )}$. Then

$$\begin{aligned} g\in \textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(T_2)=(q^{\frac{n}{2}-1}+1)\circ \textrm{GSp}_2(q). \end{aligned}$$

In particular, $|\textrm{Aut}(\textrm{PSp}_n(q)):\textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(g)|$ is relatively prime to $t_2$.

Proof

Using the information in [9, Chapter 4], we obtain

$$\begin{aligned} \textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(T_2)=(q^{\frac{n}{2}-1}+1)\circ \textrm{GSp}_2(q). \end{aligned}$$

Let T be the cyclic subgroup of order $q^{n/2-1}+1$ in $\textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(T_2)$ and observe that T is central in $\textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(T_2)$.

Suppose that $T_2\le \textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(g)$. Then $g\in \textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(T_2)$ and hence g centralizes T. Thus $T\le \textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(g)$ and hence $|\textrm{Aut}(\textrm{PSp}_n(q)):\textbf{C}_{\textrm{Aut}(\textrm{PSL}_n(q))}(g)|$ is relatively prime to $t_2$, because so is $|\textrm{PSp}_n(q):T|$.

Suppose that $T_2\nleq \textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(g)$. The usual argument using the faithful action of $\textbf{N}_{\textrm{Aut}(\textrm{PSp}_n(q))}(\langle g\rangle )/\textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(g)$ on $\langle g\rangle$ gives that $g\in \textrm{PSp}_n(q)$ and that g has order divisible by prime p with $t_2\mid p-1$. Without loss of generality, we may suppose that $|g|=p$ and hence $g\in \textrm{PSp}_n(q)$. Moreover, g is semisimple, because $t_2$ divides $p-1$ and $t_2$ cannot divide $r-1$: recall that $t_2$ is a primitive prime divisor of $r^{f(n-2)}-1$. Now, there exists $\alpha \in \mathbb {N}$ with

$$\begin{aligned} t_2=\alpha f(n-2)+1. \end{aligned}$$

If $\alpha f>1$, then $t_2>n$ and hence $T_2$ cannot permute non-trivially the eigenspaces of g. If $\alpha f=1$, then $t_2=n-1$ is prime. We show that this is impossible. Under the action of $T_2$, the vector space $V=\mathbb {F}_q^n$ decomposes as

$$\begin{aligned} V=W\perp W^\perp , \end{aligned}$$

where $\dim _{\mathbb {F}_q}(W)=n-2$ and the symplectic form induced by $\textrm{PSp}_n(q)$ on W is non-degenerate, W is an irreducible $\mathbb {F}_qT_2$-module, and $W^\perp$ is a 2-dimensional trivial module for $T_2$. Since we are supposing that $T_2$ normalizes $\langle g\rangle$, $T_2$ permutes the eigenspaces of g. The orthogonal decomposition of V above yields that g has one eigenspace of dimension $n-2$, and then another eigenspace of dimension 2, or two eigenspaces of dimension 1. In either case, since $t_2=n-1$ is prime, we see that $T_2$ fixes setwise each eigenspace of g. From this it follows that $T_2$ centralizes g.$\square$

Lemma 7.3

If $L=\textrm{PSp}_n(q)$, then (G, x, y) is in Line 10 of Table 1.

Proof

When

$$\begin{aligned} (n,q)\in \{(4,2),(4,3), (4,4),(4,8),(4,16),(4,32),(6,2),(6,3),(8,2)\}, \end{aligned}$$

the proof follows with a computation with the computer algebra system Magma. Now, by excluding these cases, the maximal factorizations of the almost simple groups with socle $L=\textrm{PSp}_n(q)$ appear in [16, Tables 1 and 2]. Moreover, by excluding these cases, we simplify some of the computations later in the proof.

Let $t_1$ be a primitive prime divisor of $r^{nf}-1$: as usual the existence of $t_1$ follows from Zsigmondy’s theorem. As $t_1$ divides |L|, without loss of generality, we may suppose that $t_1$ divides |X|. Let $T_1$ be a cyclic subgroup of X of order $t_1$, let $C_1:=\textbf{C}_{\textrm{Aut}(\textrm{PSp}_n(q))}(T_1)$ and let $T_x$ be a maximal torus of $\textrm{PGSp}_n(q)$ containing $T_1$. Thus $T_x$ is a torus of order $q^{n/2}+1$ and $T_x$ is a cyclic subgroup of $\textrm{PGSp}_n(q)$. From Lemma 7.1, we obtain

$$\begin{aligned} g\in C_1= \left\{ \begin{array}{ll} \langle T_x,\iota \rangle ,&{}\quad \text {when }n=4, r=2, f \text { is odd, } \iota \text { is a graph-field automorphism},\\ T_x,&{}\quad \text {otherwise}. \end{array}\right. \end{aligned}$$

Assume $x\in T_x$. Thus x is a semisimple element and X is a field extension subgroup of G. We now discuss the structure of X. Assume first, for simplicity, that $x\in \textrm{PSp}_n(q)$ or n/2 is odd. Thus, using [4, Section 3.4, Table B.7] and [9, Chapter 4 and Table 4.5.1], we see that the order of X divides

$$\begin{aligned} f\ell |\textrm{GU}_{n/\ell }(q^{\ell /2})|2, \end{aligned}$$

(7.1)

for some divisor $\ell$ of n with $n/\ell$ odd. When $x\in T_x\setminus \textrm{PSp}_{n}(q)$ and n/2 is even, there are a few more cases to consider. As $T_x\nleq \textrm{PSp}_{n}(q)$, q is odd. As n/2 is even, $q^{n/2}+1\equiv 2\pmod 4$ and hence $T_x=\langle a\rangle \times \langle b\rangle$, where a has order 2 and $\langle b\rangle =T_x\cap \textrm{PSp}_n(q)$ has order $(q^{n/2}+1)/2$. Using the references above, we obtain that

$$\begin{aligned} \textbf{N}_{\textrm{PGSp}_n(q)}(\langle a\rangle )=\textbf{C}_{\textrm{PGSp}_n(q)}(a)\cong \textrm{Sp}_{n/2}(q^2).2, \end{aligned}$$

where the “2” on top acts as a field automorphism. Therefore, when q is odd and n/2 is even, we have the following possibilities for the order of X:

$$\begin{aligned} 2f|\textrm{Sp}_{n/2}(q^2)|\quad \text {and}\quad f\ell |\textrm{GU}_{n/\ell }(q^{\ell /2})|, \end{aligned}$$

(7.2)

for each divisor $\ell$ of n with $n/\ell$ odd.

Before considering the element y in general, we first consider the case $n=4$. In this case, from (7.1) and (7.2), |X| divides either $4f(q^2+1)$ or $2f|\textrm{SL}_2(q^2)|=2fq^2(q^4-1)$. In the first possibility, we have

$$\begin{aligned} m(\textrm{PSp}_4(q))\le |G:Y|\le |X|\le 4f(q^2+1), \end{aligned}$$

(7.3)

where $m(\textrm{PSp}_4(q))$ is the minimal degree of a faithful permutation representation of $\textrm{PSp}_4(q)$. Now, $m(\textrm{PSp}_4(q))=(q^4-1)/(q-1)$, except when $q\in \{2,3\}$. (This information is tabulated, for instance, in [10, Table 4].) The inequality (7.3) is satisfied only when $r=2$ and $f\le 3$, or $q=r=3$. However, these cases have been checked with the help of a computer. Therefore, we may suppose that $\textbf{C}_{\textrm{Aut}(L)}(x)\cong \textrm{SL}_2(q^2).2f$. In particular, q is odd and x is an involution. Observe that $q^2\equiv 1\pmod 4$ and hence $(q^2+1)/2\equiv 1\pmod 2$. Therefore $x\in \textrm{PGSp}_4(q)\setminus \textrm{PSp}_4(q)$ and hence $\textrm{PGSp}_4(q)\le G$. In particular, $|G:X|=q^2(q^2-1)/2$.

From [3, Table 8.12], we see that $X=\textbf{C}_G(x)$ is a maximal subgroup of G. From the classification of the maximal factorizations of almost simple groups [16, 17], we deduce that Y is contained in a parabolic subgroup P of G whose unipotent radical Q is a non-abelian group of order $q^3$ (this information is in [16, (3.2.1a)]). Furthermore, from [3, Table 8.12], we get that the shape of $P\cap L$ is

$$\begin{aligned} E_q^{1+2}:((q-1)\circ \textrm{Sp}_2(q)). \end{aligned}$$

As $G=XY$, we deduce that $|G:X|=q^2(q^2-1)/2$ divides |Y|. Using this description of P it is not hard to see that the only elements $y\in P$ with the property that $\textbf{N}_P(\langle y\rangle )$ has order divisible by $q^2(q^2-1)/2$ are the elements in $\textbf{Z} Q\cong E_q$. Therefore, y is a transvection of $\textrm{PSp}_4(q)$. In particular, we find the examples in Line 10 of Table 1 (for $n=4$).

Suppose now that $n\ge 6$. Let $t_2$ be a primitive prime divisor of $r^{f(n-2)}-1$. Observe that $t_2$ does exist because $n\ge 6$ and because we are excluding the case $(n,q)=(8,2)$ from our analysis here. From (7.1) and (7.2), |X| is relatively prime to $t_2$ and hence $t_2$ divides |Y|. Let $T_2$ be a cyclic subgroup of Y of order $t_2$ and set $C_2:=\textbf{C}_{\textrm{Aut}(L)}(T_2)$. From Lemma 7.2, we obtain

$$\begin{aligned} y\in C_2=(q^{\frac{n}{2}-1}+1)\circ \textrm{GSp}_2(q). \end{aligned}$$

Write

$$\begin{aligned} y=y_{n-2}y_2, \end{aligned}$$

where $y_{n-2}$ belongs to the torus of cardinality $q^{n/2-1}+1$ and $y_2$ belongs to $\textrm{GSp}_2(q)$. Now, this decomposition of y induces a direct sum decomposition of the underlying vector space $V=V_{n-2}\perp V_2$, where y induces $y_{n-2}$ on $V_{n-2}$ and induces $y_2$ on $V_2$. Since $y_{n-2}$ is semisimple, the $\langle y_{n-2}\rangle$-module $V_{n-2}$ is the direct sum of pair-wise isomorphic irreducible modules.

Suppose that none of these modules is isomorphic to any of the irreducible $\langle y_2\rangle$ -submodules of $V_2$ (here, we are including the possibility that $y_2$ is a non-identity unipotent element and hence $V_2$ is indecomposable with a unique irreducible submodule, which is the trivial module).

In this case, $\textbf{N}_G(\langle y\rangle )=Y$ preserves the direct sum decomposition $V_2\perp V_{n-2}$ and hence Y is contained in the stabilizer in G of a 2-dimensional non-degenerate subspace of V. Now, by checking the maximal factorizations of the almost simple group G in [16, Tables 1 and 2], we see that one of the following holds:

• (i) $X\cap L\le \textrm{Sp}_{n/2}(4).2$, $q=2$, n/2 is even,

• (ii) $X\cap L\le \textrm{Sp}_{n/2}(16).2$, $q=4$, n/2 is even, $G=\textrm{Aut}(L)=L.2$,

• (iii) $X\cap L\le G_2(q)$, q is even and $n=6$.

All of these three cases can be eliminated with a computation. Indeed, since $\textrm{Sp}_{n/2}(4).2$, $\textrm{Sp}_{n/2}(16).2$ and $G_2(q)$ normalize no non-identity cyclic subgroup, we deduce that $X\cap L$ must be strictly contained in this embedding. However, by comparing the order of X (see (7.1)), Y and G, we see that the equality $G=XY$ cannot be satisfied. These computations can be performed in the same spirit as the analogous computations for almost simple groups having socle $\textrm{PSL}_n(q)$ and $\textrm{PSU}_n(q)$. For instance, in Case (iii), we have $n=6$ and hence, from (7.1), we have that |X| divides $2f|\textrm{GU}_3(q)|$; moreover, |Y| divides $|\textrm{Sp}_2(q)\perp \textrm{Sp}_4(q)|f$. A computation shows that $|X|_2|Y|_2<|G|_2$, contradicting $G=XY$. We omit the details of the remaining computations for Cases (i) and (ii).

This means that, in order to have a factorization $G=XY=\textbf{N}_G(\langle x\rangle )\textbf{N}_G(\langle y\rangle )$, some of the $\langle y_{n-2}\rangle$-irreducible submodules of $V_{n-2}$ are isomorphic to some of the irreducible submodules of $V_2$. As $y_2\in \textrm{GSp}_2(q)$, this implies that the irreducible $\langle y_{n-2}\rangle$-submodules of $V_{n-2}$ have dimension at most 2 and hence $y_{n-2}$ has order a divisor of $q+1$.

Suppose that $y=y_2$ is a unipotent element. Thus y is a transvection of $\textrm{PSp}_n(q)$ and

$$\begin{aligned} \textbf{N}_{\textrm{PGSp}_n(q)}(\langle y\rangle )\cong E_{q}^{1+(n-2)}:((q-1)\times \textrm{Sp}_{n-2}(q)). \end{aligned}$$

Assume that X is of type $\textrm{Sp}_{n/2}(q^2)$ (recall from (7.1) and (7.2) that when this happens, x is an involution and n/2 is even). Since n/2 is even, $q^{n/2}+1\equiv 2\pmod 4$ and hence $x\in \textrm{PGSp}_n(q)\setminus \textrm{PSp}_n(q)$. In particular, we find one of the examples in Line 10 of Table 1. When X is not of type $\textrm{Sp}_{n/2}(q^2)$, we deduce, by consulting the factorizations of the almost simple groups with socle $L=\textrm{PSp}_n(q)$ in [16] and by consulting the structure of X and Y, that there are no triples (G, x, y) occurring in this case.

Suppose that y is not a unipotent element. If r divides |y|, then $y_2$ is a non-identity unipotent element. However, this contradicts the fact that some of the $\langle y_{n-2}\rangle$-irreducible submodules of $V_{n-2}$ are isomorphic to some of the irreducible submodules of $V_2$. Therefore r is relatively prime to |y| and hence y is a semisimple element. From the compatibility condition between the $\mathbb {F}_q\langle y_{n-2}\rangle$-submodules of $V_{n-2}$ and the $\mathbb {F}_q\langle y_2\rangle$-submodules of $V_2$, we obtain $|y|=|y_{n-2}|=|y_2|$ and that |y| is a divisor of $q+1$. This gives that $\textbf{C}_L(y)\cong {\hat{}\,}\textrm{GU}_{n/2}(q)$ or $\textbf{C}_L(y)\cong \textrm{Sp}_{n/2}(q^2)$, depending on whether n/2 is odd or even. Using this information on the structure of X and Y and consulting the list of maximal factorizations for G in [16, Tables 1 and 2], we deduce that there are no examples arising in this case.

Assume that $x\notin T_x$. This implies $n=4$, $r=2$ and f is odd. Note that $|T_x|$ is odd while |x| is even. Thus by Lemma 3.4 we may assume that x is an involution and so is a graph-field automorphism of $\textrm{PSp}_4(q)=\textrm{Sp}_4(q)$. Thus $X\ge \textbf{N}_L(\langle x\rangle )=\textbf{C}_L(x)\cong {^{2}B_2(q)}$ is a Suzuki group. From [3, Table 8.14], we deduce that X is a maximal subgroup of G. Using the classification of the maximal factorizations of the almost simple groups having socle $\textrm{Sp}_4(q)$ [16, Table 2], we deduce that

$$\begin{aligned} Y\cap L\le \textrm{O}_4^+(q)=\textrm{SL}_2(q)\times \textrm{SL}_2(q). \end{aligned}$$

Suppose that $Y\cap L$ does not contain any of the two simple direct factors of $\textrm{O}_4^+(q)$. Then $|\textrm{O}_4^+(q):Y\cap L|\ge (q+1)^2$ because the minimal degree of a permutation representation of $\textrm{SL}_2(q)$ is $q+1$. Therefore $|Y\cap L|\le |\textrm{SL}_2(q)|^2/(q+1)^2=q^2(q-1)^2$ and hence

$$\begin{aligned} |G|\le |X||Y|\le 2f(q^2+1)q^2(q-1)\cdot q^2(q-1)^2|Y:L\cap Y|. \end{aligned}$$

As $|G|=|G:L|q^4(q^4-1)(q^2-1)$ and $|G:L|\ge |Y:Y\cap L|$, we deduce

$$\begin{aligned} (q+1)^2\le 2f(q-1), \end{aligned}$$

which is impossible. Therefore, Y contains at least one of the two simple direct factors of $\textrm{O}_4^+(q)$. Let us denote by $S_1$ and $S_2$ the two simple direct factors of $\textrm{O}_4^+(q)$. Without loss of generality we assume $S_1\le Y$. Since $\textbf{N}_G(\langle y\rangle )/\textbf{C}_G(y)$ is soluble and since $\textrm{SL}_2(q)$ is simple, we deduce $S_1\le \textbf{C}_G(y)$ and hence $y\in \textbf{C}_G(S_1)$. Using the action of the outer automorphism group of $L=\textrm{Sp}_4(q)$, we deduce $\textbf{C}_G(S_1)\le L$ and hence $y\in \textbf{C}_L(S_1)$. As $\textbf{C}_L(S_1)\le \textbf{N}_L(S_1)\le \textrm{O}_4^+(q)$, we have $\textbf{C}_L(S_1)=S_2$ and hence $y\in S_2$. Since the normalizers of the non-identity elements of $S_2\cong \textrm{SL}_2(q)$ have order q, $2(q-1)$ or $2(q+1)$, we deduce

$$\begin{aligned} |L\cap Y|\le |\textrm{O}_4^+(q)\cap Y|=|\textbf{N}_{\textrm{O}_4^+(q)}(\langle y\rangle )|\le |S_1|2(q+1)=2q(q^2-1)(q+1). \end{aligned}$$

Now, as

$$\begin{aligned} |G:L|q^4(q^4-1)(q^2-1)\!=\!|G|\le |X||Y|\!\le \! 2f(q^2+1)q^2(q-1)\cdot 2q(q^2-1)(q+1)|Y\!:\!L\cap Y| \end{aligned}$$

and $|G:L|\ge |Y:L\cap Y|$, we get $q\le 2f$, which is a contradiction. Hence, no triple arises in this case.

Using the explicit description of $\textbf{N}_G(\langle x\rangle )$, $\textbf{N}_G(\langle y\rangle )$ in Line 10, it is readily seen that $G=\textbf{C}_G(x)\textbf{C}_G(y)$. Thus we have the $\surd$ symbol in Line 10. $\square$

8 Classical Groups: Odd Dimensional Orthogonal Groups

The analysis in this section is similar to the work in Section 7; indeed, from one side, we use the classification of Liebeck, Praeger and Saxl [16, 17] and, from the other side, the factorizations arising in the context of odd dimensional orthogonal groups resemble the factorizations for symplectic groups.

Lemma 8.1

If $L=\textrm{P}\Omega _n(q)$ with n odd, then (G, x, y) is in Line 14 of Table 1.

Proof

When $(n,q)=(7,3)$, the proof follows with a computer computation: no example arises. Thus we assume that $(n,q)\ne (7,3)$. Set $m:=(n-1)/2$. We start by summarizing the maximal factorizations

$$\begin{aligned} G=AB \end{aligned}$$

of almost simple groups with socle $L=\Omega _n(q)$ (as usual we use the notation from [16, 17]):

• 1. labelcasee1 $A\cap L=N_1^-$ and $B\cap L=P_m$,

• 2. $n=7$, $A\cap L=G_2(q)$ and $B\cap L$ is either $P_1$, or $N_1^\varepsilon$, or $N_2^{\varepsilon }$, with $\varepsilon \in \{+,-\}$,

• 3. $n=13$, $q=3^f$, $A\cap L=\textrm{PSp}_6(3^f).a$ with $a\le 2$ and $B\cap L=N_1^-$,

• 4. $n=25$, $q=3^f$, $A\cap L=F_4(3^f)$ and $B\cap L=N_1^-$.

Replacing X by Y if necessary, we may suppose that

$$\begin{aligned} X\le A\quad \hbox { and }\quad Y\le B. \end{aligned}$$

(8.1)

Cases 2, 3 and 4.

Here A is an almost simple subgroup of G and hence X is a core-free proper subgroup of A. By Lemma 3.1, the factorization $G=XY$ gives rise to the factorization

$$\begin{aligned} A=X(Y\cap A) \end{aligned}$$

(8.2)

of A.

Now, from [16, Table 5], we see that $F_4(3^f)$ admits no proper factorizations. Hence $A\le X$ or $A\le Y\cap A$, which are both impossible. Therefore Case 4 does not arise.

Assume Case 2. From [16, Table 5], we see that $G_2(q)$ admits proper factorizations with q odd only when $q=3^{f}$. Observe that $f\ge 2$, because we have dealt with $\Omega _7(3)$ above. Let $A=A'B'$ be a maximal factorization of A with $X\le A'$ and with $Y\cap A\le B'$. Using the maximal factorizations of $G_2(q)$, we see that $A'\cap G_2(q)$ is one of the following groups

$$\begin{aligned} \textrm{SL}_3(q), \quad \textrm{SL}_3(q).2,\quad \textrm{SU}_3(q),\quad \textrm{SU}_3(q).2,\quad ^2G_2(q), \end{aligned}$$

where, for the last case, we require f odd, but we do not need this information here. From $G=AB=XB$ and (8.2), we deduce $|G:B|=|A:A\cap B|=|X:X\cap B|$ and hence |G : B| divides |X|. Thus

$$\begin{aligned} |G:B| ~\hbox { divides }~|A'|. \end{aligned}$$

(8.3)

Now,

$$\begin{aligned} |G:B|= \left\{ \begin{array}{ll} q^3\frac{q^3-1}{2}&{}\quad \text {when }L\cap B=N_1^-,\\ q^3\frac{q^3+1}{2}&{}\quad \text {when }L\cap B=N_1^+,\\ q^5\frac{q^6-1}{2(q-1)}&{}\quad \text {when }L\cap B=N_2^+,\\ q^5\frac{q^6-1}{2(q+1)}&{}\quad \text {when }L\cap B=N_2^-,\\ \frac{q^6-1}{q-1}&{}\quad \text {when }L\cap B=P_1. \end{array}\right. \end{aligned}$$

Using this explicit value of |G : B| and using (8.3), we deduce that

• either $L\cap B=N_1^-$ and $A'\cap G_2(q)\in \{\textrm{SL}_3(q),\textrm{SL}_3(q).2\}$, or

• $L\cap B=N_1^+$ and $A'\cap G_2(q)\in \{\textrm{SU}_3(q),\textrm{SU}_3(q).2\}$.

As $q=3^f$, we have $\gcd (3,q-1)=\gcd (3,q+1)=1$ and hence $A'$ is an almost simple group with socle $\textrm{SL}_3(q)$ or $\textrm{SU}_3(q)$. Recall now that, since $X\le A'$, we have $X=\textbf{N}_G(\langle x\rangle )=\textbf{N}_{A'}(\langle x\rangle )$. Now, it is not hard to verify that $\textrm{Aut}(\textrm{SL}_3(q))$ and $\textrm{Aut}(\textrm{SU}_3(q))$ do not contain a non-identity cyclic subgroup $\langle x\rangle$ whose normalizer has order divisible by $|G:B|\in \{(q^3-1)q^3/2,(q^3+1)q^3/2\}$. Hence, Case 2 does not arise.

The analysis for Case 3 is similar to Case 2, but simpler. The factorization for $\Omega _{13}(3^f)$ arising in Case 3 is described in detail in [16, 4.6.3, Lemma A]. Observe that A is an almost simple group with socle $\textrm{PSp}_6(q)$. As $\textbf{N}_G(\langle x\rangle )=X\le A$, we have $X=\textbf{N}_A(\langle x\rangle )$. As $G=XB$ and $G=LB$, we deduce that

$$\begin{aligned} |L:L\cap B|=|G:B|=|X:X\cap B|. \end{aligned}$$

Since $L\cap B=N_1^-$ in Case 3, we have

$$\begin{aligned} |L:L\cap B|=|\Omega _{13}(q):N_1^-|=\frac{(q^6-1)q^6}{2} \end{aligned}$$

and hence $(q^6-1)q^6/2$ divides |X|. Now, it is not hard to verify, using [3, Tables 8.28 and 8.29] that $\textrm{Aut}(\textrm{PSp}_6(q))$ contains no non-identity group elements $g\ne 1$ with $|\textbf{N}_{\textrm{Aut}(\textrm{PSp}_n(q))}(\langle g\rangle )|$ divisible by $q^6(q^6-1)/2$. Hence, Case 3 does not arise.

Replacing X with Y if necessary, $X\cap L\le N_1^{-}$ and $Y\cap L\le P_{m}$, where $N_1^{-}$ is the stabilizer in L of a 1-dimensional non-degenerate subspace of “minus type” and $P_{m}$ is the stabilizer in L of a totally isotropic subspace of dimension m; in particular, $P_m$ is a parabolic subgroup.

For simplicity, let $A_X$ be the stabilizer in G of a 1-dimensional non-degenerate subspace of “minus type” with $X\le A_X$ and let $B_Y$ be a parabolic subgroup of G with $Y\le B_Y$. It will also be convenient to let $\hat{P}_m$ be a parabolic subgroup of $\textrm{SO}_n(q)$ with $B_Y\cap L\le \hat{P}_m$. Thus

$$\begin{aligned} A_X\cap L\cong \Omega _{2m}^-(q).2\quad \text { and }\quad B_Y\cap L\cong E_q^{\frac{m(m-1)}{2}+m}:\tfrac{1}{2}\textrm{GL}_{m}(q). \end{aligned}$$

Moreover,

$$\begin{aligned} |G:A_X|=|L:L\cap A_X|=|\Omega _{2m+1}(q):\textrm{P}\Omega _{2m}^-(q).2|=\frac{q^m(q^m-1)}{2}. \end{aligned}$$

(8.4)

Now $G=XP_m$, so X acts transitively on the set of all totally isotropic subspaces of dimension m. Since $(m,q)\ne (7,3)$ we have from [7, Theorem 7.1] that $\Omega _{2m}^-(q)\unlhd X$. Since, $\Omega _{2m}^-(q)$ has trivial centre and is insoluble, it follows that $\Omega _{2m}^-(q)\le \textbf{C}_G(x)$. Thus $\textrm{SO}_n(q)\le G$ and $x\in \textrm{SO}_n(q)\backslash \Omega _n(q)$ is an involution.

We now fix an $\mathbb {F}_q$-basis $e_1,\ldots ,e_m,w,f_1,\ldots ,f_m$ of V such that the symmetric matrix defining $L=\Omega _{2m+1}(q)$ with respect to this ordered basis is the matrix

$$\begin{aligned} J= \begin{pmatrix} 0&{}0&{}I\\ 0&{}1&{}0\\ I&{}0&{}0 \end{pmatrix}, \end{aligned}$$

where we use I to denote the $m\times m$ identity matrix. Using this matrix representation, $\hat{P}_m$ has unipotent radical subgroup

$$\begin{aligned} Q=\left\{ \begin{pmatrix} I&{}v&{}B\\ 0&{}1&{}-v^t\\ 0&{}0&{}I \end{pmatrix}~|~ v\in \mathbb {F}_q^m, B\in \textrm{Mat}_{m\times m}(\mathbb {F}_q), B+B^t+vv^t=0\right\} \cong E_q^{\frac{m(m-1)}{2}+m}. \end{aligned}$$

Moreover, the Levi complement of Q in $\hat{P}_m$ is

$$\begin{aligned} \mathcal {L}=\left\{ \begin{pmatrix} A&{}0&{}0\\ 0&{}1&{}0\\ 0&{}0&{}(A^{-1})^t \end{pmatrix}~|~ A\in \textrm{GL}_m(q)\right\} \cong \textrm{GL}_m(q). \end{aligned}$$

In what follows we also need the following subgroup

$$\begin{aligned} Z=\left\{ \begin{pmatrix} I&{}0&{}B\\ 0&{}1&{}0\\ 0&{}0&{}I \end{pmatrix}~|~ B\in \textrm{Mat}_{m\times m}(\mathbb {F}_q), B^t=-B \right\} \cong E_q^{\frac{m(m-1)}{2}}. \end{aligned}$$

A simple computation yields that $Z\le \textbf{Z}(Q)$.

Let t be a primitive prime divisor of $r^{fm}-1$ and observe that the existence of t is guaranteed by Zsigmondy’s theorem, because r is odd and $m\ge 3$. By (8.4), t divides $|G:A_X|$ and hence t divides |Y|. Let T be a cyclic subgroup of Y having order t. We claim that

$$\begin{aligned} y\in \textbf{C}_G(T). \end{aligned}$$

(8.5)

Since t divides |Y| and since t is a primitive prime divisor of $r^{fm}-1$, we deduce that $T\le L$ and hence $T\le L\cap B_Y\cong P_m$. Suppose that t divides $|\textbf{N}_G(\langle y\rangle ):\textbf{C}_G(y)|$. Then, from the faithful action of T on $\langle y\rangle$, we deduce that |y| is divisible by a prime number p with $t\,|\, p-1$. Set $y':=y^{|y|/p}$. Now, $y'$ is an element of prime order p. Moreover, using the fact that t is a primitive prime divisor of $r^{fm}-1$ and that $t\,|\, (p-1)$, we have $y'\in L\cap Y\le P_m$ and $y'$ is semisimple. Furthermore, t divides $|\textbf{N}_G(\langle y'\rangle ):\textbf{C}_G(y')|$. In particular, T and $y'$ are semisimple elements in $\hat{P}_m$ and hence, using the explicit description of $\hat{P}_m$ above, we deduce that T and $y'$ are both in a Levi complement (which is isomorphic to $\textrm{GL}_m(q)$) of $\hat{P}_m$. Applying Lemma 5.1 to $\textrm{GL}_m(q)$ yields that it is impossible to have t divides $|\textbf{N}_{\textrm{GL}_m(q)}(\langle y'\rangle ):\textbf{C}_{\textrm{GL}_m(q)}(y')|$. Since t divides $|Y|=|\textbf{N}_G(\langle y\rangle )|$, we get $T\le \textbf{C}_G(g)$. Therefore, (8.5) holds true.

From (8.5), we deduce $y\in \textrm{SO}_n(q)$ and hence

$$\begin{aligned} y\in \hat{P}_m. \end{aligned}$$

(8.6)

Let

$$\begin{aligned} m_t= \begin{pmatrix} \lambda &{}0&{}0\\ 0&{}1&{}0\\ 0&{}0&{}(\lambda ^{-1})^t \end{pmatrix} \end{aligned}$$

be a generator of T. Now, using the explicit description of Q and $\mathcal {L}$ above, we see that $\textbf{C}_{\hat{P}_m}(T)=T_m\ltimes W$, where $T_m$ is a torus in $\mathcal {L}\cong \textrm{GL}_m(q)$ of cardinality $q^m-1$ and

$$\begin{aligned} W=\left\{ \begin{pmatrix} I&{}0&{}B\\ 0&{}1&{}0\\ 0&{}0&{}I \end{pmatrix}~|~ B\in \textrm{Mat}_{m\times m}(\mathbb {F}_q), B+B^t=0, \lambda B\lambda ^t=B\right\} . \end{aligned}$$

From (8.5) and (8.6), we have $y\in T_m\ltimes W$.

We claim that

$$\begin{aligned} y\in W. \end{aligned}$$

(8.7)

We argue by contradiction and we suppose that $y\notin W$. Then, replacing y by a suitable power, we may assume that y has prime order and $y\in T_m$. Using the explicit description of $\hat{P}_m$, it can be deduced that

$$\begin{aligned} Y\cap \textrm{SO}_n(q)=\textbf{N}_{\textrm{SO}_n(q)}(\langle y\rangle )=\textbf{N}_{\hat{P}_m}(\langle y\rangle )\subseteq Z\mathcal {L}. \end{aligned}$$

However, $Z\mathcal {L}$ is not transitive on the non-degenerate 1-dimensional subspaces of “minus type” and hence $A_XY\ne G$, which is a contradiction because in Case 1 Y does act transitively on the set of 1-dimensional non-degenerate subspaces of “minus type”. Therefore, we must have $y\in W$ and y is a unipotent element of order r. Therefore, (8.7) holds true.

Now, it can be shown that the set

$$\begin{aligned} \{B\in \textrm{Mat}_{m\times m}(q)\mid B^t=-B, \lambda B\lambda ^t=B\} \end{aligned}$$

contains a non-zero matrix only when m is even. Thus m is even. Moreover, from [4, Table B.12], we have

$$\begin{aligned} \textbf{C}_{L}(y)\cong E_q^{\frac{m(m-1)}{2}+m}: \textrm{Sp}_m(q). \end{aligned}$$

(8.8)

Thus we obtain the examples in Line 14 of Table 1.

Using the explicit description of $\textbf{N}_G(\langle x\rangle )$, $\textbf{N}_G(\langle y\rangle )$ in Line 14, it is readily seen that $G=\textbf{C}_G(x)\textbf{C}_G(y)$. Thus we have the $\surd$ symbol in Line 14. $\square$

9 Classical Groups: Even Dimensional Orthogonal Groups Having Witt Defect 1

We begin with the following lemma.

Lemma 9.1

Let $m\ge 5$ be odd, let $n:=2m$ and let $g\in \textrm{Aut}(\textrm{P}\Omega _n^-(q))$ with $g\ne 1$ and with $|\textbf{N}_{\textrm{Aut}(\textrm{P}\Omega _n^-(q))}(\langle g\rangle )|$ divisible by

$$\begin{aligned} q^{\frac{m(m-1)}{2}}(q^{m-1}+1)(q^{m-2}-1)\cdots (q^2+1)(q-1). \end{aligned}$$

Then g is an involution not in $\textrm{P}\Omega _n^-(q)$ and

$$\begin{aligned} \textbf{C}_{\textrm{P}\Omega _n^-(q)}(g)\cong \left\{ \begin{array}{ll} \textrm{Sp}_{n-2}(q)&{}\quad \text {when }q \text { is even},\\ \Omega _{n-1}(q)&{}\quad \text {when }q \text { is odd}.\\ \end{array}\right. \end{aligned}$$

Proof

Set $v:=q^{m(m-1)/2}(q^{m-1}+1)(q^{m-2}-1)\cdots (q^2+1)(q-1)$, $L:=\textrm{P}\Omega _n^-(q)$ and $A:=\textrm{Aut}(\textrm{P}\Omega _n^-(q))$. The proof follows by an inspection of Section 3.5 and Tables B.11, B.12 in [4]. We give some details to make this inspection more elementary.

Suppose first that g has prime order. Assume also that $g\in L$. Now,

$$\begin{aligned} |\textbf{N}_A(\langle g\rangle ):\textbf{C}_L(g)|=|\textbf{N}_A(\langle g\rangle ):\textbf{N}_L(\langle g\rangle )||\textbf{N}_L(\langle g\rangle ):\textbf{C}_L(g)|. \end{aligned}$$

(9.1)

The first factor on the right hand side of (9.1) divides $|\textrm{Out}(L)|$. Observe that the second factor on the right hand side of (9.1) divides $r-1$ when g is unipotent (because $|g|=r$ and $\varphi (r)=r-1$) and divides n when g is semisimple (because $\textbf{N}_L(\langle g\rangle )/\textbf{C}_L(g)$ acts by permuting the eigenspaces of g). Therefore, $|\textbf{C}_L(g)|$ is divisible by $v/\ell$, where $\ell :=\ell _1\ell _2$, $\ell _1:=\gcd (v,|\textrm{Out}(L)|)\le 8f$ and $\ell _2\le r-1$ when g is unipotent and $\ell _2\le 2m$ when g is semisimple. Now, using [4, Section 3.5], a case-by-case analysis shows that there is no g having centralizer divisible by such a large number.

Assume that $g\notin L$. Let $h\in L\cap \textbf{N}_A(\langle g\rangle )$. Then, $g^h=g^i$, for some $1\le i\le |g|-1$. Now,

$$h^{-1}ghg^{-1}=g^{i-1}\in L\cap \langle g\rangle =1$$

and hence $i=1$. This shows that $L\cap \textbf{N}_A(\langle g\rangle )=\textbf{C}_L(g)$. Therefore

$$\begin{aligned} |\textbf{N}_A(\langle g\rangle ):\textbf{C}_L(g)|=|\textbf{N}_A(\langle g\rangle )L:L| \end{aligned}$$

and hence $|\textbf{N}_A(\langle g\rangle ):\textbf{C}_L(g)|$ divides $|\textrm{Out}(L)|$. Therefore, $|\textbf{C}_L(g)|$ is divisible by $v/\ell$, where $\ell :=\gcd (v,|\textrm{Out}(L)|)\le 8f$. Now, using [4, Section 3.5] and the notation therein, we see that the only elements having prime order with $g\notin L$ and having centralizer divisible by such a large number are conjugate to $\gamma _1$ when q is odd and to $b_1$ when q is even. Moreover, the structure of $\textbf{C}_L(g)$ is discussed in [4, Section 3.5.2] when q is odd and in [4, Section 3.5.4] when q is even. The proof of the lemma follows in this case.

Suppose now that g does not have prime order. We need to show that no extra case arises. Observe that from the previous part of the proof, g has order a power of 2. Without loss of generality, replacing g by $g^{|g|/4}$ if necessary, we may suppose that g has order 4. Observe that $g^2$ is A-conjugate to $\gamma _1$ when q is odd and to $b_1$ when q is even. Set $\bar{A}:=\textbf{C}_A(g^2)/\langle g^2\rangle$ and adopt the “bar” notation for the projection of $\textbf{C}_A(g^2)$ onto $\bar{A}$. We have

$$\begin{aligned} \bar{A}\cong \left\{ \begin{array}{ll} \textrm{Aut}(\textrm{Sp}_{n-2}(q))&{}\quad \text {when }q \text { is even},\\ \textrm{Aut}(\Omega _{n-1}(q))&{}\quad \text {when }q \text { is odd}.\\ \end{array}\right. \end{aligned}$$

Moreover, $\overline{\textbf{N}_A(\langle g\rangle )}=\textbf{C}_{\bar{A}}(\bar{g})$. This shows that $\textbf{C}_{\bar{A}}(\bar{g})$ has order divisible by v/2. When q is odd, we may apply [4, Section 3.5] to the odd dimensional orthogonal group $\Omega _{n-1}(q)$ and we see that $\textrm{Aut}(\Omega _{n-1}(q))$ contains no involutions whose centralizer has order divisible by v/2. Similarly, when q is even, we may apply [4, Section 3.4] to the symplectic group $\textrm{Sp}_{n-2}(q)$ and we see that $\textrm{Aut}(\textrm{Sp}_{n-2}(q))$ contains no involutions whose centralizer has order divisible by v/2 (to check this it is useful to recall that $n-2=2m-2\ge 8$).$\square$

We are also going to need [18, Lemma 4.4] that lists all possibilities of $\Gamma \textrm{O}^-_{2m}(q)$ that act transitively on an orbit of $\Omega _{2m}(q)$ on nonsingular 1-subspaces. However, it is not claimed there that all groups listed are actually transitive. We rule out two possibilities with the following lemma.

Lemma 9.2

Let $Y\le \Gamma \textrm{O}_{2m}^-(2)$ such that $m\equiv 2\pmod 4$ and either $\textrm{SU}_{m/4}(2^4)$ or $\Omega _{m/2}^-(2^4)$ is normal in Y. Then Y does not act transitively on the set of nonsingular 1-subspaces.

Proof

Note that $\textrm{SU}_{m/4}(2^4)\le \Omega _{m/2}^-(2^4)\le \Gamma \textrm{O}_{2m}^-(2)$ and so it suffices to show that $Y:=N_{\Gamma \textrm{O}_{2m}^-(2)}(\Omega _{m/2}^-(2^4))$ is not transitive on the set of nonsingular 1-spaces. Let $k=\textrm{GF}(2)$, $V=k^{2m}$ and Q be a nondegenerate quadratic form on V of “minus type”. Let $\Delta =\{v\in V~|~ Q(v)=1\}$, which corresponds to the set of all nonsingular 1-subspaces of V. Consider V as a m/2-dimensional vector space over $K=\textrm{GF}(2^4)$. Following [16, p. 59], let $P:V\rightarrow K$ be a nondegenerate quadratic form on V of “minus type” such that $Q=\textrm{Tr}_{K\rightarrow k} \circ P$, that is, $Q(v)=P(v)+P(v)^2+P(v)^{4}+P(v)^{8}$ for each $v\in V$. Note that $\Delta =\{v\in V~|~ P(v)+P(v)^2+P(v)^{4}+P(v)^{8}=1\}$. Now arguing as in [16, p. 59] we have that $Y=\langle \Omega ^-_{m/2}(2^4), \phi \rangle$, where $\phi :V\rightarrow V$ has order 8 and $P(v^\phi )=P(v)^\tau$, where $\tau$ is a generator of $\textrm{Aut}(\textrm{GF}(2^4))$.

Let $v\in V$ such that $P(v)\ne 0$. Then $P(\langle v\rangle _K)=K$. Since $\textrm{Tr}_{K\rightarrow k}$ is k-linear, its kernel has size 8 and so there are precisely 8 elements $w\in \langle v\rangle _K$ such that $Q(w)=1$. Since the isometry group of P has index 4 in Y, it follows that Y is not transitive on $\Delta$. $\square$

Lemma 9.3

If $L=\textrm{P}\Omega _n^-(q)$, then (G, x, y) is in Lines 11, 12 or 13 of Table 1.

Proof

When $(n,q)=(8,2)$, the proof follows with a computer computation with the computer algebra system Magma: there are no triples in this case. Set $m:=n/2$. From [16, 17], there exist two core-free maximal subgroups A and B of G, with $X\le A$ and with $Y\le B$. Moreover, replacing x by y if necessary, from [16], we see that one of the following holds:

• 1. $L=\textrm{P}\Omega _{10}^-(2)$, $A\cap L=\textrm{Alt}(12)$ and $B\cap L=P_1$,

• 2. $A\cap L=N_1$, $B\cap L= {\hat{}\,{\textrm{GU}}}_m(q)$ and m is odd,

• 3. $A\cap L=P_1$, $B\cap L= {\hat{}\,{\textrm{GU}}}_m(q)$ and m is odd,

• 4. $A\cap L=N_1$, $B\cap L= \Omega _m^-(q^2).2$, $q\in \{2,4\}$, m is even and $G=\textrm{Aut}(L)$,

• 5. $A\cap L=N_2^+$, $B\cap L= \textrm{GU}_m(4)$, $q=4$, m is odd and $G=\textrm{Aut}(L)$.

Case 1.

This case can be dealt with a computer computation and we obtain one of the examples in Line 11 of Table 1.

Case 5.

The factorization $G=XY$ of G gives rise to the factorization

$$\begin{aligned} B=G\cap B=XY\cap B=(X\cap B)Y \end{aligned}$$

of B, via Lemma 3.1. Let us denote by $g\mapsto \bar{g}$ the natural projection from B to $\bar{B}=B/\textbf{Z}(B\cap L)=B/\textbf{Z}(\textrm{GU}_m(4))$ and observe that $\textbf{Z}(\textrm{GU}_m(q))$ has order $\gcd (m,q+1)=\gcd (m,5)$. Now, $\bar{B}$ is an almost simple group with socle $\textrm{PSU}_m(4)$ with m odd. An inspection of [16] reveals that this group $\bar{B}$ has no maximal factorizations and hence the factorization $B=(X\cap B)Y$ implies $\bar{B}=\bar{Y}$, or $\bar{B}=\overline{X\cap B}$. The second option is absurd because, by hypothesis, $X\cap L\le A\cap L=N_2^+$ and $N_2^+\cap B$ cannot project surjectively to B. Therefore $\bar{B}=\bar{Y}$ and hence $y\in \textbf{Z}(\textrm{GU}_m(4))$. In particular, $\textbf{Z}(\textrm{GU}_m(4))\ne 1$ and hence 5 divides m. Moreover,

$$\begin{aligned} Y=\textbf{N}_G(\textbf{Z}(\textrm{GU}_m(4))), \end{aligned}$$

(9.2)

$|y|=5$ and y is a semisimple element having no eigenvalue in $\mathbb {F}_{q}$.

Now, by [12, Lemma 4.1.1], $\Omega _2^+(4)\times \Omega _{n-2}^-(4)\le A\cap L=N_2^+=\textrm{O}_2^+(4)\times \textrm{O}_{n-2}^-(4)$ and $A\cap L \cap \textrm{O}^+_2(4)=\Omega ^+_2(4)$. Hence $A=\textbf{N}_G(\Omega _2^+(4))$ and $|\Omega _2^+(4)|=3$. If $\langle x\rangle =\Omega _2^+(4)$, we obtain the examples in Line 12 of Table 1. Suppose then $\langle x\rangle \ne \Omega _2^+(4)$. Let us denote by $g\mapsto \bar{g}$ the natural projection from A to $\bar{A}=A/Z$. Now, $\bar{A}$ is an almost simple group with socle $\textrm{P}\Omega _{2m-2}^-(4)$. As usual, from Lemma 3.1, the factorization $G=XY$ gives rise to a factorization $A=X(A\cap Y)$ of A and hence to the factorization $\bar{A}=\bar{X}\overline{A\cap Y}$ of the almost simple group $\bar{A}$ having socle $\textrm{P}\Omega _{2m-2}^-(q)$. As $(2m-2)/2=m-1$ is even, Case 4 holds for the factorization, $\bar{A}=\bar{X}\overline{A\cap Y}$, that is, $\bar{X}$ is contained in a subgroup of type $N_1$ of $\bar{A}$, or of type $\Omega _{m-1}^-(q^2).2=\Omega _{m-1}^-(16).2$. However, the first possibility is impossible, otherwise we would have a factorization of G where one of the two factors (namely X) is contained in the stabilizer of a 3-dimensional non-degenerate subspace of $V=\mathbb {F}_q^n$, contradicting [16]. In the second case, we claim that |X| is not divisible by |G : B|. Indeed, from (9.2), we have

$$\begin{aligned} |G:B|=|\textrm{P}\Omega _n^-(q):\textrm{GU}_m(q)| \end{aligned}$$

is divisible by $q^{m-2}-1$, because m is odd. However, if t is a primitive prime divisor of $q^{m-2}-1$, then it is readily seen that t is relatively prime to $|\Omega _{m-1}^-(q^2).2|$ and hence to |X|.

Cases 2, 3 and 4.

Suppose first that $B\cap L={\hat{}}\textrm{GU}_m(q))$. From the factorization $G=XB$, we deduce that |X| is divisible by |G : B| and hence by

$$\begin{aligned} |LB:B|=|L:L\cap B|=q^{\frac{m(m-1)}{2}}(q^{m-1}+1)(q^{m-2}-1)\cdots (q^2+1)(q-1). \end{aligned}$$

Now, Lemma 9.1 yields that x is an involution whose centraliser is $N_1$. Thus $A\cap L= N_1$. Hence, Case 3 does not occur and we only need to consider Cases 2 and 4. In both cases we have $A\cap L=N_1$. Hence $X\le N_1$ and so Y acts transitively on an L-orbit of nonsingular 1-subspaces. Thus by [18, Lemma 4.4] and Lemma 9.2 we have that one of the following is a normal subgroup $Y_0$ of Y:

• 1. $\textrm{SU}_m(q)$ and m odd;

• 2. $\textrm{SU}_{m/2}(q^2)$, with $m\equiv 2\pmod 4$, $m\ge 6$, and $q=2$ or 4;

• 3. $\Omega _m^-(q^2)$ with m even and $q=2$ or 4.

The last possibility is that $Y_0$ has trivial centre and is insoluble, so must lie in $\textbf{C}_{G}(y)$. However, inspecting [4, Section 3.5] we see that this is not possible (note that $Y_0$ is irreducible). When $Y_0=\textrm{SU}_m(q)$ we have that $B=Y=\textbf{N}_{G}(\langle y\rangle )$ for some $y\in \textbf{Z}(B\cap L)=\textbf{Z}(\,{\hat{}}\textrm{GU}_m(q))$. Thus y is semisimple of order a divisor of $q+1$ and y has no eigenvalue in $\mathbb {F}_{q}$. Moreover, the argument at the start of the paragraph yields that x is an involution whose centraliser is $N_1$ and so we have Line 11.

It remains to consider the case where $Y_0=\textrm{SU}_{m/2}(q^2)$ when $q=2$ or 4. Then $Y=\textbf{N}_{G}(\langle y\rangle )$ for some $y\in \textbf{Z}(B\cap L)=\textbf{Z}(\,{\hat{}}\textrm{GU}_{m/2}(q^2))$. Thus y is semisimple of order $q^2+1$ and y has no eigenvalue in $\mathbb {F}_{q^2}$. We also have that $B\cap L=\textrm{O}_m^-(q^2)$ and so for the maximal factorisation to exist we need $G=\textrm{Aut}(L)$. Using the argument in [16, p. 59], let $Q:V\rightarrow \textrm{GF}(q)$ and $P:V\rightarrow \textrm{GF}(q^2)$ be nondegenerate quadratic forms of “minus type” such that $Q=\textrm{Tr}_{q^2\rightarrow q}\circ P$. Let $v\in V$ such that $Q(v)=1$. As the elements of $\langle v\rangle _{\textrm{GF}(q^2)}$ have distinct P-values, we have that $B\cap N_1= \textrm{O}^-_{2m}(q^2)_v=\textrm{O}^-_{2m}(q^2)_{\langle v\rangle _{\textrm{GF}(q)^2}}=\textrm{Sp}_{m-2}(q^2)\times C_2$. By [16, Table 1] we have that $\textrm{O}^-_m(q^2)=(\textrm{Sp}_{m-2}(q^2)\times C_2)(\textrm{GU}_{m/2}(q^2).2)$ and so $B=(\textrm{Sp}_{m-2}(q^2)\times C_2)(\textrm{GU}_{m/2}(q^2).4f)$, where $f=2$ if $q=4$ and $f=1$ otherwise. Thus $B=(B\cap N_1)Y$ and so by Lemma 3.1$G=N_1Y$. Note that $N_1$ is the centraliser in G of an involution in $\textrm{SO}_{2m}^-(q)\backslash \Omega _{2m}^-(q)$ whose centraliser in L is $\textrm{Sp}_{n-2}(q)$, and so we have the factorisation in Line 13 of Table 1. It remains to show that it is not possible to have $X<N_1$. Note that $N_1=\textrm{Sp}_{n-2}(2)\times C_2$ when $q=2$, while when $q=4$ we have $N_1=(\textrm{Sp}_{n-2}(4)\rtimes \langle \phi \rangle ) \times C_2$, where $\phi$ is a field automorphism. Also if $x=(x_1,x_2)\in N_1$, then $X\le N_{\textrm{Sp}_{n-2}(4)\rtimes \langle \phi \rangle }(\langle x_1\rangle )\times C_2$. Note that it remains to consider the case where $x_1\ne 1$. Looking at |X : B| we deduce that a primitive prime divisor of $r^{f(n-2)}-1$ divides |X| and so, by Lemma 7.1, we deduce that $x_1$ lies in a maximal torus of $\textrm{PGSp}_{n-2}(q)$ of order $q^{(n-2)/2}+1$. Then looking at the possible orders of $N_{\textrm{Sp}_{n-2}(4)\rtimes \langle \phi \rangle }(\langle x_1\rangle )$ we deduce that no factorisation arises.

In Line 11, we see from [16, 3.5.2(b)] that $\hat{}\,\textrm{SU}_m(q)\le \textbf{C}_G(y)$ acts transitively on an L-orbit of nonsingular 1-spaces and so we get $G=\textbf{C}_G(x)\textbf{C}_G(y)$. However, for Lines 12 and 13 we see in [16, 3.5.1 and 3.5.2(c)] that $\textbf{C}_G(y)$ needs to contain field automorphisms to be transitive on the conjugacy class $\langle x\rangle ^G$. Since such elements of $\textbf{N}_G(\langle y\rangle )$ do not centralise y, it follows that $\textbf{C}_G(x)\textbf{C}_G(y)<G$. Thus we have the $\surd$ symbol in Line 11, whereas $\surd$ is omitted in Lines 12 and 13.