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This article is cited in 5 scientific papers (total in 5 papers) On the sharp Baer-Suzuki theorem for the N. Yanga, Zh. Wua, D. O. Revinbc, E. P. Vdovinbc a Jiangnan University, Wuxi, P. R. China b Sobolev Institute of Mathematics, Siberian Branch of the Russian Academy of Sciences, Novosibirsk, Russia c Novosibirsk State University, Novosibirsk, Russia
Russian version:
Matematicheskii Sbornik, 2023, Volume 214, Number 1, DOI: https://doi.org/10.4213/sm9698 
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    § 1. IntroductionOnly finite groups are considered in this paper, and the term ‘group’ means a finite group. Throughout, $\pi$ denotes a subset of the set $\mathbb{P}$ of prime numbers. A finite group is said to be a $\pi$-group if all prime divisors of its order belong to $\pi$. The standard notation is used: we denote by $\mathrm{O}_\pi(G)$ the $\pi$-radical of the group $G$, that is, its largest normal $\pi$-subgroup; if $M$ is a subset of $G$, then $\langle M\rangle$ denotes the subgroup generated by $M$. The Baer-Suzuki theorem is a classical result in the theory of finite groups which is included in many monographs and textbooks, for example, [1], Theorem 39.6, [2], Ch. A, (14.11), [3], Ch. 3, Theorem 8.2, [4], Theorem 2.12, and [5], Theorem 6.7.6. Baer-Suzuki Theorem. Let $p$ be a prime number, $G$ be a finite group, and let $x\in G$. Then $x\in \mathrm{O}_p(G)$ if and only if $\langle x_1,x_2 \rangle$ is a $p$-group for any elements $x_1,x_2\in G$ conjugate to $x$. This theorem was first proved by Baer [6] and then reproved by Suzuki [7]. In [8] Alperin and Lyons found a short proof based on Sylow’s theorem. Another short proof, which follows from general facts about subnormal subgroups of finite groups, was proposed by Wielandt [9]. The nontrivial part of this theorem is the ‘if’ part. Its value is clear from the following considerations. For a conjugacy class $D$ of a group $G$ the condition $D\subseteq \mathrm{O}_p(G)$ is equivalent to the condition that $\langle D\rangle$ is a $p$-subgroup. Since $D$ can be arbitrarily large, the verification of the latter condition can be difficult. The Baer-Suzuki theorem shows that it suffices to test the subgroups generated by subsets of $D$ having bounded cardinality, namely, cardinality 2. This theorem both presents a test for a group not to be simple and serves as a tool for local analysis. It is used in the theory of solvable groups (see [2]) and has also played an important role in the classification of finite simple groups.1 The Baer-Suzuki theorem has several equivalent formulations. Along with the above, the most well-known formulation gives a completely similar criterion for an element to belong to the nilpotent radical (Fitting subgroup) of a group. Generalizations and analogues of this theorem, including those for infinite groups, were considered by many authors; see, for example, [8], [9] and [12]–[29]. Analogues of this kind are often also called Baer-Suzuki theorems or theorems of Baer-Suzuki type. In particular, Gordeev, Grunewald, Kunyavskii and Plotkin [14]–[16] and, independently, Flavell, Guest and Guralnick [17], [12] proved the Baer-Suzuki theorem for the solvable radical, which states that, if any four elements in a conjugacy class in a finite group generate a solvable subgroup, then the whole class is contained in the solvable radical of this group. Moreover, it is impossible to reduce this number not only to two, but even to three elements, and, in this sense, the Baer-Suzuki theorem for the solvable radical established in the works cited is sharp. The question about other radicals of a finite group for which the Baer-Suzuki theorem holds was posed in [15], Problem 1.16. In group theory important results can often be obtained if, instead of a prime number $p$, one considers some set $\pi$ of primes and instead of $p$-subgroups, considers $\pi$-subgroups (cf., for example, the classical Sylow theorem [30] and Hall theorem [31]; also see [3], Theorems 1.2.9 and 6.4.3). Therefore, it is natural to try to obtain the Baer-Suzuki theorem for a $\pi$-radical by increasing the number of conjugate elements generating the subgroups tested in comparison to the original Baer-Suzuki theorem if necessary. In [32], Theorem 1.2, the authors proved that for any $\pi\subseteq \mathbb{P}$ there always exists a nonnegative integer $m=m(\pi)$ such that for an arbitrary group $G$
$$
\begin{equation*}
\mathrm{O}_\pi(G)=\{x\in G\mid \langle x^{g_1}, \dots, x^{g_m}\rangle\text{ is a } \pi\text{-group for any } g_1,\dots, g_m\in G\}.
\end{equation*}
\notag
$$
By Definition 1.15 in [15] the least $m$ of this kind is called the Baer-Suzuki width of the class of $\pi$-groups and is denoted by $\operatorname{BS}(\pi)$. Moreover, it was proved (see [32], Theorem 1.3) that
$$
\begin{equation*}
\begin{gathered} \, \textit{if }\ \varnothing\ne\pi\ne \mathbb{P} \ \textit{ and }\ r=r(\pi)=\min\mathbb{P}\setminus\pi, \\ \textit{then }\ r-1\leqslant \operatorname{BS}(\pi)\leqslant \max\{11, 2(r-2)\}. \end{gathered}
\end{equation*}
\notag
$$
The lower bound for $\operatorname{BS}(\pi)$ in the last inequality can be obtained from the following assertion:2 if $r\geqslant 3$, then any $r-2$ transpositions in the symmetric group $S_r$ generate a $\pi$-subgroup, whereas $\mathrm{O}_\pi(S_r)=1$ (see [32], Proposition 1.1). This estimate shows that the Baer-Suzuki width of the class of $\pi$-groups can be arbitrarily large for suitable $\pi$. At the same time, this width is finite for any $\pi$,3 and the following question arises: what is the precise value of the width? A conjecture has been made (see [32], Conjecture 1) that $\operatorname{BS}(\pi)$ in most cases coincides with the lower bound $r-1$. More precisely, there is a question of whether or not the following assertion holds. Conjecture 1 (see [32]). Let $\pi$ be a proper subset of the set of primes and $r$ be the least prime number not in $\pi$. Then Were Conjecture 1 confirmed, it could be viewed as a sharp Baer-Suzuki theorem for the $\pi$-radical: it would mean that $\operatorname{BS}(\pi)=r-1$ for $r\geqslant 5$. For $r=2$ the conjecture holds, as follows from a result due to Tyutyanov [29] (also see [24], Theorem 1); here, if $\pi\ne\varnothing$, then $\operatorname{BS}(\pi)=2$. We can see how accurate could be the estimate in Conjecture 1 for $r=3$ from the following: it is clear that $\operatorname{BS}(\pi)\geqslant 2$; for $\pi=\{2\}$ we have $r=3$ and $\operatorname{BS}(\pi)=2<3$ by the Baer-Suzuki theorem; at the same time, there are examples of sets $\pi$ with $r=3$ such that $\operatorname{BS}(\pi)\ne 2$ (see [24], Example 2). The results in [24] reduce the conjecture to the investigation of so-called almost simple groups. To formulate a statement about almost simple groups which would ensure Conjecture 1, recall the notation introduced by Guralnick and Saxl [33] and the authors of [32]. Definition 1. Let $L$ be a non-Abelian simple group, let $r$ be a prime divisor of its order, and let $x\in\operatorname{Aut}(L)$ be a nontrivial automorphism of $L$. We identify $L$ with the subgroup $\operatorname{Inn}(L)$ in $\operatorname{Aut}(L)$. It can be seen from the definition that, if $\alpha(x,L)$ and $\beta_r(x,L)$ are defined, then Taking the known reduction to almost simple groups and the results for $r=2$ (see [29] and [24]) into account, we see that Conjecture 1 would be established if we could prove the following. Conjecture 2 (see [32]). Let $r$ be an odd prime number. Let $s$ be a prime divisor of the order of a finite non-Abelian simple group $L$ such that $s=r$ if $r$ divides $|L|$ and $s>r$ otherwise. Then for any automorphism $x$ of $L$ of prime order Here the simplicity of the order of $x$ replaces in an equivalent way the a priori more general assumption $x\ne1$. Conjecture 2 is confirmed for $L$ an alternating group (see [32], Proposition 1.5) or a sporadic group (see [34], Theorem 1). The main result of this paper is that Conjecture 2 holds for two series of simple classical groups of Lie type, namely, the simple linear and simple unitary groups. Theorem 1. Let $r$ be an odd prime number. Let $L=L_n(q)$ or $L=U_n(q)$ be a finite simple linear group or a simple unitary group, and let $x\in \operatorname{Aut}(L)$ be an element of prime order. Let $s$ be a prime divisor of the order of the group $L$ such that $s=r$ if $r$ divides $|L|$ and $s> r$ otherwise. Then for every automorphism $x$ of $L$ of prime order Conjecture 2 has not yet been proved or disproved for symplectic groups, orthogonal groups, or exceptional groups of Lie type. However, it is known (see [32], Theorem 1.4) that for every non-Abelian finite simple group $L$ whose order is divisible by $r\in\mathbb{P}$, the quantity $\beta_r(x,L)$ is bounded by the number $\max\{11,2(r- 2)\}$, which depends only on $r$ and is independent of the group $L$ itself (in contrast to $\alpha(x,L)$, which can be arbitrarily large for a suitable group $L$ and an automorphism $x$ of $L$). In the proof of Theorem 1 we follow in general the scheme of reasoning used for classical groups in [32]. The proof is carried out by simultaneous induction for $L_n(q)$ and $U_n(q)$ on the dimension $n$ of the vector space associated with the group. First, the cases when $n\leqslant 3$ or $n=4$ and $q=2,3,5$ are considered. The general case is divided into two subcases. Either it can be shown that an element $x$ of $L$ induces a nonidentity automorphism on some $x$-invariant section isomorphic to $L_k(q')$ or $U_k( q')$ for $k<n$ such that its order is divisible by $s$, after which the induction assumption is applied, or it turns out that the dimension $n$ itself is bounded in terms of $r$. In the latter case the estimate in [33] (also see Table 1), bounding $\alpha(x,L)$, and therefore also $\beta_s(x,L)$, by a function of $n$, often, although by no means always, turns out to be sufficient. The situations occurring for an automorphism $x$ fall into two large blocks: either $x$ is an inner diagonal automorphism (the possible cases are listed in Lemma 10) or $x$ turns out to be a field, a graph, or a graph-field automorphism modulo the group of inner-diagonal automorphisms. Unfortunately, the results of [33] are often insufficient for a sharper estimate in comparison with Theorem 1.4 in [32], which is required in Conjecture 2, and one has to analyze a large number of quite difficult exceptional cases. Lemmas 13 and 14 are devoted to the analysis of the most fundamental cases for an inner-diagonal automorphism $x$. It is also difficult to analyze the cases in Lemma 16, when $x$ is a graph involution, especially for a relatively small dimension, when $\alpha(x,L)$ is too large in comparison with $n$. Here it turns out that the groups $L_4(q)$ and $U_4(q)$ can conveniently be viewed as the orthogonal groups $O^\pm_6(q)$, and the sections on which $x$ induces a nonidentity automorphism turn out to be isomorphic to $O_4^- (q)\cong L_2(q^2)$ and, in the most delicate case, to $O_5(q)\cong S_4(q)$ (Lemmas 17 and 26). Using Theorem 1 and results of [24], [32] and [34], we obtain the following assertion, which confirms Conjecture 1 in part. Theorem 2. Let $\pi$ be a set of primes, and $r$ be the least prime number outside $\pi$. Set Then for every finite group $G$ each non-Abelian composition factor of which is isomorphic to a sporadic, an alternating, a linear, or a unitary group. The authors hope that a further study of Conjecture 2 will make it possible to prove the sharp Baer-Suzuki theorem for the $\pi$-radical in its entirety. One can also expect that sharp bounds for $\beta_r(x,L)$ will be also of interest outside the framework of theorems of Baer-Suzuki type. Anyway, the estimates for $\alpha(x,L)$ obtained in [33] and their subsequent refinements find very wide applications. § 2. Preliminaries2.1. Group-theoretic notation, a reduction to almost simple groups and other general lemmasWe use the standard notation of group theory (see [1], [3] and [35]), as well as the notation from the Atlas of finite groups [36]. Thus, if $G$ is a group and $x,g\in G$, then $x^g=g^{-1}xg$ and $[x,g]=x^{-1}x^ g$. Also, given a group $G$, we denote by $\mathrm{O}_\infty(G)$, $\operatorname{Z}(G)$, $\operatorname{F}(G)$, $\operatorname{F}^*(G)$, $\Phi(G)$, $G'=[G,G]$ and $G^{\infty}=\mathrm{O}^{\infty}(G)$ the solvable radical, centre, Fitting subgroup, generalized Fitting subgroup, Frattini subgroup, commutator subgroup and the last member of the series of commutator subgroups (solvable residual) of $G$, respectively. Also, given a positive integer $n$ and a group $G$, we denote the sets of prime divisors of $n$ and $|G|$ by $\pi(n)$ and $\pi(G)$, respectively. For a subset $\pi$ of the set $\mathbb{P}$ of all prime numbers we put $\pi'=\mathbb{P}\setminus\pi$. Let $\mathrm{O}^\pi(G)$ be the $\pi$-residual of $G$, that is, the minimal normal subgroup with respect to inclusion such that the corresponding quotient group is a $\pi$-group. It can also be viewed as the subgroup generated by all $\pi'$-subgroups. The set of nontrivial elements of $G$ is denoted by $G^\sharp$. In accordance with [24], we use the following notation. Let $\pi$ be some set of primes and let $m$ be a nonnegative integer. For a group $G$ we write $G\in{{\mathscr B}{\mathscr S}}_{\pi}^{m}$ if
$$
\begin{equation*}
\mathrm{O}_\pi(G)=\{x\in G\mid \langle x^{g_1}, \dots, x^{g_m}\rangle\text{ is a } \pi\text{-group for every } g_1,\dots, g_m\in G\}.
\end{equation*}
\notag
$$
The following lemma describes the structure of a minimal counterexample to Conjecture 1. Lemma 1 (see [24], Lemma 7). Let $\mathscr{X}$ be a class of finite groups which is closed with respect to taking normal subgroups, homomorphic images and extensions and contains all $\pi$-groups. Assume that $\mathscr{X}\nsubseteq\mathscr{BS}_{\pi}^{m}$ for some integer $m\geqslant 2$, and let $G\in\mathscr{X}\setminus\mathscr{BS}_{\pi}^{m}$ be a group such that its order is the least possible. Then $G$ contains a subgroup $L$ and an element $x$ such that 1) $L\trianglelefteq G$; 2) $L$ is a non-Abelian simple group; 3) $L$ is not a $\pi$- or $\pi'$-group; 4) $\operatorname{C}_G(L)=1$; 5) any $m$ elements conjugate to $x$ generate a $\pi$-group; 6) $x$ has a prime order which belongs to $\pi$; 7) $G=\langle x, L\rangle$. Lemma 2 (see [24], Theorem 1). If $2\not\in\pi$, then $\mathscr{BS}_\pi^2$ coincides with the class of all finite groups. Lemma 3 (see [37], Lemma 15). Let $G$ be a group and let $x\in\operatorname{Aut}(G)$ be an automorphism whose order is equal to a power of a prime number $p$. Set $M=\operatorname{C}_G(x)$ and assume that $p$ divides $|G:M|$ and either $M=\operatorname{N}_G(M)$ or $\operatorname{Z}(M) =1$. Then $x$ normalizes, but does not centralize, some subgroup of $G$ conjugate to $M$. The following two obvious lemmas are used in the inductive reasoning. Lemma 4. Let $L$ be a non-Abelian simple group, and let $x\in \operatorname{Aut}(L)^\sharp$. Assume that $x$ preserves some subgroup $H$ and a normal subgroup $N$ of $H$ and induces a nonidentity automorphism $\overline{x}$ of $\overline{H}=H/N$. Let the group $\overline{H}$ be a non-Abelian simple group, and let its order be divisible by a prime number $r$. Then Lemma 5. Let $L$ be a non-Abelian finite simple group, and let $x,y\in \operatorname{Aut}(L)^\sharp$. Assume that $y\in \langle x^{g_1},\dots,x^{g_k}\rangle$ for some $g_1,\dots,g_k\in L$. Then for every prime divisor $r$ of the order of $L$. 2.2. Notation and preliminaries concerning classical groupsFor classical groups we use the notation from [38] and [36]. In using the Lie notation we follow [39]. Throughout what follows we assume that $q$ is a power of a fixed prime $p$ and $F=\mathbb{F}_q$ is a finite field of order $q$. We recall that for a vector space $V$ over a field $F$ the symbol $V^\sharp$ denotes the set of all nonzero vectors in this space. For a vector $v\in V$ and an element $g\in \operatorname{GL}(V)$, the image of $v$ under the action of $g$ of $v$ is denoted by $vg$. If ${v\in V}$ and $g\in \operatorname{GL}(V)$, then $[v,g]=vg-v$. For a subgroup $G\leqslant\operatorname{GL}(V)$ we set $[v,G]=\langle [v,g]\mid g\in G\rangle_F$. We note that, since $[v,G]=\langle vg-vh\mid g,h\in G\rangle_F$, the subspace $[v,G]$ of $V$ is $G$-invariant. If a space $V$ is equipped with a nondegenerate bilinear, Hermitian, or quadratic form, then we say that a subspace $U$ of $V$ is nondegenerate if the restriction of this form to $U$ is nondegenerate, and we say that $U$ is totally isotropicc if the restriction of this form to $U$ vanishes identically. To denote the image in the group $\operatorname{PGL}_n(q)$ of a matrix $(a_{ij})\in\operatorname{GL}_n(q)$ defined in terms of its entries we use the same entries enclosed in square brackets: $[a_{ij}]$. In what follows we denote by $\tau$ the automorphism of the group $\operatorname{GL}_n(q)$ acting by the rule where ${}^\top$ denotes matrix transposition. We denote the automorphism of the group $\operatorname{GL}_n(q)$ acting according to the rule by $\varphi_{p^m}$. We denote the automorphisms of the groups $\operatorname{SL}_n(q)$ and $\operatorname{PGL}_n(q)$ and $\operatorname{PSL}_n(q)$ induced by $\tau$ and $\varphi_{p^m}$ by the same symbols $\tau$ and $\varphi_{p^m}$. In particular, if $q=p^k$ and $r$ divides $k$, then $\varphi_{q^{1/r}}$ is a field automorphism of order $r$.As a rule, we identify the group $\operatorname{PSU}_n(q)$ with $\mathrm{O}^{p'}(\operatorname{C}_{\operatorname{PGL}_n(q^2)}(\tau\varphi_{q}))$. As usual, in our considerations of linear and unitary groups we often use the unified notation $\operatorname{PSL}_n^\varepsilon(q)$, where $\varepsilon\in\{+,-\}$, by setting
$$
\begin{equation*}
\operatorname{PSL}_n^{+}(q)=\operatorname{PSL}_n(q)\quad\text{and} \quad \operatorname{PSL}_n^{-}(q)=\operatorname{PSU}_n(q).
\end{equation*}
\notag
$$
The groups $\operatorname{PGL}^\varepsilon_n(q)$, $\operatorname{GL}^\varepsilon_n(q)$ and $\operatorname{SL}^\varepsilon_n(q)$ are defined similarly. Following [36], we also use the shorthand notation
$$
\begin{equation*}
L_n(q)=L_n^+(q)=\operatorname{PSL}_n(q)\quad\text{and} \quad U_n(q)=L_n^-(q)=\operatorname{PSU}_n(q).
\end{equation*}
\notag
$$
In using the Lie notation we set
$$
\begin{equation*}
A_{n-1}(q)=A_{n-1}^+(q)=\operatorname{PSL}_n(q)\quad\text{and} \quad {}^2A_{n-1}(q)=A_{n-1}^-(q)=\operatorname{PSU}_n(q).
\end{equation*}
\notag
$$
The symbol $E_t$ denotes the $ t\times t $ identity matrix, and the symbol $A\otimes B$ denotes the Kronecker product of the matrices $A$ and $B$. As applied to automorphisms of groups of Lie type, we use the following terminology, which is close to that in [39] and slightly different from that in [40]. The notions of an inner diagonal automorphism in [40] and [39] coincide and do not differ from the one we use. In [39], Definition 2.5.10, the subgroups $\Phi_K$ and $\Gamma_K$ were introduced in the automorphism group of an arbitrary group of Lie type $K$. For groups of Lie type we usually use the letter $L$, so the corresponding subgroups are denoted by $\Phi_L$ and $\Gamma_L$. These subgroups can be identified with the groups of field and graph automorphisms of $L$ in the sense of [40], respectively. We denote the group of inner diagonal automorphisms of the group $L$ by $\widehat{L}$. As concerns linear and unitary groups, let us say more about their automorphisms. In the case of $L=\operatorname{PSL}^\varepsilon_n(q)$ we have $\widehat{L}=\operatorname{PGL}^\varepsilon_n(q)$. Lemma 6 (see [39], Theorem 2.5.12). Let ${\mathbb F}_q$ be a field of order $q=p^k$, let $n\geqslant 2$ and $L=A_{n-1}^\varepsilon(q)$. Then $\operatorname{Aut}(L)$ is a split extension of the group $\widehat{L}$ by means of the Abelian group $\Phi_L\Gamma_L$. Here $\Phi_L\Gamma_L\cong \Phi_L\times \Gamma_L$ and the groups $\Phi_L$ and $\Gamma_L$ are cyclic. If $\varepsilon=+$, then $|\Phi_L|=k$, and $|\Gamma_L|=2$ for $n>2$ and $|\Gamma_L|=1$ for $n=1$. If $\varepsilon=-$, then $|\Phi_L|=2k$ and $|\Gamma_L|=1$. For a linear simple group $L$ an automorphism $x\in\operatorname{Aut}(L)\setminus\widehat{L}$ is said to be
Let $L$ be a unitary simple group. Let $x\in\operatorname{Aut}(L)\setminus\widehat{L}$. An automorphism $x$ is said to be
We note that the concepts of field and graph-field automorphisms $x$ of the group $L$ modulo $\widehat{L}$ defined in this way coincide with the concepts of field and graph-field automorphisms in [39], Definition 2.5.13, respectively, in the case when $\langle x\rangle\cap \widehat{L}=1$ (in particular, when $x$ has a prime order). The following lemma contains some information from [33] about the parameter $\alpha(x,L)$, where $x$ is an automorphism of prime order of a classical simple group $L$. Lemma 7. Let $L$ be a simple classical group and $x\in \operatorname{Aut}L$ be an element of prime order. Then $\alpha(x,L)$ is as specified in the last column of Table 1. The proof of the following assertion can be extracted from the general proof of Theorems 4.1–4.4 in [33]. More precisely, this assertion was proved in [33] for linear groups on p. 534, for unitary groups on p. 536, for symplectic ones on p. 538, and for orthogonal ones on p. 539. Lemma 8. Let $L$ be a simple classical group and $x\in \widehat{L}$ be a nontrivial element of prime order induced by an irreducible semisimple element of the similarity group corresponding to $L$. Then $\alpha(x,L)\leqslant 3$. Lemma 9. Let $L=L_2(q)$ and let $r\in \pi(L)$ be an odd number not dividing $q$. Then $\beta_r(x,L)=2$ for any involution $x\in\widehat{L}$. Proof. It follows from the hypotheses that $r$ divides $q-\varepsilon$ for some $\varepsilon=\pm1$. We assume first that $x\in L$. Since all involutions in $L$ are conjugate (if $q$ is even, then this follows from Sylow’s theorem, the equation $L=\widehat{L}$ and the well-known fact that the Sylow normalizer of a Sylow 2-subgroup of $L$ is a Frobenius group with kernel of order $q$ and the complement of order $q-1$ which acts regularly on the nonidentity 2-elements of the Sylow 2-subgroups; for odd $q$, see [39], Theorem 4.5.1 and Table 4.5.1), we may assume that $x$ is contained in a dihedral subgroup $D$ of order $q-\varepsilon$ or $2(q-\varepsilon)$, depending on the parity of $q$, and inverts all elements of the unique subgroup $\langle y\rangle$ of order $r$ in $D$ (see [5], Assertion 1.6.9). Then the element has order $r$ and $\beta_r(x,L)=2$, as stated.
If $x\notin L$, then $q$ is odd. All involutions in $\widehat{L}\setminus L$ are conjugate by elements of $L$, as follows from [39], Theorem 4.5.3 and Table 4.5.2, and are conjugate to the involution in the dihedral subgroup $\widehat{D}$ of order $2(q-\varepsilon)$ that inverts the elements of the unique subgroup $\langle y\rangle$ of order $r$ in $\widehat{D}$. Arguing as in the previous case, we obtain the required assertion. This completes the proof of the lemma. Lemma 10. Let $\Delta=\Delta(V)$ be the group of similarities of a finite-dimensional vector space $V$ equipped with a nondegenerate or trivial bilinear or Hermitian form. Let $x\in\Delta$ be an element of primary order whose image in the group $\Delta(V)/\operatorname{Z}(\Delta(V))$ has a prime order. Then one of the following cases holds: 1) $x$ is unipotent and stabilizes a subspace of dimension $1$; 2) the form is trivial, and $x$ is semisimple and preserves two complementary nontrivial subspaces; 3) the form is nondegenerate, and $x$ is semisimple and preserves a proper nondegenerate nontrivial subspace and its orthogonal complement; 4) the form is nondegenerate, its Witt index is $\frac{1}{2}\dim V$ (in particular, the dimension of $V$ is even), and $x$ is semisimple and preserves a maximal totally isotropic subspace; 5) $x$ is semisimple and acts irreducibly on $V$. Proof. The element $x$ is either unipotent or semisimple. In the first case let $l$ be the least positive integer such that $(x-1)^l=0$. Then $(x-1)^{l-1}\ne 0$, and there exists a vector $v\in V$ such that Thus, $ux=u$ for the nonzero vector $u=v(x-1)^{l-1}$. Hence $\langle u\rangle$ is a one-dimensional $x$-invariant subspace, that is, 1) holds.
Now let the element $x$ be semisimple. If it is irreducible, then 5) holds. If the form is trivial, then 2) holds by Maschke’s theorem (see [35], Theorem (1.9)). Therefore, we assume that $x$ is reducible and the form is nondegenerate. Let $U$ be a proper nontrivial $x$-invariant subspace. Then $x$ stabilizes the orthogonal complement $U^\perp$. If $U$ is nondegenerate, then $U\cap U^\perp=0$ and $U^\perp$ is the complement to $U$ in $V$, and thus 3) holds. If $U$ is not nondegenerate, then its radical ${\operatorname{rad}U=U\cap U^\perp}$ is a proper nontrivial totally isotropic $x$-invariant subspace. Thus, $U$ can be assumed to be totally isotropic. If $\dim U<\frac{1}{2}\dim V$, then $U<U^\perp$, $U=\operatorname{rad}U^\perp$, and by Maschke’s theorem $U^\perp$ has an $x$-invariant complement $W$ to $U$, and $W\cong U^\perp/U$ is nondegenerate, that is, 3) holds. On the other hand, if $\dim U=\frac{1}{2}\dim V$, then 4) holds. This completes the proof of the lemma. Lemma 11 (see [33], Lemma 2.2). Let $L$ be a simple group of Lie type and let $G = \widehat{L}$ and $x\in G^\sharp$. Then the following assertions hold. 1) If the element $x$ is unipotent, then let $P_1$ and $P_2$ be different parabolic maximal subgroups of $G$ containing a common Borel subgroup, and let $U_1$ and $U_2$ be the unipotent radicals of the subgroups $P_1$ and $P_2$, respectively. Then the element $x$ is conjugate to an element of $P_i\setminus U_i$ for $i = 1$ or $2$. 2) If $x$ is semisimple, then assume that $x$ is contained in some parabolic subgroup of the group $G$. If the rank of $L$ is at least 2, then there is a parabolic maximal subgroup $P$ with Levi complement $J$ such that $x$ is conjugate to some element of $J$ which does not centralize any of the (possibly solvable) Levi components of $J$. Lemma 12. Let $\mathbb{F}_q$ be a field of odd characteristic $p$, and let $\beta\in\mathbb{F}^*_q$. Let $\mathbb{F}_{q_0}$ denote the subfield of $\mathbb{F}_q$ generated by $\beta^3$. Consider the elements of $\operatorname{SL}_2(q)$. Then either the subgroup $H=\langle x,y \rangle$ of $\operatorname{SL}_2(q)$ is isomorphic to $\operatorname{SL}_2(q_0)$ or $q_0=9$ and $H$ is isomorphic to a subgroup of $\operatorname{SL}_2(q_0)$ whose image in $\operatorname{PSL}_2(q_0)$ is isomorphic to $A_5$, and $H$ itself contains a subgroup isomorphic to $\operatorname{SL}_2(3)$. In particular, $H$ always contains a subgroup isomorphic to $\operatorname{SL}_2(p)$, which in turn contains the matrix which is the unique element of $\operatorname{SL}_2(q)$ of order $2$. Proof. Consider the matrix
$$
\begin{equation*}
g=\begin{pmatrix} 1 & {} \\ {} & \beta^2 \end{pmatrix}\in \operatorname{GL}_2(q)
\end{equation*}
\notag
$$
and the subgroup $H^g=\langle x^g,y^g\rangle\cong H$. From direct calculations we see that
$$
\begin{equation*}
x^g=\begin{pmatrix} 1 & {} \\ 1 & 1 \end{pmatrix} \quad\text{and}\quad y^g=\begin{pmatrix} 1 & \beta^3 \\ {} & 1 \end{pmatrix}.
\end{equation*}
\notag
$$
Now it follows from [3], Ch. 2, Theorem 8.4, that the subgroup $H^g$, and so also $H$, is as required in the lemma.
This completes the proof of the lemma. Lemma 13. Let $V$ be a space with nondegenerate Hermitian form over the field $F=\mathbb{F}_{q^2}$, which has an odd dimension $\dim V=n\geqslant 5$. Then for a unipotent element $x\in\operatorname{SU}(V)$ of prime order one of the following cases holds: 1) $x$ stabilizes a nondegenerate subspace of dimension $1$; 2) $x$ stabilizes a maximal totally isotropic subspace and induces a nonidentity transformation of it; 3) $q$ is odd, and there is an element $x^g$ conjugate to $x$ such that, in the subgroup $\langle x, x^g\rangle$, some involution stabilizes a maximal totally isotropic subspace and induces a nonscalar transformation of it. Proof. For $\alpha\in F=\mathbb{F}_{q^2}$ we write $\overline{\alpha}=\alpha^q$.
Representatives of the conjugacy classes of unipotent elements of the group $\operatorname{SU}(V)\cong\operatorname{SU}_n(q)$ were found in [41], Proposition 2.2. As is known, up to conjugation by an element of $\operatorname{GU}(V)$, to every unipotent element $x$ there corresponds a unique (up to a permutation of the summands) partition and a decomposition of the space $V$ into the sum of pairwise orthogonal nondegenerate $x$-invariant subspaces $V_1,\dots,V_s$ such that $\dim V_t=n_t$ for $t\in\{1,\dots,s\}$ and the action of $x$ on $V_t$ is defined as follows.We fix $\beta,\gamma\in F^*$ in such a way that $\beta+\overline{\beta}=0$ and $\gamma+\overline{\gamma}=-1$. If $n_t=2k$ for some $k=k(t)$, then there is an ordered basis in $V_t$, such that
$$
\begin{equation*}
(e_i^t,e_j^t)=(f_i^t,f_j^t)=0, \quad (e_i^t,f_j^t)=\delta_{ij} \quad\text{for all } i,j=1,\dots,k,
\end{equation*}
\notag
$$
and the following equations hold:
$$
\begin{equation*}
\begin{aligned} \, e^t_ix&=e^t_i+\dots+e^t_k+\beta f^t_k \quad\text{for all } i=1,\dots,k, \\ f^t_ix&=f^t_i- f^t_{i-1} \quad\text{for all } i=2,\dots,k, \\ f^t_1x&=f^t_{1}. \end{aligned}
\end{equation*}
\notag
$$
If $n_t=2k+1$, $k=k(t)$, then there is an ordered basis in $V_t$, such that
$$
\begin{equation*}
\begin{gathered} \, (e_i^t,e_j^t)=(f_i^t,f_j^t)=(e_i^t,d^t)=(f_i^t,d^t)=0, \\ (d^t,d^t)=1,\qquad (e_i^t,f_j^t)=\delta_{ij} \quad\text{for all } i,j=1,\dots,k, \end{gathered}
\end{equation*}
\notag
$$
and the following equations hold:
$$
\begin{equation*}
\begin{aligned} \, e^t_ix&= e^t_i+\dots+e^t_k+d^t+\gamma f^t_k \quad\text{for all } i=1,\dots,k, \\ d^tx&=d^t- f^t_{k}, \\ f^t_ix&=f^t_i- f^t_{i-1} \quad\text{for all } i=2,\dots,k, \\ f^t_1x&=f^t_{1}. \end{aligned}
\end{equation*}
\notag
$$
It is clear that case 1), that is, the existence of a one-dimensional $x$-invariant nondegenerate subspace, is equivalent to the equality $n_t=1$ for some $t\in\{1,\dots,s\}$. Therefore, we assume that $n_t>1$ for all $t$. Since $n$ is odd, some $n_t$ is too, and it satisfies $n_t\geqslant 3$. Now it follows from the fact that $x$ is of order $p$, where $p$ is the characteristic of the field, that $p>2$. In fact, for $p=2$ the element $x^2$ acts nontrivially on $V_t$:
$$
\begin{equation*}
\begin{aligned} \, e^t_kx^2 &=(e^t_k+d^t+\gamma f_k^t)x=e_k^t+d^t+\gamma f_k^t+d^t+f_k^t+\gamma(f_k^t+f) \\ &=e_k^t+f_k^t+\gamma f\ne e_k^t, \end{aligned}
\end{equation*}
\notag
$$
where $f=0$ for $k=1$ and $f=f^t_{k-1}$ otherwise.
Denote by $U$ the subspace spanned by all the $f^t_i$, where $t=1,\dots,s$ and $i=1,\dots, k(t)$. It is clear that $U$ is totally isotropic and $x$-invariant. Assume that exactly one summand in the sum $n_1+\dots+n_s$ is odd. Then $U$ is a maximal totally isotropic subspace. In this case, if $k=k(t)>1$ for some $t\in\{1,\dots,s\}$, then it follows from the equality that $x$ acts nontrivially on $U$, that is, case 2) of the lemma holds. Therefore, we can assume that Since $n\geqslant 5$, we see that $s\geqslant 2$ and $n_1=2$. Consider an element $g\in \operatorname{GL}(V)$ such that
$$
\begin{equation*}
\begin{gathered} \, e^1_1g=\beta^{-1}f^1_1, \qquad f^1_1g=-\beta e_1^1, \\ e^t_1g=e^t_1, \quad f^t_1g=f^t_1 \quad\text{for } t>1, \\ d^sg=d^s, \end{gathered}
\end{equation*}
\notag
$$
where, as above, $\beta\in F^*$ and $\beta+\overline{\beta}=0$. Then $g\in\operatorname{SU}(V)$, as can readily be seen, and the matrices of the elements $x$ and $x^g$ in the basis
$$
\begin{equation*}
e^1_1,f^1_1,e^2_1,f^2_1,\dots, e^{s-1}_1,f^{s-1}_1,e^{s}_1,d^s,f^{s}_1
\end{equation*}
\notag
$$
have the form
$$
\begin{equation*}
{ \begin{pmatrix} 1 & \beta & & & & & & & & \\ & 1 & & & & & & & & \\ & & 1 & \beta & & & & & & \\ & & & 1 & & & & & & \\ & & & & \ddots & & & & & \\ & & & & & 1 & \beta & & & \\ & & & & & & 1 & & & \\ & & & & & & & 1 & 1 & \gamma \\ & & & & & & & & 1 & -1 \\ & & & & & & & & & 1 \end{pmatrix}}\ \ \text{and} \ \ { \begin{pmatrix} 1 & & & & & & & & & \\ -\beta^2 & 1 & & & & & & & & \\ & & 1 & \beta & & & & & & \\ & & & 1 & & & & & & \\ & & & & \ddots & & & & & \\ & & & & & 1 & \beta & & & \\ & & & & & & 1 & & & \\ & & & & & & & 1 & 1 & \gamma \\ & & & & & & & & 1 & -1 \\ & & & & & & & & & 1 \end{pmatrix}, }
\end{equation*}
\notag
$$
respectively, where the elements $\beta$ and $\gamma$ of $F$ are chosen as described above. Further, consider the subgroup Then $H$ is a subgroup of the group $K$ of block-diagonal matrices of the form where $A\in\operatorname{SL}_2(q^2)$ and $B$ is an arbitrary unitriangular matrix of order $n-2$ over $\mathbb{F}_{q^2}$. Consider also the epimorphism
$$
\begin{equation*}
\overline{\phantom{G}}\colon K\to \operatorname{SL}_2(q^2)
\end{equation*}
\notag
$$
acting by the formula
$$
\begin{equation*}
\begin{pmatrix} A & {} \\ {} & B \end{pmatrix}\mapsto A.
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
\overline{(x^g)}{}^{-1}= \begin{pmatrix} 1 & {} \\ \beta^2 & 1 \end{pmatrix}, \qquad \overline{x}= \begin{pmatrix} 1 & \beta \\ {} & 1 \end{pmatrix}.
\end{equation*}
\notag
$$
Therefore, as follows from Lemma 12, the subgroup
$$
\begin{equation*}
\overline{H}=\overline{\langle(x^g)^{-1},{x} \rangle} \leqslant\overline{K}=\operatorname{SL}_2(q^{{2}})
\end{equation*}
\notag
$$
contains a unique element $\overline{y}$ of order 2 of the group $\operatorname{SL}_2(q^{{2}})$, and this is the matrix Consider its pre-image $y$ in $H$, which is also of order $2$. Then the matrix of $y$, in the basis
$$
\begin{equation*}
e^1_1,f^1_1,e^2_1,f^2_1,\dots, e^{s-1}_1,f^{s-1}_1,e^{s}_1,d^s,f^{s}_1,
\end{equation*}
\notag
$$
has the form5
$$
\begin{equation*}
\begin{pmatrix} -1 & & & & & & & & & \\ & -1 & & & & & & & & \\ & & 1 & * & & & & & & \\ & & & 1 & & & & & & \\ & & & & \ddots & & & & & \\ & & & & & 1 & * & & & \\ & & & & & & 1 & & & \\ & & & & & & & 1 & * & * \\ & & & & & & & & 1 & * \\ & & & & & & & & & 1 \end{pmatrix},
\end{equation*}
\notag
$$
and $f_1^1,f_1^2,\dots, f_1^{s-1}, f^s_1$ are eigenvectors of $y$:
$$
\begin{equation*}
f_1^1y=-f^1_1,\qquad f_1^ty=f_1^t \quad\text{for } t>1.
\end{equation*}
\notag
$$
Therefore, the maximal totally isotropic subspace is invariant under the action of $y$, and $y$ induces a nonscalar transformation of $U$. Thus, the case 3) holds.
Finally, let us show that, if there is more than one odd integer among $n_1,\dots, n_s$, then the case 2) holds again. Without loss of generality we can assume that $n_1,\dots, n_{s'}$ are odd and $n_{s'+1},\dots, n_s$ are even and, since $n$ is odd, it follows that $s'$ is also odd. Consider the orthogonal complement $U^\perp$ of $U$. It is $x$-invariant, and because $\operatorname{codim} U^\perp=\dim U=n-\dim U^\perp$ coincides with the number of vectors $e_i^t$, $t=1,\dots,s$, $i=1,\dots, k_t$, we have
$$
\begin{equation*}
U^\perp=\langle d^{t'}, f_i^t\mid t'=1,\dots,s', \ t=1,\dots,s,\ i=1,\dots, k_t\rangle_F.
\end{equation*}
\notag
$$
Consider the $(s'-1)/2$ pairs of vectors $(d^1,d^2),\dots,(d^{s'-2},d^{s'-1})$. Fix a pair $(d^{t'},d^{t'+1})=(d,d')$ and construct a pair $(a,b)=(a^{t'},b^{t'})$ according to the following rule. We take $\mu\in \mathbb{F}_{q^2}$ such that $\mu\overline{\mu}=-1$ and write Since $(d,d')= 0$ and $(d,d)=(d',d')=1$, we have and similarly $(b,b)=0$. We note that that is, $x$ induces the identity transformation of the quotient space
$$
\begin{equation*}
(\langle d,d'\rangle_F+U)/U=(\langle a,b\rangle_F+U)/U.
\end{equation*}
\notag
$$
It is now clear that is a maximal totally isotropic subspace of the space $V$, which is also $x$-invariant. Here
$$
\begin{equation*}
a^1x=(d^1+\mu d^2)x=d^1+\mu d^2 - f^1_{k(1)}-\mu f^1_{k(2)}\ne d^1+\mu d^2= a^1,
\end{equation*}
\notag
$$
that is, $x$ acts nontrivially on $W$ and case 2) holds.
This completes the proof of Lemma 13. Lemma 14. Let one of the following statements hold: Let the sign $\varepsilon\in\{+,-\}$ indicate which of these assertions holds, let $F=\mathbb{F}_q$ for $\varepsilon=+$ and $F=\mathbb{F}_{q^2}$ for $\varepsilon=-$, and let $n=\dim V$. Assume that the subgroups $L_U\leqslant \operatorname{GL}^\varepsilon(U)$ and $L_W\leqslant \operatorname{GL}^\varepsilon(W)$ are irreducible, and define $L=L_U\times L_W$ as a subgroup of $\operatorname{GL}^\varepsilon(V)$ in the natural way. Assume that an element $x\in L$ is such that its natural projection onto $L_U$ acts irreducibly on $U$. Then the following assertions hold. 1) There is an element $g\in \operatorname{SL}^\varepsilon(V)$ such that the subgroup $G=\langle L,x^g\rangle$ acts irreducibly on $V$. 2) Assume that $\dim U\leqslant\dim W$ and $\operatorname{SL}^\varepsilon(W)\leqslant L_W\leqslant\operatorname{GL}^\varepsilon(W)$. Then either an element $g\in \operatorname{SL}^\varepsilon(V)$ can be chosen so that the subgroup $G\leqslant \operatorname{GL}^\varepsilon(V)$ defined in $1)$ is primitive, or $(n,q)\in\{(4,2),(4,3)\}$ and any imprimitivity system of the group $G$ defined in $1)$ has cardinality $4$. 3) Assume that $(n,q)\notin\{(4,2),(4,3),(4,5)\}$, $|x|$ is a prime number, and the element $x$ has no eigenvectors for $\varepsilon=+$ or has no nondegenerate eigenvectors for $\varepsilon=-$. Let $U$ be an $x$-invariant subspace of minimum dimension, and let $t=\operatorname{codim} U$. Set Then some $m+1$ elements conjugate to $x$ by elements of $\operatorname{SL}^\varepsilon(V)$ generate a subgroup which contains one of the following subgroups: Proof. First we show that
In fact, suppose that a proper subspace $X$ of the space $V$ is $L$-invariant. Since $2\leqslant \dim U\leqslant \dim W$ and the subgroups $L_U$ and $L_W$ are irreducible on the corresponding subspaces, the equations $[u,L_U]=U$ and $[w,L_W]=W$ hold for any $u\in U^\sharp $ and $w\in W^\sharp$. Let $v\in X^\sharp$. Then the vector $v$ can uniquely be represented in the form $v=u+w$, where $u\in U$ and $w\in W$. If $u\ne 0$, then, as the group $L_U$ is irreducible, we have Therefore, if $u\ne 0$, then $X$ contains $U$. A similar reasoning shows that if $w\ne 0$, then $X$ contains $W$. Assertion $(*)$ is proved.Further, we claim that In fact, choose bases $e_1,\dots,e_s$ and $f_1,\dots,f_t$ of $U$ and $W$, respectively, that are orthonormal in case $(-)$. Since $s=\dim U\geqslant 2$, it follows that assertion $(**)$ is satisfied for the element $g\in\operatorname{SL}^\varepsilon(V)$ defined by the equalities
$$
\begin{equation*}
e_ig=e_i \quad\text{for } i<s, \qquad f_ig= f_i \quad\text{for } i<t, \qquad e_sg=- f_t\quad\text{and} \quad f_tg= e_s.
\end{equation*}
\notag
$$
We claim that the subgroup $G=\langle L,x^g\rangle$ is irreducible. Assume that $X$ is a proper nontrivial $G$-invariant subspace. Then it is $L$-invariant, and so either $X=U$ or $X=W$. On the other hand $Ug$ is a $x^g$-invariant $x^g$-irreducible subspace, and thus $Ug\cap X$ is $x^g$-invariant too, which contradicts the $x^g$-irreducibility of the subspace $Ug$. Assertion 1) is proved. Before proving 2) we make an obvious remark: the choice of an element $g\in \operatorname{SL}^\varepsilon(V)$ in the proof of assertion 1) can be corrected if necessary in such a way that the pair of subspaces $\{U,W\}$ is not an imprimitivity system for the group $G$. Assume that where $m>1$, and $V_i^y\in\Omega:=\{V_1,\dots, V_m\}$ for any $i\in\{1,\dots,m\}$ and $y\in G$, that is, $\Omega$ is an imprimitivity system for the group $G$. Our objective is to show that $m=n=4$ and $q\leqslant 3$.We also conclude from the irreducibility of the group $G$ that $G$ acts transitively on $\Omega$; in particular, all the $V_i$ are isometric and their dimensions are equal. Let $\dim V_i=k$. Since the sum of the elements of every $L$-orbit on $\Omega$ is invariant under $L$, it follows from assertion $(*)$ that either $L$ acts transitively on $\Omega$ or there are exactly two $L$-orbits on this set, the sum of the elements of one of them is $U$ and that of the other orbit is $W$. It can readily be seen that the case of a nontransitive action of $L$ on $\Omega$ is equivalent to the existence of $V_i\in\Omega$ such that $V_i\leqslant W$. We note that in this case $\{V_j\mid V_j\leqslant W\}$ is a system of imprimitivity for the groups $L$ and $L_W$ on $W$. Let
$$
\begin{equation*}
\operatorname{SL}^\varepsilon(W)\leqslant L_W\leqslant\operatorname{GL}^\varepsilon(W),
\end{equation*}
\notag
$$
as in assertion 2). Then the group $L_W$ is primitive, and therefore $W$ coincides with $V_i$. In this case as follows from $(*)$. Taking the relation into account we obtain $m=2$ and $\Omega =\{U,W\}$. However, we have assumed that $\{U,W\}$ is not an imprimitivity system for the group $G$.
Consider the case when $L$ is transitive on $\Omega$. Let denote the projection onto $W$ parallel to $U$. Then $vy\omega=v\omega y$ for any $v\in V$ and $y\in L$. In particular, $L_W$ acts transitively on $\Omega\omega=\{V_i\omega\mid i=1,\dots,m\}$, and the dimensions of all projections $V_i\omega$ are the same. We denote them by $k'$.First consider case $(+)$. The group $\operatorname{SL}(W)$ acts transitively on the subspaces of the same dimension of $W$. Since $\operatorname{SL}(W)\leqslant L_W$, the set $\Omega\omega$ coincides with the set of $k'$-dimensional subspaces of $W$, whose number is certainly not less than that of the set of one-dimensional subspaces. Thus, if $\dim W=t$, then
$$
\begin{equation*}
m=|\Omega|\geqslant |\Omega\omega|\geqslant \frac{q^t-1}{q-1}\geqslant 2^t-1\geqslant 2^{n/2}-1.
\end{equation*}
\notag
$$
If $m<n$, then $m=n/k\leqslant n/2$. It follows from the assumptions that $n\geqslant 4$; however, the inequality $n/2\geqslant 2^{n/2}-1$ fails to hold for such $n$. Therefore, $m=n$. The following inequality fails for $n>4$:
$$
\begin{equation*}
n\geqslant 2^{t}-1, \quad\text{where } t\geqslant \frac n2.
\end{equation*}
\notag
$$
Therefore, $n=4$. Finally, the inequality
$$
\begin{equation*}
4=n\geqslant \frac{q^t-1}{q-1}\geqslant \frac{q^{n/2}-1}{q-1}=q+1
\end{equation*}
\notag
$$
means that $q\leqslant 3$.
Thus, assertion 2) is proved in case $(+)$. We note that the number $(q^t-1)/(q-1)$ of one-dimensional subspaces of a $t$-dimensional space $W$ over $\mathbb{F}_q$ coincides as a rule with the degree $\mu(\operatorname{SL}_t(q))$ of the minimal permutation representation of the group $\operatorname{SL}(W)=\operatorname{SL}_t(q)$: see [42]. In dealing with case $(-)$, first we rule out the case when $t=3$. To do this we consider in detail the action of $\operatorname{SU}(W)$ on $\Omega\omega$ for $t=3$ directly. For all other $t$ we will use the information about the degrees of the minimal permutation representations $\mu(\operatorname{SU}_t(q))$ of the group $\operatorname{SU}_t(q)$ from the same paper [42]. We also use the chain of inequalities
$$
\begin{equation*}
2t\geqslant n\geqslant m=|\Omega|\geqslant |\Omega\omega|\geqslant \mu(\operatorname{SU}_t(q)),
\end{equation*}
\notag
$$
which hold similarly to case $(+)$.
Let $t=3$. By Witt’s lemma any two isometric subspaces of $W$ are taken to each other by some element of $\operatorname{SU}(W)$. The dimensions of totally isotropic subspaces in $W$ do not exceed $[3/2]=1$. Computing the number of the corresponding subspaces as the index of the stabilizer of one of them in the group $\operatorname{SU}(W)$, we conclude that, contrary to the inequality $2t\geqslant |\Omega\omega|$, one of the following cases occurs: $\bullet$ the $V_i\omega$ are totally isotropic, their dimension is 1, and $\bullet$ the $V_i\omega$ are nondegenerate of dimension 1 and
$$
\begin{equation*}
|\Omega\omega|=q^2(q^2+q+1)\geqslant 2^2(2^2+2+1)= 28>6=2t;
\end{equation*}
\notag
$$
$\bullet$ the $V_i\omega$ are nondegenerate of dimension 2, there are as many of them as there are (one-dimensional nondegenerate) orthogonal complements to them, and thus the bound from the previous case holds; $\bullet$ the $V_i\omega$ are degenerate but not isotropic, of dimension 2, these are exactly the orthogonal complements to their (one-dimensional isotropic) radicals, there are as many of them as there are one-dimensional isotropic subspaces, that is, the estimate from the first case holds. From [42], Table 1, taking the isomorphism $\operatorname{SU}_2(q)\cong \operatorname{SL}_2(q)$ into account, we obtain
$$
\begin{equation*}
\mu(\operatorname{SU}_t(q)) =\begin{cases} 2 &\text{for } (t,q)=(2,2), \\ 3 &\text{for } (t,q)=(2,3), \\ 5 &\text{for } (t,q)=(2,5), \\ 7 &\text{for } (t,q)=(2,7), \\ 6 &\text{for } (t,q)=(2,9), \\ 11 &\text{for } (t,q)=(2,11), \\ q+1 &\text{for } t=2, \ q\ne 2,3,5,7,9, 11, \\ 2 &\text{for } (t,q)=(3,2), \\ 50 &\text{for } (t,q)=(3,5), \\ q^3+1 &\text{for } t=3, \ q\ne 2,5, \\ (q+1)(q^3+1) &\text{for } t=4, \\ \displaystyle\frac{(q^{t-1}-(-1)^{t-1})(q^{t}-(-1)^{t})}{(q^{2}-1)} &\text{for } t>4. \end{cases}
\end{equation*}
\notag
$$
We assume that $(t,q)\notin\{(2,2),(2,3)\}$, since otherwise $(n,q)\in\{(4,2),(4,3)\}$ and $m=n$, as in the conclusion of assertion 2). If $t=2$ and $q\in\{5,7,9,11\}$, then $\mu(\operatorname{SU}_t(q))>4=2t$, contrary to the inequality $2t\geqslant \mu(\operatorname{SU}_t(q))$. If $q\ne 5,7,9,11$, then again, The case $t=3$ has been excluded above. For $t=4$ we have again
$$
\begin{equation*}
\mu(\operatorname{SU}_t(q))=(q+1)(q^3+1)\geqslant (2+1)(2^3+1)=27>8=2t.
\end{equation*}
\notag
$$
Finally, let us exclude the case $t>4$. For even $t=2d>4$ we have
$$
\begin{equation*}
\begin{aligned} \, \mu(\operatorname{SU}_t(q)) &=\frac{(q^{t-1}-(-1)^{t-1})(q^{t}-(-1)^{t})}{(q^{2}-1)} \\ &=(q^{t-1}+1)\sum_{i=0}^{d-1}q^{2i}\geqslant d(2^{t-1}+1)>t\cdot 2^{t-2}>2t. \end{aligned}
\end{equation*}
\notag
$$
For odd $t=2d+1>4$ we have
$$
\begin{equation*}
\begin{aligned} \, \mu(\operatorname{SU}_t(q)) &=\frac{(q^{t-1}-(-1)^{t-1})(q^{t}-(-1)^{t})}{(q^{2}-1)} \\ &=(q^{t}+1)\sum_{i=0}^{d-1}q^{2i}\geqslant d(2^t+1)>(t-1)\cdot 2^{t-1}>2t. \end{aligned}
\end{equation*}
\notag
$$
In all cases we arrive at a contradiction with the inequality $2t\geqslant \mu(\operatorname{SU}_t(q))$. Assertion 2) is completely proved.
Let us prove 3). Let $x=x_Ux_W$, where $x_U\in \operatorname{GL}^\varepsilon(U)$ and $x_W\in \operatorname{GL}^\varepsilon(W)$. Since the element $x$ is semisimple, has a prime order and has no eigenvectors for $\varepsilon=+$ and no nondegenerate eigenvectors in $V$ for $\varepsilon=-$, we obtain We do not assume that the group $L$ has already been fixed, and we construct it for the element $x$ in a special way, after which we verify that it satisfies the conditions in the lemma.It is clear from the assumptions that $t\geqslant n/2\geqslant 2$. Since $x$ is a semisimple element and $(n,q)\notin\{(4,2),(4,3),(4,5) \}$, it follows from Lemma 7 that there exist $m$ elements $g_1,\dots,g_m\in \operatorname{SL}^\varepsilon(W)$ for which
$$
\begin{equation*}
L_W:=\langle x_W^{g_1}, \dots, x_W^{g_m}\rangle\geqslant \operatorname{SL}^\varepsilon(W).
\end{equation*}
\notag
$$
Here $L_W=\langle \operatorname{SL}^\varepsilon(W), x_W\rangle$. We set
$$
\begin{equation*}
L_U:=\langle x_U\rangle\quad\text{and} \quad L:=\langle x^{g_1}, \dots, x^{g_m}\rangle.
\end{equation*}
\notag
$$
Then $L\leqslant L_U\times L_W$, and the projections of $L$ onto the factors $L_U$ and $L_W$ are surjective. To show that $L$ satisfies the conditions in the lemma, it remains to establish that $L$ coincides with $L_U\times L_W$.
We note first that $x_W\in \operatorname{SL}^\varepsilon(W)$; in particular, $L_W=\operatorname{SL}^\varepsilon(W)$. This follows from the fact that the order of the image of the element $x_W$ in the quotient group $\operatorname{GL}^\varepsilon(W)/\operatorname{SL}^\varepsilon(W)$ divides $(|x|, q-\varepsilon)=1.$ Now, consider the canonical epimorphism
$$
\begin{equation*}
\phi\colon {L_U\times L_W}\to ({L_U\times L_W})\,/\,\Phi({L_U\times L_W}).
\end{equation*}
\notag
$$
Taking into account that $L_U$ is a group of prime order and $L_W\cong \operatorname{SL}^\varepsilon(W)$, and using known properties of the Frattini subgroup (see [2]. Ch. A, Lemma (9.4)) we have
$$
\begin{equation*}
\Phi({L_U\times L_W})=\Phi(L_U)\times \Phi(L_W)=\Phi(L_W)=\operatorname{Z}(L_W).
\end{equation*}
\notag
$$
Therefore, $L^\phi$ is isomorphic to a subgroup of $\mathbb{Z}_{|x|}\times L_n^\varepsilon(q)$ whose projections onto all factors are surjective. However, $L^\phi$ has composition factors isomorphic to $\mathbb{Z}_{|x|}$ and $L_n^\varepsilon(q)$, and thus so that $L=L_U\times L_W$.
Now, by assertions 1) and 2) of the lemma, which we have proved, there exists $g\in\operatorname{SL}^\varepsilon(V)$ such that the group
$$
\begin{equation*}
G:=\langle L, x^{g}\rangle=\langle x^{g_1}, \dots, x^{g_m},x^{g}\rangle
\end{equation*}
\notag
$$
is irreducible and primitive as a subgroup of $\operatorname{GL}^\varepsilon(V)$. Since $\operatorname{SL}^\varepsilon(W)\leqslant G$, it follows that $G$ contains a long root subgroup of $\operatorname{SL}^\varepsilon(V)$. Let $R$ be the normal closure in $G$ of this root subgroup. Since $G$ is primitive, $R$ is irreducible. In particular, $\mathrm{O}_p(R)=1$. Now, as follows from Theorem II in [43], the group $R$ is isomorphic to one of groups specified in assertion 3) (for $\varepsilon=+$ one can also use the results of [44] and [45]).
This completes the proof of Lemma 14. Lemma 15 ([39], Proposition 4.9.1). Let $L={}^d\Sigma(q)$ be a simple group of Lie type over the field ${\mathbb F}_q$, where $\Sigma$ is an irreducible root system and $d$ is either an empty symbol or $2$ (that is, ${}^d\Sigma(q)\ne{}^3D_4(q)$). Let $x$ and $y$ be automorphisms of the group $L$ having the same prime order, and assume that $x$ and $y$ are both field or graph-field automorphisms modulo the group $\widehat{L}$. Then the subgroups $\langle x\rangle$ and $\langle y\rangle$ in $\operatorname{Aut}(L)$ are conjugate by an element of $\widehat{L}$. If $x$ is a field automorphism, then If $x$ is a graph-field automorphism and ${}^d\Sigma\in\{A_{n-1},D_{n}\}$, then $|x|=2$ and ${\operatorname{C}_{L}(x)={}^2\Sigma(q^{1/2})}$. Lemma 16. Let $L=L_n^\varepsilon(q)$ be a simple projective special linear or unitary group, and let $n\geqslant 4$. Consider graph involutions modulo $\widehat{L}$ in the automorphism group of $L$. Then the following assertions hold. 1) If $n$ is odd, then all graph involutions modulo $\widehat{L}$ are conjugate with respect to $\widehat{L}$, and every such involution normalizes some subgroup $K$ in $L$ isomorphic to $\operatorname{SL}^\varepsilon_{n-1}(q)$ and induces a nontrivial automorphism on $K/\operatorname{Z}(K)$. 2) If $n$ is even and $q$ is odd, then there are three classes of $\widehat{L}$-conjugacy of graph involutions modulo $\widehat{L}$. For representatives $x_0$, $x_+$ and $x_-$ of these classes the following holds:
$$
\begin{equation*}
\operatorname{F}^*(C_L(x_\delta))\cong \begin{cases} S_n(q) &\textit{for } \delta=0, \\ O^+_n(q) &\textit{for } \delta=+, \\ O^-_n(q) &\textit{for } \delta=-. \end{cases}
\end{equation*}
\notag
$$
The element $x_\delta$ normalizes a subgroup $K_{\delta}$ of $L$ that is the image, under the natural homomorphism, of a subgroup
$$
\begin{equation*}
\bigl(\operatorname{GL}_m^\varepsilon(q)\times \operatorname{GL}_{n-m}^\varepsilon(q)\bigr) \cap \operatorname{SL}_n^\varepsilon(q)
\end{equation*}
\notag
$$
of $\operatorname{SL}_n^\varepsilon(q)$, where
$$
\begin{equation*}
m=\begin{cases} 2 & \textit{for } (\varepsilon,\delta)\in \{(+,0),(+,+),(-,+),(-,-)\}, \\ 1 & \textit{for } (\varepsilon,\delta)\in \{(+,-),(-,0)\}, \end{cases}
\end{equation*}
\notag
$$
and $x_\delta$ induces a nontrivial automorphism on the component of the quotient group $K_{\delta}/\operatorname{Z}(K_{\delta})$ that is isomorphic to $\operatorname{PSL}_{n-m}^\varepsilon(q)$. 3) If both $n$ and $q$ are even, then there are two $\widehat{L}$-conjugacy classes of graph involutions modulo $\widehat{L}$. For any involution $x$ the group $L$ has an $x$-invariant subgroup $K$ for which $K/\operatorname{Z}(K)\cong\operatorname{PSL}_n^\varepsilon(q)$, and $x$ induces a nontrivial automorphism of $K/\operatorname{Z}(K)$. Proof. The graph involutions modulo $\widehat{L}$ are contained in the coset $\widehat{L}\tau$ (see the beginning of § 2.2). Some information about $\widehat{L}$-conjugacy classes of these involutions, their number, representatives and centralizers can be found in [37], Lemma 10.
If $n$ is odd, then all involutions in $\widehat{L}\tau$ are conjugate to $\tau$ by elements in $\widehat{L}$. Therefore, we can assume that $x=\tau$. Then $x$ normalizes the subgroup $K$ of $L$ that is the image in $\operatorname{PSL}_n^\varepsilon(q)$ of a subgroup
$$
\begin{equation*}
\bigl(\operatorname{GL}_{n-1}^\varepsilon(q)\times \operatorname{GL}_1^\varepsilon(q)\bigr)\cap \operatorname{SL}_n^\varepsilon(q)
\end{equation*}
\notag
$$
and consists of the images of matrices of the form
$$
\begin{equation*}
\begin{pmatrix} A&0 \\ 0&(\det A)^{-1} \end{pmatrix},
\end{equation*}
\notag
$$
where $A$ ranges over $ \operatorname{GL}^\varepsilon_{n-1}(q)$, and normalizes the subgroup $K^{\infty}\cong \operatorname{SL}_{n-1}^\varepsilon(q)$. It is clear that $x=\tau$ induces a nontrivial automorphism of $K^{\infty}/Z(K^{\infty})\cong \operatorname{PSL}^\varepsilon_{n-1}(q)$. This completes the proof of the assertion 1).
Consider the case when $n$ is even and $q$ is odd. If $\varepsilon=-$, then see [46], p. 288, and [47], p. 43, for the proof of the assertion about the existence of the required subgroup $K_\delta$. Therefore, we assume that $\varepsilon=+$. In this case (see [46], p. 285), the element $x$ is conjugate by an element of $\widehat{L}$ to one of the three pairwise nonconjugate involutions $x_0$, $x_+$ and $x_-$ induced on $L$ by the images of the products
$$
\begin{equation*}
J^0\tau,J^+\tau,J^-\tau\in \langle\operatorname{GL}_n(q),\tau\rangle,
\end{equation*}
\notag
$$
respectively, where
Here the centralizers of the elements $x_\delta$ are as indicated in assertion 2). We show that $x_\delta$ normalizes a subgroup $K_\delta$ as specified in 2). The elements $J^-$ and $\tau$, and therefore also their product $J^-\tau$, normalize a subgroup
$$
\begin{equation*}
(\operatorname{GL}_{n-1}(q)\times \operatorname{GL}_1(q))\cap \operatorname{SL}_n(q)
\end{equation*}
\notag
$$
of the group $\operatorname{SL}_n(q)$, and thus $x_-$ normalizes the subgroup $K_-$ defined just as $K$ in the proof of assertion 1); it is clear that $x_-$ induces a nonidentity automorphism of $K_-$. In the case when $x$ is conjugate to $x_-$ assertion 2) is proved.
The elements $J^0\tau$ and $J^+\tau$ normalize the subgroup of $\operatorname{GL}_n(q)$ consisting of the block-diagonal matrices of the form where $A\in \operatorname{GL}_{n-2}(q)$ and $B\in\operatorname{GL}_2(q)$. Intersecting this subgroup with $\operatorname{SL}_n(q)$ and considering the image of this intersection in $\operatorname{PSL}_n(q)$ we obtain a group $K_0=K_+$ which is invariant under $x_0$ and $x_+$ and such as indicated in assertion 2) for $m=2$. This completes the proof of 2).Finally, consider the case when $q$ and $n$ are even. Here, in accordance with [37], Remark 11, and [47], Lemma 3.7, for $\varepsilon=-$ we treat $\operatorname{GL}^\varepsilon_n(q)=\operatorname{GU}_n(q)$ as the group of matrices of the linear transformations of a vector space of dimension $n=2m$ over the field $\mathbb{F}_{q^2}$ that preserve the Hermitian form $(\,\cdot\,,\,\cdot\,)$ defined by the equalities
$$
\begin{equation*}
(e_i, e_j) =(f_i, f_j) =0, \quad (e_i, f_j) =\delta_{ij} \quad\text{for all } i,j=1,\dots,m,
\end{equation*}
\notag
$$
with respect to some ordered basis $e_1, \dots, e_m, f_m, \dots, f_1$.
Then (see [37], Lemma 10) the group $L$ has exactly two $\widehat{L}$-conjugacy classes of graph automorphisms modulo $\widehat{L}$ of order $2$, and some of their representatives are induced by the elements
First let $\varepsilon=+$. Then $\tau$, $J^0$ and $t$ normalize the subgroup of all block-diagonal matrices of the form
$$
\begin{equation*}
\begin{pmatrix} A & {} \\ {}& B \end{pmatrix}, \quad \text{where } A\in \operatorname{GL}_2(q), \ \ B\in\operatorname{GL}_{n-2}(q) \quad\text{and} \quad \det A\cdot\det B=1,
\end{equation*}
\notag
$$
in $\operatorname{SL}_n(q)$. The image of this subgroup in $\operatorname{PSL}_n(q)$ contains the normal subgroup $K\cong \operatorname{SL}_{n-2}(q)$, which is invariant under the automorphisms induced by $\tau$ and $J^0t\tau$, and both automorphisms are nontrivial on $K/\operatorname{Z}(K)$.
Now let $\varepsilon=-$. The element $t_0$ of $\operatorname{GL}_n(q^2)$ centralizes the $\langle\tau,\varphi_q\rangle$-invariant subgroup $G^*$ of matrices of the form
$$
\begin{equation*}
\begin{pmatrix} 1 & & \\ {}& A & \\ {} & & 1 \end{pmatrix}, \quad \text{where } A\in \operatorname{SL}_{n-2}(q^2),
\end{equation*}
\notag
$$
and centralizes every subgroup of this subgroup. In it, the subgroup $K^*=C_{G^*}(\tau\varphi_q)\cong \operatorname{SU}_{n-2}(q)$ is invariant with respect to $\varphi_q$ and $t_0$ and is contained in $\operatorname{SU}_n(q)$. The elements $\varphi_q$ and $t_0\varphi_q$ normalize the subgroup
$$
\begin{equation*}
N_{\operatorname{SU}_n(q)}(K^*)\cong (\operatorname{GU}_{n-2}(q) \times\operatorname{GU}_2(q))\cap\operatorname{SU}_n(q)
\end{equation*}
\notag
$$
of $\operatorname{SU}_n(q)$. We denote the image of the subgroup $K^*$ in $\operatorname{PSL}_n(q^2)$ by $K$. It is clear that $K\leqslant \operatorname{PSU}_n(q)$. The elements $\varphi_q$ and $t_0\varphi_q$ induce nontrivial automorphisms with consistent action on the isomorphic groups $K^*/\operatorname{Z}(K^*)$ and $K/\operatorname{Z}(K)\cong \operatorname{PSU}_{n-2}(q)$.
This completes the proof of Lemma 16. Since for a graph automorphism $x$ modulo $\widehat{L}$ of the group $L=L_n^\pm(q)$, where $n=4$, the bound for $\alpha(x,L)$ is irregular (see Lemma 7), for $n=4$ we need additional information about subgroups of $L$ normalized but not centralized by $x$. This information is contained in the following lemma, in which we use the well-known isomorphisms $L_4^\pm(q)\cong O_6^\pm(q)$: see [38], Proposition 2.9.1. Lemma 17. Let $V$ be a vector space over the field $\mathbb{F}_q$ of odd order, let $\dim V=6$, and let $V$ be equipped with a nondegenerate symmetric bilinear form of sign $\varepsilon\in\{+,-\}$. Let $\mathrm{O}$ be the group of isometries of space $V$, let $\Delta$ be its similarity group, let $\operatorname{SO}$ be the group of elements with determinant $1$ of $\mathrm{O}$, and let $\Omega=\mathrm{O}'=\Delta'$. Let denote the canonical epimorphism. Let $L=\overline{\Omega}=O_{2n}^\varepsilon(q)$. Then the following assertions hold. 1) The canonical graph automorphism $\overline{\gamma}$ of the group $L$ is contained in $\overline{\mathrm{O}}\setminus \overline{\operatorname{SO}}$, and the group $\overline{\Delta}$ coincides with $\langle\widehat{L},\overline{\gamma}\rangle$. 2) All graph involutions modulo $\widehat{L}$ are the images of involutions in $\Delta$. 3) There are three $\widehat{L}$-conjugacy classes of graph involutions modulo $\widehat{L}$ with representatives $\overline{\gamma}_1=\overline{\gamma}$, $\overline{\gamma}_2$ and $\overline{\gamma}_{2}'$, where each $\gamma_i$, $i=1,2$, is an involution in $\mathrm{O}$ whose eigenvalue $-1$ has multiplicity $2i-1$, and $\gamma_{2}'$ is an involution in $\Delta\setminus \mathrm{O}$ whose eigenvalue $-1$ has multiplicity $3$. Here every involution $\overline{\gamma}_1$ and $\overline{\gamma}_2$ normalizes but does not centralize a subgroup of $L$ isomorphic to $O_{5}(q)\cong S_4(q)$, and the involution $\overline{\gamma}_{2}'$ normalizes but does not centralize a subgroup in $L$ isomorphic to $O_{4}^-(q)\cong L_2(q^2)$. Proof. All assertions of the lemma, except for the existence of subgroups normalized but not centralized by relevant involutions, follow6 from [39], Theorems 4.5.1 and 4.5.2, Tables 4.5.1 and 4.5.2, and Remark 4.5.4.
We claim that the involutions $\gamma_i$, $i=1,2$, act nonscalarly on some nondegenerate subspace $U$ of $V$ such that $\dim U=5$. Since $\gamma_i$ is an isometry, the subspaces $V_+$ and $V_-$ of eigenvectors with eigenvalues $1$ and $-1$, respectively, are orthogonal to each other, and $V= V_+\oplus V_-$. Therefore, $V_+$ and $V_-$ are nondegenerate $\gamma_i$-invariant subspaces and $\gamma_i$ acts scalarly on each of them. Let $u\in V_+$ be a nondegenerate vector. Then $W=u^\perp$ is a $\gamma_i$-invariant nondegenerate complement to $\langle u\rangle$. Since $\dim V_+=6 - 2i + 1>1$, the transformation $\gamma_i$ has both eigenvalues $1$ and $-1$ on $W$ and acts nonscalarly on $W$. Thus, $\overline{\gamma}_i$ normalizes but does not centralize the image of the commutator subgroup of the isometry group of $W$, which is isomorphic to $O_{5}(q)$. By [39], Table 4.5.1, the centralizer of $\overline{\gamma}_{2}'$ in $\widehat{L}$ is isomorphic to some group of automorphisms of the group $O_{4}^-(q)\cong L_2(q^2)\cong O_3(q^2)$ which contains the latter; this centralizer has the trivial centre and an even index in $\widehat{L}$. We conclude from Lemma 3 that $\overline{\gamma}_{2}'$ normalizes but does not centralize a subgroup of $L$ isomorphic to $O_{4}^-(q)$. This completes the proof of the lemma. Lemma 18 ([32], Lemma 2.18). Let $L$ be a simple classical group or an alternating group and let $r$ be an odd prime number not dividing the order of the group $L$. Then the following assertions hold. 1) If $L=A_n$, then $r\geqslant n+1$ and $n\leqslant r-1$. 2) If $L=L_n(q)$, then $ r\geqslant n+2$ and $n\leqslant r-2$. 3) If $L=U_n(q)$, then $ r\geqslant n+2$ and $n\leqslant r-2$. 4) If $L=S_{n}(q)$, then $r\geqslant n+3$ and $n\leqslant r-3$. 5) If $S=O_{n}(q)$ and $n$ is odd, then $r\geqslant n+2$ and $n\leqslant r-2$. 6) If $S=O^+_{n}(q)$ or $S=O^-_{n}(q)$ and $n$ is even, then $r\geqslant n+1$ and $n\leqslant r-1$, respectively. 7) If $L=L_n(q^2)$, then $r\geqslant 2n+3$ and $n\leqslant (r-3)/2$. Lemma 19. Let $L=L_n^\varepsilon(q)$ and let $x$ be an automorphism of odd prime order of the group $L$. Then one of the following assertions holds. 1) There exists an element $g\in L$ such that the subgroup $\langle x,x^g\rangle$ is not solvable. 2) $q=3$, $x$ is a transvection, and there exist $g_1,g_2\in L$ such that the subgroup $\langle x,x^{g_1},x^{g_2}\rangle$ is not solvable. 3) $\varepsilon=-$, $q=2$, $x$ is a pseudoreflection of order $3$, and there exist $g_1,g_2,g_3\in L$ such that the subgroup $\langle x,x^{g_1},x^{g_2},x^{g_3}\rangle$ is not solvable. The proof follows from Theorem A* in [17]. Lemma 20 ([32], Proposition 2). Let $L=A_n$, $n\geqslant 5$, let $r\leqslant n$ be an odd prime, and let $x\in \operatorname{Aut} (L)$ be an element of prime order. Then 1) $\beta_{r}(x,L)=r-1$ if $x$ is a transposition; 2) one of the following statements holds: (a) $\beta_{r}(x,L)\leqslant r-1$, (b) $r=3$, $n=6$, $x$ is an involution not belonging to $S_6$, and $\beta_{r,L}(x) = 3$. Lemma 21 (see [34], Theorem 1). Let $r$ be an odd prime number. Let $L$ be one of $26$ sporadic simple groups. Also let $s$ be a prime divisor of the order of the group $L$ such that $s=r$ if $r$ divides $|L|$ and $s> r$ otherwise. Then for any automorphism $x$ of $L$ of prime order. § 3. Proof of Theorem 1We divide the proof of Theorem 1 into several cases. We start from cases of low dimension, which are considered in a series of successive lemmas. Then we consider inner-diagonal automorphisms $x$ which admit a natural action on the associated vector space. We consider separately the situations when $x$ stabilizes a one-dimensional subspace of the so-called inherited type (in particular, the case when the automorphism $x$ is induced by a unipotent element is fully considered there) and when $x$ is a semisimple element that does not stabilize such one-dimensional subspaces. Finally, we consider situations when $x$ is a field, a graph-field, or a graph automorphism modulo the group of inner-diagonal automorphisms. 3.1. Cases of small dimensionLemma 22. Theorem 1 holds for $L\in\{L^\pm_4(2),L^\pm_4(3),L^\pm_4(5)\}$. Proof. Let $L=L_4^\varepsilon(q)$, where $\varepsilon\in\{+,-\}$ and $q\in\{2,3,5\}$. Theorem 1 holds for the group $L=L_4(2)\cong A_8$ by Lemma 20; therefore, we assume that $(\varepsilon,q)\ne(+,2)$.
Consider the case $r=3$. Then $s=r=3$. For $|x|=2$, see the remark at the end of the proof of the lemma, and here we assume that $|x|>2$. By Lemma 19, except in the case when $L=U_4(2)$ and $x$ is a pseudoreflection of order $3$ (and we can leave out this case) some two or three elements conjugate to $x$ generate a nonsolvable group. The group $L$ under consideration has no sections isomorphic to simple Suzuki groups; therefore, it follows from the Glauberman-Thompson theorem (see [48], Ch. II, Corollary 7.3) that the order of this nonsolvable subgroup of $L$ is divisible by 3. Thus, in all cases $\beta_3(x,L)\leqslant 3$. Now we assume that $r\geqslant 5$. By Lemma 7 we have $\alpha(x,L)\leqslant 4$, except in the following cases:
This yields the lemma for $r>5$, and also for $r=5$ apart from the above exceptional cases. Consider the remaining exceptions, assuming that $r=5$. Since $5$ divides $|L|$, we have $s=5$. Let $L=L_4^\pm(5)$ and let $x$ be a graph involution modulo $\widehat{L}$. Then, by Lemma 16, 2), the element $x$ normalizes some subgroup $K$ of $L$ such that $K/\operatorname{Z}(K)\cong L_2^\pm(5)\cong A_5$ or $K/\operatorname{Z}(K)\cong L_3^\pm(5)$, and $x$ induces a nontrivial automorphism $y$ on $K/\operatorname{Z}(K)$. Hence we conclude from Lemmas 4, 7 and 20 that
$$
\begin{equation*}
\beta_5(x,L)\leqslant\beta_5(y,K/\operatorname{Z}(K))\leqslant 4=5-1.
\end{equation*}
\notag
$$
Let $x$ be a graph automorphism of $L_4(3)$ modulo inner-diagonal automorphisms and let $|x|=2$. In the table of characters of the group $\operatorname{Aut}(L_4(3))$ in [36], corresponding to this case are involutions in the conjugacy classes $2D$, $2E$, $2F$, and $2G$; moreover, the classes $2D$ and $2E$ are interchanged by a diagonal automorphism so that $2E$ can be ignored. Let us use the well-known fact from the character theory that, given elements $a,b$ and $c$ of a group $G$, the number $\operatorname{m}(a,b,c)$ of pairs $(u,v)\in a^G\times b^G$ such that $uv=c$ can be found from the character table by the formula
$$
\begin{equation*}
\operatorname{m}(a,b,c)=\frac{|a^G|\,|b^G|}{|G|}\sum_{\chi\in\operatorname{Irr}(G)} \frac{\chi(a)\chi(b)\overline{\chi(c)}}{\chi(1)};
\end{equation*}
\notag
$$
see [35], Exercise (3.9). Using the character table in [36] of the corresponding extension of $L_4(3)$, we see that7 in both $2F$ and $2G$, there is a pair of elements whose product belongs to the class $5A$ and has order $5$ (more precisely, $\operatorname{m}(2F,2F,5A)=\operatorname{m}(2G,2G,5A)=20$). Therefore, $\beta_5(x,L)=2$ if $x$ is an involution in the class $2F$ or $2G$. Also, the class $2A$, which consists of inner involutions, contains a pair of elements whose product belongs to $5A$, since $\operatorname{m}(2A,2A,5A)=5$. Moreover, every element of $2A$ is a product of two involutions in $2D$, because $\operatorname{m}(2D,2D,2A)=2$. Therefore, for an involution $x$ in $2D\cup 2E$ we have $\beta_5(x,L)\leqslant 4=5-1$.
The other cases are considered similarly. $\bullet$ $L=U_4(2)$ and $x$ is a transvection (the class $2A$):
$$
\begin{equation*}
\operatorname{m}(2A,2A,2B)=2, \quad \operatorname{m}(2B,2B,5A)=5 \quad\Longrightarrow \quad \beta_5(x, U_4(2))\leqslant 4 \quad\text{for } x\in 2A.
\end{equation*}
\notag
$$
$\bullet$ $L=U_4(2)$ and $x$ is a graph involution modulo $\widehat{L}$ (the classes $2C$ and $2D$):
$$
\begin{equation*}
\begin{gathered} \, \operatorname{m}(2C,2C,5A)=5 \quad \Longrightarrow\quad \beta_5(x, U_4(2))= 2 \quad\text{for } x\in 2C; \\ \operatorname{m}(2D,2D,2B)=2, \quad\operatorname{m}(2B,2B,5A) =5\quad\Longrightarrow \quad \beta_5(x, U_4(2))\,{\leqslant}\, 4,\quad\text{for } x\,{\in}\, 2D. \end{gathered}
\end{equation*}
\notag
$$
$\bullet$ $L=U_4(3)$ and $x$ is a graph involution modulo $\widehat{L}$ (the classes $2D$, $2E$ and $2F$):
$$
\begin{equation*}
\begin{gathered} \, \operatorname{m}(2D,2D,2A)=2, \quad \operatorname{m}(2A,2A,5A)=5\quad\Longrightarrow\quad \beta_5(x, U_4(3))\,{\leqslant}\, 4 \quad\text{for } x\,{\in}\, 2D; \\ \operatorname{m}(2E,2E,5A)=5 \quad \Longrightarrow\quad \beta_5(x, U_4(3))= 2 \quad\text{for } x\in 2E; \\ \operatorname{m}(2F,2F,5A)=5 \quad \Longrightarrow\quad \beta_5(x, U_4(3))= 2 \quad\text{for } x\in 2F. \end{gathered}
\end{equation*}
\notag
$$
In a similar way, with the help of the character tables in [49] we consider the inner involutions $x$ for $s=3$ and prove that $\beta_3(x,L)=2.$
This completes the proof of Lemma 22. Lemma 23. Theorem 1 holds for $L= L_2(q)$. Proof. We perform induction on $q$. The validity of Theorem 1 for the groups $L_2(4)$, $L_2(5)$ and $L_2(9)$ follows from Lemma 20 and the isomorphisms $L_2(4)\cong L_2(5)\cong A_5$ and $L_2(9)\cong A_6$. Therefore, we assume that $q\ne 4,5,9$.
If $x$ is not a field automorphism of order $2$, then by Lemma 7, for any prime divisor $s$ of the order of the group $L$ we have which yields the assertion of the lemma. In particular, the lemma holds if $q$ is a prime.Now assume that $|x|=2$ and $x$ is a field automorphism modulo $\widehat{L}$. Consider $\varphi=\varphi_{q^{1/2}}$, which is the canonical field automorphism of order 2 induced by the mapping
$$
\begin{equation*}
\begin{pmatrix} u&v \\ w&z \end{pmatrix}\mapsto \begin{pmatrix} u^{q^{1/2}}&v^{q^{1/2}} \\ w^{q^{1/2}}&z^{q^{1/2}} \end{pmatrix}.
\end{equation*}
\notag
$$
Since the subgroups $\langle x\rangle$ and $\langle \varphi\rangle$ of $\langle \widehat{L},\varphi\rangle$ are conjugate with respect to $\widehat{L}$ by Lemma 15, we can assume that $x=\varphi$. Since $q\ne 9$, we have $\beta_s(x,L)\leqslant\alpha(x,L)\leqslant 4$ by Lemma 7. Therefore, Theorem 1 holds for any $r\geqslant 5$.
Assume that $r=3$. Then $r$ divides the order of $L$ and $s=r=3$. Moreover, by Lemma 15
$$
\begin{equation*}
\operatorname{PSL}_2(q^{1/2})\leqslant \operatorname{C}_L(x)\leqslant \operatorname{PGL}_2(q^{1/2})
\end{equation*}
\notag
$$
and $\operatorname{C}_L(x)$ is a subgroup of even index in $L$. This subgroup is almost simple since $q\ne 4,9$, and therefore $\operatorname{Z}(\operatorname{C}_L(x))=1$. By Lemma 3 the element $x$ normalizes but does not centralize a subgroup conjugate to $\mathrm{O}^{p'}(\operatorname{C}_L(x))\cong L_2(q^{1/2})$. It remains to use the induction assumption.
This completes the proof of the lemma. Lemma 24. Theorem 1 holds for $L= U_3(q)$. Proof. Lemma 7 implies that $\alpha(x,L)\leqslant 3$, except in the case when $q=3$, $|x|=2$ and $\alpha(x,L)=4$. Therefore, which yields the lemma for $r>3$; moreover, except in the case when $s=r=3$, $L=U_3(3)$ and $|x|=2$.
To analyze the last case we use the character table of the group $U_3(3)$ and its automorphism group which is presented in [36]. In $U_3(3)$ all inner involutions form the single conjugacy class $2A$, and all noninner ones form the class $2B$. Arguing as in the final part of Lemma 22 and using GAP (see [49]) we see that
$$
\begin{equation*}
\begin{gathered} \, \operatorname{m}(2A,2A,3B)=3 \quad \Longrightarrow\quad \beta_3(y, U_3(3))= 2 \quad\text{for } y\in 2A; \\ \operatorname{m}(2B,2B,3A)=36 \quad \Longrightarrow\quad\beta_3(y, U_4(3))= 2 \quad\text{for } y\in 2B. \end{gathered}
\end{equation*}
\notag
$$
This completes the proof of the lemma. Lemma 25. Theorem 1 holds for $L=L_3(q)$. Proof. First assume that an automorphism $x$ is not graph-field modulo $\widehat{L}$. By Lemma 7 we have $\alpha(x,L)\leqslant 3$. Thus, the following inequality holds for any prime divisor $s$ of the order of the group $L$: $\beta_s(x,L)\leqslant \alpha(x,L)\leqslant 3$, which implies the assertion of the lemma.
Now let $x$ be a graph-field automorphism modulo $\widehat{L}$, and let $|x|=2$, that is, $\widehat{L}x=\widehat{L} \tau\varphi$, where
$$
\begin{equation*}
\tau\colon A\mapsto (A^{-1})^\top, \qquad \varphi=\varphi_{q^{1/2}}\colon (a_{ij})\mapsto (a_{ij}^{q^{1/2}}).
\end{equation*}
\notag
$$
By Lemma 15 we can assume that $x=\tau\varphi$. In this case we have $\alpha(x,L)\leqslant 4$ by Lemma 7, that is, the assertion of the lemma holds for any $r\geqslant 5$. Assume that $r=3$. Then $s=r=3$ and, as follows from Lemma 15,
$$
\begin{equation*}
\operatorname{C}_{\widehat{L}}(x)\cong \operatorname{PGU}_3(q^{1/2})\quad\text{and} \quad\mathrm{O}^{p'}(\operatorname{C}_L(x))\cong U_3(q^{1/2}).
\end{equation*}
\notag
$$
Since
$$
\begin{equation*}
\mathrm{O}^{p'}(\operatorname{C}_L(x)) \leqslant \operatorname{C}_{L}(x)\leqslant C_{\widehat{L}}(x),
\end{equation*}
\notag
$$
we can readily see that the index $|L:\operatorname{C}_L(x)|$ is even. Moreover, $\operatorname{Z}(\operatorname{C}_L(x))=1$. By Lemma 3 there exists a subgroup $M$ of $L$ such that $M$ is conjugate to $\operatorname{C}_L(x)$ and $x$ normalizes $M$ but does not centralize it. Let $y$ be an automorphism of the group $\mathrm{O}^{p'}(M)\cong U_3(q^{1/2})$ induced by $x$. It remains to use Lemma 24 and the fact that
$$
\begin{equation*}
\beta_3(x,L)\leqslant \beta_3(y,\mathrm{O}^{p'}(M))\leqslant 3.
\end{equation*}
\notag
$$
This completes the proof of the lemma. Lemma 26. If $L=S_4(q)$, where $q$ is odd and $x\in \widehat{L}$ is an involution, then $ \beta_5(x,L)\leqslant 4$. Proof. We use the isomorphisms
$$
\begin{equation*}
L\cong O_5(q)\cong \Omega_5(q), \qquad \widehat{L} \cong \operatorname{SO}_5(q)\quad\text{and} \quad \operatorname{GO}_5(q)=\langle -E\rangle\times \operatorname{SO}_5(q),
\end{equation*}
\notag
$$
where $E$ is the $5\times 5$ identity matrix; see [36], Table 2, and [38], Proposition 2.9.1. In accordance with [38], Proposition 2.6.1, consider a five-dimensional vector space $V$ over $F=\mathbb{F}_q$ with a nondegenerate symmetric bilinear form $(\,\cdot\,,\,\cdot\,)$ defined on $V$ and a basis $e_1,\dots,e_5$ such that the following equations hold for some $\zeta\in\mathbb{F}_q^*$:
Let $G$ be the group of all linear transformations $g$ of the space $V$ such that and let us identify $\widehat{L}$ with the subgroup $\{g\in G\mid \det g=1\}$ of index 2 of the group $G$. Then $L=G'$. The discriminant of the restriction of the form to $\langle e_1,e_5\rangle_F$ is a square in the field $F$. We choose mutually orthogonal nonzero vectors $e_1'$ and $e_5'$ in the subspace $\langle e_1,e_5\rangle_F$ such that
$$
\begin{equation*}
(e_1',e_1')=(e_5',e_5')\notin (\mathbb{F}_q^*)^2\zeta
\end{equation*}
\notag
$$
(this is possible; see [38], Remark on p. 27 and Proposition 2.5.12). Then the restriction of the form to has sign $+$, and its restriction to has sign $-$ (see [38], Propositions 2.5.12 and 2.5.13). For brevity we denote the resulting ordered bases by
$$
\begin{equation*}
\mathscr{E}\colon e_1,e_2,e_3,e_4,e_5\quad\text{and} \quad\mathscr{E}'\colon e_1',e_2,e_3,e_4,e_5'.
\end{equation*}
\notag
$$
We denote the matrices of a linear transformation $g$ of the space $V$ written in the bases $\mathscr{E}$ and $\mathscr{E}'$, respectively, by the symbols
As is known (see [39], Table 4.5.1), the group $L$ has two conjugacy classes of involutions, and there are two other classes in $\widehat{L}\setminus L$. Let us indicate representatives of these classes. These are $x_1^\square$, $x_1^\boxtimes$, $x_2^\square$ and $x_2^\boxtimes$ such that
$$
\begin{equation*}
[x_1^\square]=[x_1^\boxtimes]'=\operatorname{diag}(-1,-1,1,1,1)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
[x_2^\square]=[x_2^\boxtimes]'=\operatorname{diag}(-1,-1,-1,-1,1).
\end{equation*}
\notag
$$
The elements defined in this way are readily decomposed into products of reflections with respect to the vectors $e_i$ and $e_i'$, and their spinor norms are readily determined; from this we conclude that
$$
\begin{equation*}
x_1^\square,x_2^\square\in L\quad\text{and} \quad x_1^\boxtimes,x_2^\boxtimes\in\widehat{L}\setminus L.
\end{equation*}
\notag
$$
It follows from the definition that the elements $x_1^\boxtimes$ and $x_2^\boxtimes$ stabilize the nondegenerate subspace such that the restriction of the form to $W'$ has sign $-$. Moreover, these elements induce nonscalar transformations $y_1$ and $y_2$ of $W'$, which preserve the form on $W'$. Therefore, if $H$ is the group of isometries of $W'$, then $y_1$ and $y_2$ induce nontrivial automorphisms of The order of the group $L_2(q^2)$ is divisible by 5, and by Lemma 23
$$
\begin{equation*}
\beta_5(x_i^\boxtimes, L)\leqslant \beta_5(y_i, H'/\operatorname{Z}(H'))\leqslant 4 \quad\text{for } i=1,2.
\end{equation*}
\notag
$$
Further, consider the embedding $S_5\hookrightarrow G$ given by the action of the group $S_5$ on the indices of the vectors $e_1,\dots,e_5$. We denote the image of a substitution $\sigma\in S_5$ by $\sigma^*$. The involution $\tau=(15)(24)\in A_5$ inverts the element $\sigma=(12345)$ of order $5$ in $A_5$, and therefore $\tau\tau^\sigma=\sigma^2$ is an element of order $5$ and $\beta_5(\tau, A_5)=2$. It follows from the simplicity of $A_5$ that $\sigma^*,\tau^*\in L$ and $\beta_5(\tau^*,L)=2$. Since $\tau^*\in L$, the involution $\tau^*$ is conjugate to $x_1^\square$ or $x_2^\square$, and since the eigenvalue $-1$ of $\tau^*$ has multiplicity $2$, $\tau^*$ is conjugate to $x_1^\square$. Thus, Finally, for the elements $g=(15)(34)^*$ and $h=(25)(34)^*$ in $L$ we have so that
$$
\begin{equation*}
\beta_5(x_2^\square,L)\leqslant 2\beta_5(x_1^\square,L)= 4
\end{equation*}
\notag
$$
by Lemma 5.
This completes the proof of Lemma 26. 3.2. Proof of Theorem 1: the structure and general remarksWe begin the proof of Theorem 1. As in the assumptions of the theorem, let $L=L^\varepsilon_n(q)$ and let $x\in\operatorname{Aut}(L)$ be an automorphism of prime order. We recall that $r$ is an odd prime number and $s\in\pi(L)$ is such that $s=r$ if $r\in\pi(L)$ and $s>r$ if $r\notin\pi(L)$. Our objective is to prove the inequality
$$
\begin{equation}
\beta_s(x,L)\leqslant \begin{cases} r & \text{for } r=3, \\ r-1 & \text{for } r>3. \end{cases}
\end{equation}
\tag{3.1}
$$
We argue using induction on $|L|$. By Lemmas 22–25 we can assume that $n\geqslant 4$ and for $n=4$ the number $q$ is distinct from $2$, $3$ and $5$. It can also be assumed that (except when $n=4$ and $x$ is a graph automorphism modulo $\widehat{L}$) the inequality holds, because otherwise
$$
\begin{equation*}
\beta_s(x, L)\leqslant\alpha(x,L)\leqslant n\leqslant r-1
\end{equation*}
\notag
$$
by Lemma 7, and (3.1) holds. This and Lemma 18 imply that In §§ 3.3 and 3.4 we consider all cases when $x\in\widehat{L}$, and in § 3.5, all cases when $x\in\operatorname{Aut}(L)\setminus \widehat{L}$. We assume that $F=\mathbb{F}_q$ if $\varepsilon=+$ and $F=\mathbb{F}_{q_{}^2}$ if $\varepsilon=-$. Since $x$ has a prime order, in the case when $x\in\widehat{L}$ the element $x$ is either unipotent or semisimple. We identify the group $L^\varepsilon_n(q)$ with $\operatorname{PSL}^\varepsilon(V)$, where $V$ is an $n$-dimensional vector space over the field $F$, which, in case $\varepsilon=-$, is equipped with a nondegenerate Hermitian form. Then $\widehat{L}=\operatorname{PGL}^\varepsilon(V)$, so that $\widehat{L}$ acts on the set of subspaces of the space $V$. Considering an element $x\in\operatorname{PGL}^\varepsilon(V)$ we say that an $x$-invariant subspace $U\leqslant V$ has the inherited type if $U$ is an arbitrary subspace for $\varepsilon=+$ or for unipotent $x$, and $U$ is a nondegenerate space for $\varepsilon=-$ and semisimple $x$. Subsections 3.3 and 3.4 correspond to the cases when there exists or does not exist a one-dimensional subspace of inherited type for $x$. We also say that $x\in\operatorname{PGL}^\varepsilon(V)$ acts scalarly on an $x$-invariant subspace $U$ if the restriction to $U$ is proportional to the identity transformation for some pre-image of the element $x$ in $\operatorname{GL}^\varepsilon(V)$ (equivalently, each one-dimensional subspace of $U$ is $x$-invariant). 3.3. Inner-diagonal automorphism stabilizing a one-dimensional subspace of inherited typeIn this part of the proof we consider all cases when $x$ preserves some one-dimensional subspace $U$ of inherited type of the space $V$ (in particular, this assumption covers the case when $x$ is unipotent). The following subcases are possible: (a) $x$ is unipotent and $\varepsilon=+$; (b) $x$ is unipotent and $\varepsilon=-$; (c) $x$ is semisimple and either $\varepsilon=+$ or $U$ is nondegenerate. Consider case (a). Let $P_1$ be the stabilizer of some line in the natural module, and let $P_2$ be the stabilizer of a hyperplane. By Lemma 11 we can assume that $x\in P_i\setminus \mathrm{O}_p(P_i)$ for $i=1$ or $i=2$, and $x$ acts nonscalarly on the components of the quotient $P_i/\mathrm{O}_p(P_i)$. In particular, this means that $x\notin \mathrm{O}_\infty(P_i)$. Consider the canonical epimorphism
$$
\begin{equation*}
\overline{\phantom{G}}\colon P_i\to P_i/\mathrm{O}_\infty(P_i).
\end{equation*}
\notag
$$
Then $\overline{x}\ne1$ and $\operatorname{F}^*(\overline{P_i})\cong L_{n-1}(q)$. It follows from Fermat’s little theorem and (3.2) that $r$ divides $|L_{n-1}(q)|$ and, by Lemma 4,
$$
\begin{equation*}
\beta_r(x, L)\leqslant\beta_r\bigl(\overline{x}, \operatorname{F}^*(\overline{P_i})\bigr).
\end{equation*}
\notag
$$
Using the induction assumption we obtain (3.1). The theorem is proved in case (a). Let case (b) hold. If $x$ also stabilizes some nondegenerate one-dimensional subspace of $V$, then Theorem 1 is established by repeating the reasoning used in case (a), after making the natural replacement of ${L^+_{n-1}(q)=L_{n-1}(q)}$ by ${L^-_{n-1}(q)=U_{n-1}(q)}$ and the parabolic subgroup $P_i$ by the stabilizer of a nondegenerate one-dimensional subspace, which is isomorphic to the image in $\operatorname{PGU}_n(q)$ of the subgroup
$$
\begin{equation*}
\operatorname{GU}_1(q)\times \operatorname{GU}_{n-1}(q)\leqslant \operatorname{GU}_n(q).
\end{equation*}
\notag
$$
Therefore, in case (b) we assume that $x$ does not stabilize any nondegenerate one-dimensional subspace. Consider two maximal parabolic subgroups:
By Lemma 11, up to conjugation by an element in $L$ we have $x\in P_i\setminus \mathrm{O}_p(P_i)$ for $i=1$ or $2$. As in case (a), we have $x\notin \mathrm{O}_\infty(P_i)$; consider the canonical epimorphism
$$
\begin{equation*}
\overline{\phantom{G}}\colon P_i\to P_i/\mathrm{O}_\infty(P_i).
\end{equation*}
\notag
$$
We assume that $i=1$, that is, $x$ stabilizes a maximal totally isotropic subspace $W$ and induces a nonidentity transformation of it. Then
$$
\begin{equation*}
\operatorname{F}^*(\overline{P_1})\cong L_{[n/2]}(q^2) \quad\text{and}\quad \overline{x}\in \operatorname{F}^*(\overline{P_1})^\sharp.
\end{equation*}
\notag
$$
Inequality (3.2) and Fermat’s little theorem show that $r$ divides $|L_{[n/2]}(q^2)|$ because
$$
\begin{equation*}
2\biggl[\frac{n}{2}\biggr]\geqslant n-1\geqslant r-1.
\end{equation*}
\notag
$$
Now Lemma 4, in combination with the induction assumption, gives
$$
\begin{equation*}
\beta_r(x,L)\leqslant \beta_r\bigl(\overline{x},\operatorname{F}^*(\overline{P_1})\bigr) \leqslant \begin{cases} r & \text{for } r=3, \\ r-1 & \text{for } r>3. \end{cases}
\end{equation*}
\notag
$$
We assume that $i=2$. Then $\operatorname{F}^*(\overline{P_2})\cong U_{n-2}(q)$, where $\overline{x}\in \operatorname{F}^*(\overline{P_1})^\sharp$. If $r$ divides $|U_{n-2}(q)|$, then again, according to Lemma 4 we have
$$
\begin{equation*}
\beta_r(x,L)\leqslant \beta_r\bigl(\overline{x},\operatorname{F}^*(\overline{P_2})\bigr);
\end{equation*}
\notag
$$
hence we obtain (3.1) by the induction assumption. Therefore, analyzing the remaining subcases of case (b) we assume that $r$ does not divide $|U_{n-2}(q)|$, which implies that $r\geqslant n$ by Lemma 18. Therefore, $n=r$ by (3.2). In particular, $n$ is odd and $n\geqslant 5$. According to Lemma 13, one of the following subcases holds:
As above, let $P_1$ be the stabilizer of a maximal totally isotropic subspace of $W$ and consider the canonical epimorphism
$$
\begin{equation*}
\overline{\phantom{G}}\colon P_1\to P_1/\mathrm{O}_\infty(P_1).
\end{equation*}
\notag
$$
Here
$$
\begin{equation*}
\operatorname{F}^*(\overline{P_1})\cong L_{[n/2]}(q^2)=L_{(r-1)/2}(q^2).
\end{equation*}
\notag
$$
The involution $\overline{y}$ normalizes but does not centralize this subgroup, inducing an inner-diagonal automorphism on $L_{(r-1)/2}(q^2)$. It follows from Fermat’s little theorem that $r$ divides and thus it divides $|\operatorname{F}^*(\overline{P_1})|$. If $n=r=5$, then $\dim W=(r-1)/2=2$. By Lemma 9, in this case
$$
\begin{equation*}
\beta_r(y,L)\leqslant\beta_r(\overline{y}, \operatorname{F}^*(\overline{P_1}))=2,
\end{equation*}
\notag
$$
so that, taking the inclusion $y\in\langle x,x^g\rangle$ into account, from Lemma 5 we obtain that is, the theorem in question holds. Let $n=r>5$. Then $r\geqslant 7$, $\dim W=(r-1)/2\geqslant 3$, and by Lemma 7 and taking the fact that $\overline{y}$ is an inner-diagonal involution of the group $L_{(r-1)/2}(q^2)$ into account we have
$$
\begin{equation*}
\beta_r(y,L)\leqslant\beta_r\bigl(\overline{y}, \operatorname{F}^*(\overline{P_1})\bigr)\leqslant \alpha\bigl(\overline{y}, \operatorname{F}^*(\overline{P_1})\bigr)\leqslant \frac{r-1}{2},
\end{equation*}
\notag
$$
so that Case (b) has been treated completely. Consider case (c) when a semisimple element $x$ stabilizes a one-dimensional subspace $U$ which is also nondegenerate for $\varepsilon=-$. Let $W$ be an $x$-invariant complement to $U$, and let $W\perp U$ if $\varepsilon=-$. Without loss of generality we can assume that a pre-image of $x$ in $\operatorname{GL}^\varepsilon(V)$ acts nonscalarly on $W$. In fact, if $x$ acts scalarly on $W$, then $W$ consists of eigenvectors of a pre-image $x^*$ of $x$ in $\operatorname{GL}^\varepsilon(V)$ and, since $x\ne 1$, the eigenvalue $\lambda$ corresponding to vectors in $W$ differs from the eigenvalue $\mu$ corresponding to vectors in $U$. We choose a one-dimensional subspace $U_0$ of $W$ (nondegenerate for $\varepsilon=-$; the possibility to choose $U_0$ in this way is ensured by Witt’s lemma) and consider the $x$-invariant (orthogonal for $\varepsilon=-$) complement $W_0$ to $U_0$. Since $n\geqslant 4$, among the eigenvalues of the restriction of $x^*$ to $W_0$ are both $\lambda$ and $\mu$, and thus $x^*$ acts nonscalarly on $W_0$, and we can replace $U$ by $U_0$ and $W$ by $W_0$. Now $x$ is contained in the image in $\operatorname{PGL}^\varepsilon_n(q)$ of a subgroup of the form
$$
\begin{equation*}
\operatorname{GL}^\varepsilon_1(q)\times \operatorname{GL}_{n-1}^\varepsilon(q)
\end{equation*}
\notag
$$
and induces a nontrivial automorphism $\overline{x}$ on the unique non-Abelian composition factor $L^\varepsilon_{n-1}(q)$ of this image. As in case (a), using Lemma 18 and (3.2) we conclude that $r$ divides $|L^\varepsilon_{n-1}(q)|$. From this, by Lemma 4 and the induction assumption we derive inequality (3.1). The theorem in case (c) is proved. 3.4. An inner-diagonal automorphism that does not stabilize one-dimensional subspaces of inherited typeHere we consider all cases when a semisimple element $x\in\widehat{L}=\operatorname{PGL}^\varepsilon(V)$ has no invariant one-dimensional subspaces of inherited type. If $x$ acts irreducibly on $V$, then
$$
\begin{equation*}
\beta_r(x,L)\leqslant\alpha(x,L)\leqslant 3\leqslant \begin{cases} r & \text{for } r=3, \\ r-1 & \text{for } r>3, \end{cases}
\end{equation*}
\notag
$$
by Lemma 8. Therefore, we assume that $V$ has a proper nontrivial $x$-invariant subspace. Let us reduce the situation to the case when this subspace is of inherited type and hence has dimension $\geqslant 2$, in order to use then Lemma 14. By Lemma 10, if $x$ has no proper invariant subspaces of inherited type, then $\varepsilon=-$, $n$ is even and $x$ stabilizes a totally isotropic subspace $U$ of dimension $n/2$. Since $x$ is semisimple, it follows from Maschke’s theorem that $x$ also stabilizes some totally isotropic subspace $W$ of the same dimension such that If $x$ acts scalarly on both $U$ and $W$, then we consider nonzero vectors $u\in U$ and $w\in W$ such that $(u,w)\ne 0$. For a pre-image $x^*\in \operatorname{GL}^-(V)$ of $x$ the subspaces $U$ and $W$ consist of eigenvectors; moreover, since $x^*\notin \operatorname{Z}(\operatorname{GL}^-(V))$, the vectors $u$ and $w$ correspond to different eigenvalues. Thus, $\langle u,w\rangle_F$ is a proper $x$-invariant subspace of inherited type. Assume that $x$ acts nonscalarly on $U$ or $W$. Then $x^*$ is contained in the stabilizer of this subspace in $\operatorname{GL}^-(V)$; this stabilizer is isomorphic to $\operatorname{GL}_{n/2}(q^2)$, $x$ is contained in the image $H$ of this stabilizer in $\widehat{L}=\operatorname{PGU}(V)$, and $x$ induces a nontrivial automorphism $\overline{x}$ of $H^\infty/\operatorname{Z}(H^\infty)\cong L_{n/2}(q^2)$. Further, as usual, it follows from (3.2) and Lemma 18 that $r$ divides $|L_{n/2}(q^2)|$, and relation (3.1) holds by the induction assumption and the inequality
$$
\begin{equation*}
\beta_r(x,L)\leqslant\beta_r(\overline{x},H^\infty/\operatorname{Z}(H^\infty)).
\end{equation*}
\notag
$$
If $x$ acts scalarly on both $U$ and $W$, then $x$ acts nonscalarly on $\langle u,w\rangle_F$, as required. Thus, throughout the rest of this subsection we assume that $x\in \widehat{L}$ is a semisimple element, $U$ is a nonzero $x$-invariant subspace of inherited type of minimum dimension (in particular, $x$ acts irreducibly on $U$), and $W$ is an $x$-invariant subspace, also of inherited type, complementing $U$ to $V$. Here
$$
\begin{equation*}
\dim U\geqslant 2\quad\text{and} \quad t=\dim W=n-\dim U\geqslant \dim U.
\end{equation*}
\notag
$$
The element $x$ acts nonscalarly on $W$ and therefore induces a nontrivial automorphism $\overline{x}$ of the group $\operatorname{PSL}^\varepsilon(W)\cong L^\varepsilon_t(q)$. If $r$ divides $|L^\varepsilon_t(q)|$, then by the induction assumption the inequality
$$
\begin{equation*}
\beta_r(x,L)\leqslant\beta_r(\overline{x},L^\varepsilon_t(q))
\end{equation*}
\notag
$$
implies (3.1). Therefore, we assume that $r$ does not divide $|L^\varepsilon_t(q)|$ and, in particular, $r\geqslant 5$. By Lemma 18 the following inequality holds: Lemma 14 implies that, for
$$
\begin{equation*}
m=\begin{cases} t&\text{for }t>2, \\ 3&\text{for }t=2, \end{cases}
\end{equation*}
\notag
$$
some $m+1$ elements conjugate to $x$ by means of elements of $L$ generate a subgroup $H$ of $L$ that contains a normal subgroup from the following list:
Hence, taking (3.2) and Lemma 18 into account we conclude that the order of the subgroup $H$ is divisible by $r$. Therefore,
$$
\begin{equation*}
\beta_r(x,L)\leqslant m+1= \begin{cases} t+1\leqslant r-1 & \text{for } t>2, \\ 4=5-1\leqslant r-1 & \text{for } t=2. \end{cases}
\end{equation*}
\notag
$$
The case under consideration has been analyzed completely. 3.5. Field, graph-field and graph automorphismsIn the group $\operatorname{SL}_n^\varepsilon(q)$ consider the subgroup $H$ of matrices where $A$ ranges over the group $\operatorname{SL}_{n-1}^\varepsilon(q)$, and consider the image $K$ of $H$ in $L_n^\varepsilon(q)$. It is clear that $K/\operatorname{Z}(K)\cong L_{n-1}^\varepsilon(q)$. It follows from (3.2) and Lemma 18 that $r$ divides $|L_{n-1}^\varepsilon(q)|$. It is also clear that $H$ and $K$ are invariant with respect to the automorphisms $\varphi_{p^m}$ and $\tau$ and, moreover, $y$ induces a nonidentity automorphism $\overline{y}$ on $K/\operatorname{Z}(K)$ in each of the following cases:
By Lemmas 15 and 16, if $x$ is a field automorphism or a graph-field automorphism modulo $\widehat{L}$, or if $n$ is odd and $x$ is a graph automorphism modulo $\widehat{L}$, then the subgroup $\langle x\rangle$ is conjugate to $\langle y\rangle$ for some $y$ as above with respect to $\widehat{L}$. By the induction assumption, from the relations
$$
\begin{equation*}
\beta_r(x,L)=\beta_r(y,L)\leqslant \beta(\overline{y}, K/\operatorname{Z}(K))
\end{equation*}
\notag
$$
we can derive (3.1) in these cases. Therefore, it remains to consider the case when $n$ is even and $x$ is a graph automorphism modulo $\widehat{L}$. For even $n>4$ it follows from (3.2) that The same inequality holds for $n=4$ and $r=3$. In these cases Lemma 18 implies that the orders of the groups $L_{n-1}^\varepsilon(q)$ and even of the $L_{n-2}^\varepsilon(q)$ are divisible by $r$. Lemma 16 implies that any graph automorphism $x$ modulo $\widehat{L}$ of the group $L$ normalizes but does not centralize a subgroup $H$ of $L$ such that $H^\infty/\operatorname{Z}(H^\infty)$ is isomorphic to $L_{n-1}^\varepsilon(q)$ or $L_{n-2}^\varepsilon(q)$. Applying the induction assumption to an automorphism induced by $x$ on $H^\infty/\operatorname{Z}(H^\infty)$ we obtain inequality (3.1).It remains to consider the case when $n=4$ and $r>3$. Since $\alpha(x,L)\leqslant 6$, by Lemma 7, for $r\geqslant 7$ we have
$$
\begin{equation*}
\beta_s(x,L)\leqslant \alpha(x,L)\leqslant 6=7-1\leqslant r-1.
\end{equation*}
\notag
$$
Therefore, we assume that $r=5$. Then we see that $r$ divides $|L|$ and $s=r$. Except when $q$ is odd and the graph involution $x$ modulo $\widehat{L}$ is conjugate to $x_-$ (see Lemma 16), we see that the representatives of conjugacy classes of graph involutions that are indicated in the proof of Lemma 16 centralize the element $\varphi=\varphi_p$ in $\operatorname{Aut}(L)$ (see the beginning of § 2.2), and we can replace $L$ by the simple group $\mathrm{O}^{p'}(\operatorname{C}_L(\varphi))\cong L^\varepsilon_4(p)$ on which these representatives induce a graph involution of the same type modulo inner-diagonal automorphisms. Therefore, we may assume that Since the cases $q=2,3,5$ have already been considered in Lemma 22, we assume that, in any case, $q$ is odd and $q\geqslant 7$. Let us use the isomorphism and regard $L$ as a projective orthogonal group and the element $x$ as a graph involution in this group modulo the group of inner-diagonal automorphisms. By Lemma 17 the involution $x$ normalizes but does not centralize a subgroup $H$ of $L$ which is isomorphic to either $O_5(q)\cong S_4(q)$ or $O_4^-(q)\cong O_3(q^2)\cong L_2(q^2)$ and, in either case, $|H|$ is divisible by 5.If $H\cong S_4(q)$, then $\operatorname{Aut}(H)=\widehat{H}$, and therefore $x$ induces an inner-diagonal automorphism $\overline{x}$ of $H$, and by Lemmas 4 and 26 we have
$$
\begin{equation*}
\beta_5(x,L)\leqslant \beta_5(\overline{x},H)\leqslant 4=5-1.
\end{equation*}
\notag
$$
In case $H\cong L_2(q^2)$, the same reasoning with a reference to Lemma 9 yields the required result. This completes the proof of Theorem 1. § 4. Proof of Theorem 2Let $\pi$ be a proper subset of the set of primes that contains at least two distinct elements. Let $r$ be the least prime outside $\pi$, and let
$$
\begin{equation*}
m= \begin{cases} r & \text{for } r\in\{2,3\}, \\ r-1 & \text{for } r>3. \end{cases}
\end{equation*}
\notag
$$
Suppose that Theorem 2 is false. Then $r\geqslant 3$ by Lemma 2. We write
$$
\begin{equation*}
\begin{aligned} \, \mathscr{F}=\mathscr{S}\cup \{A_n\mid n\geqslant 5\} &\cup \bigl\{L^\varepsilon_n(q)\mid n\geqslant 2, \ \varepsilon=\pm,\ q\text{ is a power of a prime number} \\ &\qquad \text{and } (\varepsilon, n,q)\ne(\pm,2,2), (\pm,2,3), (-,3,2) \bigr\}, \end{aligned}
\end{equation*}
\notag
$$
where $\mathscr{S}$ is the set of 26 sporadic groups. As the class $\mathscr{X}$, we take the class of all finite groups for which any non-Abelian composition factor is either isomorphic to a group in $\mathscr{F}$ or is a $\pi$-group. Then $\mathscr{X}\setminus \mathscr{BS}_\pi^m\ne\varnothing$, since the inclusion $\mathscr{X}\subseteq \mathscr{BS}_\pi^m$ would mean that the theorem under consideration is true. In accordance with Lemma 1, a group $G$ of the least order in $\mathscr{X}\setminus \mathscr{BS}_\pi^m$ contains a normal non-Abelian simple subgroup $L$ and an element of prime order $x$ with the following properties:
Let $s$ be the least prime divisor of $|L|$ not belonging to $\pi$. Then either $r$ divides $|L|$ and $s=r$ or $r$ does not divide $|L|$ and $s>r$. According to Theorem 1 and Lemmas 20 and 21, that is, there are elements $g_1,\dots,g_m\in L$ such that $|\langle{ x^{g_1},\dots,x^{g_m}}\rangle|$ is divisible by $s$. Then $\langle{ x^{g_1},\dots,x^{g_m}}\rangle$ is not a $\pi$-subgroup, contrary to the fact that any $m$ elements conjugate to $x$ generate a $\pi$-subgroup.This completes the proof of the theorem.
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