Locally Finite Periodic Groups Saturated with Finite Simple Orthogonal Groups of Odd Dimension

Abstract

Suppose that $n$ is an odd integer, $n\geq 5$. We prove that a periodic group $G$, saturated with finite simple orthogonal groups $O_{n}(q)$ of odd dimension over fields of odd characteristic, is isomorphic to $O_{n}(F)$ for some locally finite field $F$ of odd characteristic. In particular, $G$ is locally finite and countable.

Introduction

Let $M$ be some set of finite groups. A group $G$ is said to be saturated with groups from $M$, if each finite subgroup of $G$ lies in a subgroup isomorphic to some element of $M$.

The main goal of the paper is to prove the following result:

Theorem

Suppose that $m$ is an integer, $m\geq 2$, and $M$ is a set whose elements are finite simple orthogonal groups of dimension $n=2m+1$ over fields of odd characteristic. If $G$ is a periodic group saturated with groups from $M$, then $G$ is isomorphic to a simple orthogonal group $O_{n}(F)$ for some locally finite field $F$.

A particular case of this theorem was proved in [1].

1. Preliminary Facts

We will be using the notation and results of [2, 3]. Recall some of them.

Suppose that $F$ is a field of odd characteristic, $n=2m+1$ is an odd integer, $n\geq 5$, and $V$ is a vector space of dimension $n$ over $F$, while $e=\{e_{1},e_{2},\dots,e_{n}\}$ is a basis for $V$ over $F$. Let $f$ be a symmetric bilinear form on $V$ such that $f(e_{i},e_{i})=1$ and $f(e_{i},e_{j})=0$ for all $i,j\in\{1,2,\dots,n\}$, $i\neq j$. The basis $e$ is called the standard basis for $f$. Suppose that $f_{1}$ is a symmetric bilinear form on $V$ such that $f_{1}(e_{i},e_{i})=\mu$, where $\mu$ is an element of $F$ which is not a square, while $f_{1}(e_{i},e_{j})=0$ for all $i$ and $j$ satisfying $i\neq j$. Then $f$ and $f_{1}$ are not isometric, and every nondegenerate form on $V$ is isometric to $f$ or $f_{1}$. The group of linear transformations of $V$ that preserve $f$, preserves $f_{1}$ as well and is denoted by $GO(V)$. The subgroup $SO(V)=GO(V)\cap SL(V)$ has index 2 in $GO(V)$ and is called the special orthogonal group $V$. The group $\Omega_{n}(F)=\Omega(V)=[SO(V),SO(V)]$ is simple and $GO(V)/\Omega(V)$ is an elementary abelian group of order 4. The group $\Omega_{n}(F)=O_{n}(F)$ is isomorphic to a simple group $B_{m}(F)$ of Lie type $B$.

Suppose that $t$ is an involution (i.e. an element of order 2) from $L=\Omega(V)$. Then $V$ is an orthogonal direct sum of the subspaces $V^{+}(t)=\{v\in V\mid vt=v\}$ and $V^{-}(t)=\{v\in V\mid vt=-v\}$. Denote the dimensions of $V^{+}(t)$ and $V^{-}(t)$ by $d(t)$ and $r(t)$. It is clear that $d(t)$ and $r(t)$ are the defect and the rank of the transformation $t-1$ respectively. Two involutions $t,t_{1}\in\Omega(V)$ conjugate if and only if $d(t)=d(t_{1})$. Therefore, $\Omega_{n}(F)$ contains exactly $m=(n-1)/2$ classes of conjugated involutions.

Lemma 1

Let $A$ be a maximal elementary abelian subgroup of $L=\Omega(V)$. Then $|A|=2^{n-1}$ or $|A|=2^{n-2}$, $C_{L}(A)=A$, and each involution from $L$ is conjugate to an involution from $A$. If involutions $t$ and $t_{1}$ from $L$ are contained in $A$ and conjugate in $L$; i.e., $d(t)=d(t_{1})$; then $t$ and $t_{1}$ are conjugate in $N_{L}(A)$.

Let $t$ be an involution from $A$. If $d(t)=1$, then $|t^{N_{L}(A)}|$ is equal to $n$ if $|A|=2^{n-1}$; and $1$, if $|A|=2^{n-2}$. If $d(t)\neq 1$, then $|t^{N_{L}(A)}|>n$.

The normalizer $N_{L}(A)$ in $L$ acts transitively by conjugation on each of the sets $D_{s}=\{t\in A\mid d(t)=s\}$, $s=1,3,\dots$. In the case when $|A|=2^{n-1}$, the subgroup $N_{L}(A)$ has a subgroup $N_{0}\geq A$ such that $|N_{L}(A):N_{0}|\leq 2$, $N_{0}\simeq A:\operatorname{Alt}(n)$ and $N_{0}$ acts transitively by conjugation on each of the sets $D_{s}$.

Proof

The subgroup $A$ is diagonal in some orthogonal basis $e=\{e_{1},\dots,e_{n}\}$ for the space $V$; therefore, $A=A^{*}\cap\Omega(V)$, where $A^{*}=\{a^{*}\in GO(V)\mid e_{i}a^{*}=\pm e_{i}\}$. Since $A^{*}$ contains $-1\not\in\Omega(V)$, the order of $A$ is at most $2^{n-1}$, and we may assume that either the basis $e$ is standard or $(e_{i},e_{i})=1$ for $i>1$ and $(e_{1},e_{1})=\mu$, where $\mu$ is not a square in $F$. We start with the second case. Thus, the transformation $c$, satisfying

$$ e_{1}c=-e_{1},\quad e_{2}c=-e_{2},\quad e_{i}c=e_{i}\text{ for }i>2, $$

does not belong to $\Omega(V)$; i.e., $\langle c,-1\rangle\cap A=1$. On the other hand, every transformation $a\in A^{*}$ such that $e_{1}a=e_{1}$ and $d(a)$ is odd, is contained in $A$. Hence, $|A|=2^{n-2}$ and

$$ A=\{a=[1,\alpha_{2},\dots,\alpha_{n}]\mid\alpha_{i}=\pm 1,i=2,\dots,n;\ d(a)\text{ is odd}\}, $$

where $[\alpha_{1},\alpha_{2},\dots,\alpha_{n}]$ denotes the diagonal matrix with the element $\alpha_{i}$ in the entry $(i,i)$, $i=1,2,\dots,n$. It is clear that the number of elements $a\in A$ such that $d(a)=d$ is equal to 1, if $d=1$; and

$$ C_{n-1}^{d-1}=\frac{(n-1)(n-2)\dots(n-d+1)}{(d-1)!}, $$

if $d>1$. If $2\leq i<j<k\leq n$ and $c=c(i,j,k)$ is a transformation such that $e_{i}c=e_{j}$, $e_{j}c=e_{k}$, $e_{k}c=e_{i}$, and $e_{l}c=e_{l}$, given $l\not\in\{i,j,k\}$; then $[1,\alpha_{2},\dots,\alpha_{n}]^{c}=[1,\beta_{2},\dots,\beta_{n}]$, where $\beta_{i}=\alpha_{k}$, $\beta_{j}=\alpha_{i}$, $\beta_{k}=\alpha_{j}$, and $\beta_{l}=\alpha_{l}$, given $l\not\in\{i,j,k\}$. Obviously, $c\in N_{GO(V)}(A)$ and the order of $c$ is equal to 3; therefore, $c\in\Omega(V)$. Conjugating $a=[1,\alpha_{2},\dots,\alpha_{n}]$ by an element $g$, equal to the product $c(i_{1},j_{1},k_{1})\dots c(i_{l},j_{l},k_{l})$ for a suitable $i_{s}$, $j_{s}$, $k_{s}$, there is no difficulty in obtaining the element $a^{g}=[1,1,\dots,1,-1,\dots,-1]$, where the first $d(a)$ of the diagonal elements equal 1, and the rest of them equal $-1$. This shows that every two involutions $t,t_{1}\in A$, that are conjugate in $L$, i.e. involutions with condition $d(t)=d(t_{1})$, are conjugate in $N_{L}(A)$. Moreover, there exists only one involution $t$ in $A$ such that $d(t)=1$, and the number of involutions $t$ with condition $d(t)=d>1$ is equal to

$$ C_{n-1}^{d-1}=\frac{(n-1)(n-2)\dots(n-d+1)}{(d-1)!}=\frac{(n-1)(n-2)}{2}\prod\limits_{i=3}^{d-1}\frac{n-i}{i}. $$

We will show that this number is more than $n$ by induction on $d\geq 3$ (recall that $d$ is odd and $n\geq 5$). For $d=3$,

$$ C_{n-1}^{d-1}=\frac{(n-1)(n-2)}{2}=\frac{n^{2}-3n+2}{2}\geq n+1>n. $$

Moreover, $C_{n-1}^{d-1}=C_{n-1}^{n-d}$; therefore, we may assume that $d-1\leq n-d$, i.e. $d\leq\frac{n+1}{2}$. Now, given $i-1\leq d-1$, we have $\frac{n-i}{i}>1$; i.e., $\prod\nolimits_{i=3}^{n-1}\frac{n-i}{i}>1$, which implies that $C_{n-1}^{d-1}>n$.

Direct checking shows that $C_{GO(V)}(A)=A^{*}$, and hence $C_{L}(A)=A$.

Consider the standard basis $e$. Then

$$ A=\{a=[\alpha_{1},\alpha_{2},\dots,\alpha_{n}]\mid\alpha_{1},\alpha_{2},\dots,\alpha_{n}\in\{1,-1\};\ d(a)\text{ is odd}\}. $$

In particular, $C_{L}(A)=A$. Clearly, $N^{*}=N_{GO(V)}(A^{*})=A^{*}:\operatorname{Sym}(n)$ and $N=N^{*}\cap L=A:H$, where $H\simeq\operatorname{Sym}(n)$ or $\operatorname{Alt}(n)$, acts transitively on the set of involutions $t\in A$ with the common parameter $d(t)$. Thus, if $t\in A$ and $d(t)=d$, then $|t^{N}|=C_{n}^{d}$. As before, it is an easy check that $|t^{N}|>n$ if $d(t)>1$, and $|t^{N}|=n$ if $d(t)=1$.

Lemma 2

Suppose that $|F|=q$, while $A$ is a maximal elementary abelian subgroup $L=\Omega(V)$ with order $2^{n-1}$, and $a$ and $b$ are two distinct involutions from $A$ such that $d(a)=d(b)=1$ and $K=\langle a,b\rangle$. Then $C_{L}(a)=L_{1}:\langle b\rangle$, where $L_{1}\simeq\Omega^{\varepsilon}_{n-1}(V)$, $\varepsilon 1\equiv q(\operatorname{mod}4)$, and $C_{L}(K)=L_{0}\times K$, where $L_{0}\simeq\Omega_{n-2}(q)$. Moreover, $C_{L}(a)$ is maximal in $L$, and $C_{L}(K)$ is maximal in $C_{L}(a)$.

Proof

By Lemma 1, $A$ contains $n$ involutions $t$ for which $d(t)=1$ and all of them are conjugate in $N_{L}(A)$. Moreover, $N_{L}(A)$ acts double transitively on the set of such involutions; hence, all subgroups $K$ of the statement of Lemma 2 are conjugate in $N_{L}(A)$.

Since $C_{GO(V)}(a)=GO(V^{+})\times GO(V^{-})$, where

$$ V^{+}=\{v\in V\mid va=v\},\quad V^{-}=\{v\in V\mid va=-a\}; $$

therefore, $C_{L}(a)=C_{GO(V)}(a)\cap\Omega(V)$ includes $\Omega(V^{-})$ as a subgroup of index 2. The proof of Lemma 11.53 in [2] implies that $\Omega(V^{-})=\Omega^{\varepsilon}(V^{-})$, where $\varepsilon$ is defined by a congruence $q\equiv\varepsilon 1(\operatorname{mod}4)$ and $a$ is the only involution in $\Omega(V^{-})$ such that $d(a)=1$; hence, $b\not\in\Omega(V^{-})$ and $C_{L}(a)=\Omega(V^{-})\langle b\rangle$. Further,

$$ C_{L}(K)=C_{C_{L}(a)}(b)=\langle b\rangle\times C_{\Omega(V^{-})}(b)=L_{0}\times K, $$

where $L_{0}\simeq\Omega_{n-2}(q)$. The maximality of $C_{L}(K)$ in $C_{C_{L}(a)}(b)$ and of $C_{L}(a)$ in $L$ follows from the maximality of geometric subgroups of $\Omega_{n}(q)$ and $\Omega_{n-1}^{\varepsilon}(q)$ (see Tables 3.5.D, E, and F in [4] and Tables 8.31, 33, 39, 50, 52, 58, 66, 68, 74, 82, 84, and 85 in [3].

2. Proof of the Theorem

Let $n$ be an odd integer, $n\geq 5$, while $G$ is a periodic group such that its every subgroup lies in a subgroup isomorphic to $\Omega_{n}(q)$ for some odd $q$ which is a power of a prime. Our goal is to show that $G$ is isomorphic to $\Omega_{n}(F)$ for a suitable locally finite field $F$.

Because $\Omega_{5}(q)\simeq S_{4}(q)$, according to [5] it is true for $n=5$. By induction we may assume that $n\geq 7$, and the claim is true when $n$ is replaced with $n-1$.

Let $M(G)$ be the set of all subgroups of $G$ isomorphic to elements of $M$, while $L=\{\Omega_{n}(q)\mid q\text{ is odd}\}$. If $M(G)$ has a subgroup isomorphic to $\Omega_{n}(q)$, where $q\equiv 1(\operatorname{mod}4)$, then we fix and denote by $L=L(q)$ one of such subgroups. If there are no such subgroups, then we fix and denote as $L=L(q)$ some (arbitrary) element of $M(G)$. In both cases we will identify $L$ with $\Omega(V)$, where $V$ is an orthogonal space of dimension $n$ over a field of order $q$.

We will also fix an elementary abelian 2-subgroup $A$ of order $2^{n-1}$ from $L$, elements $a$ and $b$ from $A$, and a subgroup $K$ as they are defined in Lemma 2. If $L_{1}$ is a subgroup of $G$ isomorphic to $\Omega(V_{1})$ for some space $V_{1}$ of dimension $n$ over a field of order $q_{1}$, and $t$ is an involution from $L_{1}$; then denote by $d_{L_{1}}(t)$ the dimension of the space of fixed points of $t$ in $V_{1}$.

Lemma 3

$C_{G}(A)=A$, $N_{G}(A)$ contains a subgroup of index $1$ or $2$ coinciding with $N_{0}\simeq A:\operatorname{Alt}(n)$ from Lemma $1$.

Proof

Suppose that $c\in C_{G}(A)$. Then $C=\langle c,A\rangle$ is a finite subgroup lying in some element $L_{1}\in M(G)$. Applying Lemma 1 with $L_{1}$ instead of $L$, we get that $c\in C_{L_{1}}(A)=A$. So, $C_{G}(A)=A$, and therefore $N_{G}(A)$ is a finite subgroup lying in some $L_{2}\in M(G)$. Now we use Lemma 1 with $L_{2}$ in place of $L$ and derive that $N_{G}(A)$ includes $N_{0}$ as a subgroup of index 1 or 2.

Lemma 4

If $K\leq L_{1}\in M(G)$, then $d_{L_{1}}(a)=d_{L_{1}}(b)=d(a)=1$.

Proof

Let $A_{1}$ be a maximal elementary abelian 2-subgroup of $L_{1}$ including $K$.

If $A_{1}=A$, Lemma 1 with $L_{1}$ in place of $L$ implies that the subgroup $N_{0}$ lies in $L_{1}$. Because $|a^{N_{0}}|=|b^{N_{0}}|=n$, we have $d_{L_{1}}(a)=d_{L_{1}}(b)=1=d_{L}(a)$. In this case the claim of the lemma is true.

Suppose that $A_{1}\leq A$ and $A_{1}\neq A$. Then $|A:A_{1}|=2$ by Lemma 1. Set $C=C_{G}(A_{1})$. It is clear that $A\leq C$; and, if $t\in A\setminus A_{1}$, then

$$ C_{C}(t)=C_{G}(t)\cap C=C_{G}\left(\langle t,A_{1}\rangle\right)=C_{G}(A)=A. $$

By Shunkov’s Theorem [6], $C$ is locally finite. Let $N_{1}=N_{L_{1}}(A_{1})\leq N_{G}(A_{1})$. Since $A_{1}$ is finite, $N_{G}(A_{1})/C$ is finite, and so $N_{G}(A_{1})$ is a locally finite group. It follows that $N_{G}(A_{1})=CH$, where $H$ is a finite subgroup including $A$. Suppose that $H\leq L_{2}\in M(G)$.

Because $A\leq L_{2}$, we have $d_{L_{2}}(a)=d_{L_{2}}(b)=1$. Lemma 1 implies that $1=|a^{N_{G}(A_{1})}|=|a^{H}|$, which yields that $d_{L_{1}}(a)=d_{L_{1}}(b)=1$. On the other hand, by Lemma 1$A_{1}$ has the only involution $i$ with condition $d(i)=1$; therefore, the case under consideration when $A\neq A_{1}\leq A$ is impossible.

By induction on $s=|A:(A_{1}\cap A)|+|A_{1}:(A_{1}\cap A)|$, we will show that $|A_{1}|=|A|=2^{n-1}$ and $d_{L_{1}}(a)=d_{L_{1}}(b)=1$.

If $s\leq 3$, then either $A=A_{1}\cap A$, or $A_{1}=A_{1}\cap A$, and these cases are already done. Hence, we can assume that $A_{1}\neq A_{1}\cap A\neq A$.

Let $t\in A\setminus(A_{1}\cap A)$, $t_{1}\in A_{1}\setminus(A_{1}\cap A)$. Then $R=\langle t,t_{1},A_{1}\cap A\rangle$ is finite, and so $R\leq L_{2}\in M(G)$. Suppose that $\langle t,A_{1}\cap A\rangle\leq A_{2}$, where $A_{2}$ is a maximal elementary abelian subgroup in $L_{2}$. Then

$$ |A_{2}|\geq|A_{2}\cap A|>|A_{1}\cap A|. $$

By the inductive hypothesis, $|A_{2}|=|A|$ and $d_{L_{2}}(a)=d_{L_{2}}(b)=d(a)=1$. Further, $\langle t_{1},A_{1}\cap A\rangle\leq L_{1}\cap L_{2}$ and $\langle t_{1},A_{1}\cap A\rangle\leq A_{3}$, where $A_{3}$ is a maximal elementary abelian 2-subgroup of $L_{2}$. Because $a,b\in A_{3}$, we have $|A_{3}|=|A|$. Moreover, $|A_{3}\cap A_{1}|>|A\cap A_{1}|$. By the inductive hypothesis, $d_{L_{1}}(a)=d_{L_{1}}(b)=d_{L_{2}}(a)=1$. The lemma is proved.

Lemma 5

$C_{G}(K)=K\times R$, where $R\simeq\Omega_{n-2}(F)$ for some locally finite field $F$ of odd characteristic.

Proof

Let $\overline{C}=C_{G}(K)/K$. We want to show that $\overline{C}$ is saturated with groups from the set $M_{1}=\{\Omega_{n-2}(q)\mid q\text{ is odd}\}$.

Suppose that $\overline{X}$ is a finite group from $\overline{C}$, while $X$ is the full preimage of $\overline{X}$ in $G$. By condition, $K\leq X\leq L_{1}\in M(G)$; and by Lemma 2$C_{L_{1}}(K)=K\times R_{1}$, where $R_{1}\simeq\Omega_{n-2}(q_{1})$ for odd $q_{1}$. Thus, $\overline{C}$ is saturated with groups from the set $M_{1}=\{\Omega_{n-2}(q)\mid q\text{ is odd}\}$. By the inductive hypothesis, $\overline{C}\simeq\Omega_{n-2}(F)$ for some locally finite field $F$ of odd characteristic. In particular, $C$ is a locally finite group. We will show that $[C,C]\cap K=1$. Let $c\in[C,C]$. Then $c=[c_{1},c_{2}][c_{3},c_{4}]\dots[c_{p-1}c_{p}]$ for some $p$ and suitable elements $c_{1},\dots,c_{p}\in C$. The subgroup $\langle K,c_{1},\dots,c_{p}\rangle$ is finite and lies in $K\times Y$, where $Y\simeq\Omega_{n-2}(q_{2})$ for some $q_{2}$. It is clear that $c\in Y$ and $c\not\in K$. Since $C=K[C,C]$; therefore, $C=K\times[C,C]$, and the lemma is proved.

Lemma 6

$C_{G}(K)$ lies in a subgroup $P$ of $C_{G}(a)$ which is the union of an ascending sequence of subgroups $P_{i}$, $i=1,2,\dots$, isomorphic to $\Omega_{n-1}^{\lambda}(q_{i}).2$, with $q_{i}=\lambda 1\ (\operatorname{mod}4)$, $\lambda\in\{+,-\}$, and $\lambda$ depends on the choice of $L$ and is common for all $i$.

Proof

By Lemma 5, $C_{G}(K)$ is locally finite and countable. If $C_{G}(K)$ is finite, then we may assume that $C_{G}(K)=C_{L}(K)$, and the lemma is true by Lemma 2. Suppose that $C_{G}(K)$ is infinite and $C_{G}(K)=\{g_{i}\mid i\in{𝕅}\}$. Put $P_{0}=C_{L}(a)$. Let $g_{i_{1}}$ be an element of $C_{G}(K)$ not belonging to $P_{1}$, and the number $i_{1}$ is the smallest of those subject to that condition. The subgroup $\langle C_{P_{0}}(K),N_{0}\rangle$ coincides with $P_{0}$ by Lemma 2. Let $L_{1}$ be an element of $M(G)$ containing $C_{P_{0}}(K)$ and let $g_{i}$ be the first element in order not belonging to $C_{P_{0}}(K)$. By condition, $L_{1}\simeq\Omega_{n}(q_{1})$ for some $q_{1}$. The subgroup $C_{L_{1}}(K)$ is maximal in $C_{L_{1}}(a)$; and, because $N_{0}\not\leq C(K)$, the subgroup $\langle C_{L_{1}}(K),N_{0}\rangle$ coincides with $C_{L_{1}}(a)\simeq\Omega_{n-1}^{\lambda}(q_{1}).2$. Since $C_{L_{0}}(K)<C_{L_{1}}(K)$; therefore, $P_{0}=C_{L_{0}}(a)<C_{L_{1}}(a)=P_{1}$.

Similarly, let $L_{2}$ be an element of $M(G)$ including $C_{P_{1}}(K)$ and let $g_{i_{2}}$ be the first element in order not belonging to $C_{P_{1}}(K)$. As before, $P_{2}=C_{L_{2}}(a)$, and $P_{2}$ includes $C_{L_{1}}(a)$. Proceeding this construction of the subgroups $P_{i}$’s in a similar way, we will get an ascending sequence of subgroups $P_{i}\simeq\Omega_{n-1}^{\lambda}(q_{i})$ whose union $P$ includes $C_{G}(a)$. The lemma is proved.

Lemma 7

$P=C_{G}(a)$.

Proof

Suppose the contrary. Let $t\in C_{G}(a)\setminus P$. The subgroup $\langle K,K^{t}\rangle$ is generated by the elements $a$, $b$, and $b^{t}$. Since $\langle b,b^{t}\rangle$ is a finite group, so is $\langle K,K^{t}\rangle$. Because $\langle K,K^{t}\rangle$ lies in the subgroup $L^{*}$ isomorphic to $\Omega_{n}(q^{*})$ for some $q^{*}$; therefore,

$$ \langle K,K^{t}\rangle\leq D=C_{L^{*}}(a)\simeq\Omega_{n-1}^{\lambda}(q^{*})\cdot 2. $$

Now, $K$ and $K^{t}$ are conjugate in $D$, because $A$ and $A^{t}$ are conjugate in $D$ and $N_{D}(A)$ acts double transitively on the set of involutions $a^{*}\in A$ with condition $\alpha_{D}(a^{*})=1$. Let $\Gamma$ be a graph with vertex set $\Sigma=\{K^{d}\mid d\in D\}$ such that two vertices $K^{r}$ and $K^{s}$ are adjacent if and only if $[K^{r},K^{s}]=1$. Suppose that $\Delta$ is a connected component in $\Gamma$ that includes $K$. Since $K\leq A$ and $C_{N}(K)\neq C_{N}(a)$, we have $K^{C_{N}(a)}\neq\{K\}$, and so $|\Delta|\geq 2$. Thus, if $\Delta\neq\Sigma$, then $D$ acts on $\Sigma$ by conjugation transitively and imprimitively. Hence, the stabilizer of the vertex $K$ in $D$ equal to $N_{D}(K)$ is not maximal in $D$, which is untrue. Therefore, $\Delta=\Sigma$ and there is a sequence $t_{1},t_{2},\dots,t_{r}=t$ of elements from $C_{L^{*}}(a)$ such that $1=[K,K^{t_{1}}]=[K^{t_{i}},K^{t_{i+1}}]$, $i=1,2,\dots,r-1$. By induction on $r$, we will show that $K^{t_{r}}\leq P$. If $r=1$, then $K^{t}\leq C_{G}(K)$; and by Lemma 6$K^{t}\leq P$. Suppose that $r>1$ and $K^{t_{r-1}}\leq P$. There exists $u\in P$ such that $K^{t_{1}u}=K$ and $1=[K,K^{t_{2}u}]=\dots=[K^{t_{r-1}u},K^{t_{r}u}]$, $i=2,\dots,r$. By the inductive hypothesis, $K^{t_{r}u}\leq P$ and $K^{t_{r}}\leq P$.

So, $K^{t}\leq P$ for every $t\in C_{G}(a)$. The subgroup $P$ is locally finite; therefore, $\langle K,t\rangle$ is finite and lies in $L^{*}\in M(G)$. Suppose that $H=C_{L^{*}}(a)$. Then

$$ H=\langle C_{L^{*}}(K),N_{G}(A)\cap C_{L^{*}}(a)\rangle\leq P. $$

Because $t\in H$, we have $t\in P$, and the lemma is proved.

Lemma 8

$C_{G}(a)$ lies in a subgroup $Z\simeq\Omega_{n}(F)$ of $G$ for some locally finite field $F$.

Proof

By Lemma 7, $C_{G}(a)$ is countable and locally finite. Let $L_{0}=L$ and

$$ C_{G}(a)=C_{L_{0}}(a)\cup\{g_{i}\in C_{G}(a)\mid i\in{𝕅}\}. $$

The subgroup $C_{1}=\langle C_{L}(a),g_{i_{1}}\rangle$, with $g_{i_{1}}$ the first element in order not belonging to $L_{0}$, is finite and hence lies in $L_{1}\in M(G)$. Because $L_{0}$ includes $N_{G}(A)$ and $C_{L_{1}}(a)$ is maximal in $L_{1}$, the subgroup $\langle C_{1},N_{G}(A)\rangle$ coincides with $L_{1}$.

Let $C_{2}=\langle C_{1},g_{i_{2}}\rangle$, where $g_{i_{2}}$ is the first element in order which is not contained in $L_{1}$. By condition, $C_{2}\leq L_{2}\in M(G)$. It is clear that $C_{2}\leq C_{L_{2}}(a)$ and $L_{2}$ includes $N_{L}(A)=N_{L_{2}}(A)$. Since $C_{L_{2}}(a)$ is maximal in $L_{2}$ and $N_{L_{2}}(A)\not\leq C_{L_{2}}(a)$, the subgroup $L_{2}$ coincides with $\langle C_{L_{2}}(a),N_{L_{2}}(A)\rangle$ and includes $L_{1}$.

Reasoning similarly, we construct subgroups $L_{3},L_{4},\dots\in M(G)$ with condition $L_{i}\leq L_{i+1}$, $i=3,4,\dots$. The union $Z$ of the so-obtained sequence includes $C_{G}(a)$. By the main result of each of the papers [7,8,9,10,11], $Z\simeq\Omega_{n}(F)$ for some locally finite field $F$, and the lemma is proved.

Lemma 9

$Z=G$.

Proof

By Lemma 8, $Z$ is countable and locally finite. Suppose that $g\in G$ and $a^{g}\neq a$. The group $\langle a,a^{g}\rangle$ is finite. Therefore, $\langle a,a^{g}\rangle$ lies in such a subgroup $R$ of the group $G$ that is isomorphic to $\Omega_{n}(r)$ for some $r$. Let $\Delta$ be a set of involutions belonging to $R$ and conjugate to $a$ in $G$. Let $\Gamma$ be a graph with vertex set $\Delta$ such that two vertices $a^{g_{1}}$ and $a^{g_{2}}$ are adjacent if and only if $[a^{g_{1}},a^{g_{2}}]=1$. Since $C_{R}(a)$ is a maximal subgroup of $R$ and $|C_{R}(a)\cap\Delta|\geq 2$; we have by analogy to Lemma 8 that the graph $\Gamma$ is connected. This implies as in Lemma 8 that $a^{G}\subseteq Z$, $\langle a^{G}\rangle=Z$, and $Z\trianglelefteq G$. Because $Z$ is locally finite, $\langle a,g\rangle$ is finite, and we may assume that $\langle a,g\rangle\leq R$. Since $\langle a^{R}\rangle=R$; therefore, $g\in\langle c^{R}\rangle\leq Z$. The lemma is proved, which completes the proof of the theorem.