# Sharply doubly transitive groups with saturation conditions

Anatoliy I.Sozutov, Evgeny B. Durakov

Annotation: A number of conditions were found under which a sharply doubly
transitive permutation group has an abelian normal divider.

Keywords: exactly doubly transitive group, Frobenius groups, saturation
condition, finite and generalized finite the elements.

Recall that the permutation group $G$ of the set $F$ ($|F|\geq k$) is called
is exactly $k$ -transitive on $F$ if for any two ordered sets
$(\alpha_{1},\,...,\,\alpha_{k})$ and $(\beta_{1},\,...,\,\beta_{k})$ elements
from $F$ such that $\alpha_{i}\neq\alpha_{j}$ and $\beta_{i}\neq\beta_{j}$ for
$i\neq j$, there is exactly one element of the group $G$ taking $\alpha_{i}$
to $\beta_{i}$ ($i=1,\,...,\,k$).

As K. Jordan [9] proved, in a finite sharply doubly transitive group $T$
regular permutations (moving each element of the set $F$) together with the
identity substitution constitute an abelian normal subgroup. Is Jordan’s
theorem true for infinite groups, in the general case is unknown (see, for
example, [3, questions 11.52, 12.48]). To particular solutions of this central
question in the theory of near-fields and near-domains [10] dozens of works by
famous authors are devoted. We especially note that in 2014 in [4, 8] exactly
doubly transitive groups without abelian normal subgroups, characteristics 2.
[5] and [6] constructed examples of exactly 2-transitive groups and simple
exactly 2-transitive groups, respectively, that have characteristic 0 and do
not contain non-trivial Abelian normal subgroups. And in 2008 [7] studied the
question for exactly 2-transitive groups of characteristic 3. For other
characteristics the question remains open. Sharply 2-transitive groups are
closely related to algebraic structures such as near-fields, near-areas, $KT$
-fields (Kerby-Tits fields), projective planes, etc. (see [10, Ch. V], [12,
chap. 20]).

We continue to investigate infinite exactly doubly transitive groups and
related near-domains [1, 2]. In this paper, a number of conditions are found
under which a group has an abelian normal divisor (see [3, questions 11.52,
12.48]), and the corresponding near-domain [10] is a near-field. All necessary
definitions are given in paragraph 1. We only recall that the group $G$ is
saturated with groups of some set of finite groups $\mathfrak{X}$ if every
finite subgroup of $G$ is contained in a subgroup of the group $G$ isomorphic
to some group from $\mathfrak{X}$.

This work was supported by the Russian Foundation for Basic Research (project
19-01-00566 A) and Krasnoyarsk mathematical center, funded by the Ministry of
Education and Science of the Russian Federation in the framework of activities
for the creation and development of regional NMCs (Agreement 075-02-2020-1534
/ 1).

######  Theorem 1.

Sharply doubly transitive group $T$ saturated with finite Frobenius groups of
substitutions, the set $F$ of odd characteristic has a regular abelian normal
subgroup and the almost domain $F(+,\cdot)$ is a near-field if at least one of
the following properties of the stabilizer $T_{\alpha}$ of the point
$\alpha\in F$:

  1. (1)

$T_{\alpha}$ — Shunkov group;

  2. (2)

$T_{\alpha}$ is a periodic group and $T_{\alpha}$ does not contain conjugate
dense subgroups;

  3. (3)

$T_{\alpha}$ is (locally) a finite normal subgroup of order more than 2;

  4. (4)

$T_{\alpha}$ contains a finite element $a$ of prime odd order not equal to
five.

######  Theorem 2.

Let in a sharply triply transitive permutation group $G$ odd characteristic
point stabilizer has a regular abelian normal subgroup and each of its finite
subgroups is contained in a finite unsolvable subgroup of the group $G$. Then
$G$ is locally finite.

##  1\. Results and concepts used

Let $T$ be a sharply doubly transitive group permutations of the set $F$.
According to M. Hall’s theorem [10], we assume that $F$ is almost a domain
with operations of addition $+$ and multiplication $\cdot$, and $T$ is its
affine transformation group $x\to a+bx$ $(b\neq 0)$, and $T_{\alpha}\simeq
F^{*}(\cdot)$ is the stabilizer of the point $\alpha\in F$. Let $T_{\alpha}$
be the stabilizer of the point $\alpha\in F$, $J$ denotes the set of
involutions of the group $T$. For each involution $k\in J$ by $N_{k}$ we
denote the set $kJ=\\{kj\mid j\in J\\}$. The result of a substitution action
$t\in T$ to an element (point) $\gamma\in F$ is denoted by $\gamma^{t}$. The
following statements are true.

######  1.

$(T,T_{\alpha})$ — a Frobenius pair, i.e. $T_{\alpha}\cap T^{t}_{\alpha}=1$
for anyone $t\in T\setminus T_{\alpha}$.

######  2.

The group $T$ contains involutions and all involutions in $T$ are conjugate.
The set $J\setminus T_{\alpha}$ is nonempty and $T_{\alpha}$ is transitive on
$J\setminus T_{\alpha}$. Product of two different involutions from $T$ is a
regular substitution, i.e. acting on $F$ without fixed points.

######  3.

If $T_{\alpha}$ contains an involution $j$, then it is unique in $T_{\alpha}$,
$T=T_{\alpha}N_{j}$ and $T$ acts by conjugation on the set $J$ of all its
involutions exactly twice transitively. All products $kv$, where $k,v\in J$,
$k\neq v$ in the group $T$ are conjugate and have the same or a simple odd
order $p$, or infinite order. In the first case ${\rm Char}\,T=p$, in the
second case ${\rm Char}\,T=0$ (if there are no involutions in $T_{\alpha}$,
then ${\rm Char}\,T=2$ by definition).

######  4.

The following result was proved in [1] Let ${\rm Char}\,T=p>2$, $b$ is
strictly real with respect to $j$ is an element from $T\smallsetminus
T_{\alpha}$, $A=C_{T}(b)$ and $V=N_{T}(A)$. Then:

1) the subgroup $A$ is periodic, abelian is inverted by the involution $j$ and
is strongly isolated in $T$;

2)the subgroup $V$ acts exactly twice transitively on the orbit
$\Delta=\alpha^{A}$, moreover, $A$ is an elementary abelian regular normal
subgroup of $V$, $H=V\cap T_{\alpha}$ is a point stabilizer and $V=A\lambda
H$. If in addition $|\bigcap_{x\in T_{\alpha}}H^{x}|>2,$ then $A$ is a regular
abelian normal subgroup of $T$, and near-domain $F(+,\cdot)$ is a near-field.

The next proposition shows that the saturation condition finite Frobenius
groups partially holds in an arbitrary exactly doubly transitive group of odd
characteristic.

######  5.

When ${\rm Char}\,T=p>2$, each dihedral subgroup of $T$ is contained in a
finite Frobenius subgroup of $T$ with kernel of order $p$ and a cyclic
complement of order $p-1$.

Recall that $(a,b)$ \- finiteness condition in the group $G$ means that $a$
and $b$ — its nontrivial elements, and in the group $G$ all subgroups $\langle
a,b^{x}\rangle$ ($x\in G$). The elements $a$ and $b$ are called is generalized
finite, and if $a=b$, then $a$ is called finite elements in the group $G$. The
following proposition was proved in [2, Lemmas 1, 2 and Theorem]:

######  6.

Let $T$ be a sharply doubly transitive group characteristics ${\rm
Char}\,T\neq 2$ contains elements $a,b$ with $(a,b)$ \- condition limbs. Then

  1. (1)

If for any $x\in T$ the subgroup $\langle a^{x}\rangle$ and $\langle b\rangle$
are not incident, then at least one of the elements $a,b$ belongs to the
stabilizer of some point of the near-domain $F$;

  2. (2)

If $|a|=|b|>2$, $T$ has a regular abelian normal subgroup;

  3. (3)

If $|a|\cdot|b|=2k>4$, $T$ has a regular abelian normal subgroup.

######  7.

[13, Theorem 2] If the group $T$ has is a finite element of order $>2$, then
$T$ has a regular abelian normal subgroup and near-domain $F$ is a near-field.

######  8.

If $T$ contains a locally finite subgroup containing a regular substitution
and intersecting $T_{\alpha}$ by a normal subgroup in it, consisting of more
than two elements, then $T$ has a regular abelian normal subgroup, and the
near-domain $F(+,\cdot)$ is near-field.

######  9.

If ${\rm Char\,}T\neq 2$, $T$ contains the Frobenius group $V$ with involution
and complement $H=V\cap T_{\alpha}$, where $T_{\alpha}$ is the stabilizer of
the point $\alpha$, and in $H$ there is a normal in $T_{\alpha}$ is a subgroup
of order $>2$, then the group $T$ has a regular abelian normal subgroup.

######  10.

In the complement of a finite Frobenius group, each cyclic subgroup of prime
order $q>5$ is normal.

##  2\. Proof of Theorem 1

Let the group $T$ and the near-domain $F$ satisfy the conditions of Theorem 1.
The set of Frobenius subgroups of the group $T$ containing finite subgroup
$L$, we denote by $\mathfrak{X}(L)$. For any Frobenius group $M\leq T$, we
agree to its kernel denote by $F_{M}$, and the complement suitable for the
meaning of the text — through $H_{M}$. The notation $L\leq H_{M}$ often used
in the following text for the subgroup $L\leq M\in\mathfrak{X}(L)$ will be
mean that $L$ is contained in some complement $H_{M}$ group $M$.

Comment. It is well known that a near-domain $F(+,\cdot)$ is a near-field if
and only if when $T$ has a regular abelian normal subgroup. Therefore, to
prove Theorem 1 it suffices to prove the existence in the infinite group $T$
of a regular abelian normal subgroup.

######  Lemma 1.

We can assume what ${\rm Char}\,T\neq 3$.

Proof In the case of ${\rm Char}\,T=3$, the group $T$ has a regular abelian
normal subgroup and without the additional saturation condition (see, for
example, [20, Lemma 2.7]. The lemma is proved.

######  Lemma 2.

When $T_{\alpha}$ has is a finite element of prime order $q>5$, the theorem is
true.

Proof Let ${\rm Char}\,T=p>3$, $j$ be an involution from $T_{\alpha}$ and $a$
— final in $T_{\alpha}$ is an element of prime order $q>5$. Let’s choose an
arbitrary element $t\in T_{\alpha}$. Due to the finiteness of the element $a$
and item 3, the subgroup $L_{t}=\langle a,a^{t},j\rangle$ is finite, and
$L_{t}\leq M\in\mathfrak{X}(K)$. In view of item 1 and the properties of
finite Frobenius groups $M\cap T_{\alpha}\leq H_{M}$, $F_{M}\cap T_{\alpha}=1$
and $F_{M}$ is an elementary abelian $p$ -group, consisting of all products
$jk$, where $k\in J\cap M$. According to the structure of complements in
finite Frobenius groups $L_{t}=\langle a,a^{t}\rangle=\langle a\rangle$. Since
the element $t\in T_{\alpha}$ is arbitrary, we conclude that that the subgroup
$\langle a\rangle$ is normal in $T_{\alpha}$. Therefore, the subgroup
$L=\langle a,j\rangle$ is contained in the appendix finite Frobenius group
$M^{x}\in\mathfrak{X}(A)$ for any $x\in T_{\alpha}$, and the set-theoretic
union of the kernels of all such groups (for $x\in T_{\alpha}$), contains the
set $N_{j}$. From this and the equality $T=T_{\alpha}N_{j}$ (item 3) it
follows that all subgroups $L_{g}=\langle a,a^{g}\rangle$ are finite for any
$g\in T$, and by item 6 $T$ possesses a regular abelian normal subgroup. The
lemma is proved.

######  Lemma 3.

If $T_{\alpha}$ is a Shunkov group and ${\rm Char}\,T\neq 2$, then
$T_{\alpha}$ has a local finite periodic part and subgroup
$\Omega_{1}(T_{\alpha})$ generated by all elements of prime orders from
$T_{\alpha}$, there is a group of one of the following types: 1) (locally)
cyclic group; $\Omega_{1}(T_{\alpha})=C\times L$, where $C$ — (locally) cyclic
$\\{2,3\\}^{\prime}$ – group, and $L\simeq SL_{2}(3)$; 3)
$\Omega_{1}(T_{\alpha})=\times L$, where $C$ is (locally) cyclic
$\\{2,3,5\\}^{\prime}$ – group, $L\simeq SL_{2}(5)$. In any case, in
$T_{\alpha}$ is a finite normal subgroup of order greater than 2.

Proof By virtue of the saturation condition and the condition ${\rm
Char}\,T\neq 2$, each finite the subgroup $K$ of $T_{\alpha}$ is contained in
the complement of the finite group Frobenius $M\in\mathfrak{X}(K)$, and
according to To the proposition [14, Proposition 11], its subgroup
$\Omega_{1}(K)$ has the structure indicated in the lemma. By [13, Theorem 1],
the same structure has subgroup $\Omega_{1}(T_{\alpha})$, and $T_{\alpha}$
possesses a local finite periodic part, we denote it by $S$. If $S$ not
2-group, then $\Omega_{1}(T_{\alpha})$ obviously contains a finite normal
subgroup of $T_{\alpha}$ of order, greater 2. If $S$ is a 2-group, then by
[21, Theorem 2] either a quasicyclic group, either locally quaternionic and
also contains finite normal in $T_{\alpha}$ a subgroup of order $2^{n}$ for
any $n>1$. The lemma is proved.

######  Lemma 4.

The theorem is true if $T_{\alpha}$ contains a finite normal subgroup $L$,
$|L|>2$.

Proof Since the involution of $j$ in $T_{\alpha}$ is unique, we can assume
that $j\in L$. By the saturation condition $M\in\mathfrak{X}(L)$, and the
subgroup $M\cap T_{\alpha}$ is strongly isolated in $M$. Therefore, the
subgroup $L$ is the complement of some finite group Frobenius
$M\in\mathfrak{X}(L)$. From the normality of $L$ to $T_{\alpha}$ it follows
that $M^{t}\in\mathfrak{X}(L)$ and $H_{M}=L$. Hence it follows that set-
theoretic union of the kernels of all such groups (for $t\in T_{\alpha}$),
contains the set $N_{j}$. Let $a$ be an element of order greater than 2 from
$L$ (since $|L|>2$, such an element exists by virtue of item 3). It follows
from what has been proved that all subgroups $L_{g}=\langle a,a^{g}\rangle$
are finite for any $g\in T$, and by item 6 $T$ possesses a regular abelian
normal subgroup. The lemma is proved.

######  Lemma 5.

When $T_{\alpha}$ contains a finite element $a$ of order $3$, the theorem is
also true.

Proof As in the lemma 2, it is proved that for any $t\in T_{\alpha}$ the
finite subgroup $L_{t}=\langle a,a^{t}\rangle$ is contained in the appendix of
a finite Frobenius group $M$ from $\mathfrak{X}(L)$. As in the lemma 3, we
conclude that either $L_{t}=\langle a\rangle$, or $L_{t}$ is isomorphic to one
of the groups $SL_{2}(3)$, $SL_{2}(5)$. By the main theorem in [16], the
normal closure $L=\langle a^{T_{\alpha}}\rangle$ of $a$ in $T_{\alpha}$ is
locally finite. As follows from the proof of the lemma 3, or $L=\langle
a\rangle$, or $L$ is isomorphic to one of the groups $SL_{2}(3)$, $SL_{2}(5)$.
By lemma 4, the theorem is true. The lemma is proved.

A proper subgroup $H$ of a group $G$ is called conjugate dense, if $H$ has a
non-empty intersection with every conjugacy class in $G$ elements.

######  Lemma 6.

The theorem is true when $T$ is a periodic group and $T_{\alpha}$ has no
conjugate dense subgroups.

Proof By item 3, $T_{\alpha}$ has a unique involution $j$, therefore, for an
arbitrary element $a$ of finite order from $T_{\alpha}$ the subgroup
$L=\langle a,j\rangle$ is finite. Saturation condition $L\leq
M\in\mathfrak{X}(L)$, and we can assume that $H_{M}=L$, $F_{M}\subseteq
N_{j}$. By item 4, for any nonidentity element $b\in N_{j}$, the subgroup
$A=C_{T}(b)$ is periodic, contained in $N_{j}$ and strongly isolated in $T$,
in this case, $N_{T}(A)=A\leftthreetimes H$, where $H=T_{\alpha}\cap
N_{T}(A)$. Since in view of paragraphs. 2, 3 $T_{\alpha}$ acts transitively by
conjugation on the set $N_{j}$, then $F_{M}^{x}\leq A$ for some $x\in
T_{\alpha}$. This implies that $a^{x}\in H$ and $H$ is the conjugate dense
subgroup group $T_{\alpha}$. According to the conditions of the lemma,
$H=T_{\alpha}$, by item 3, $A=N_{j}$, and the lemma is proved.

We now complete the proof of the theorem. The first statement of the theorem
follows from Lemmas 3, 4. Statement 2 is proved in Lemma 6. Statement 3 of
Theorem coincides with Lemma 4. Statement 4 follows from Lemmas 2 and 5. The
theorem is proved.

##  3\. Proof of Theorem 2

Let $G$ be an infinite sharply triply transitive permutation group multitudes
$X=F\cup\\{\infty\\}$ satisfies the conditions of Theorem 2. As in [22], by
$B$ we denote the stabilizer $G_{\alpha}$ of the point $\alpha\in X$ and
through $H$ — stabilizer $G_{\alpha\,\beta}=G_{\alpha}\cap G_{\beta}$ of two
points $\alpha=\infty\in X$, $\beta\in F$. Let also $J$ be the set of
involutions of the group $G$, and $J_{m}$ be the set involutions stabilizing
exactly $m$ points, $m=0,\,1,\,2$.

Proof of the theorem. By the lemma, 1 $B=U\leftthreetimes H$ is a Frobenius
group, $H$ contains an involution $z$ and $N=N_{G}(H)=H\leftthreetimes\langle
v\rangle$, where $v\in J$. For $b\in U^{\\#}$ there is an element $a\in H$ of
order $p-1$ such that $\langle a,b\rangle=\langle
b\rangle\leftthreetimes\langle a\rangle$ — exactly twice transitive group of
order $p(p-1)$. Let $S$ be an arbitrary finite subgroup from $U$ containing an
arbitrary element $c$ from $U\setminus\langle b\rangle$ and $K=\langle
b,S,a\rangle$. The subgroup $K$ is obviously finite and by the saturation
condition $K\leq M$, where $M$ is a finite unsolvable group. Let $L=L(M)$ be
the layer of the group $M$ [11, Proposition 1.4, p. 53]. It is clear that
$Z(L)=1$ and since the 2-rank of the group $G$ is 2, then $L$ is a simple
group. By virtue of lemma 6 [22] $L$ is isomorphic to $L_{2}(p^{n})$ and
$P=U\cap L$ is a Sylow $p$ -subgroup of the group $L$. As known, all cyclic
subgroups of $P$ are conjugate in the subgroup $N_{L}(P)$, and $P^{\\#}$
splits into two conjugate classes. Since $\langle a,b\rangle\leq M$ and
$\langle a\rangle$ acts transitively to $\langle b\rangle^{\\#}$, then
obviously $|M:L|=2$ and $M\simeq PGL_{2}(p^{n})$. The subgroup $N_{M}(P)$ acts
transitively on $P^{\\#}$, therefore $c=b^{h}$ for some $h\in H\cap M$. Since
the element $c\in P^{\\#}$ we conclude that $H$ is a periodic group. By [15,
Theorem 2], the group $G$ is locally finite. The theorem is proved.

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