# Criterion of nonsolvability of a finite group
and recognition of direct squares of simple groups

Zh. Wang, A. V. Vasil’ev, M. A. Grechkoseeva, and A. Kh. Zhurtov

###### Abstract.

The spectrum $\omega(G)$ of a finite group $G$ is the set of orders of its
elements. The following sufficient criterion of nonsolvability is proved: if
among the prime divisors of the order of a group $G$, there are four different
primes such that $\omega(G)$ contains all their pairwise products but not a
product of any three of these numbers, then $G$ is nonsolvable. Using this
result, we show that for $q\geqslant 8$ and $q\neq 32$, the direct square
$Sz(q)\times Sz(q)$ of the simple exceptional Suzuki group $Sz(q)$ is uniquely
characterized by its spectrum in the class of finite groups, while for
$Sz(32)\times Sz(32)$, there are exactly four finite groups with the same
spectrum.

Key words: criterion of nonsolvability, simple exceptional group, element
orders, recognition by spectrum.

MSC: 20D60, 20D06.

A. V. Vasil’ev was supported by National Natural Science Foundation of China
(No. 12171126). A. V. Vasil’ev and M. A. Grechkoseeva were supported by the
Program of fundamental scientific researches of the Siberian Branch of Russian
Academy of Sciences, I.1.1, project 0314-2019-0001.

## Introduction

Let $G$ be a finite group (in what follows, all groups are assumed to be
finite). The set of prime divisors of its order and the set of its element
orders are denoted by $\pi(G)$ and $\omega(G)$, respectively, the latter set
for brevity we call the _spectrum_ of the group $G$. Groups $G$ and $H$ are
called _isospectral_ if $\omega(G)=\omega(H)$. A group $G$ is called
_recognizable by spectrum_ if any group isospectral to $G$ is isomorphic to
$G$. More details about recognition of simple and related groups by spectrum
can be found in the recent survey [1]. The present paper concerns the
recognition of direct squares of finite simple groups.

As a rule, the first step in proving the recognizability of nonsolvable groups
is to establish that the groups isospectral to them are also nonsolvable, so
we start with the following sufficient criterion of nonsolvability of a finite
group, which may be of independent interest.

######  Theorem 1.

Let $G$ be a finite group. Suppose that $\pi(G)$ includes a subset

$\sigma(G)=\\{p_{i}\mid i\in I\\},\,|I|\geqslant 4,$

satisfying the following:

  1. (1)

$p_{i}p_{j}\in\omega(G)$ for all distinct $i,j\in I;$

  2. (2)

$p_{i}p_{j}p_{k}\not\in\omega(G)$ for all pairwise distinct $i,j,k\in I.$

Then $G$ is nonsolvable. In particular, there is no such group of odd order.

######  Remark 1.

Theorem 1 is wrong for $|I|=3$. It suffices to consider a $3$-primary group
$G$ with $\pi(G)=\\{p,q,r\\}$ which is the direct product of two Frobenius
groups whose kernels are a $p$-group and $q$-group, and the complements are
$r$-groups.

######  Remark 2.

If $|I|\geqslant 6$, then Theorem 1 follows from [2, Theorem 1]. Indeed, let
$G$ satisfy the conditions of Theorem 1 and $\sigma=\sigma(G)$. If $G$ is
solvable, then it includes a Hall $\sigma$-subgroup $H$ and, by Condition 2,
the inequality $\alpha(H)\leqslant 2$ holds. Here, as in [2], we denote by
$\alpha(H)$ the maximum of $|\pi(\langle x\rangle)|$ for all elements $x$ of
$H$. Applying [2, Theorem 1], we obtain that

$|\pi(H)|\leq\alpha(H)(\alpha(H)+3)/2\leqslant 5;$

a contradiction. If $|I|\leqslant 5$, then, as examples from [3, 2] show, it
is impossible to derive the conclusion of Theorem 1 applying Condition 2 only.

######  Remark 3.

If $\sigma(G)$ satisfies the additional condition: $p_{i}$ does not divide
$p_{j}-1$ for all $i,j\in I$, then the conclusion of Theorem 1 was proved in
[4, Proposition 1].

Before returning to recognition, we introduce for convenience one more concept
related to element orders: the prime graph $GK(G)$ of a group $G$ is a graph
with vertex set $\pi(G)$, in which two vertices $p$ and $q$ are adjacent if
and only if $p\neq q$ and $pq\in\omega(G)$. Denote by $t(G)$ the maximum
number of pairwise nonadjacent vertices in $GK(G)$. If the group $L$ satisfies
the condition $t(L)\geqslant 4$, then it is easy to see that the direct square
$L\times L$ satisfies the conditions of Theorem 1, and we get

######  Corollary 2.

Let $L$ be a finite group and $t(L)\geqslant 4$. If $G$ is a finite group with
$\omega(G)=\omega(L\times L)$, then $G$ is nonsolvable.

######  Remark 4.

The condition $t(L)\geqslant 4$ is satisfied for the vast majority of simple
groups $L$, namely, for alternating groups of degree at least $19$, classical
groups of dimension at least $11$, and all exceptional groups except
$G_{2}(q)$, ${}^{3}D_{4}(q)$, and ${}^{2}F_{4}(2)^{\prime}$ (see more in [5]).

Despite the fact that there are a huge number of simple groups recognizable by
spectrum (see [1, Theorem 1]), there are only two examples of a recognizable
group that is a direct square of a simple group: $Sz(2^{7})\times Sz(2^{7})$
[6] and $J_{4}\times J_{4}$ [4]. The reason for this is that the standard
methods of proving the nonsolvability of groups isospectral to a simple group
are based on properties of its prime graph. Clearly, they are not applicable
to proving the nonsolvability of a group isospectral to the square of a simple
group, since in this case the corresponding graph is always complete.
Proposition 1 of [4], mentioned in Remark 3, is a significant advance in this
direction, but it is difficult to verify the conditions of this proposition
for squares of arbitrary simple groups. As Corollary 2 shows, Theorem 1 of the
present paper does not have this disadvantage and therefore allows us to hope
for new examples of recognizable groups that are the squares of simple groups.

In this paper we consider the direct squares of simple Suzuki groups $Sz(q)$.
Recall that $Sz(q)={}^{2}B_{2}(q)$, where $q=2^{\alpha}$ with odd $\alpha$, is
a twisted exceptional group of Lie type. It is well known that $Sz(q)$ is a
nonabelian simple group if $q\geqslant 8$ and a Frobenius group of order
$5\cdot 4$ if $q=2$. As mentioned above, $Sz(2^{7})\times Sz(2^{7})$ is
uniquely (up to isomorphism) characterized by its spectrum in the class of
finite groups [6]. We prove the following generalization of this result.

######  Theorem 3.

If $q\geqslant 8$ and $q\neq 32$, then the group $Sz(q)\times Sz(q)$ is
recognizable by spectrum.

######  Theorem 4.

Let $L=Sz(32)$. The finite groups isospectral to $L\times L$ are exactly the
groups of the form $(L\times L)\rtimes\langle\varphi\rangle$, where either
$\varphi=1$, or $\varphi$ normalizes each direct factor and induces a field
automorphism of order $5$ on it. In particular, up to isomorphism, there are
four finite groups isospectral to $L\times L$.

Note that the recognizability by spectrum of the simple Suzuki groups
themselves was established by W. Shi [7] about 30 years ago. It is also worth
noting that the proof of Theorem 1 does not depend on the classification of
finite simple groups, while the proof of Theorems 3 and 4 uses only the
assertion proved back in 1972 in [8] that nonabelian simple groups whose
orders are not divisible by $3$ are exactly the Suzuki groups.

The article is organized as follows. In the first section, the necessary
preliminary information and results are collected, in the second, the
nonsolvability criterion from Theorem 1 is proved, and in the third, the
recognizability of the squares of Suzuki groups is established.

Concluding the introduction, the authors would like to express their deep
gratitude and admiration to Viktor Danilovich Mazurov whose article [6],
published in Algebra and Logic 25 years ago, served as a source of inspiration
for this paper. Also the authors are grateful to A. A. Buturlakin for valuable
comments.

##  1\. Preliminaries

For a natural number $n$, $\pi(n)$ denotes the set of prime divisors of $n$.
If $\pi$ is the set of primes, then $\pi^{\prime}$ is the set of primes not in
$\pi$.

Let $G$ be a group. The set $\omega(G)$ is closed with respect to taking
divisors; therefore, it is uniquely defined by its subset $\mu(G)$ consisting
of elements maximal with respect to divisibility relation. The exponent of $G$
is denoted by $\operatorname{exp}(G)$. Also, $\operatorname{Aut}G$ and
$\operatorname{Out}G$ are respectively the automorphism group and the outer
automorphism group of $G$, $F(G)$ is its Fitting subgroup, and
$\operatorname{Soc}(G)$ is its socle, i.e., the product of the minimal normal
subgroups. If $G$ is a $p$-group, then $\Omega_{1}(G)$ is its subgroup
generated by elements of order $p$. If a group $B$ acts on a group $A$, then
$A\rtimes B$ denotes their natural semidirect product.

######  Lemma 1.1 (Bang [9], Zsigmondy [10]).

Let $q,n\geqslant 2$ be integers. Then either there is a prime number $r$ that
divides $q^{n}-1$ and does not divide $q^{i}-1$ for all $i<n$, or one of the
following conditions is satisfied:

  1. (1)

$q=2$ and $n=6$;

  2. (2)

$q$ is a Mersenne prime and $n=2$.

A prime number $r$ from Lemma 1.1 is called a _primitive prime divisor_ of
$q^{n}-1$.

######  Lemma 1.2.

If $G$ is a solvable group, then $t(G)\leqslant 2$.

###### Proof.

Otherwise, the spectrum of a Hall $\sigma$-subgroup of $G$, where $\sigma$ is
the set of three vertices of the graph $GK(G)$ that are not pairwise adjacent
to each other, consists only of powers of primes, which contradicts [11,
Theorem 1]. ∎

A group $G$ is called a cover of a group $L$ if there exists an epimorphism
from $G$ onto $L$. The following four lemmas give known facts about the orders
of elements in covers of finite groups. Note that the references are given to
those papers in which these statements were formulated in a form convenient
for further discussion. Say, Lemma 1.6 is a variation of the celebrated Hall –
Higman theorem [12].

######  Lemma 1.3.

[13, Lemma 10] Let $K$ be a normal elementary abelian subgroup of a group $G$,
$G/K\simeq L$ and $G_{1}=K\rtimes L$ is the natural semidirect product. Then
$\omega(G_{1})\subseteq\omega(G)$.

######  Lemma 1.4.

[6, Lemma 4] Let $G=P\rtimes(R_{1}\times\dots\times R_{k})$, where $P$ is a
nontrivial $p$-subgroup, $R_{i}$ is an elementary abelian group of order
$r_{i}^{2}$, $1\leqslant i\leqslant k$, and primes $p$, $r_{1}$, …, $r_{k}$
are pairwise distinct. Then $G$ contains an element of order $pr_{1}\dots
r_{k}$.

######  Lemma 1.5.

[14, Lemma 1] Let $P$ be a normal $p$-subgroup of a finite group $G$ and let
$G/P$ be a Frobenius group with kernel $F$ and cyclic complement $C$. If $p$
does not divide $F$ and $F\not\subseteq PC_{G}(P)/P$, then $G$ contains an
element of order $p|C|$.

######  Lemma 1.6.

[15, Lemma 3.6] Let $G=R\rtimes\langle g\rangle$, where $R$ is an $r$-group,
$g$ is of prime order $s$, $s\neq r$, and $[R,g]\neq 1$. Suppose that $G$ acts
faithfully on an elementary abelian $p$-group $P$, where $p\neq r$. Then
either the natural semidirect product $P\rtimes G$ contains an element of
order $ps$, or the following conditions are satisfied:

  1. (1)

$C_{R}(g)\neq 1$;

  2. (2)

$R$ is nonabelian;

  3. (3)

$r=2$ and $s$ is a Fermat prime.

Lemma 1.7 collects well-known facts about Suzuki groups, and the next two
lemmas deal with element orders in covers and automorphic extensions of these
groups.

######  Lemma 1.7.

Let $G=Sz(q)$, where $q=2^{\alpha}\geqslant 8$. Then

  1. (1)

$|G|=q^{2}(q-1)(q^{2}+1);$

  2. (2)

$\mu(G)=\\{4,q-1,q+\sqrt{2q}+1,q-\sqrt{2q}+1\\};$

  3. (3)

$G$ includes Frobenius groups of orders $q^{2}(q-1)$, $4(q+\sqrt{2q}+1)4$ and
$4(q-\sqrt{2q}+1)$ with cyclic complements of orders $(q-1)$, $4$ and $4$,
respectively;

  4. (4)

$\operatorname{Aut}G=G\rtimes\langle\varphi\rangle$, where $\varphi$ is a
field automorphism of $G$ of order $\alpha$, and if
$\psi\in\langle\varphi\rangle$, then $C_{G}(\psi)\simeq Sz(q^{1/|\psi|});$

  5. (5)

the Schur multiplier of $G$ is trivial for $q>8$ and is an elementary abelian
group of order $4$ for $q=8$.

###### Proof.

Items 1–4 were established in [16], Item 5 — in [17]. ∎

######  Lemma 1.8.

Let $G=Sz(q)$, where $q\geqslant 8$, and $g\in G$ an $r$-element for an odd
prime $r$. If $G$ acts faithfully on a $p$-group $V$ and $p$ is odd, then the
coset $Vg$ of the natural semidirect product $V\rtimes G$ contains an element
of order $p|g|$.

###### Proof.

We can assume that $V$ is an elementary abelian $p$-group and consider $V$ as
a $G$-module. We may also assume that $G$ acts irreducibly on $V$. By [18,
Theorem 1.1], the minimal polynomial of $g$ in this representation is equal to
$x^{|g|}-1$. Hence, there exists $v\in V$ such that
$v(1+g+g^{2}+\dots+g^{|g|-1})\neq 0$. Then $(vg)^{|g|}\neq 0$, hence
$|vg|=p|g|$. ∎

######  Lemma 1.9.

Let $G=Sz(q)$, where $q\geqslant 8$.

  1. (1)

If $\psi$ is a field automorphism of $G$, then the set of the orders of
elements in the coset $G\psi$ is equal to
$|\psi|\cdot\omega(Sz(q^{1/|\psi|}))$.

  2. (2)

If $g\in\operatorname{Aut}G$, then $|g|<2q$.

  3. (3)

If $q\geqslant 32$ and $g\in\operatorname{Aut}G\setminus G$, then $|g|<q$.

###### Proof.

Let $q=2^{\alpha}$. The set of element orders of $\operatorname{Aut}G\setminus
G$ is the union of the sets $\gamma\cdot\omega(Sz(q^{1/\gamma}))$ over all
nonidentity divisors $\gamma$ of $\alpha$ [19, Theorem 2]. It is clear from
Item 2 of Lemma 1.7 that each number in $\omega(Sz(q^{1/\gamma}))$ is less
than $2q^{1/\gamma}$. Since $2\gamma q^{1/\gamma}\leqslant 2q$ for $q\geqslant
8$ and $2\gamma q^{1/\gamma}\leqslant 6q^{1/3}<q$ for $q\geqslant 32$, we
obtain the required estimates. ∎

##  2\. Proof of Theorem 1

Let $G$ be a minimal (by order) counterexample to the assertion of the
theorem. Since $G$ is a solvable group, Hall’s theorem ensures the existence
of a Hall $\sigma$-subgroup for any $\sigma\subseteq\pi(G)$. If we take
$\sigma$ consisting of four primes from the subset $\sigma(G)$ given in the
theorem, then the corresponding Hall $\sigma$-subgroup is solvable and
satisfies all conditions of the theorem, therefore, by the assumption on the
minimality of counterexample, it must coincide with $G$.

Thus, in what follows, $G$ is a $4$-primary solvable group satisfying the
hypotheses of the theorem, i.e., $\pi(G)=\\{p,q,r,s\\}$, and the spectrum of
$G$ contains all pairwise products of numbers from $\pi(G)$ (Condition 1 of
the theorem), but does not contain any one of the triple products (Condition
2).

######  Lemma 2.1.

For each nontrivial element $x\in G$, the inequality $|\pi(C_{G}(x))|<4$
holds.

###### Proof.

Suppose that $G$ has a nontrivial element $x$ with $|\pi(C_{G}(x))|=4$. We may
assume that $x$ has prime order. By Lemma 1.2, $C_{G}(x)$ contains an element
whose order is divisible by three different primes from $\pi(G)$, and this
contradicts Condition 2 of the theorem. ∎

######  Lemma 2.2.

If $p\in\pi(G)$, then $G$ contains a noncyclic subgroup of order $p^{2}$.

###### Proof.

Otherwise, $G$ contains, up to conjugation, a unique subgroup $P$ of order
$p$. Condition 1 of the theorem yields $|\pi(C_{G}(P))|=4$, where $C_{G}(P)$
is the centralizer of $P$ in $G$, which contradicts Lemma 2.1. ∎

######  Lemma 2.3.

Let $H$ be a Hall subgroup of $G$. Then the Fitting subgroup $F(H)$ of $H$ is
not cyclic and $|\pi(F(H))|\leqslant 2$.

###### Proof.

Denote $F(H)$ by $F$. The inequality $|\pi(F)|\leqslant 2$ follows from
Condition 2 of the theorem.

Assume that $F$ is cyclic. Then $H\neq F$ by Lemma 2.2. Since the group $H$ is
solvable, we have $C_{H}(F)\subseteq F$ [20, Theorem 6.1.3]. Therefore, the
quotient group $H/F$ embeds in the group $\operatorname{Aut}F$ and, in
particular, is abelian. If $F$ is a $p$-group, then by Lemma 2.2, $H/F$
contains a noncyclic $p^{\prime}$-group, which is impossible because the
$p^{\prime}$-part of $\operatorname{Aut}F$ is cyclic. Let $\pi(F)=\\{p,q\\}$,
where $p>q$, and let $W$ be a Sylow $q$-subgroup of $F$. Since $p>q$, a Sylow
$p$-subgroup $P$ of $H$ centralizes $W$. The group $H/F$ is abelian, so the
image of $P$ in $H/F$ is normal in $H/F$, and hence $P\times W$ is normal in
$H$. It follows that $P\leq F$, which contradicts Lemma 2.2. ∎

######  Lemma 2.4.

Every minimal normal subgroup of $G$ is a Sylow subgroup of $G$.

###### Proof.

Let $N$ be a minimal normal $p$-subgroup of $G$, and let $H$ be a Hall
$p^{\prime}$-subgroup of $G$. By Lemma 2.2, the group $H$ contains an
elementary abelian $q$-subgroup of order $q^{2}$ for every $q\in\pi(H)$, and
by Lemma 1.4, the subgroup $NH$ contains an element of order $pq$. Thus, the
group $NH$ satisfies the hypotheses of the theorem, and so $G=NH$. ∎

We fix until the end of the proof the notation: $F$ is the Fitting subgroup
$F(G)$ of $G$ and $H$ is a complement of $F$ in $G$, i.e., a Hall
$\pi(F)^{\prime}$-subgroup of $G$. By Lemma 2.4, the group $F$ is a direct
product of Sylow noncyclic elementary abelian subgroups of $G$.

######  Lemma 2.5.

Let $N$ be a normal $r$-subgroup of $H$. Then either $N$ is a Sylow
$r$-subgroup of $G$ and $\operatorname{exp}(N)=r$, or $N$ is a cyclic group of
odd order, or $N$ has a characteristic subgroup of order $2$.

###### Proof.

If $N$ is a cyclic group of even order or $N$ is not cyclic but does not
contain elementary abelian subgroups of order $r^{2}$, then $r=2$ and $N$ has
a characteristic subgroup of order $2$. Hence, we may assume that $N$ includes
an elementary abelian subgroup of order $r^{2}$. By Lemma 1.4, the subgroup
$FN$ contains an element of order $rp$ for every $p\in\pi(F)$. Arguing further
as in Lemma 2.4, we obtain that $N$ is a Sylow subgroup of $H$, and so of $G$.

By the minimality of $G$, there are no proper noncyclic characteristic abelian
subgroups in $N$. If $N$ is abelian, then $N=\Omega_{1}(N)$ and
$\operatorname{exp}(N)=r$, as required. Let $N$ be nonabelian. Then $Z(N)$ is
a cyclic group, and for $r=2$, $N$ has a characteristic subgroup of order $2$.
If $r\neq 2$, then in virtue of [20, Theorem 5.5.3] the group $N$ is the
central product of a cyclic group and an extraspecial group $M$ of period $r$.
Then $N=\Omega_{1}(N)=M$, and we are done. ∎

Recall that $|\pi(F)|\leqslant 2$ by Lemma 2.3. Therefore, Theorem 1 will
follow from the next two lemmas.

######  Lemma 2.6.

$|\pi(F)|\neq 1$.

###### Proof.

Assume the opposite. Then $F$ is a noncyclic elementary abelian $p$-group
which is a Sylow subgroup of $G$, and $H$ is a Hall $p^{\prime}$-subgroup of
$G$.

Let $N$ be a normal primary subgroup of $H$. If $N$ has a characteristic
subgroup of order $2$, then this subgroup is central in $H$ and, therefore,
$G$ contains an element of order $2rs$ forbidden by Condition 2, where
$\\{r,s\\}=\pi(G)\setminus\\{2,p\\}$. Applying Lemma 2.5, we obtain that
either $N$ is a Sylow subgroup of $G$ of prime exponent, or $N$ is a cyclic
group of odd order.

Consider now the Fitting subgroup $U=F(H)$ of $H$. By Lemma 2.3, the group $U$
is not cyclic. Hence, one of the Sylow subgroups of $U$, say, a $q$-subgroup,
is not cyclic. Thus, by the previous paragraph, $U$ contains a Sylow
$q$-subgroup $Q$ of $G$ and $Q$ is of prime exponent. Below we consider
separately the cases when $|\pi(U)|=2$ and $|\pi(U)|=1$.

Let $\pi(U)=\\{q,r\\}$ and $V$ be a Sylow $r$-subgroup of $U$. Then $U=Q\times
V$. If $V$ is also a Sylow subgroup of $G$, then, applying Lemma 1.4, we see
that $pqr\in\omega(G)$. Hence $V$ is a cyclic group and $r\neq 2$.

Let $S$ be a Sylow $s$-subgroup of $G$. If $C_{S}(V)$ contains a noncyclic
elementary abelian subgroup, then $C_{S}(V)$ contains an element that
centralizes nontrivial elements in both $Q$ and $V$, and so $H$ contains an
element of forbidden order $qrs$. Therefore, $C_{S}(V)$ has at most one
subgroup of order $s$. In particular, $C_{S}(V)<S$ due to Lemma 2.2. On the
other hand, $S/C_{S}(V)=N_{S}(V)/C_{S}(V)$ is a cyclic group of order dividing
$r-1$. Hence $C_{S}(V)\neq 1$ and $s<r$. Denote by $Z$ the unique subgroup of
order $s$ in $C_{S}(V)$. It is clear that $Z\unlhd S$ and, therefore, $Z\leq
Z(S)$.

Let $J$ be a Hall $\\{r,s\\}$-subgroup of $H$ containing $S$ and
$\overline{J}=J/V$. The images of other subgroups of $J$ in $\overline{J}$
will also be denoted by a bar.

Let $\overline{K}=F(\overline{J})$ and
$\overline{C}=\overline{C_{S}(V)}\cap\overline{K}$. Then $\overline{C}$ is
normal in $\overline{J}$, because

$\overline{C}=\overline{C_{J}(V)}\cap\left(\overline{S}\cap\overline{K}\right),$

while $\overline{C_{J}(V)}$ and $\overline{S}\cap\overline{K}$ are normal
subgroups of $\overline{J}$. Assume that $\overline{Z}\not\leq\overline{K}$.
Since $C_{\overline{J}}(\overline{K})\leq\overline{K}$, $\overline{K}$ must
contain a Sylow $r$-subgroup $\overline{T}$ on which $\overline{Z}$ acts
nontrivial. Applying Lemma 1.6 to the action of
$\overline{T}\rtimes\overline{Z}$ on $Q$ and noting that $r\neq 2$, we obtain
an element of order $sq$ in $QZ$. However, $QZ$ centralizes $V$, so $G$
contains an element of forbidden order $sqr$. Thus,
$\overline{Z}\leq\overline{K}$, whence $\overline{Z}\leq\overline{C}$, and
$\overline{Z}$ is normal in $\overline{J}$ as the only subgroup of
$\overline{C}$ of order $s$.

Let $R$ be a Sylow $r$-subgroup of $J$ (and $G$). Then
$[\overline{R},\overline{Z}]=1$ since $r>s$. Therefore, $Z$ stabilizes the
normal series $1<V<R$ and so acts trivially on $R$ [20, Lemma 5.3.2]. Consider
the action of $R\times Z$ on $QF$. The group $Z$ acts on $Q$ without fixed
points, otherwise $G$ contains an element of order $sqr$. Hence $QZ$ is a
Frobenius group, and since $C_{G}(F)=F$, it acts on $F$ faithfully. Therefore,
$C_{F}(Z)\neq 1$ in view of Lemma 1.5. The group $R$ acts on $C_{F}(Z)$ and,
applying Lemmas 2.2 and 1.4, we obtain an element of order $rpq$; a
contradiction.

Let now $\pi(U)=\\{q\\}$, i.e., $U=Q$ is a Sylow $q$-subgroup of $G$. Let, as
above, $J$ be a Hall $\\{r,s\\}$-subgroup of $H$. Then $G=FUJ$.

Put $V=F(J)$. By Lemma 2.3, the group $V$ is not cyclic. Hence one of its
Sylow subgroups, say, the $r$-subgroup $W$ is not cyclic. Denote by $S$ a
Sylow $s$-subgroup of $J$ (and $G$).

Assume that $r=2$ and $W$ contains a unique involution $z$. Then the whole
group $S$ centralizes $z$. If $C_{U}(z)=1$, then $U\rtimes\langle z\rangle$ is
a Frobenius group and $C_{F}(z)\neq 1$ by Lemma 1.5. Hence, at least one of
the centralizers $C_{U}(z)$ and $C_{F}(z)$ is nontrivial. This centralizer is
invariant under the action of $S$, so applying Lemma 1.4, we obtain an element
of order $2sq$ or $2sp$; a contradiction. Arguing in the same way as in Lemma
2.5, we conclude that $W$ is a Sylow $r$-subgroup of $G$ and either $W$ is
abelian or has an odd order.

Let $\pi(V)=\\{r,s\\}$ and $T$ a Sylow $s$-subgroup of $V$, i.e., $V=W\times
T$. Repeating the argument of the previous paragraph about the action of
$S\times\langle z\rangle$ on $UF$ with $W$ instead of $S$ and any element
$x\in T$ instead of $z$, we arrive at a contradiction.

Thus, $V=W$. There is a nontrivial element $x\in S$ of order $s$ such that
$C_{V}(x)\neq 1$. On the other hand, $[V,x]\neq 1$. The group $V\rtimes\langle
x\rangle$ acts faithfully on $U$ and on $F$. By Lemma 1.6, we get that
$C_{U}(x)\neq 1$ and $C_{F}(x)\neq 1$. Hence $|\pi(C_{G}(x))|=4$. This
contradiction completes the proof of the lemma. ∎

######  Lemma 2.7.

$|\pi(F)|\neq 2$.

###### Proof.

Again, suppose the contrary. Then $F=P\times Q$, where $P$ and $Q$ are
noncyclic elementary abelian Sylow $p$-subgroups and $q$-subgroups of $G$, and
$H$ is a Hall $\\{r,s\\}$-subgroup of $G$.

Assume first that $H$ is of odd order, and let $U=F(H)$ be the Fitting
subgroup of $H$. By Lemma 2.3, the group $U$ is not cyclic, therefore,
applying Lemma 2.5, we obtain that $U$ contains a Sylow subgroup of $G$, say,
an $r$-subgroup $R$, and $R$ has exponent $r$. Denote by $S$ a Sylow
$s$-subgroup of $H$ (and $G$). By the minimality of $G$, the group $S$ is an
elementary abelian group of order $s^{2}$.

By Lemma 1.4, there are $x,y\in S$ such that $C_{P}(x)\neq 1$ and
$C_{Q}(y)\neq 1$. Condition 2 implies that $C_{Q}(x)=C_{P}(y)=1$. In
particular, $x\neq y$. If one of these elements, say $x$, centralizes $R$,
then $R$ acts on $C_{P}(x)$ and we get an element of order $prs$. Hence
$[R,x]\neq 1$ and $[R,y]\neq 1$.

If $[R,x]\cap C_{R}(Q)=1$, then by Lemma 1.6, we have $C_{Q}(x)\neq 1$, but
this is not the case. In particular, $C_{R}(Q)>1$. The group $C_{R}(Q)$ is
cyclic, otherwise by Lemma 1.4, $G$ contains an element of order $rqp$. Thus,
$C_{R}(Q)$ is of order $r$ and $[R,x]=C_{R}(Q)$. Similarly, $[R,y]=C_{R}(P)$.
It is clear that $C_{R}(Q)\cap C_{R}(P)=C_{R}(U)=1$, so $R$ includes a
subgroup $[R,x]\times[R,y]$ of order $r^{2}$. Let $g,h\in R$ be such that
$[g,x]\neq 1$, $[h,y]\neq 1$. Then $[g,xy]=[g,y][g,x]^{y}\not\in[R,y]$ and
$[h,xy]=[h,y][h,x]^{y}\not\in[R,x]$. Consequently, $[R,xy]$ lies neither in
$C_{R}(P)$ nor in $C_{R}(Q)$, and hence there is an element of order $pqr$ in
$G$; a contradiction.

Now let $2\in\pi(H)$ and let $s$ be the only odd prime divisor of the order of
$H$. By Lemma 2.2, the group $H$ contains a noncyclic subgroup $T$ of order
$4$. If $t$ is an involution from $T$, then Condition 2 of the theorem yields
$C_{P}(t)=1$ or $C_{Q}(t)=1$. Choose from $P$ and $Q$ the group on which two
different involutions $t$ and $u$ of $T$ act fixed-point-freely. Let this be
$P$. Then the involution $v=tu$ centralizes $P$ and inverts $Q$. The
involution $v^{h}$ has the same property for every $h\in H$. If $v^{h}\neq v$,
we get an element of forbidden order in $G$: $2pq$ in the case when the
dihedral group $D=\langle v,v^{h}\rangle$ contains an elementary abelian
subgroup of order $4$, and $spq$ otherwise. Thus, the involution $v$ lies in
the center of $H$. The group $PH$ contains an element $x$ of order $pr$, so
$xv$ is an element of order $2pr$. This contradiction proves the lemma, and
the theorem as well. ∎

##  3\. Recognition of squares of Suzuki groups

In this section, we will prove Theorems 3 and 4. We will use the main
properties of Suzuki groups listed in Lemma 1.7 without explicit reference to
this lemma.

Let $L=Sz(q)$, where $q=2^{\alpha}\geqslant 8$. The set $\mu(L)$ consists of
the numbers $m_{1}(q)=4$, $m_{2}(q)=q-1$, $m_{3}(q)=q-\sqrt{2q}+1$, and
$m_{4}(q)=q+\sqrt{2q}+1$. Note that $m_{3}(q)m_{4}(q)=q^{2}+1$,
$q/2<m_{i}(q)<2q$ for $i\neq 1$, and $(m_{i}(q),m_{j}(q))=1$ for $i\neq j$. In
particular, $t(L)=4$.

Let $G$ be a finite group with $\omega(G)=\omega(L\times L)$. Then

$\mu(G)=\\{m_{i}(q)\cdot m_{j}(q)\mid 1\leqslant i,j\leqslant 4\text{ and
}i\neq j\\}.$

By Corollary 2, the group $G$ is nonsolvable. Let $K$ be the solvable radical
of $G$ and $\overline{G}=G/K$. Since $3\not\in\pi(G)$, it follows from [8]
that $\operatorname{Soc}(\overline{G})=L_{1}\times L_{2}\times\dots\times
L_{k}$, where $L_{i}=Sz(2^{\alpha_{i}})$ and $\alpha_{i}\geqslant 3$ for all
$1\leqslant i\leqslant k$.

For every $i$, the number $2^{\alpha_{i}}-1$ must divide the exponent of $G$,
which in turn divides $4(2^{4\alpha}-1)$, so all $\alpha_{i}$ divide $\alpha$.
Hence $m_{2}(2^{\alpha_{i}})$ divides $m_{2}(q)$. Furthermore, one of the
numbers $m_{3}(2^{\alpha_{i}})$ and $m_{4}(2^{\alpha_{i}})$ divides
$m_{3}(q)$, and the other divides $m_{4}(q)$ (see, e.g., [21, p. 18]). This
immediately implies that $k\leqslant 2$: otherwise,
$4ps\in\omega(G)\setminus\omega(L\times L)$ for some prime divisors $p$ and
$s$ of $m_{2}(2^{\alpha_{2}})$ and $m_{3}(2^{\alpha_{3}})$.

In what follows, we fix some primitive prime divisors $r_{2}$ and $r_{4}$ of
the numbers $2^{\alpha}-1$ and $2^{4\alpha}-1$, respectively, and note that
each of them is greater than $\alpha$. In particular, none of them divides
$|\operatorname{Aut}L_{i}/L_{i}|$ for any $i$.

Let $k=1$. Suppose $\alpha_{1}\neq\alpha$. Then $r_{2}$ and $r_{4}$ do not
divide $|\overline{G}|$ and so $r_{2}r_{4}\in\omega(K)$. It is clear that
$r_{4}$ divides $m_{i}(q)$ for $i\in\\{3,4\\}$. Let $j\in\\{3,4\\}$ and $j\neq
i$. Since $m_{j}(q)>q/2=2^{\alpha-1}\geqslant 2^{\alpha_{1}+1}$, it follows
from Item 2 of Lemma 1.9 that $m_{j}(q)\not\in\omega(\overline{G})$. On the
other hand, $r_{2}m_{j}(q),r_{4}m_{j}(q)\in\omega(G)$, so
$r_{2}s,r_{4}s\in\omega(K)$ for some prime divisor $s$ of
$m_{j}(q)/(m_{j}(q),\exp(\overline{G}))$. Similarly, if $T$ is a Sylow
$2$-subgroup of $G$, then $2s\in\omega(KT)$. Thus, the solvable group $KT$
satisfies the hypothesis of Theorem 1 with $\sigma(KT)=\\{r_{2},r_{4},2,s\\}$;
a contradiction.

Therefore, $\alpha_{1}=\alpha$ and $L_{1}=L$. Let $\alpha\geqslant 5$. Denote
by $G_{1}$ the preimage of $L$ in $G$ and show by induction on the order of
$K$ that $L$ is a direct factor in $G_{1}$. Let $V$ be a minimal normal in
$G_{1}$ subgroup of $K$. Then $V$ is an elementary abelian $p$-group for some
prime $p$. By induction, we may assume that $G_{1}/V=K/V\times H/V$, where
$H/V\simeq L$. Suppose $C_{H}(V)=V$. If $p=2$, then applying Lemma 1.5 to a
Frobenius subgroup of order $4m_{3}(q)$, we obtain that $8\in\omega(G)$. If
$p$ is odd, then by Lemma 1.8, $G$ contains an element of order $p^{l+1}$,
where $p^{l}$ is the greatest power of $p$ in $\omega(L)$. Therefore,
$C_{H}(V)=H$. In view of $\alpha\geqslant 5$, the Schur multiplier of $L$ is
trivial, so $H=V\times L$. Since $[K,L]\leq V$, the equalities $[K,L,L]=1$ and
$[L,K,L]=1$ hold, and hence $[L,K]=[L,L,K]=1$.

Thus, $K\times L=G_{1}\leq G$ and hence $\omega(K)\subseteq\omega(L)$. As
above, choose $j\in\\{3,4\\}$ so that $r_{4}$ does not divide $m_{j}(q)$, and
let $s\in\pi(m_{j}(q))$. By virtue of Lemma 1.2 and the inclusion
$\omega(K)\subseteq\omega(L)$, the intersection
$\pi(K)\cap\\{2,r_{2},r_{4},s\\}$ does not contain more than two numbers. On
the other hand, the numbers $2$, $r_{2}$, $r_{4}$ do not divide $|G/(K\times
L)|$, so two of them must divide $|K|$. Denote by $p$ the third one. Then
$pm_{j}(q)$ is coprime to $|K|$, so $pm_{j}(q)\in\omega(\overline{G})$.
Applying Item 3 of Lemma 1.9, we get that $q>pm_{j}(q)>pq/2$; a contradiction.

Let $\alpha=3$. Then the Schur multiplier of $L$ has order $4$, $m_{2}(q)=7$,
$m_{3}(q)=5$, and $m_{4}(q)=13$. It is clear that
$\overline{G}=\operatorname{Soc}(\overline{G})$. Suppose $7\not\in\pi(K)$ and
let $x\in G$ be an element of order $7$. Since $\langle x\rangle$ is a Sylow
subgroup of $G$, we have $\pi(C_{K}(x))=\\{2,5,13\\}$. By Lemma 1.2,
$C_{K}(x)$ contains an element of order $pr$ for $p\neq r$ and hence
$7pr\in\omega(G)$; a contradiction. Hence $7\in\pi(K)$. Similarly,
$5,13\in\pi(K)$.

Arguing in the same way as in the case of $\alpha\geqslant 5$, we conclude
that either $L$ is again a direct factor, i.e., in this case $G=K\times L$, or
there is a normal solvable subgroup $W$ of $G$ such that the quotient group
$G/W$ is the central product of a solvable group $N$ and a perfect central
extension $M$ of $L$, and the center $Z(M)$ is nontrivial. In the first case,
applying Lemma 1.2, we see that the product of two of the numbers $5,7,13$
lies in $\omega(K)$, and hence the product of all three numbers lies in
$\omega(G)$; a contradiction. Let us consider the second case. Since $Z(M)\neq
1$, there are no odd numbers in $\pi(N)$; in particular, $7\in\pi(W)$. Denote
by $J$ the complete preimage of $M$ in $G$. By the Frattini argument, we may
assume that a Sylow $7$-subgroup $P$ of the group $W$ is normal in $J$. If
$C_{J}(P)W/W=M$, then $2\cdot 5\cdot 7\in\omega(G)$. Let $C_{J}(P)W/W\leqslant
Z(M)$. A Borel subgroup of $L$ is a Frobenius group with kernel of order
$2^{6}$ and complement of order $7$. Hence, if $T$ is its preimage in $M$,
then $T=R\rtimes\langle g\rangle$, where $R$ is a $2$-group, $|g|=7$, and
$[R,g]\neq 1$. Applying Lemma 1.6 to the action of $T/C_{T}(P)$ on $P$ and
noting that $7$ is not a Fermat prime, we obtain an element of order $49$ in
$G$. This contradiction completes the proof that $k\neq 1$.

Thus $k=2$. Suppose $K\neq 1$ and show that
$\omega(G)\not\subseteq\omega(L\times L)$. Without loss of generality, we may
assume that $K$ is an elementary abelian $p$-group. By Lemma 1.3, we may also
assume that $L_{1}\times L_{2}$ is a subgroup of $\overline{G}$. Suppose that
$p\neq 2$ or $L_{1}\times L_{2}$ centralizes $K$. There are odd
$r\in\pi(L_{1})$ and $s\in\pi(L_{2})$ such that $prs\not\in\omega(L\times L)$,
and let $x\in L_{1}$ and $y\in L_{2}$ be elements of orders $r$ and $s$,
respectively. Applying Lemma 1.8 to $L_{1}$, we get that
$C_{K}(x)\neq\leavevmode\nobreak\ 1$. The group $L_{2}$ acts on $C_{K}(x)$,
and applying Lemma 1.8 again, we see that $y$ has a fixed point in $C_{K}(x)$.
Thus, $prs\in\omega(G)$. Hence $p=2$ and, say, $L_{1}$ acts on $K$ faithfully.
Since $L_{1}$ has Frobenius subgroups with kernels of orders $m_{3}(q)$,
$m_{4}(q)$ and cyclic complements of order $4$, it follows by Lemma 1.5 that
$G$ contains an element of order $8$.

Therefore, $K=1$. If at least one of the numbers $\alpha_{1}$ and $\alpha_{2}$
is distinct from $\alpha$, then $r_{2}r_{4}\not\in\omega(G)$. Hence,
$L_{1}\simeq L_{2}\simeq L$ and $\operatorname{Aut}(L_{1}\times
L_{2})=(L_{1}\times
L_{2})\rtimes(\langle\varphi\rangle\wr\langle\tau\rangle)$, where $\varphi$ is
some fixed automorphism of $L_{1}$ of odd order $\alpha$ and $\tau$ is the
automorphism of $L_{1}\times L_{2}$ of order $2$ interchanging the components.
If $2\in\pi(G/\operatorname{Soc}(G))$, then we may assume that $\tau\in G$,
and taking $g$ to be an element of $L_{1}$ of order $4$, we see that
$8=|g\tau|\in\omega(G)$. This is false, and so
$G\leq\operatorname{Aut}L_{1}\times\operatorname{Aut}L_{2}$.

Suppose $G\neq L_{1}\times L_{2}$. Then $G$ contains
$\psi=(\psi_{1},\psi_{2})$, where $\psi_{i}$ is a field automorphism of
$L_{i}$, $i=1,2$, and $|\psi|=p$ is a prime. It is clear that

$C_{L\times L}(\psi)=C_{L_{1}}(\psi_{1})\times C_{L_{2}}(\psi_{2})\simeq
Sz(q^{1/|\psi_{1}|})\times Sz(q^{1/|\psi_{2}|}).$

Assume that $q^{1/|\psi_{i}|}>2$ for some $i$. Then there is
$s\in\pi(C_{L_{i}}(\psi_{i}))$ such $s\neq 2$, $s\neq p$ and
$ps\not\in\omega(L)$. For this prime $s$, we have
$2ps\in\omega(G)\setminus\omega(L\times L)$, a contradiction. It follows that
$|\psi_{1}|=|\psi_{2}|=p$ and $q=2^{p}$. On the other hand,
$p\in\pi(G)=\pi(L)$, and so $p$ divides $(2^{p}-1)(2^{2p}+1)$. This yields
$p=5$. Thus if $q\neq 2^{5}$, then $G\simeq L\times L$, and this completes the
proof of Theorem 3.

Let $q=2^{5}$ and $M=(L_{1}\times L_{2})\rtimes\langle\psi\rangle$, where
$\psi$ induces on each factor $L_{i}$ a field automorphism of order $5$. We
claim that $\omega(M)=\omega(L\times L)$. Let $g\in M\setminus(L_{1}\times
L_{2})$. Then $g=(g_{1}\psi_{1},g_{2}\psi_{2})$, where
$|\psi_{1}|=|\psi_{2}|=5$. By Item 1 of Lemma 1.9, for every $i=1,2$, the
order of $g_{i}\psi_{i}$ lies in $5\cdot\omega(Sz(2))$ and hence divides $20$
or $25$. Since the order of $g$ is equal to the least common multiple of
$|g_{1}\psi_{1}|$ and $|g_{2}\psi_{2}|$, this order divides $100$ and,
therefore, lies in $\omega(L\times L)$.

It remains to find the number of pairwise nonisomorphic groups $M$. Groups $A$
and $B$ such that $L_{1}\times L_{2}\leq A,B\leq\operatorname{Aut}(L_{1}\times
L_{2})$ are isomorphic if and only if their images in
$\operatorname{Out}(L_{1}\times L_{2})$ are conjugate. Recall that
$\operatorname{Aut}(L_{1}\times L_{2})=(L_{1}\times
L_{2})\rtimes(\langle\varphi\rangle\wr\langle\tau\rangle)$. We denote the
images of $\varphi$ and $\tau$ in $\operatorname{Out}(L_{1}\times L_{2})$ by
the same symbols. Then the image of $M$ in $\operatorname{Out}(L_{1}\times
L_{2})$ is the cyclic group $X_{l}$ generated by
$(\varphi^{l},\varphi^{\tau})$ for some $1\leqslant l\leqslant 4$. It is clear
that $X_{l}$ and $X_{m}$, with $l\neq m$, are conjugate if and only if
$X_{l}^{\tau}=X_{m}$. It follows that $X_{2}$ and $X_{3}$ are conjugate, while
$X_{1}$, $X_{2}$ and $X_{4}$ are pairwise not conjugate. Thus, up to
isomorphism, there are four groups isospectral to $L\times L$: $L_{1}\times
L_{2}$ and the full preimages of $X_{l}$, with $l=1,2,4$, in
$\operatorname{Aut}(L_{1}\times L_{2})$. This completes the proof of Theorem
4.

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Zhigang Wang,

School of Science, Hainan University, Haikou, Hainan, P. R. China;

<EMAIL_ADDRESS>
Andrey V. Vasil’ev, Sobolev Institute of Mathematics, Novosibirsk, Russia;

School of Science, Hainan University, Haikou, Hainan, P. R. China;

<EMAIL_ADDRESS>
Maria A. Grechkoseeva,

Sobolev Institute of Mathematics, Novosibirsk, Russia;

<EMAIL_ADDRESS>
Archil Kh. Zhurtov,

Kabardino-Balkarian State University, Nalchik, Russia;

<EMAIL_ADDRESS>