# On characterization by Gruenberg–Kegel graph of finite simple exceptional
groups of Lie type

Natalia V. Maslova, Viktor V. Panshin, and Alexey M. Staroletov  Natalia
Vladimirovna Maslova
Krasovskii Institute of Mathematics and Mechanics UB RAS,
16, S. Kovalevskaja str., Yekaterinburg, 620108, Russia
Ural Federal University,
19, Mira str., Yekaterinburg, 620002, Russia<EMAIL_ADDRESS>Viktor
Vladimirovich Panshin
Novosibirsk State University,
1, Pirogova str., Novosibirsk, 630090, Russia
Sobolev Institute of Mathematics,
4, Acad. Koptyug ave., Novosibirsk, 630090, Russia<EMAIL_ADDRESS>Alexey
Mikhailovich Staroletov
Sobolev Institute of Mathematics,
4, Acad. Koptyug ave., Novosibirsk, 630090, Russia<EMAIL_ADDRESS>
###### Abstract.

The Gruenberg–Kegel graph $\Gamma(G)$ of a finite group $G$ is the graph whose
vertex set is the set of prime divisors of $|G|$ and in which two distinct
vertices $r$ and $s$ are adjacent if and only if there exists an element of
order $rs$ in $G$.

A finite group $G$ is called almost recognizable (by Gruenberg–Kegel graph) if
there is only finite number of pairwise non-isomorphic finite groups having
Gruenberg–Kegel graph as $G$. If $G$ is not almost recognizable, then it is
called unrecognizable (by Gruenberg–Kegel graph).

Recently P. J. Cameron and the first author have proved that if a finite group
is almost recognizable, then the group is almost simple. Thus, the question of
which almost simple groups (in particular, finite simple groups) are almost
recognizable is of prime interest. We prove that every finite simple
exceptional group of Lie type, which is isomorphic to neither
${{}^{2}}B_{2}(2^{2n+1})$ with $n\geq 1$ nor $G_{2}(3)$ and whose
Gruenberg–Kegel graph has at least three connected components, is almost
recognizable. Moreover, groups ${{}^{2}}B_{2}(2^{2n+1})$, where $n\geq 1$, and
$G_{2}(3)$ are unrecognizable.

In memory of Irina Dmitrievna Suprunenko

Throughout the paper we consider only finite groups and simple graphs, and
henceforth the term group means finite group and the term graph means simple
graph, that is undirected graph without loops and multiple edges.

Let $G$ be a group. The spectrum $\omega(G)$ is the set of all element orders
of $G$. The prime spectrum $\pi(G)$ is the set of all primes belonging to
$\omega(G)$. A graph $\Gamma(G)$ whose vertex set is $\pi(G)$ and in which two
distinct vertices $r$ and $s$ are adjacent if and only if $rs\in\omega(G)$ is
called the Gruenberg–Kegel graph or the prime graph of $G$.

We say that the group $G$ is

  * •

recognizable (by Gruenberg–Kegel graph) if for every group $H$ the equality
$\Gamma(H)=\Gamma(G)$ implies that $G\cong H$;

  * •

$k$-recognizable (by Gruenberg–Kegel graph), where $k$ is a positive integer,
if there are exactly $k$ pairwise non-isomorphic groups having the same
Gruenberg–Kegel graph as $G$;

  * •

almost recognizable (by Gruenberg–Kegel graph) if it is $k$-recognizable by
Gruenberg–Kegel graph for a positive integer $k$;

  * •

unrecognizable (by Gruenberg–Kegel graph) if there are infinitely many
pairwise non-isomorphic groups having the same Gruenberg–Kegel graph as $G$.

Note that groups can be characterized by various numerical sets. For example,
if we replace $\Gamma(G)$ in these definitions with $\omega(G)$, then we
obtain the corresponding definitions for recognizability by spectrum.
Nevertheless, if the characterization set is not specified, we suppose that it
is the Grunberg-Kegel graph. It is easy to see that if $\omega(G)=\omega(H)$
for groups $G$ and $H$, then $\Gamma(G)=\Gamma(H)$, however, the converse
implication is not true in general. Consider alternating groups $A_{5}$ and
$A_{6}$. Clearly, $\Gamma(A_{5})=\Gamma(A_{6})$, where both graphs are empty
graphs on three vertices $2$, $3$, and $5$, and, on the other hand, we see
that $4\in\omega(A_{6})\setminus\omega(A_{5})$.

Recently P. J. Cameron and the first author have proved [7] that a group $G$
is almost recognizable if and only if each group $H$ with
$\Gamma(G)=\Gamma(H)$ is almost simple. Thus, the question of which almost
simple groups are almost recognizable is of prime interest. At the same paper
[7], a survey of known results on recognition of simple groups has been
presented. Note that there are not many completed results at the moment. The
situation is much better in the case of the characterization problem by
spectrum, where recognition is established for many nonabelian simple groups
[15].

In [36], V. D. Mazurov conjectured that if a simple group $G$ is not
isomorphic to $A_{6}$ and $\Gamma(G)$ has at least three connected components,
then $G$ is recognizable by its spectrum. This conjecture was proved in a
series of papers, the final result was obtained in [28], where the author
proved that groups $E_{7}(2)$ and $E_{7}(3)$ are recognizable by
Gruenberg–Kegel graph, therefore, these groups are recognizable by spectrum.

Connected components of Gruenberg–Kegel graphs of simple groups were described
in [46, 27]. A complete result after corrections of mistakes can be found, for
example, in [2, Tables 1–3]. If $G$ is a simple group, then $\Gamma(G)$ has at
least three connected components if and only if one of the following
statements holds:

$(1)$ $G\cong G_{2}(q)$, where $q$ is a power of $3$, or
$G\cong{{}^{2}}G_{2}(q)$, where $q=3^{2n+1}>3$;

$(2)$ $G\cong{{}^{2}}B_{2}(q)\cong Sz(q)$, where $q=2^{2n+1}>2$;

$(3)$ $G\cong F_{4}(q)$, where $q$ is even, or $G\cong{{}^{2}}F_{4}(q)$ for
$q=2^{2n+1}>2$;

$(4)$ $G\cong E_{8}(q)$;

$(5)$ $G\cong A_{1}(q)\cong PSL_{2}(q)$, where $q>3$;

$(6)$ $G\cong{{}^{2}}D_{n}(3)\cong P\Omega_{2n}^{-}(3)$, where $n=2^{m}+1\geq
3$ is a prime;

$(7)$ $G$ is one of the following finite simple groups of Lie type:
${{}^{2}}A_{5}(2)\cong PSU_{6}(2)$, $E_{7}(2)$, $E_{7}(3)$, $A_{2}(4)\cong
PSL_{3}(4)$, ${{}^{2}}E_{6}(2)$;

$(8)$ $G$ is one of the following finite simple sporadic groups: $M_{11}$,
$M_{23}$, $M_{24}$, $J_{3}$, $HiS$, $Suz$, $Co_{2}$, $Fi_{23}$, $F_{3}$,
$F_{2}$, $M_{22}$, $J_{1}$, $O^{\prime}N$, $LyS$, $Fi^{\prime}_{24}$, $F_{1}$,
$J_{4}$;

$(9)$ $G\cong A_{n}$, where $n>6$ and both $n$ and $n-2$ are primes.

We consider the recognition problem of the simple exceptional groups of Lie
type whose Gruenberg–Kegel graphs have at least three connected components.
The main result of this paper is the following statement.

######  Main Theorem.

Every finite simple exceptional group of Lie type, which is isomorphic to
neither ${{}^{2}}B_{2}(2^{2n+1})$ with $n\geq 1$ nor $G_{2}(3)$ and whose
Gruenberg–Kegel graph has at least three connected components, is almost
recognizable by Grunberg–Kegel graph. Moreover, groups
${{}^{2}}B_{2}(2^{2n+1})$, where $n\geq 1$, and $G_{2}(3)$ are unrecognizable
by Gruenberg–Kegel graph.

In fact, much has been known about the groups in the theorem before in the
context of the recognition problem. The groups $E_{7}(2)$, $E_{7}(3)$, and
${{}^{2}}E_{6}(2)$ are known to be recognizable (see [28] and [29]). The
recognizability of simple groups ${{}^{2}}G_{2}(q)$ with $q>3$ was established
in [48].

If $G$ is a nonabelian simple group, then we say that $G$ is quasirecognizable
by its Gruenberg–Kegel graph if every group $H$ with $\Gamma(G)=\Gamma(H)$ has
a unique nonablelian composition factor $S$ and $S\cong G$. In [50], it was
proved that each group $G_{2}(q)$, where $q$ is an odd power of $3$, is
quasirecognizable; however, this result contains an error since
$\Gamma(G_{2}(3))=\Gamma(PSL_{2}(13))$. Quasirecognizability of the simple
groups $F_{4}(q)$ and ${{}^{2}}F_{4}(q)$, where $q>2$ is even, was proved in
[24] and [1], respectively. We prove Main Theorem for groups $G_{2}(q)$, where
$q>3$, $F_{4}(q)$, and ${}^{2}F_{4}(q)$ in Section 3. The quasirecognizability
of groups ${{}^{2}}B_{2}(q)$ was proved in [50]. We consider the groups
${{}^{2}}B_{2}(q)$ and $G_{2}(3)$ in Section 4.

In [47], A. V. Zavarnitsine proved that if $G=E_{8}(q)$, where $q\equiv 0,\pm
1\pmod{5}$, and $H$ is a group such that $\Gamma(H)=\Gamma(G)$, then $H\cong
E_{8}(u)$ for a prime power $u\equiv 0,\pm 1\pmod{5}$. In Section 5, we prove
a similar result for groups $E_{8}(q)$, where $q\equiv\pm 2\pmod{5}$.
Therefore, as a corollary, Main Theorem holds for the groups $E_{8}(q)$.
However, the question of quasirecognizability for the groups $E_{8}(q)$ is
still open (see Section 5 for details).

Provide a mini-survey of known results on characterization by Gruenberg–Kegel
graph of the remaining simple groups whose Gruenberg–Kegel graphs have at
least three connected components.

First consider groups $PSL_{2}(q)$, where $q=p^{k}$ and $p$ is a prime. The
groups $PSL_{2}(4)\cong PSL_{2}(5)$, $PSL_{2}(7)$, $PSL_{2}(8)$, and
$PSL_{2}(9)\cong A_{6}$ are unrecognizable while the groups $PSL_{2}(11)$ and
$PSL_{2}(25)$ are $2$-recognizable (see, for example, [7, Theorem 1.7], [31,
Theorem 5] and [17]). The group $PSL_{2}(49)$ is $5$-recognizable (see [34],
this result is obtained as a consequence of [31, Theorem 5] and direct
calculations with using Lemma 1.15). Let $k=1$ and $p>11$. In [25], it was
proved that if $p\not\equiv 1\pmod{12}$, then $G$ is recognizable, and if
$p\equiv 1\pmod{12}$, then $G$ is quasirecognizable. However, this result has
at least one mistake since the group $PSL_{2}(13)$ is not quasirecognizable.
If $p$ is odd and $k>2$ or $k=2$ and $p>7$, then $G$ is $(k,2)$-recognizable
(see [22] and [21]). If $q>8$ is even, then $G$ is quasirecognizable (see [26,
Theorem 3.3]). In general, we suggest that the results on recognition of the
groups $PSL_{2}(q)$ by Gruenberg–Kegel graph need a careful revision since the
authors in their proofs refer to the paper [18] which contains a number of
rather serious inaccuracies.

If $G\cong P\Omega_{2n}^{-}(3)$, where $n\geq 3$ is a prime, then $G$ is
recognizable (see [11]). The group $PSU_{6}(2)$ is $2$-recognizable (see [30,
Theorem 2]), the group $PSL_{3}(4)$ is $5$-recognizable since
$\Gamma(PSL_{3}(4))=\Gamma(PSL_{2}(49))$.

The problem of recognition by Gruenberg–Kegel graph has been solved for all
finite simple sporadic groups. In particular, the groups $J_{1}$, $M_{22}$,
$M_{23}$, $M_{24}$, $Co_{2}$, $J_{4}$, $J_{3}$, $Suz$, $O^{\prime}N$, $LyS$,
$F_{3}$, $Fi_{23}$, $Fi^{\prime}_{24}$, $F_{2}$, and $F_{1}$ are recognizable
(see [17, Theorem 3], [48, Theorem B], [30, Theorem 1], and [32]) while the
groups $M_{11}$ and $HiS$ are $2$-recognizable (see [17, Theorem 3] and [30,
Theorem 2], respectively).

Let $G\cong A_{n}$, where $n>6$ and both $n$ and $n-2$ are primes. If $n=7$,
then $\Gamma(G)=\Gamma(PSL_{2}(49))$ and, therefore, $G$ is $5$-recognizable.
If $n=13$, then $G$ is recognizable (see [39, Lemma 24]), and if $n\geq 19$,
then $G$ is known to be quasirecognizable (see [39, Theorem 1]). The question
of recognizability by Gruenberg–Kegel graph of the groups $A_{n}$, where
$n\geq 19$ is a prime and $n-2$ is also a prime, is still open, and the
conjecture [39, Conjecture 1] is that these groups are recognizable.

##  1\. Preliminaries

Let $n$ be an integer. Denote by $\pi(n)$ the set of all prime divisors of
$n$. Let $\pi$ be a set of primes. The largest divisor $m$ of $n$ such that
$\pi(m)\subseteq\pi$ is called a $\pi$-part of $n$ and is denoted by
$n_{\pi}$. By $\pi^{\prime}$ we denote the set of primes which do not belong
to $\pi$. If $\pi$ consists of a unique element $p$, then we will write
$n_{p}$ and $n_{p^{\prime}}$ instead of $n_{\\{p\\}}$ and
$n_{\\{p\\}^{\prime}}$, respectively.

If $n$ is an integer and $r$ is an odd prime with $(r,n)=1$, then $e(r,n)$
denotes the multiplicative order of $n$ modulo $r$. Given an odd integer $n$,
we put $e(2,n)=1$ if $n\equiv 1\pmod{4}$, and $e(2,n)=2$ otherwise.

The following lemma is proved in [3], and also in [51].

######  Lemma 1.1 (Bang–Zsigmondy).

Let $q$ be an integer greater than $1$. For every positive integer $m$ there
exists a prime $r$ with $e(r,q)=m$ but for the cases $q=2$ and $m=1$, $q=3$
and $m=1$, and $q=2$ and $m=6$.

Fix an integer $a$ with $|a|>1$. A prime $r$ is said to be a primitive prime
divisor of $a^{i}-1$ if $e(r,a)=i$. We write $r_{i}(a)$ to denote some
primitive prime divisor of $a^{i}-1$ if such a prime exists, and $R_{i}(a)$ to
denote the set of all such divisors.

######  Lemma 1.2 ([14, Lemma 6]).

Let $q$, $m$, and $k$ be positive integers. Then $R_{mk}(q)\subseteq
R_{m}(q^{k})$. If, in addition, $(m,k)=1$, then $R_{m}(q)\subset
R_{m}(q^{k})$.

######  Lemma 1.3.

If $r\in R_{i}(q)$, then $r=ik+1$, where $k$ is a positive integer.

###### Proof.

This is a consequence of Fermat’s little theorem. ∎

Given a positive integer $i\neq 2$, denote by $k_{i}(a)$ the product of all
primitive prime divisors of $a^{i}-1$ with multiplicities counted. In case
$i=2$ put $k_{2}(a)=k_{1}(-a)$. It is not difficult to verify that if $i$ is
divisible by 4 then $k_{i}(a)=k_{i}(-a)$ and if $i$ is odd then
$k_{i}(a)=k_{2i}(-a)$. The following general formula [38] expresses
$k_{i}(a)$, where $i>2$, in terms of cyclotomic polynomials:

(1) $k_{i}(a)=\frac{\Phi_{i}(a)}{(r,\Phi_{(i)_{r^{\prime}}}(a))},$

where $r$ is the greatest prime divisor of $i$.

The following assertion is well-known and its proof is elementary.

######  Lemma 1.4.

Suppose that $q>1$ is an integer. For a positive integer $i$, an odd prime $r$
divides $q^{i}-1$ if and only if $e(r,q)$ divides $i$.

The following lemma is a particular case of the well-known Nagell–Ljunggren
equation.

######  Lemma 1.5 ([37]).

Suppose that $x$, $y$, and $k$ are positive integers. If $x^{2}+x+1=y^{k}$,
then either $k=1$ or $k=3$, $x=18$, and $y=7$.

The following technical lemma follows from Lemma 1.5, but we give an
independent proof. This statement will be needed in the proof of Main Theorem
for the group $G_{2}(q)$.

######  Lemma 1.6.

Suppose that $q$ is an integer greater than $2$ and
$\pi(q^{2}+\varepsilon{q}+1)=\\{r\\}$, where $\varepsilon\in\\{+,-\\}$ and $r$
is a prime. Then $r\equiv 1\pmod{6}$. Moreover, if $q\equiv 1\pmod{8}$ then
either $r\equiv 1\pmod{8}$ or $r\equiv 3\pmod{8}$.

###### Proof.

By assumption, there exists a positive integer $n$ such that
$q^{2}+\varepsilon{q}+1=r^{n}$. Clearly, $r$ is odd. Note that
$q^{2}+\varepsilon{q}+1$ is not divisible by 9, so if $r=3$, then $n=1$ and
$q\in\\{1,2\\}$. If $r\neq 3$, then $r$ divides
$k_{3}(\varepsilon{q})=\frac{q^{2}+\varepsilon{q}+1}{(q-\varepsilon 1,3)}$ and
hence $r\in R_{3}(\varepsilon{q})$. It follows from Lemma 1.3 that $r\equiv
1\pmod{6}$.

Suppose that $q\equiv 1\pmod{8}$. Then either $q^{2}+\varepsilon{q}+1\equiv
1\pmod{8}$ or $q^{2}+\varepsilon{q}+1\equiv 3\pmod{8}$. Therefore, if $r\equiv
5\pmod{8}$ or $r\equiv 7\pmod{8}$, then $n$ is even. On the other hand,
$(q-1)<q^{2}-q+1<q^{2}$ and $q^{2}<q^{2}+q+1<(q+1)^{2}$, so $n$ cannot be
even. This implies that either $r\equiv 1\pmod{8}$ or $r\equiv 3\pmod{8}$. ∎

Let $G$ be a finite group. Denote the number of connected components of
$\Gamma(G)$ by $s(G)$, and the set of connected components of $\Gamma(G)$ by
$\\{\pi_{i}(G)\mid 1\leq i\leq s(G)\\}$; for a group $G$ of even order, we
assume that $2\in\pi_{1}(G)$. Denote by $t(G)$ the _independence number_ of
$\Gamma(G)$, that is the greatest size of a coclique (i. e. induced subgraph
with no edges) in $\Gamma(G)$. If $r\in\pi(G)$, then denote by $t(r,G)$ the
greatest size of a coclique in $\Gamma(G)$ containing $r$.

######  Lemma 1.7.

Let $A$ and $B$ be normal subgroups of a group $G$ such that $A\leq B$. If
$r,s\in\pi(B/A)\setminus(\pi(A)\cup\pi(G/B))$, then $r$ and $s$ are adjacent
in $\Gamma(G)$ if and only if $r$ and $s$ are adjacent in $\Gamma(B/A)$.

###### Proof.

The proof of this lemma is elementary. ∎

######  Lemma 1.8 ([42]).

Let $G$ be a finite group with $t(G)\geq 3$ and $t(2,G)\geq 2$. Then the
following statements hold.

$(1)$ There exists a nonabelian simple group $S$ such that
$S\unlhd\overline{G}=G/K\leq\operatorname{Aut}(S)$, where $K$ is the solvable
radical of $G$ (i. e., the largest solvable normal subgroup of $G$).

$(2)$ For every coclique $\rho$ of $\Gamma(G)$ of size at least three, at most
one prime in $\rho$ divides the product $|K|\cdot|\overline{G}/S|$. In
particular, $t(S)\geq t(G)-1$.

$(3)$ One of the following two conditions holds:

$(3.1)$ $S\cong A_{7}$ or $L_{2}(q)$ for some odd $q$, and $t(S)=t(2,S)=3$.

$(3.2)$ Every prime $p\in\pi(G)$ nonadjacent to $2$ in $\Gamma(G)$ does not
divide the product $|K|\cdot|\overline{G}/S|$. In particular, $t(2,S)\geq
t(2,G)$.

######  Lemma 1.9 ([39, Lemma 1]).

Let $N\unlhd G$ be an elementary abelian subgroup and $H=G/N$. Define a
homomorphism $\phi:H\rightarrow\operatorname{Aut}(N)$ as follows
$n^{\phi(gN)}=n^{g}$. Then $\Gamma(G)=\Gamma(N\rtimes_{\phi}H)$.

######  Lemma 1.10 ([7, Proposition 3.1]).

Let $\pi$ be a finite set of primes. The number of pairwise nonisomorphic
nonabelian simple groups $S$ with $\pi(S)\subseteq\pi$ is finite, and is at
most $O(|\pi|^{3})$.

Let $S$ be a finite simple group of Lie type in characteristic $p$. Let $A$ be
any abelian $p$-group with an $S$-action. Any element $s\in S$ is said to be
unisingular on $A$ if $s$ has a (nonzero) fixed point on $A$. The group $S$ is
said to be unisingular if every element $s\in S$ acts unisingularly on every
finite abelian $p$-group $A$ with an $S$-action. Denote by
$PSL^{\varepsilon}_{n}(q)$, where $\varepsilon\in\\{+,-\\}$, the group
$PSL_{n}(q)$ if $\varepsilon=1$ and $PSU_{n}(q)$ if $\varepsilon=-1$.
Similarly, $E^{\varepsilon}_{6}(q)$ denotes the simple group $E_{6}(q)$ if
$\varepsilon=1$ and ${}^{2}E_{6}(q)$ if $\varepsilon=-1$.

######  Lemma 1.11 ([16, Theorem 1.3]).

A finite simple group $S$ of Lie type of characteristic $p$ is unisingular if
and only if $S$ is one of the following:

$(i)$ $PSL_{n}^{\varepsilon}(p)$ with $\varepsilon\in\\{+,-\\}$ and $n$
divides $p-\varepsilon 1$;

$(ii)$ $P\Omega_{2n+1}(p)$, $PSp_{2n}(p)$ with $p$ odd;

$(iii)$ $P\Omega_{2n}^{\varepsilon}(p)$ with $\varepsilon\in\\{+,-\\}$, $p$
odd, and $\varepsilon=(-1)^{n(p-1)/2}$;

$(iv)$ ${{}^{2}}G_{2}(q)$, $F_{4}(q)$, ${{}^{2}}F_{4}(q)$, $E_{8}(q)$ with $q$
arbitrary;

$(v)$ $G_{2}(q)$ with $q$ odd;

$(vi)$ $E_{6}^{\varepsilon}(p)$ with $\varepsilon\in\\{+,-\\}$ and $3$ divides
$p-\varepsilon 1$;

$(vii)$ $E_{7}(p)$ with $p$ odd.

######  Lemma 1.12 ([47, Proposition 2]).

Let $G={}^{3}D_{4}(q)$ act on a nonzero vector space $V$ over a field of
characteristic not dividing $q$ (possibly, zero). Then each element of $G$ of
order $q^{4}-q^{2}+1$ fixes on $V$ a nonzero vector.

The following lemma is also well-known, but we provide its proof for
completeness.

######  Lemma 1.13.

Let $G$ be a group, $g\in G$ an element of order $r$, and $\phi$ a non-trivial
irreducible representation of $G$ on a nonzero vector space $V$. If the
minimum polynomial degree of $\phi(g)$ equals to $r$, then $g$ fixes on $V$ a
nonzero vector.

###### Proof.

Let $A=\phi(g)$. Since $g^{r}=1$, we have $A^{r}=1$, and, therefore, the
minimal polynomial for $A$ divides the polynomial $x^{r}-1$. Since the minimum
polynomial degree of $A$ equals to $r$, we have that the minimum polynomial
for $A$ is $x^{r}-1$.

By the Cayley–Hamilton theorem, $A$ is a root of its characteristic
polynomial. In particular, $1$ is an eigenvalue of $A$, and, therefore, each
eigenvector of $A$ which corresponds to the eigenvalue $1$, is fixed by
$A=\phi(g)$. ∎

######  Lemma 1.14 ([41, Theorem 1.1]).

Let $G$ be one of the groups ${{}^{2}}B_{2}(q)$, where $q>2$,
${{}^{2}}G_{2}(q)$, where $q>3$, ${{}^{2}}F_{4}(q)$, $G_{2}(q)$,
${{}^{3}}D_{4}(q)$. Let $g\in G$ an element of prime power order coprime to
$q$. Let $\phi$ be a non-trivial irreducible representation of $G$ over a
field $F$ of characteristic $l$ coprime to $q$. Then the minimum polynomial
degree of $\phi(g)$ equals $|g|$, unless possibly when $G={{}^{2}}F_{4}(8)$,
$l=3$, $p=109$ and $\phi(1)<64692$.

######  Lemma 1.15 ([10, Lemma 4]).

Let $G$ be a finite simple group, $F$ a field of characteristic $p>0$, $V$ an
absolute irreducible $GF$-module, and $\beta$ a Brauer character of $V$. If
$g\in G$ is an element of prime order distinct from $p$, then

$dimC_{V}(g)=(\beta_{\langle g\rangle},1_{\langle
g\rangle})=\frac{1}{|g|}\sum_{x\in{\langle g\rangle}}\beta(x).$

######  Lemma 1.16.

Let $G$ be a group with a non-trivial nilpotent normal subgroup $K$ such that
$G/K$ has a subgroup $H$ isomorphic to $E_{8}(q)$, where $q$ is a prime power.
Then $R_{24}(q)\subset\pi_{1}(G)$.

###### Proof.

Note that $R_{24}(q)\subset\pi(H)$.

Let $q=p^{l}$, where $p$ is a prime. Suppose that $K$ is a $p$-group.
Factoring $G$ and $K$ by $K^{\prime}$, we can assume that $K$ is abelian.
According to [2, Table 3], $p$ lies in $\pi_{1}(G)$. By Lemma 1.11, $H$ is
unisingular and hence $p$ is adjacent to each element of $\pi(H)$. Therefore,
$R_{24}(q)\subset\pi_{1}(G)$.

Take any $r\in\pi(K)$. Since Sylow $2$-subgroups of $H$ are non-cyclic, we
infer that either $r=2$ or $2$ and $r$ are adjacent in $\Gamma(G)$ (see, for
example, [12, Theorem 10.3.1]). Therefore, $r\in\pi_{1}(G)$.

Suppose that there exists $r\in\pi(K)\cap R_{24}(q)$. Since $\pi(K)$ is a
clique in $\Gamma(G)$ and $r\in\pi_{1}(G)$, we get that
$R_{24}(q)\subset\pi_{1}(G)$.

If $K$ is not a $p$-group and $\pi(K)\cap R_{24}(q)=\varnothing$, then we can
choose $r\in\pi(K)\setminus(\\{p\\}\cup R_{24}(q))$. By Lemma 1.12, $r$ is
adjacent in $\Gamma(G)$ to any prime from $R_{24}(q)$. This implies that
$R_{24}(q)\subset\pi_{1}(G)$. ∎

##  2\. The Grunberg–Kegel graphs of some exceptional groups of Lie type

A criterion for the adjacency of vertices in the Gruenberg–Kegel graph for all
finite nonabelian simple groups was obtained in [44]. Based on this paper and
[45], in this section, we collect the necessary information for exceptional
groups of Lie type from Main Theorem.

By the compact form of the Gruenberg–Kegel graph $\Gamma(G)$ for a group $G$
we mean a graph whose vertices are labeled with sets of primes. A vertex
labeled by a set $\tau$ represents the clique of $\Gamma(G)$ such that every
vertex in this clique labeled by a prime from $\tau$. An edge connecting two
sets represents the set of edges of $\Gamma(G)$ that connect each vertex in
the first set with each vertex in the second.

######  Lemma 2.1 ([45, Proposition 2.7] and [44, Proposition 3.2]).

Let $G\cong F_{4}(q)$, where $q=2^{n}$, $r,s\in\pi(G)$ and $r\neq s$. Then $r$
and $s$ are nonadjacent if and only if one of the following conditions holds
(up to permutation):

  1. (1)

$2=r$, and $e(s,q)\in\\{8,12\\}$.

  2. (2)

$s,r\neq 2$, $k=e(r,q),\leavevmode\nobreak\ l=e(s,q)$, $1\leq k<l$ and either
$l\in\\{8,12\\}$, or $l=6$ and $k\in\\{3,4\\}$, or $l=4$ and $k=3$.

In particular, the compact form for $\Gamma(F_{4}(q))$ is the following.

$R_{1}$$R_{2}$$R_{6}$$R_{3}$$R_{4}$$2$$R_{12}$$R_{8}$

######  Remark.

Note that $R_{6}(2)$ and $R_{1}(2)$ are empty sets.

######  Lemma 2.2 ([44, Proposition 3.3] and [45, Proposition 2.9]; see also
[9, Lemma 3]).

Let $G\cong{}^{2}F_{4}(q)$, where $q=2^{2n+1}$, $r,s\in\pi(G)$ and $r\neq s$.
Put $m_{1}(n)=q-1$, $m_{2}(n)=q+1$, $m_{3}(n)=q^{2}+1$, $m_{4}(n)=q^{2}-q+1$,
$m_{5}(n)=q^{2}-\sqrt{2q^{3}}+q-\sqrt{2q}+1$,
$m_{6}(n)=q^{2}+\sqrt{2q^{3}}+q+\sqrt{2q}+1$. Then $r$ and $s$ are nonadjacent
in $\Gamma(G)$ if an only if one of the following conditions holds:

  1. (1)

$2=r$, $s$ divides $m_{k}(n)$, $s\neq 3$, and $k>3$.

  2. (2)

$2\neq s,r$; either $3\not=r\in\pi(m_{k}(n))$, $3\not=s\in\pi(m_{l}(n))$ for
$k\neq l$, and $\\{k,l\\}\neq\\{1,2\\},\leavevmode\nobreak\ \\{1,3\\}$; or
$r=3$ and $s\in\pi(m_{l}(n))$, where $l\in\\{3,5,6\\}$.

In particular, the compact form for $\Gamma({}^{2}F_{4}(q))$ is the following.

$2$$\pi(q+1)\\!\setminus\\!\\{3\\}$$3$$\pi(q-1)$$\pi(q^{2}-q+1)\\!\setminus\\!\\{3\\}$$\pi(q^{2}+1)$$\pi(q^{2}+\sqrt{2q^{3}}+q+\sqrt{2q}+1)$$\pi(q^{2}-\sqrt{2q^{3}}+q-\sqrt{2q}+1)$

######  Lemma 2.3 ([45, Proposition 2.7] and [44, Propositions 3.2, 4.5]).

Let $G\cong G_{2}(q)$, where $q=3^{k}$, $r,s\in\pi(G)$ and $r\neq s$. Then $r$
and $s$ are nonadjacent in $\Gamma(G)$ if and only if $e(r,q)$ or
$e(s,q)\in\\{3,6\\}$.

In particular, the compact form for $\Gamma(G_{2}(q))$ is the following.

$R_{1}$$R_{2}$$\\{3\\}$$R_{3}$$R_{6}$

######  Remark.

Note that the set $R_{1}(3)$ is empty.

######  Lemma 2.4 ([45, Proposition 2.7] and [44, Propositions 3.2, 4.5]).

Let $G\cong E_{8}(q)$, where $q$ is a power of a prime $p$. Suppose that
$r,s\in\pi(G)$ with $r\neq s$. Then $r$ and $s$ are nonadjacent in $\Gamma(G)$
if and only if one of the following conditions holds:

  1. (1)

$r\in\\{2,p\\}$, $s\neq p$ and $e(s,q)\in\\{15,20,24,30\\}$.

  2. (2)

$s,r\not\in\\{2,p\\}$, $k=e(r,q),\leavevmode\nobreak\ l=e(s,q)$, $1\leq k<l$,
and either $l=6$ and $k=5$, or $l\in\\{7,14\\}$ and $k\geq 3$, or $l=9$ and
$k\geq 4$, or $l\in\\{8,12\\}$ and $k\geq 5$, $k\neq 6$, or $l=10$ and $k\geq
3$, $k\not\in\\{4,6\\}$, or $l=18$ and $k\not\in\\{1,2,6\\}$, or $l=20$ and
$r\cdot k\neq 20$, or $l\in\\{15,24,30\\}$.

In particular, the compact form for $\Gamma(E_{8}(q))$ is the following. Here,
$R(q)=R_{1}(q)\cup R_{2}(q)\cup\\{p\\}$ and the vector from $5$ to $R_{4}(q)$
and the dotted edge $\\{5,R_{20}(q)\\}$ indicate that $R_{4}(q)$ and
$R_{20}(q)$ are not adjacent, but if $5\in R_{4}(q)$ (i.e.,
$q^{2}\equiv-1\pmod{5}$), then there exist edges between $5$ and the primes
from $R_{20}(q)$.
$R$$R_{18}$$R_{5}$$R_{3}$$R_{8}$$R_{12}$$R_{6}$$R_{10}$$R_{9}$$R_{14}$$R_{7}$$R_{4}$$5$$R_{20}$$R_{15}$$R_{24}$$R_{30}$

##  3\. Almost recognizability of groups $G_{2}(q)$, $F_{4}(q)$, and
${{}^{2}}F_{4}(q)$ by Gruenberg–Kegel graph

In this section, we prove Main Theorem for groups $L=F_{4}(q)$, where $q\geq
2$ is a power of $2$, $L={{}^{2}}F_{4}(q)$, where $q=2^{2m+1}>2$, and
$L=G_{2}(q)$, where $q>3$ is a power of $3$.

######  Theorem 1.

If $G$ is a group with $\Gamma(G)=\Gamma(L)$, then
$L\cong\operatorname{Inn}(L)\unlhd G\leq\operatorname{Aut}(L)$. In particular,
$L$ is almost recognizable by Gruenberg–Kegel graph.

By Lemmas 2.1, 2.2, and 2.3, we see that $t(L)\geq 3$ and $t(2,L)\geq 2$. It
follows from Lemma 1.8 that there exists a nonabelian simple group $S$ such
that $S\cong\operatorname{Inn}(S)\leq G/K\leq\operatorname{Aut}(S)$, where $K$
is the solvable radical of $G$. Moreover, by the Thompson theorem on finite
groups with fixed-point-free automorphisms of prime order [40, Theorem 1], $K$
is nilpotent. To prove Theorem 1, it suffices to show that $S\cong L$ and
$K=1$. These two facts are established in the following four lemmas.

######  Lemma 3.1.

$S\cong L$.

###### Proof.

If $q>2$ is even and $L={{}^{2}}F_{4}(q)$ or $L=F_{4}(q)$, then Lemma follows
from [1, Theorem 3.4] and [24, Theorem 3.3], respectively.

Suppose that $L=F_{4}(2)$. Then $\pi(S)\subseteq\pi(L)=\\{2,3,5,7,13,17\\}$.
Note that $13\in R_{12}(2)$ and $17\in R_{8}(2)$. By Lemma 2.1, we find that
$13$ and $17$ are nonadjacent to all vertices in $\Gamma(G)$. Therefore,
$\\{13,17\\}\subset\pi(S)$ by Lemma 1.8. Inspecting [49, Table 1], we see that
$S\in\\{PSU_{4}(4)$, $PSU_{3}(17)$, $PSL_{2}(13^{2})$, $PSp_{4}(13)$,
$PSL_{3}(16)$, $PSp_{6}(4)$, $P\Omega_{8}^{+}(4)$, $F_{4}(2)\\}$. Using [4,
Corollary 3], we find that $5$ and $17$ are adjacent in
$\Gamma(PSL_{2}(13^{2}))$ and $\Gamma(PSL_{3}(16))$, while $3$ and $17$ are
adjacent in $\Gamma(PSU_{3}(17))$ and $\Gamma(PSU_{4}(4))$. According to [5,
Corollaries 2–4], 5 and 17 are adjacent in $\Gamma(PSp_{4}(13))$,
$\Gamma(PSp_{6}(4))$, and $\Gamma(P\Omega_{8}^{+}(4))$. This implies that
$S\cong F_{4}(2)$, as claimed.

It remains to consider the case $L\cong G_{2}(q)$, where $q>3$ is a power of
3. If $q>3$ is an odd power of $3$, then $S\cong L$ by [50, Theorem 1.1].
Suppose that $L\cong G_{2}(q)$, where $q=3^{2k}$ for a positive integer $k$.
By Lemma 2.3, sets $R_{3}(q)$ and $R_{6}(q)$ are connected components in
$\Gamma(L)$. It follows from Lemma 1.8 that sets $R_{3}(q)$ and $R_{6}(q)$ are
connected components in $\Gamma(S)$ and hence $\Gamma(S)$ has at least three
connected components. Therefore, $S$ isomorphic to a group listed in
Introduction before Main Theorem. We show that $S\cong L$ considering each
case for $S$ separately. We will extensively use that
$k_{3}(q)=\frac{q^{2}+q+1}{(q-1,3)}=q^{2}+q+1$ and
$k_{6}(q)=\frac{q^{2}-q+1}{(q+1,3)}=q^{2}-q+1$ (see equation (1)).

Case $S\cong A_{p}$, where $p>6$ and both $p$ and $p-2$ are primes. The
connected components of $\Gamma(S)$ are $\pi_{1}(S)$, $\\{p\\}$, and
$\\{p-2\\}$. We know that $R_{3}(q)$ and $R_{6}(q)$ are connected components
in $\Gamma(S)$, so either $p-2\in R_{3}(q)$ or $p-2\in R_{6}(q)$. It follows
from Lemma 1.3 that $p-3$ is divisible by $3$; a contradiction since $p\neq
3$.

Case $S\cong E_{8}(u)$, where $u$ is a prime power. Since
$3^{2k}\equiv(-1)^{k}\pmod{5}$, we infer that $q\equiv\pm 1\pmod{5}$. This
implies that there exists $\varepsilon\in\\{+,-\\}$ such that
$q^{2}+\varepsilon{q}+1\equiv 3\pmod{5}$. Take any prime divisor $r$ of
$k_{3}(\varepsilon{q})=q^{2}+\varepsilon{q}+1$ such that $r\not\equiv
1\pmod{5}$. Since $r\in R_{3}(\varepsilon{q})$, Lemma 2.4 implies that $r\in
R_{j}(u)$, where $j\in\\{15,20,24,30\\}$. Then $r-1$ is divisible by $j$ and
hence $j=24$. Since $R_{24}(u)$ is a connected component in $\Gamma(S)$, we
find that $R_{3}(\varepsilon{q})=R_{24}(u)$. If $K\neq 1$, then Lemma 1.16
implies that $R_{24}(u)\subseteq\pi_{1}(G)$; a contradiction since
$r\not\in\pi_{1}(G)$. Assume that there exists $s\in\pi(\overline{G}/S)$ such
that $s>3$. It follows from Lemma 1.8 that $s\in\pi_{1}(G)$. By [13, Theorem
2.5.12, Definition 2.5.13], there exists an element $g\in G\setminus S$ such
that $|g|=s$ and $g$ acts on $S$ as a field automorphism. It is known that
$C_{S}(g)\cong E_{8}(u^{1/s})$ (see, e.g., [13, Proposition 4.9.1]). By Lemma
1.2, we see that $R_{24}(u^{1/s})\subseteq R_{24}(u)$ and hence $s$ is
adjacent to $r$ in $\Gamma(G)$; we arrive at a contradiction since
$s\in\pi_{1}(G)$ and $r\not\in\pi_{1}(G)$. Now consider a coclique
$\rho=\\{s_{1},s_{2},\ldots,s_{12}\\}$ of maximal size in $\Gamma(S)$. By
Lemma 2.4, we have $(s_{i},6)=1$ for every $i\in\\{1,\ldots,12\\}$. Applying
Lemma 1.7, we find that $t(G)\geq 12$; a contradiction with Lemma 2.3.

Cases
$S\in\\{PSU_{6}(2),PSL_{3}(4),M_{11},M_{23},M_{24},J_{3},HiS,Suz,Co_{2},Fi_{23},F_{2},M_{22},Fi_{24}^{\prime},F_{1}\\}$.
By Lemma 1.8, we infer that $\\{2,r_{3}(q),r_{6}(q)\\}$ is a coclique in
$\Gamma(S)$. It follows from Lemma 1.3 that $r_{3}(q)\equiv 1\pmod{6}$ and
$r_{6}(q)\equiv 1\pmod{6}$. Using [8], we see that there is no such a coclique
in $\Gamma(S)$; contradiction.

Case $S\cong F_{4}(u)$, where $u$ is even. Since $q^{2}+q+1\equiv 3\pmod{8}$,
there exists $r\in R_{3}(q)$ such that $r\not\equiv 1\pmod{8}$. By Lemmas 1.8,
2.1, and 1.3, we infer that $r\in R_{12}(u)$. Then $\pi(q^{2}+q+1)=R_{12}(u)$.
On the other hand, we have $q^{2}+q+1\equiv 7\pmod{12}$ and hence there exists
$s\in\pi(q^{2}+q+1)$ such that $s\not\equiv 1\pmod{12}$; a contradiction with
Lemma 1.3.

Case $S\cong{}^{2}F_{4}(u)$, where $u=2^{2m+1}>2$. By Lemma 2.2, it is true
that $\pi_{2}(S)\cup\pi_{3}(S)=\pi(u^{4}-u^{2}+1)\subseteq\pi(u^{6}+1)$.
Consider any prime $r\in R_{3}(q)$. Then $r\in\pi_{2}(S)\cup\pi_{3}(S)$ and
$r$ divides $(u^{3})^{2}+1$. Since $(u^{3})^{2}\equiv-1\pmod{r}$, we find that
$-1$ is a quadratic residue modulo $r$. This implies that $r\equiv 1\pmod{4}$.
Since $r$ is an arbitrary element of $R_{3}(q)$, we infer that
$q^{2}+q+1\equiv 1\pmod{4}$. On the other hand, $q^{2}+q+1\equiv
3^{4k}+3^{2k}+1\equiv 3\pmod{4}$; a contradiction.

Case $S\cong{}^{2}B_{2}(u)$, where $u=2^{2m+1}>2$. According to [2, Table 3],
we can assume that $\pi_{2}(S)=\pi(u-1)$, $\pi_{3}(S)=\pi(u-\sqrt{2u}+1)$, and
$\pi_{4}(S)=\pi(u+\sqrt{2u}+1)$. Note that
$\pi(u-\sqrt{2u}+1)\cup\pi(u+\sqrt{2u}+1)=\pi(u^{2}+1)$. If
$R_{3}(q)\neq\pi_{2}(S)$, then $R_{3}(q)\subseteq\pi(u^{2}+1)$ and we get a
contradiction arguing as in the case $S\cong{}^{2}F_{4}(u)$. Therefore, we can
assume that $R_{3}(q)=\pi(u-1)$. Suppose that $r\leavevmode\nobreak\
|\leavevmode\nobreak\ u-1=2^{2m+1}-1$. Then $r\leavevmode\nobreak\
|\leavevmode\nobreak\ 2^{2n+2}-2$ and hence $2$ is a quadratic residue modulo
$r$. This implies that $r\equiv\pm 1\pmod{8}$. It follows that
$q^{2}+q+1\equiv\pm 1\pmod{8}$; a contradiction since $q^{2}+q+1\equiv
3^{4k}+3^{2k}+1\equiv 3\pmod{8}$.

Case $S\cong{{}^{2}}G_{2}(u)$, where $u=3^{2m+1}>3$. According to [2, Table
2], we can assume that $\pi_{2}(S)=\pi(u-\sqrt{3u}+1)$ and
$\pi_{3}(S)=\pi(u+\sqrt{3u}+1)$. Take any prime $r\in R_{12k}(3)$. By Lemma
1.2, we infer that $r\in R_{6}(q)$. We know that
$R_{6}(q)\subseteq\pi_{2}(S)\cup\pi_{3}(S)=\pi(u^{2}-u+1)=R_{6}(u)$. By Lemma
1.4, we conclude that $12k\leavevmode\nobreak\ |\leavevmode\nobreak\
6\cdot(2m+1)$; a contradiction.

Case $S\cong{{}^{2}}D_{p}(3)$, where $p=2^{m}+1\geq 3$ is prime. According to
[2, Table 2], we can assume that $\pi_{2}(S)=\pi((3^{p-1}+1)/2)$ and
$\pi_{3}(S)=\pi((3^{p}+1)/4)$. Take any $r\in R_{12k}(3)$. By Lemma 1.2, we
infer that $r\in R_{6}(q)$. Since $r\in\pi_{2}(S)\cup\pi_{3}(S)$, we find that
$r$ divides $3^{2(p-1)}-1$ or $3^{2p}-1$. It follows from Lemma 1.4 that
$12k\leavevmode\nobreak\ |\leavevmode\nobreak\ 2(p-1)$ or
$12k\leavevmode\nobreak\ |\leavevmode\nobreak\ 2p$; a contradiction.

Case $S\cong A_{1}(u)\cong PSL_{2}(q)$, where $u=2^{m}>2$. According to [2,
Table 2], we can assume that $\pi_{2}(S)=\pi(u-1)$ and $\pi_{3}(S)=\pi(u+1)$.
Note that 3 divides $u^{2}-1$. On the other hand,
$\pi(u^{2}-1)=\pi_{2}(S)\cup\pi_{3}(S)=\pi_{2}(G)\cup\pi_{3}(G)$; a
contradiction since $3\in\pi_{1}(G)$.

Case $S\cong A_{1}(u)\cong PSL_{2}(u)$, where $3<u=v^{n}$ and $u$ is odd.
Consider $\varepsilon\in\\{+,-\\}$ such that $u\equiv\varepsilon 1\pmod{4}$.
According to [2, Table 2], we can assume that $\pi_{1}(S)=\pi(u-\varepsilon
1)$, $\pi_{2}(S)=\\{v\\}$, and $\pi_{3}(S)=\pi(\frac{u+\varepsilon 1}{2})$.
Therefore, there exists $\tau\in\\{+,-\\}$ such that
$\pi(q^{2}+\tau{q}+1)=\\{v\\}$ and
$\pi(q^{2}-\tau{q}+1)=\pi(\frac{u+\varepsilon 1}{2})$. Moreover, we know that
$\pi(u-\varepsilon 1)\subseteq\pi(3(q^{2}-1))$. Since
$q^{2}-q+1=(q-1)^{2}+(q-1)+1$, Lemma 1.5 implies that either
$q^{2}+\tau{q}+1=v$ or $q=19$ and $v=7$. By assumption, $q=3^{2k}$ and hence
$q^{2}+\tau{q}+1=v$. Then $v-1$ is divisible by $9$. Therefore, $v\geq 19$ and
$3\in\pi(\frac{u-1}{2})\setminus\pi(q^{2}-\tau{q}+1)$. This implies that
$\varepsilon=+$ since, by Lemma 2.3, $3\in\pi_{1}(G)$.

Suppose that $n$ is even. Then $v^{2}-1$ divides $u-1$, so
$\pi(v+1)\subseteq\pi(q^{2}-1)$. Take any $r\in\pi(v+1)$. Since $q\equiv\pm
1\pmod{r}$ and $r$ divides $q^{2}+\tau{q}+2$, we find that $r$ divides $3\pm
1$ and hence $r=2$. Therefore, $q^{2}+\tau{q}+2$ is a power of 2. On the other
hand, $q^{2}+\tau{q}+2\equiv 1\pm 1+2\pmod{8}$. This implies that
$q^{2}+\tau{q}+2\leq 4$; a contradiction.

We can assume that $n$ is odd. Then $\frac{v+1}{2}$ divides $\frac{u+1}{2}$.
Take any $r\in\pi(\frac{v+1}{2})$. Then $r$ divides both $q^{2}+\tau{q}+2$ and
$q^{2}-\tau{q}+1$. Therefore, $2\tau{q}\equiv-1\pmod{r}$ and hence $0\equiv
4q^{2}+4\tau{q}+8\equiv 4\tau{q}+9\equiv 7\pmod{r}$. This implies that $r=7$
and $v+1=2\cdot 7^{m}$ for a positive integer $m$. Since $q\equiv 1\pmod{8}$
and $q^{2}+\tau{q}+2=2\cdot 7^{m}$, we infer that $3+\tau 1\equiv
2\cdot(-1)^{m}\pmod{8}$. This implies that $\tau=-1$ and $m$ is even. A
straightforward calculation shows that $9^{2k}-9^{k}+2$ is not divisible by 49
for all positive integer $k$; a contradiction.

Cases
$S\in\\{F_{3},O^{\prime}N,J_{1},LyS,J_{4},{}^{2}E_{6}(2),E_{7}(2),E_{7}(3)\\}$.
According to [2, Tables 2, 3], we see that there exist primes $r$ and $s$ such
that $\pi(q^{2}+q+1)=\\{r\\}$ and $\pi(q^{2}-q+1)=\\{s\\}$. Applying Lemma
1.6, we find that $r\equiv s\equiv 1\pmod{6}$ and the remainders of $r$ and
$s$ divided by 8 belong to the set $\\{1,3\\}$. Inspecting [2, Tables 2, 3],
we see that in each case there are no two primes satisfying these
restrictions; a contradiction.

Thus, we conclude that $S\cong G_{2}(u)$, where $u$ is a power of $3$. Now
Lemma 1.1 implies immediately that $u=q$ and, therefore, $S\cong L$. This
completes the proof of the lemma.

∎

Remark. For $q=3$, the paper [50] contains a mistake, and
$S\in\\{G_{2}(3),PSL_{2}(13)\\}$ by [20, Table 1].

Show that $K=1$. Consider a minimal (by order) counterexample $G$ to this
claim.

######  Lemma 3.2.

$K$ is an elementary abelian $r$-group for some $r\in\pi_{1}(L)$.

###### Proof.

Let $r$ be a prime divisor of $|K|$. Since $K$ is nilpotent, we have a
decomposition $K=P\times U$, where $P$ is a Sylow $r$-subgroup of $K$ and $U$
is a normal subgroup of $K$ such that $r\not\in\pi(U)$. Since $U$ and
$\Phi(P)$ are characteristic subgroups of $K$, the subgroup $N=U\times\Phi(P)$
is a normal subgroup of $G$. Then $\Gamma(G/N)$ is a subgraph of $\Gamma(G)$.
On the other hand, $\Gamma(L)$ is a subgraph of $\Gamma(G/N)$ and hence
$\Gamma(G/N)=\Gamma(L)$. By the minimality of $G$, we infer that $N=1$ and,
therefore, $K$ is an elementary abelian $r$-group. ∎

By Lemmas 1.9, 3.1 and 3.2, we can assume that $G=K\rtimes X$, where $K$ is an
elementary abelian $r$-group for $r\in\pi_{1}(L)$ and $Soc(X)=S\cong L$.

######  Lemma 3.3.

$K=1$ if $L=F_{4}(q)$.

###### Proof.

According to [33, Table 5.1], $S$ has a subgroup $H$ isomorphic to
${{}^{3}}D_{4}(q)$.

Assume that $r\not\in\\{2\\}\cup R_{12}(q)$. By Lemma 1.12, $r$ is adjacent to
each element from $R_{12}(q)$ in $\Gamma(KH)$. It follows from Lemma 2.1 that
$\Gamma(G)\not=\Gamma(L)$; a contradiction.

Assume that $r=2$. By Lemma 1.11, $S$ is unisingular and, therefore, $2$ is
adjacent to all other vertices in $\Gamma(KS)$. We arrive at a contradiction
with Lemma 2.1.

Assume that $r\in R_{12}(q)$. Since $q$ is even, by [33, Table 5.1], $S$ has a
subgroup $H_{1}\cong P\Omega_{8}^{+}(q)$. Now by [6, Table 8.50], $H_{1}$ has
a subgroup $H_{2}\cong P\Omega_{4}^{-}(q)\times P\Omega_{4}^{-}(q)\cong
PSL_{2}(q^{2})\times PSL_{2}(q^{2})$. Therefore, for each $s\in R_{4}(q)$, a
Sylow $s$-subgroup of $S$ is non-cyclic. This implies that $r$ and $s$ are
adjacent in $\Gamma(KH_{2})$ (see, for example, [12, Theorem 10.3.1]).
Therefore, $r$ and $s$ are adjacent in $\Gamma(G)$; a contradiction with Lemma
2.1. Thus, $K=1$. ∎

######  Lemma 3.4.

$K=1$ if $L\cong G_{2}(q)$ for $q>3$ or $L\cong{{}^{2}}F_{4}(q)$ for $q>2$.

###### Proof.

Let $p=2$ if $L\cong{{}^{2}}F_{4}(q)$ and $p=3$ if $L\cong G_{2}(q)$. By Lemma
1.11, $L$ is unisingular and, therefore, if $r=p$, then $\Gamma(G)$ is
connected; a contradiction. Therefore, $r\in\pi_{1}(L)\setminus\\{p\\}$. By
Lemmas 1.13 and 1.14, $r$ is adjacent to a prime from $\pi_{2}(L)$; a
contradiction. Thus, $K=1$. ∎

This completes the proof of Theorem 1.

Since the group $L$ is known to be almost recognizable, the following natural
question arises.

######  Problem 1.

Suppose that $L$ is a group from the statement of Theorem 1. Find a positive
integer $k$ such that $L$ is $k$-recognizable by Gruenberg-Kegel graph.

##  4\. Unrecognizability of groups ${}^{2}B_{2}(q)$ and $G_{2}(3)$ by
Gruenberg–Kegel graphs

In this short section, we show that groups ${}^{2}B_{2}(q)$ with $q>2$ and
$G_{2}(3)$ are unrecognizable.

######  Proposition 4.1.

Let $G={{}^{2}}B_{2}(q)$, where $q>2$ is an odd power of $2$. Then $G$ is
unrecognizable by Gruenberg–Kegel graph.

###### Proof.

By [16, Lemma 3.6], there exists a $4$-dimensional module $V$ over the field
of order $q$ such that each nontrivial element from each maximal torus of $G$
acts fixed-point freely on $V$. Thus, $\Gamma(V\rtimes G)=\Gamma(G)$ and,
therefore, $G$ is unrecognizable by [7, Theorem 1.2]. ∎

######  Proposition 4.2.

Let $G\cong G_{2}(3)$. Then $G$ is unrecognizable by Gruenberg–Kegel graph.

###### Proof.

Using [8], we find that $\Gamma(G_{2}(3))$ is the following:

$2$$3$$7$$13$

Moreover, $\Gamma(G)=\Gamma(PSL_{2}(13))$. By [19, P. 9] and Lemma 1.15, there
is a $6$-dimensional irreducible $PSL_{2}(13)$-module $V$ over a field of
characteristic two such that all elements in $PSL_{2}(13)$ of orders $7$ and
$13$ act fixed-point freely on $V$. Therefore,

$\Gamma(V\rtimes PSL_{2}(13))=\Gamma(PSL_{2}(13))=\Gamma(G).$

Thus, by [7, Theorem 1.2], $G$ is unrecognizable by Gruenberg–Kegel graph. ∎

##  5\. Almost recognizability of groups $E_{8}(q)$ by Gruenberg–Kegel graph

To complete the proof of Main Theorem, it remains to consider the case of
groups $E_{8}(q)$. In [47], A. V. Zavarnitsine proved that if $G$ is a finite
group such that $\Gamma(G)=\Gamma(E_{8}(q))$, where $q\equiv 0,\pm 1\pmod{5}$,
then $G\cong E_{8}(u)$ for some $u\equiv 0,\pm 1\pmod{5}$. The aim of this
section is to prove the following similar result for the remaining cases for
$q$.

######  Theorem 2.

Let $L=E_{8}(q)$, where $q\equiv\pm 2\pmod{5}$ is a prime power, and $G$ be a
group such that $\Gamma(G)=\Gamma(L)$. Then $G\cong E_{8}(u)$ for some prime
power $u$ with $u\equiv\pm 2\pmod{5}$.

###### Proof.

Since $\Gamma(G)$ is disconnected, Lemma 1.8 implies that there exists a
nonabelian simple group $S$ such that $S\leq G/K\leq\operatorname{Aut}(S)$,
where $K$ is the solvable radical of $G$. By the Thompson theorem on finite
groups with fixed-point-free automorphisms of prime order [40, Theorem 1], $K$
is nilpotent. By Lemmas 1.8 and 2.4, we find that $s(S)\geq s(L)=4$,
$t(2,S)\geq t(2,L)=5$ and $t(S)\geq t(L)-1=11$. According to [2, Table 1] and
[44, Tables 2, 4, and 5], we find that either $S\cong F_{1}$ or $S\cong
E_{8}(u)$, where $u$ is a prime power.

Suppose that $S\cong F_{1}$. According to [2, Table 3], we see that $s(S)=4$,
for each $i\geq 2$, $|\pi_{i}(S)|=1$, and
$\pi(S)\setminus\pi_{1}(S)=\\{41,59,71\\}$. At the same time,
$\pi_{2}(G)=R_{15}(q)$, $\pi_{3}(G)=R_{24}(q)$, and $\pi_{4}(G)=R_{30}(q)$. By
Lemma 1.3, the numbers $15$, $24$, and $30$ divide $r_{i}-1$ for pairwise
distinct primes $r_{i}$ from $\pi(S)\setminus\pi_{1}(S)$; a contradiction.

Suppose that $S\cong E_{8}(u)$, where $u$ is a power of a prime $v$ and
$u\equiv 0,\pm 1\pmod{5}$. Since $q\equiv\pm 2\pmod{5}$, we find that $5\in
R_{4}(q)$. Denote

$\theta=\\{r_{9}(q),r_{14}(q),r_{7}(q),r_{18}(q),r_{15}(q),r_{24}(q),r_{30}(q)\\}.$

By Lemma 2.4, we see that $\theta\cup\\{5\\}$ is a coclique of size 8 in
$\Gamma(G)$. It follows from Lemma 1.8 that at least six elements of $\theta$
belong to $\pi(S)$. Take any $r\in\pi(S)\cap\theta$. Since $r$ and 5 are
nonadjacent in $\Gamma(G)$, they are nonadjacent in $\Gamma(S)$. We know that
$5\in\\{v\\}\cup R_{1}(u)\cup R_{2}(u)$ and hence $r\in R_{20}(u)\cup
R_{15}(u)\cup R_{24}(u)\cup R_{30}(u)$ according to Lemma 2.4. This implies
that at least two elements from $\theta\cap\pi(S)$ are adjacent in
$\Gamma(S)$; a contradiction.

Suppose that $S\cong E_{8}(u)$, where $v^{l}=u\equiv\pm 2\pmod{5}$. According
to [2, Table 3], we can assume that $\pi_{2}(G)=R_{15}(q)$,
$\pi_{3}(G)=R_{24}(q)$, and $\pi_{4}(G)=R_{30}(q)$, while
$\pi_{2}(S)=R_{15}(u)$, $\pi_{2}(S)=R_{24}(u)$, and $\pi_{4}(S)=R_{30}(u)$. If
$K\neq 1$, then Lemma 1.16 implies that $R_{24}(u)\subset\pi_{1}(G)$; a
contradiction since $R_{24}(u)$ must coincide with a connected component of
$\Gamma(G)$ not containing 2. Therefore, we can assume that $S\leq
G\leq\operatorname{Aut}(S)$.

Now prove that $G/S=1$. If a prime $r$ divides $|G:S|$, then by [13, Theorem
2.5.12, Definition 2.5.13, Proposition 4.9.1], $G\setminus S$ contains a field
automorphism $x$ of $S$ of order $r$ with $C_{S}(x)\geq E_{8}(u^{1/r})$. This
implies that $r$ is adjacent in $\Gamma(G)$ to each prime from
$\pi(E_{8}(u^{1/r}))$. Suppose that $r\not\in\\{2,3,5\\}$. By Lemma 1.2,
$\Gamma(G)$ is connected and hence $\Gamma(G)\not=\Gamma(S)$. If $r=2$, then
by Lemma 1.2, $R_{15}(u)\subset\pi_{1}(G)$; a contradiction. If $r=5$, then by
Lemma 1.2, $r$ is adjacent in $\Gamma(G)$ to some primes from $R_{24}(u)$ and
hence $R_{24}(u)\subset\pi_{1}(G)$; a contradiction. Therefore, we can assume
that $G/S$ is a 3-group. By Lemma 2.4, we see that $r_{i}(q)$ is nonadjacent
to 2 in $\Gamma(G)$ if and only if $i\in\\{15,20,24,30\\}$. Similarly, a prime
$r_{i}(u)$ is nonadjacent to 2 in $\Gamma(S)$ if and only if
$i\in\\{15,20,24,30\\}$. Since $R_{20}(q)\subset\pi_{1}(G)$ and
$R_{20}(u)\subset\pi_{1}(G)$, we infer that $R_{20}(q)=R_{20}(u)$. By Lemma
1.2, we infer that $3$ is adjacent in $\Gamma(G)$ to a prime from $R_{20}(u)$,
therefore, $\Gamma(G)\not=\Gamma(S)$.

Thus, $G=S$. The proof of Theorem 2 is complete. ∎

######  Corollary 1.

For each value of $q$, the group $E_{8}(q)$ is almost recognizable by
Gruenberg–Kegel graph.

###### Proof.

If $G$ is a group such that $\Gamma(G)=\Gamma(E_{8}(q))$, then by [47, Theorem
1] and Theorem 2, we have $G\cong E_{8}(u)$ for a prime power $u$. By Lemma
1.10, for a given $q$, the number of possibilities for $u$ is finite. This
implies that there is only the finite number of possibilities for $G$ (up to
isomorphism), in particular, $E_{8}(q)$ is almost recognizable by
Gruenberg–Kegel graph. ∎

Problem 2. Do there exist prime powers $q$ and $q_{1}$ with $q\not=q_{1}$ and
$\Gamma(E_{8}(q))=\Gamma(E_{8}(q_{1}))$?

######  Corollary 2.

If $G$ is a finite group such that $\Gamma(G)=\Gamma(E_{8}(q))$ and
$|G|=|E_{8}(q)|$ for some prime power $q$, then $G\cong E_{8}(q)$.

###### Proof.

By Theorem 2, if $\Gamma(G)=\Gamma(E_{8}(q))$, then $G\cong E_{8}(q_{1})$ for
some prime power $q_{1}$. It is clear that a function

$f(x)=x^{120}(x^{2}-1)(x^{8}-1)(x^{12}-1)(x^{14}-1)(x^{18}-1)(x^{20}-1)(x^{24}-1)(x^{30}-1)$

strictly monotonically increases if $x\geq 1$. Thus, if

$f(q_{1})=|G|=|E_{8}(q)|=f(q),$

then $q_{1}=q$, and, therefore, $G\cong E_{8}(q)$. ∎

In [43], it was proved that if $G$ is a simple group and $H$ is a group such
that $\omega(H)=\omega(G)$ and $|H|=|G|$, then $H\cong G$. Thus, each simple
group is uniquely determined by its order and spectrum. It is known that if
$q$ is odd and $n\geq 3$, then $\Gamma(P\Omega_{2n+1}(q))=\Gamma(PSp_{2n}(q))$
and $|P\Omega_{2n+1}(q)|=|PSp_{2n}(q)|$ but these groups are not isomorphic.
Therefore, it is natural to consider the following problem.

Problem 3. For which simple groups $G$ is the following true: if $H$ is a
group with ${\Gamma(H)=\Gamma(G)}$ and $|H|=|G|$, then $H$ is isomorphic to
$G$?

Problem 3 was formulated by B. Khosravi in his survey paper [23, Question
4.2], by A.S. Kondrat’ev in frame of the open problems session of the 13th
School-Conference on Group Theory Dedicated to V. A. Belonogov’s 85th Birthday
(see [35, Question 4]), and was independently formulated by W. Shi in a
personal communication with the first author. Also Problem 3 was formulated in
the paper by P. J. Cameron and the first author (see [7, Problem 2]). It is
clear that if a simple group is quasirecognizable by Gruenberg–Kegel graph,
then Problem 3 solves in the positive for this group. At the same time,
Corollary 2 gives a solution of Problem 3 for finite simple groups $E_{8}(q)$
which are not necessary quasirecognizable by Gruenberg–Kegel graph.

##  6\. Acknowledgements

The first author is supported by the Ministry of Science and Higher Education
of the Russian Federation, project 075-02-2022-877 for the development of the
regional scientific and educational mathematical center ”Ural Mathematical
Center” (for example, Section 5). The second author is supported by the
Mathematical Center in Akademgorodok under the agreement No. 075-15-2022-281
with the Ministry of Science and Higher Education of the Russian Federation
(for example, Section 3). The third author is supported by RAS Fundamental
Research Program, project FWNF-2022-0002 (for example, Section 4).

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