Characterization of some finite simple groups by the set of orders of vanishing elements and order
UDC 512.5 Характеризацiя деяких скiнченних простих груп множиною порядкiв зникаючих елементiв та порядку Let $G$ be a finite group. We say that an element $g$ of $G$ is a vanishing element if there exists an irreducible complex character $X$ of $G$ such that $X(g) = 0$. Ghasemabadi, Iranmanesh, Mava...
Saved in:
| Date: | 2021 |
|---|---|
| Main Authors: | , |
| Format: | Article |
| Language: | English |
| Published: |
Institute of Mathematics, NAS of Ukraine
2021
|
| Online Access: | https://umj.imath.kiev.ua/index.php/umj/article/view/1069 |
| Tags: |
Add Tag
No Tags, Be the first to tag this record!
|
| Journal Title: | Ukrains’kyi Matematychnyi Zhurnal |
| Download file: |  |
Institution
Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860507138472804352 |
|---|---|
| author | Askary, S. S. Askary, S. |
| author_facet | Askary, S. S. Askary, S. |
| author_sort | Askary, S. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2025-03-31T08:46:33Z |
| description | UDC 512.5
Характеризацiя деяких скiнченних простих груп множиною порядкiв зникаючих елементiв та порядку
Let $G$ be a finite group. We say that an element $g$ of $G$ is a vanishing element if there exists an irreducible complex character $X$ of $G$ such that $X(g) = 0$. Ghasemabadi, Iranmanesh, Mavadatpour (2015), present the following conjecture: Let $G$ be a finite group and $M$ a finite nonabelian simple group such that $Vo(G)=Vo(M)$ and $|G|=|M|$. Then $G \cong M $. We answer in affirmative this conjecture for $M = ^2 D_{r+1}(2)$, where $r = 2^n - 1 \geq 3$ and either $2^r+1$ or $2^{r+1}+1$ is a prime number and $M = ^2 D_{r}(3)$, where $r = 2^n + 1 \geq 5$ and either $(3^{r-1}+1)/2$ or $(3^{r}+1)/4$ is prime. |
| doi_str_mv | 10.37863/umzh.v73i11.1069 |
| first_indexed | 2026-03-24T02:04:33Z |
| format | Article |
| fulltext |
DOI: 10.37863/umzh.v73i11.1069
UDC 512.5
S. Askary (Imam-Ali Univ., Tehran, Iran)
CHARACTERIZATION OF SOME FINITE SIMPLE GROUPS
BY THE SET OF ORDERS OF VANISHING ELEMENTS AND ORDER
ХАРАКТЕРИЗАЦIЯ ДЕЯКИХ СКIНЧЕННИХ ПРОСТИХ ГРУП
МНОЖИНОЮ ПОРЯДКIВ ЗНИКАЮЧИХ ЕЛЕМЕНТIВ ТА ПОРЯДКУ
Let G be a finite group. We say that an element g of G is a vanishing element if there exists an irreducible complex
character \chi of G such that \chi (g) = 0. Ghasemabadi, Iranmanesh, Mavadatpour (2015), present the following conjecture:
Let G be a finite group and M a finite non-Abelian simple group such that Vo(G) = Vo(M) and | G| = | M | . Then
G \sim = M . We answer in affirmative this conjecture for M = 2Dr+1(2), where r = 2n - 1 \geq 3 and either 2r + 1 or
2r+1 + 1 is a prime number and M = 2Dr(3), where r = 2n + 1 \geq 5 and either (3r - 1 + 1)/2 or (3r + 1)/4 is prime.
Нехай G — скiнченна група. Елемент g \in G є зникаючим елементом, якщо iснує незвiдний комплексний характер
\chi \in G такий, що \chi (g) = 0. Гасемабадi, Iранманеш та Мавадатпур (2015) запропонували гiпотезу: якщо G —
скiнченна група, а M — скiнченна неабелева проста група, для яких Vo(G) = Vo(M) i | G| = | M | , то G \sim = M . Ми
доводимо цю гiпотезу для M = 2Dr+1(2), де r = 2n - 1 \geq 3, якщо або 2r + 1, або 2r+1 + 1 є простим числом, i
для M = 2Dr(3), де r = 2n + 1 \geq 5, якщо або (3r - 1 + 1)/2, або (3r + 1)/4 є простим.
1. Introduction. Let G be a finite group. It is well-known that the set of values cd(G) = \{ \chi (1) :
\chi \in Irr(G)\} has a strong influence on the group structure of G, where Irr(G) denotes the set of
irreducible complex characters of G. We say that an element g of G is a vanishing element if there
exists an irreducible complex character \chi of G such that \chi (g) = 0. Denote Van(G) the set \{ g \in G :
\chi (g) = 0 for some \chi \in Irr(G)\} , Vo(G) the set \{ o(g) : g \in Van(G)\} consisting of the orders of
the elements in Van(G).
In [16], it is shown that if G is a finite group such that Vo(G) = Vo(A5), then G \sim = A5. In
[17] it is proved that if G is a finite group such that Vo(G) = Vo(Sz(22m+1)), where m \geq 1,
then G \sim = Sz(22m+1). But not all finite simple groups are characterizable by the set of orders of
their vanishing elements. For example, Vo(PSL(3, 5)) = Vo(Aut(PSL(3, 5))), but PSL(3, 5) \ncong
\ncong Aut(PSL(3, 5)). The following conjecture is one of the important problem:
Conjecture. Let G be a finite group and M a finite non-Abelian simple group such that
Vo(G) = Vo(M) and | G| = | M | . Then G \sim = M. The above conjecture was proved for simple
groups PSL(2, q), where q \in \{ 5, 7, 8, 9, 17\} , PSL(3, 4), A7, Sz(8) and Sz(32). Then in [9], it is
proved that sporadic simple groups, alternating groups, projective special linear groups PSL(2, p),
where p is an odd prime, and finite simple Kn-groups where n \in \{ 3, 4\} , satisfying this conjecture.
Now, we prove this conjecture for some finite simple groups as follows:
Theorem A. If G is a finite group such that Vo(G) = Vo(2Dr+1(2)) and | G| = | 2Dr+1(2)| ,
where r = 2n - 1 \geq 3 and either 2r + 1 or 2r+1 + 1 is prime, then G \sim = 2Dr+1(2).
Theorem B. If G is a finite group such that Vo(G) = Vo(2Dr(3)) and | G| = | 2Dr(3)| , where
r = 2n + 1 \geq 5 and either (3r - 1 + 1)/2 or (3r + 1)/4 is prime, then G \sim = 2Dr(3).
c\bigcirc S. ASKARY, 2021
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 11 1443
1444 S. ASKARY
Let X be a finite set of positive integers. The prime graph \Pi (X) is a graph whose vertices are
the prime divisors of elements of X divisable by pq. For a finite group G, we denote by \omega (G) the
set of element orders of G, and by \pi (G) the set of prime divisors of | G| . The graph \Pi (\omega (G)) is
denoted by GK(G) and is called the Gruenberg – Kegel graph of G. We denote by t(G) the number
of connected components of GK(G) and by \pi i(G), i = 1, 2, . . . , t(G), the vertex set of the ith
connected components of GK(G). If 2 \in \pi (G), we always assume that 2 \in \pi 1(G). The prime
graph \Pi (Vo(G)) is denoted by \Gamma (G) and is called the vanishing prime graph of G. Obviously the
vanishing prime graph of G is a subgraph of Gruenberg – Kegel graph of G.
Throughout this paper, we denote by \pi (n) the set of prime divisors of integer n. All further
notation can be found in [4], for instance.
2. Preliminaries. A 2-Frobenius group is a group G which has a normal series 1\unlhd H\unlhd K\unlhd G,
where K and G/H are Frobenius groups with kernels H and K/H, respectively. Also, we know
that 2-Frobenius groups are solvable.
Definition 2.1 [18]. Let a and n be integers greater than 1. Then a Zsigmondy prime of an - 1
is a prime l such that l | (an - 1) but l \nmid (ai - 1) for 1 \leq i < n.
Lemma 2.1 [18]. Let a and n be integers greater than 1. Then there exists a Zsigmondy prime
of an - 1, unless (a, n) = (2, 6) or n = 2 and a = 2s - 1 for some natural number s.
Remark 2.1. If l is a Zsigmondy prime of an - 1, then Fermat’s little theorem shows that n | l - 1.
Put
Zn(a) = \{ l : l is a Zsigmondy prime of an - 1\} .
If r \in Zn(a) and r | am - 1, then we can see at once that n | m.
Lemma 2.2 [3]. Let G be a Frobenius group of even order with kernel K and complement H.
Then t(G) = 2, the prime graph components of G are \pi (H) and \pi (K) and the following assertions
hold:
(1) K is nilpotent;
(2) | K| \equiv 1 (mod | H| ).
Lemma 2.3 [3]. Let G be a 2-Frobenius group. Then:
(a) t(G) = 2, \pi 1 = \pi (G/K) \cup \pi (H) and \pi 2 = \pi (K/H);
(b) G/K and K/H are cyclic, | G/K| | (| K/H| - 1) and G/K \leq \mathrm{A}\mathrm{u}\mathrm{t}(K/H).
Lemma 2.4 [15]. If G is a finite group such that t(G) \geq 2, then G has one of the following
structures:
(a) G is a Frobenius group or 2-Frobenius group;
(b) G has a normal series 1 \unlhd H \unlhd K \unlhd G such that \pi (H) \cup \pi (G/K) \subseteq \pi 1 and K/H is a
non-Abelian simple group. In particular, H is nilpotent, G/K \lesssim \mathrm{O}\mathrm{u}\mathrm{t}(K/H) and the odd order
components of G are the odd order components of K/H.
Lemma 2.5 [7, 8]. (i) If G is a finite non-Abelian simple group except A7, then GK(G)=\Gamma (G).
(ii) If G is a solvable group, then \Gamma (G) has at most 2 connected components.
Lemma 2.6 [7]. Let G be a finite nonsolvable group. If \Gamma (G) is disconnected. Then G has
a unique non-Abelian composition factor S, and t(S) is greater than or equal to the number of
connected components of \Gamma (G), unless G is isomorphic to A7.
Lemma 2.7 [7]. Let G be a group and K a nilpotent normal subgroup of G. If K
\bigcap
Van(G) \not =
\not = 0, then there exists g \in K
\bigcap
Van(G) whose order is divisable by every prime in \pi (K).
The following lemma is an easy consequence of [12] (Corollary 22.26).
Lemma 2.8. If \chi \in Irr(G) vanishes on a p-element for some prime p, then p | \chi (1).
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 11
CHARACTERIZATION OF SOME FINITE SIMPLE GROUPS BY THE SET. . . 1445
Let p be a prime number. A character \chi \in Irr(G) is said to be of p defect zero, if p \nmid | G| /\chi (1).
Also, if \chi \in Irr(G) is of p defect zero, then for every element g \in G such that p | o(g), we have
\chi (g) = 0 [11] (Theorem 8.17).
Lemma 2.9 [6]. The equation pm - qn = 1, where p and q are primes and m,n > 1 has only
solution, namely, 32 - 23 = 1.
Lemma 2.10 [6]. With the exceptions of the relations (239)2 - 2(13)4 = - 1 and 35 - 2(11)2 = 1
every solution of the equation
pm - 2qn = \pm 1, p, q prime, m, n > 1,
has exponents m = n = 2; i.e., it comes from a unit p - q.21/2 of the quadratic field \BbbQ (21/2) for
which the coefficients p and q are primes.
3. Proofs of the main results. Proof of Theorem A. By the assumption Vo(G) =
= Vo(2Dr+1(2)), it is obvious that \Gamma (G) = \Gamma (2Dr+1(2)). By Lemma 2.6, we know that
\Gamma (2Dr+1(2)) = GK(2Dr+1(2)) has 3 connected components including an isolated vertex p, where
p \in \{ 2r + 1, 2r+1 + 1\} . Also, note that
| G| = 2r(r+1)(2r - 1)(2r + 1)(2r+1 + 1)
r - 1\prod
i=1
(22i - 1).
Since p \in Vo(2Dr+1(2)) and Vo(G) = Vo(2Dr+1(2)), so p \in Vo(G). Thus there exist an element
g \in G and irreducible character \chi \in Irr(G) such that o(g) = p and \chi (g) = 0. So p | \chi (1) and
since | G| p = p, we conclude that p \nmid | G| /\chi (1). Therefore, \chi is a p-defect zero, and, hence, for
every element h \in G such that p | o(h), we have \chi (h) = 0. So, by the fact p is an isolated vertex
in \Gamma (G), we conclude that p is an isolated vertex in GK(G). Hence, t(G) \geq 2.
Since \Gamma (G) has three connected components, Lemma 2.6 implies that G is not a solvable group
and consequently G is not a 2-Frobenius group. We also claim that G is not a Frobenius group.
Suppose that G is a Frobenius group with kernel K and complement H. So | G| = | H| | K| and
| H| | | K| - 1. Lemma 2.2 implies that GK(G) has two connected components \pi (H) and \pi (K),
and since | H| < | K| , it follows that | H| = p and | K| = | G| /p. In both cases p = 2r + 1 and
p = 2r+1+1, one can get a contradiction by the fact that | H| | | K| - 1. Therefore G is not a Frobenius
group. So, by Lemma 2.4, G has a normal 1\unlhd H\unlhd K\unlhd G such that \pi (H)\cup \pi (G/K) \subseteq \pi 1 and K/H
is a non-Abelian simple group and G/K \leq Aut(K/H). By Lemma 2.6, we have t(K/H) \geq 3.
In both cases p = 2r + 1 and p = 2r+1 + 1, we use the classification of finite non-Abelian simple
groups with more than two Gruenberg – Kegel graph connected components to prove that K/H is
isomorphic to 2Dr+1(2).
Case 1. First suppose that p = 2r + 1.
Step 1. K/H is not an sporadic simple group.
Suppose that K/H is an sporadic simple group. Then p = 2r+1 \in \{ 5, 7, 11, 13, 17, 19, 23, 29, 31,
37, 41, 43, 47, 53, 59, 61, 67, 71\} . If K/H \sim = Fi22, then p = 2r + 1 = 17, 23 or 29. The only possi-
bility is r = 4, but r = 2n - 1 \geq 3, which is impossible. For other sporadic simple groups one get
a contradiction similarly.
Step 2. K/H is not an alternating group.
Let K/H \sim = Ap\prime , where p\prime and p\prime - 2 are primes. If p\prime - 2 = p = 2r+1, then p\prime = 2r+3 is a prime
number, which is impossible. Let p\prime = p = 2r +1 and p\prime > 7. Since p\prime - 7 = 2(2r - 1 - 3) | | K/H| ,
we have 2r - 1 - 3 | | G| , which is impossible. If p\prime = 7, then p\prime = 2r + 1, which is impossible. For
p\prime = 5, we have 2r + 1 = 5 and hence r = 2, but r = 2n - 1 \geq 3, which is a contradiction.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 11
1446 S. ASKARY
Step 3. K/H is not a simple group of lie type, except 2Dr+1(2).
If K/H is isomorphic to 2A5(2), E7(2), E7(3), A2(4) or 2E6(2), then we easily get a contra-
diction similar to sporadic simple groups.
a) Let K/H \sim = A1(q
\prime ), where q\prime = 2m > 2. Therefore q\prime - 1 = p or q\prime +1 = p. If q\prime - 1 = p =
= 2r+1, then 2m - 2r = 2. Since m \geq 2 and r \geq 3, we get a contradiction. So q\prime +1 = p = 2r+1
and, hence, m = r and | K/H| = q\prime (q\prime - 1)(q\prime + 1) = 2r(2r - 1)(2r + 1). On the other hand,
G/K \leq Out(K/H), which implies that | G/K| | r. Therefore, 2r+1(2r+1+1)
\prod r - 1
i=1
(22i - 1) | | H| .
By considering \Gamma (G) we conclude that there exist g \in G and \chi \in Irr(G) such that \pi (o(g)) \subseteq
\subseteq \pi (2r+1+1) and \chi (g) = 0. Since \pi (o(g)) \subseteq \pi (2r+1+1), (2r+1+1, 2r +1) = 1 and H \unlhd G, we
conclude that g \in H. Therefore, H is a nilpotent normal subgroup of G such that H
\bigcap
Van(G) \not = \phi .
Now, Lemma 2.7 implies that there exist a vanishing element whose order is divisible by all prime
divisors of | H| . So all prime divisors of | H| are adjacent in \Gamma (G), which is a contradiction by Table
9 of [14].
b) Let K/H \sim = A1(q
\prime ), where 3 < q\prime \equiv \varepsilon (\mathrm{m}\mathrm{o}\mathrm{d} 4) for \varepsilon = \pm 1. Hence q\prime = 2r + 1 = p or
(q\prime + \varepsilon )/2 = 2r + 1 = p. First let (q\prime + \varepsilon )/2 = 2r + 1. If \varepsilon = 1, then q\prime - 2r+1 = 1, which is a
contradiction with Lemma 2.9.
If \varepsilon = - 1, then q\prime \equiv - 1(\mathrm{m}\mathrm{o}\mathrm{d} 4). Since 4 | (q\prime + 1), we can conclude that q\prime = u\alpha , where u is
odd prime. Thus p \in Z\alpha (u) and hence by Remark 2.1, \alpha | p - 1 = 2r. Therefore, \alpha = 2t, which
implies that q = u\alpha \equiv 1(\mathrm{m}\mathrm{o}\mathrm{d} 4), which is a contradiction. Now let q\prime = 2r +1 = p. So q\prime - 2r = 1
and, by Lemma 2.9, q\prime = 9, which implies that r = 3. Therefore, | G| = 212 \times 34 \times 5 \times 7 \times 17,
| K/H| = 23\times 32\times 5 and | G/K| | 2. Hence, | H| = 29\times 32\times 7\times 17. Now, similar to the above case,
we can conclude that all prime divisors of order of H are adjacent in \Gamma (G), which is impossible.
c) Let K/H \sim = E8(q
\prime ). Then p = 2r + 1 is an element of the set
\{ q\prime 8 \pm q\prime 7 \mp q\prime 5 - q\prime 4 \mp q\prime 3 \pm q\prime + 1, q\prime 8 - q\prime 6 + q\prime 4 - q\prime 2 + 1, q\prime 8 - q\prime 4 + 1\} .
So, p = 2r +1 < (q\prime 8 + q\prime 7 + q\prime 6 + q\prime 5 + q\prime 4 + q\prime 3 + q\prime 2 + q\prime +1)(q\prime - 1) = q\prime 9 - 1 < q\prime 9 +1, which
implies that 2r < q\prime 9 and, hence, | K/H| > | G| , which is impossible.
d) Let K/H \sim = Sz(q\prime ), where q\prime = 22m+1 > 2. If 22m+1 - 1 = p = 2r+1, then 22m+1 - 2r = 2,
which is impossible. If 22m+1 \pm 2m+1 +1 = 2r +1, then 2m+1(2m \pm 1) = 2r, which is impossible.
e) Let K/H \sim = 2F4(q
\prime ), where q\prime = 22m+1 > 2. Then 22(2m+1)\pm 23m+2+22m+1\pm 2m+1+1 =
= 2r + 1, which implies that 2m+1(23m+1 \pm 22m+1 + 2m \pm 1) = 2r, which is a contradiction.
f) Let K/H \sim = 2G2(q
\prime ) for q\prime = 32m+1 > 3. Therefore 32m+1 \pm 3m+1 + 1 = 2r + 1, and,
consequently, 3m+1(3m \pm 1) = 2r, which is impossible. If K/H \sim = G2(q
\prime ), where q\prime \equiv 0(\mathrm{m}\mathrm{o}\mathrm{d} 3)
and K/H \sim = 2B2(q
\prime ), one can get a contradiction similarly.
g) Let K/H be isomorphic to 2D\prime
p(3), where p\prime = 2m + 1. Then either (3p
\prime
+ 1)/4 = 2r + 1 or
(3p
\prime - 1 + 1)/2 = 2r + 1. Now, if (3p
\prime
+ 1)/4 = 2r + 1, then 3p
\prime - 3 = 2r+2, which is impossible. If
(3p
\prime - 1 + 1)/2 = 2r + 1, then 3p
\prime - 1 - 2r+1 = 1, which is impossible by Lemma 2.9.
h) Therefore K/H \sim = 2Dr\prime +1(2), where r\prime = 2m - 1 \geq 3. Obviously m \leq n. Since p \in
\in \pi (K/H), it follows that p = 2r + 1 is a divisor of
2r
\prime (r\prime +1)(2r
\prime - 1)(2r
\prime
+ 1)(2r
\prime +1 + 1)
r\prime - 1\prod
i=1
(22i - 1).
Note that p is a primitive prime divisors of 2r + 1. Now, if m < n, then p \nmid | G| , a contradiction.
Therefore m = n and, hence, r\prime = r. Thus, K/H \sim = 2Dr+1(2).
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 11
CHARACTERIZATION OF SOME FINITE SIMPLE GROUPS BY THE SET. . . 1447
Case 2. Now suppose that p = 2r+1 + 1.
Step 1. K/H is not an sporadic simple group.
Suppose that K/H is an sporadic simple group. Then p = 2r+1 + 1 \in \{ 5, 7, 11, 13, 17, 19,
23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71\} . If K/H \sim = Fi23, then p = 2r+1+1 = 17, 23 or 29. The
only possibility is r = 3. But | Fi23| \nmid | 2D4(2)| , a contradiction. For other sporadic simple groups,
one get a contradiction similarly.
Step 2. K/H is not an alternating group.
Let K/H \sim = Ap\prime , where p\prime and p\prime - 2 are primes. If p\prime = 2r+1 + 1, then p\prime - 2 = 2r+1 - 1 is a
prime number, which is a contradiction. If p\prime - 2 = 2r+1 + 1, then p\prime = 2r+1 + 3 is a divisor of
| G| = 2r(r+1)(2r - 1)(2r + 1)(2r+1 + 1)
r - 1\prod
i=1
(22i - 1),
which is impossible.
Step 3. K/H is not a simple group of lie type, except 2Dr+1(2).
If K/H is isomorphic to 2A5(2), E7(2), E7(3), A2(4) or 2E6(2), then we easily get a contra-
diction similar to sporadic simple groups.
a) Let K/H \sim = A1(q
\prime ), where q\prime = 2m > 2. Therefore q\prime - 1 = p or q\prime + 1 = p. If q\prime - 1 =
= p = 2r+1+1, then 2m - 2r+1 = 2, which is impossible. If q\prime +1 = p = 2r+1+1, then m = r+1
and | K/H| = 2r+1(2r+1 - 1)(2r+1 + 1). On the other hand, G/K \leq Out(K/H), which implies
that | G/K| | r + 1. Therefore 2(2r - 1)(2r + 1)
\prod r - 1
i=1
(22i - 1) | | H| . By considering \Gamma (G) we
conclude that there exist g \in G and \chi \in Irr(G) such that \pi (o(g)) \subseteq \pi (2r+1) and \chi (g) = 0. Since
\pi (o(g)) \subseteq \pi (2r + 1), (2r+1 + 1, 2r + 1) = 1 and H \unlhd G, we conclude that g \in H. Therefore, H is
a nilpotent normal subgroup of G such that H
\bigcap
Van(G) \not = \phi . Now, Lemma 2.7 implies that there
exist a vanishing element whose order is divisible by all prime divisors of | H| . So all prime divisors
of | H| are adjacent in \Gamma (G), which is a contradiction by Table 9 of [14].
b) Let K/H \sim = A1(q
\prime ), where 3 < q\prime \equiv \varepsilon (\mathrm{m}\mathrm{o}\mathrm{d} 4) for \varepsilon = \pm 1. Hence q\prime = 2r+1 + 1 = p or
(q\prime + \varepsilon )/2 = 2r+1 + 1 = p. First let (q\prime + \varepsilon )/2 = 2r+1 + 1. If \varepsilon = 1, then q\prime - 2r+2 = 1, which is
a contradiction with Lemma 2.9.
If \varepsilon = - 1, then q\prime \equiv - 1(\mathrm{m}\mathrm{o}\mathrm{d} 4). Since 4 | (q\prime + 1), we can conclude that q\prime = u\alpha , where u is
odd prime. Thus p \in Z\alpha (u) and hence by Remark 2.1, \alpha | p - 1 = 2r+1. Therefore, \alpha = 2t, which
implies that q = u\alpha \equiv 1(\mathrm{m}\mathrm{o}\mathrm{d} 4), which is a contradiction.
Now let q\prime = 2r+1 + 1 = p. So q\prime - 2r+1 = 1 and, by Lemma 2.9, q\prime = 9, which implies that
r = 2. Since r = 2n - 1 \geq 3, we get a contradiction.
c) Let K/H \sim = E8(q
\prime ). Then p = 2r+1 + 1 is an element of the set
\{ q\prime 8 \pm q\prime 7 \mp q\prime 5 - q\prime 4 \mp q\prime 3 \pm q\prime + 1, q\prime 8 - q\prime 6 + q\prime 4 - q\prime 2 + 1, q\prime 8 - q\prime 4 + 1\} .
So p = 2r+1 + 1 < (q\prime 8 + q\prime 7 + q\prime 6 + q\prime 5 + q\prime 4 + q\prime 3 + q\prime 2 + q\prime + 1)(q\prime - 1) = q\prime 9 - 1 < q\prime 9 + 1,
which implies that 2r+1 < q\prime 9 and hence | K/H| > | G| , which is impossible.
d) Let K/H \sim = Sz(q\prime ), where q\prime = 22m+1 > 2. If 22m+1 - 1 = p = 2r+1 + 1, then 22m+1 -
- 2r+1 = 2, which is impossible. If 22m+1 \pm 2m+1 + 1 = 2r+1 + 1, then 2m+1(2m \pm 1) = 2r+1,
which is impossible.
e) Let K/H \sim = 2F4(q
\prime ), where q\prime = 22m+1 > 2. Then 22(2m+1)\pm 23m+2+22m+1\pm 2m+1+1 =
= 2r+1 + 1, which implies that 2m+1(23m+1 \pm 22m+1 + 2m \pm 1) = 2r+1, which is a contradiction.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 11
1448 S. ASKARY
f) Let K/H \sim = 2G2(q
\prime ) for q\prime = 32m+1 > 3. Therefore 32m+1 \pm 3m+1 + 1 = 2r+1 + 1, and
consequently 3m+1(3m \pm 1) = 2r+1, which is impossible. If K/H \sim = G2(q
\prime ), where q\prime \equiv 0(\mathrm{m}\mathrm{o}\mathrm{d} 3)
and K/H \sim = 2B2(q
\prime ), one can get a contradiction similarly.
g) Let K/H be isomorphic to 2D\prime
p(3), where p\prime = 2m+1. Then either (3p
\prime
+1)/4 = 2r+1+1 or
(3p
\prime - 1+1)/2 = 2r+1+1. Now, if (3p
\prime
+1)/4 = 2r+1+1, then 3p
\prime - 3 = 2r+3, which is impossible.
If (3p
\prime - 1 + 1)/2 = 2r+1 + 1, then 3p
\prime - 1 - 2r+2 = 1, which is impossible by Lemma 2.9.
h) Therefore K/H \sim = 2Dr\prime +1(2), where r\prime = 2m - 1 \geq 3. Obviously m \leq n. Since p \in
\in \pi (K/H), it follows that p = 2r+1 + 1 is a divisor of
2r
\prime (r\prime +1)(2r
\prime - 1)(2r
\prime
+ 1)(2r
\prime +1 + 1)
r\prime - 1\prod
i=1
(22i - 1).
Note that p is a primitive prime divisors of 2r+1 + 1. Now, if m < n, then p \nmid | G| , a contradiction.
Therefore m = n and hence r\prime = r. Thus K/H \sim = 2Dr+1(2). So in both cases K/H \sim = 2Dr+1(2)
and since | G| = | 2Dr+1(2)| , it is obvious that H = 1 and G = K, hence, G \sim = 2Dr+1(2).
Theorem A is proved.
Proof of Theorem B. By the assumption Vo(G) = Vo(2Dr(3)), it is obvious that \Gamma (G) =
= \Gamma (2Dr(3)). By Lemma 2.6, we know that \Gamma (2Dr(3)) = GK(2Dr(3)) has 3 connected compo-
nents including an isolated vertex p, where p \in \{ (3r - 1 + 1)/2, (3r + 1)/4\} . Also, note that
| G| = 3r(r - 1)(3r + 1)
r - 1\prod
i=1
(32i - 1).
Since p \in Vo(2Dr(3)) and Vo(G) = Vo(2Dr(3)), so p \in Vo(G). Thus there exist an element
g \in G and irreducible character \chi \in Irr(G) such that o(g) = p and \chi (g) = 0. So p | \chi (1) and
since | G| p = p, we conclude that p \nmid | G| /\chi (1). Therefore \chi is a p-defect zero, and hence for every
element h \in G such that p | o(h), we have \chi (h) = 0. So, by the fact p is an isolated vertex in
\Gamma (G), we conclude that p is an isolated vertex in GK(G). Hence, t(G) \geq 2.
Since \Gamma (G) has three connected components, Lemma 2.6 implies that G is not a solvable group
and consequently G is not a 2-Frobenius group. We also claim that G is not a Frobenius group.
Suppose that G is a Frobenius group with kernel K and complement H. So | G| = | H| | K| and
| H| | | K| - 1. Lemma 2.2 implies that GK(G) has two connected components \pi (H) and \pi (K),
and since | H| < | K| , it follows that | H| = p and | K| = | G| /p. In both cases p = (3r - 1 + 1)/2
and p = (3r + 1)/4, one can get a contradiction by the fact that | H| | | K| - 1. Therefore G is not a
Frobenius group. So, by Lemma 2.4, G has a normal 1\unlhd H \unlhd K \unlhd G such that \pi (H)\cup \pi (G/K) \subseteq
\subseteq \pi 1 and K/H is a non-Abelian simple group and G/K \leq Aut(K/H). By Lemma 2.6, we have
t(K/H) \geq 3. In both cases p = (3r - 1 + 1)/2 and p = (3r + 1)/4, we use the classification of
finite nonabelian simple groups with more than two Gruenberg – Kegel graph connected components
to prove that K/H is isomorphic to 2Dr(3).
Case 1. First suppose that p = (3r - 1 + 1)/2.
Step 1. K/H is not an sporadic simple group.
Suppose that K/H is an sporadic simple group. Then p = (3r - 1+1)/2 \in \{ 5, 7, 11, 13, 17, 19, 23,
29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71\} . If K/H \sim = F1, then p = (3r - 1 + 1)/2 = 41. The only pos-
sibility is r = 5. But | F1| \nmid | 2D5(3)| , which is impossible. For other sporadic simple groups one get
a contradiction.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 11
CHARACTERIZATION OF SOME FINITE SIMPLE GROUPS BY THE SET. . . 1449
Step 2. K/H is not an alternating group.
Let K/H \sim = Ap\prime , where p\prime and p\prime - 2 are primes. If p\prime - 2 = p = (3r - 1 + 1)/2, then
p\prime = (3r - 1 + 5)/2 is a prime number, which is impossible. Let p\prime = p = (3r - 1 + 1)/2, then
p\prime - 2 = (3r - 1 - 3)/2 is a prime number, which is a contradiction.
Step 3. K/H is not a simple group of lie type, except 2Dr(3).
If K/H is isomorphic to 2A5(2), E7(2), E7(3), A2(4) or 2E6(2), then we easily get a contra-
diction similar to sporadic simple groups.
a) Let K/H \sim = A1(q
\prime ), where q\prime = 2m > 2. therefore q\prime - 1 = p or q\prime + 1 = p. If q\prime - 1 =
= p = (3r - 1 + 1)/2, then 2q\prime = 3r - 1 + 3 and hence 2m+1 = 3(3r - 2 + 1), which is impossible.
If q\prime + 1 = p = (3r - 1 + 1)/2, then 3r - 1 - 2m+1 = 1 and, by Lemma 2.10, r - 1 = 2. Since
r = 2n + 1 \geq 5, we get a contradiction.
b) Let K/H \sim = A1(q
\prime ), where 3 < q\prime \equiv \varepsilon (\mathrm{m}\mathrm{o}\mathrm{d} 4) for \varepsilon = \pm 1. Hence q\prime = (3r - 1 + 1)/2 = p
or (q\prime + \varepsilon )/2 = (3r - 1 + 1)/2 = p. First let (q\prime + \varepsilon )/2 = (3r - 1 + 1)/2. If \varepsilon = 1, then q\prime = 3r - 1
and | K/H| = 3r - 1(3r - 1 - 1)(3r - 1 + 1)/2. On the other hand, G/K \leq Out(K/H), which implies
that | G/K| | r - 1. Therefore 3r(3r + 1)/4 | | H| . By considering \Gamma (G) we conclude that there
exist g \in G and \chi \in Irr(G) such that \pi (o(g)) \subseteq \pi ((3r + 1)/4) and \chi (g) = 0. Since \pi (o(g)) \subseteq
\subseteq \pi ((3r + 1)/4), ((3r + 1)/4, (3r - 1 + 1)/2) = 1 and H \unlhd G, we conclude that g \in H. Therefore
H is a nilpotent normal subgroup of G such that H
\bigcap
Van(G) \not = \phi . Now, Lemma 2.7 implies that
there exist a vanishing element whose order is divisible by all prime divisors of | H| . So all prime
divisors of | H| are adjacent in \Gamma (G), which is a contradiction by Table 9 of [14].
If \varepsilon = - 1, then q\prime = 3r - 1+2 and | K/H| = (3r - 1+1)(3r - 1+2)(3r - 1+3). Since (3r - 1+2) \nmid
\nmid | G| , we get a contradiction.
If q\prime = (3r - 1 + 1)/2 = p, then | K/H| = 3/8
\bigl(
(3r - 1 - 1)(3r - 1 + 1)(3r - 2 + 1)
\bigr)
. On the other
hand, G/K \leq Out(K/H), which implies that | G/K| | 2. Now, similar to the above for \varepsilon = +1,
we can get a contradiction.
c) Let K/H \sim = E8(q
\prime ). Then (3r - 1 + 1)/2 is an element of the set
\{ q\prime 8 \pm q\prime 7 \mp q\prime 5 - q\prime 4 \mp q\prime 3 \pm q\prime + 1, q\prime 8 - q\prime 6 + q\prime 4 - q\prime 2 + 1, q\prime 8 - q\prime 4 + 1\} .
So p = (3r - 1 + 1)/2 < (q\prime 8 + q\prime 7 + q\prime 6 + q\prime 5 + q\prime 4 + q\prime 3 + q\prime 2 + q\prime +1)(q\prime - 1) = q\prime 9 - 1 < q\prime 9 +1,
which implies that 3r - 1 < q\prime 10 and hence | K/H| > | G| , which is impossible.
d) Let K/H \sim = Sz(q\prime ), where q\prime = 22m+1 > 2. If 22m+1 - 1 = p = (3r - 1 + 1)/2, then
22m+2 = 3r + 3, which is impossible.
e) Let K/H \sim = 2F4(q
\prime ), where q\prime = 22m+1 > 2. Then 22(2m+1)\pm 23m+2+22m+1\pm 2m+1+1 =
= (3r - 1+1)/2, which implies that 2m+1(23m+1\pm 22m+1+2m\pm 1) = 3r - 1, which is a contradiction.
f) Let K/H \sim = 2G2(q
\prime ) for q\prime = 32m+1 > 3. Therefore 32m+1 \pm 3m+1 + 1 = (3r - 1 + 1)/2, and
consequently 3m+1(3m\pm 1) = (3r - 1+1)/2, which is impossible. If and K/H \sim = 2B2(q
\prime ), similarly
we get a contradiction.
g) Let K/H be isomorphic to 2Dp\prime +1(2), where p\prime = 2n - 1, n \geq 2. Therefore 2p
\prime
+ 1 =
= (3r - 1+1)/2 or 2p
\prime +1+1 = (3r - 1+1)/2. If 2p
\prime
+1 = (3r - 1+1)/2, then 3r - 1 - 2p
\prime +1 = 1 and, by
Lemma 2.10, r - 1 = 2. Since r = 2n+1 \geq 5, we get a contradiction. For 2p
\prime +1+1 = (3r - 1+1)/2,
similar to the above we get a contradiction.
h) Therefore K/H \sim = 2Dr\prime (3), where r\prime = 2m +1 \geq 5. Obviously m \leq n. Since p \in \pi (K/H),
it follows that p = (3r - 1 + 1)/2 is a divisor of
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 11
1450 S. ASKARY
3r
\prime (r\prime - 1)(3r
\prime
+ 1)
r\prime - 1\prod
i=1
(32i - 1).
Note that p is a primitive prime divisors of (3r - 1 + 1)/2. Now, if m < n, then p \nmid | G| , a contradic-
tion. Therefore m = n and hence r\prime = r. Thus, K/H \sim = 2Dr(3).
Case 2. If p = (3r + 1)/4, then similar to case 1, we can conclude that K/H \sim = 2Dr(3) and by
the fact that | G| = | 2Dr(3)| , we have H = 1, G = K and G \sim = 2Dr(3) as required.
Theorem B is proved.
References
1. S. Asgary, N. Ahanjideh, Characterization of PSL(3, q) by nse, Math. Rep., 19 (60), № 4, 425 – 438 (2017).
2. S. Asgary, N. Ahanjideh, nse Characterization of some finite simple groups, Sci. Ann. Comput. Sci., 3 (2), 797 – 806
(2016).
3. G. Chen, Further reflections on Thompson’s conjecture, J. Algebra, 218, 276 – 285 (1999).
4. J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, R. A. Wilson, Atlas of finite groups, Clarendon Press, New
York (1985).
5. G. Y. Chen, On Frobenius and 2-Frobenius group, J. Southwest China Normal Univ., 20, № 5, 485 – 487 (1995).
6. P. Crescenzo, A Diophantine equation which arises in the theory of finite groups, Adv. Math., 17, 25 – 29 (1975).
7. S. Dolfi, E. Pacific, L. Sanus, P. Spiga, On the vanishing prime graph of finite groups, J. London Math. Soc., Ser. 2,
82, 167 – 183 (2010).
8. S. Dolfi, E. Pacific, L. Sanus, P. Spiga, On the vanishing prime graph of solvable groups, J. Group Theory, 13,
189 – 206 (2010).
9. M. F. Ghasemabadi, A. Iranmanesh, M. Ahanjideh, A new characterization of some families of finite simple groups,
Rend. Semin. Mat. Univ. Padova, 137, 57 – 74 (2017).
10. M. F. Ghasemabadi, A. Iranmanesh, F. Mavadatpur, A new characterization of some finite simple groups, Sib. Math.
J., 56, 78 – 82 (2015).
11. I. M. Isaacs, Character theory of finite groups, Pure and Appl. Math., vol. 69, Acad. Press, New York (1976).
12. G. James, M. Liebeck, Representations and characters of groups, Cambridge Math. Textbooks (1993).
13. H. Shi, G. Y. Chen, Relation between Bp(3) and Cp(3) with their order components where p is an odd prime, J.
Appl. Math. Inform., 27, 653 – 659 (2009).
14. A. V. Vasil’ev, E. P. Vdovin, An adjacency ceriterion for the prime graph of a finite simple group, Algebra and Logic,
44, 381 – 406 (2005).
15. J. S. Williams, Prime graph components of finite groups, J. Algebra, 69, 487 – 513 (1981).
16. J. Zhang, Z. Li, C. Shao, Finite groups whose irreducible characters vanish only on elements of prime power order,
Int. Electron. J. Algebra, 9, 114 – 123 (2011).
17. J. Zhang, C. Shao, Z. Shen, A new characterization of Suzuki,s simple groups, J. Algebra and Appl., 16, 1 – 6 (2017).
18. K. Zsigmondy, Zur Theorie der Potenzreste, Monatsh. Math. Phys., 3, 265 – 284 (1892).
Received 17.09.19
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 11
|
| id | umjimathkievua-article-1069 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:04:33Z |
| publishDate | 2021 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/1d/d3eaa9695323c33f5fa409d7b2b13a1d.pdf |
| spelling | umjimathkievua-article-10692025-03-31T08:46:33Z Characterization of some finite simple groups by the set of orders of vanishing elements and order Characterization of some finite simple groups by the set of orders of vanishing elements and order Askary, S. S. Askary, S. Vanishing element finite simple groups Vanishing prime graph Vanishing element finite simple groups Vanishing prime graph UDC 512.5 Характеризацiя деяких скiнченних простих груп множиною порядкiв зникаючих елементiв та порядку Let $G$ be a finite group. We say that an element $g$ of $G$ is a vanishing element if there exists an irreducible complex character $X$ of $G$ such that $X(g) = 0$. Ghasemabadi, Iranmanesh, Mavadatpour (2015), present the following conjecture: Let $G$ be a finite group and $M$ a finite nonabelian simple group such that $Vo(G)=Vo(M)$ and $|G|=|M|$. Then $G \cong M $. We answer in affirmative this conjecture for $M = ^2 D_{r+1}(2)$, where $r = 2^n - 1 \geq 3$ and either $2^r+1$ or $2^{r+1}+1$ is a prime number and $M = ^2 D_{r}(3)$, where $r = 2^n + 1 \geq 5$ and either $(3^{r-1}+1)/2$ or $(3^{r}+1)/4$ is prime. УДК 512.5 Нехай $G$ — скiнченна група. Елемент $g \in G$ є зникаючим елементом, якщо iснує незвiдний комплексний характер $\chi \in G$ такий, що $\chi (g) = 0$. Гасемабадi, Iранманеш та Мавадатпур (2015) запропонували гiпотезу: якщо $G$ — скiнченна група, а $M$ — скiнченна неабелева проста група, для яких $\mathrm{V}\mathrm{o} (G) = \mathrm{V}\mathrm{o} (M)$ та $| G| = | M|$ , тодi $G \cong M $.Ми доводимо цю гiпотезу для $M = ^2 D_{r+1}(2)$, де $r = 2n - 1 \geq 3$, якщо або $2^r + 1$, або $2^{r+1}+1$ є простим числом та для $M = ^2 D_{r}(3)$, де $r = 2^n + 1 \geq 5$, якщо або $(3^{r-1}+1)/2$ , або $(3^{r}+1)/4$ є простим. Institute of Mathematics, NAS of Ukraine 2021-11-23 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1069 10.37863/umzh.v73i11.1069 Ukrains’kyi Matematychnyi Zhurnal; Vol. 73 No. 11 (2021); 1443 - 1450 Український математичний журнал; Том 73 № 11 (2021); 1443 - 1450 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1069/9156 Copyright (c) 2021 S. Asgary |
| spellingShingle | Askary, S. S. Askary, S. Characterization of some finite simple groups by the set of orders of vanishing elements and order |
| title | Characterization of some finite simple groups by the set of orders of vanishing elements and order |
| title_alt | Characterization of some finite simple groups by the set of orders of vanishing elements and order |
| title_full | Characterization of some finite simple groups by the set of orders of vanishing elements and order |
| title_fullStr | Characterization of some finite simple groups by the set of orders of vanishing elements and order |
| title_full_unstemmed | Characterization of some finite simple groups by the set of orders of vanishing elements and order |
| title_short | Characterization of some finite simple groups by the set of orders of vanishing elements and order |
| title_sort | characterization of some finite simple groups by the set of orders of vanishing elements and order |
| topic_facet | Vanishing element finite simple groups Vanishing prime graph Vanishing element finite simple groups Vanishing prime graph |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1069 |
| work_keys_str_mv | AT askarys characterizationofsomefinitesimplegroupsbythesetofordersofvanishingelementsandorder AT s characterizationofsomefinitesimplegroupsbythesetofordersofvanishingelementsandorder AT askarys characterizationofsomefinitesimplegroupsbythesetofordersofvanishingelementsandorder |