Groups with the same prime graph as the simple group Dn(5)
Let G be a finite group. The prime graph of G is denoted by Γ(G). Let G be a finite group such that Γ(G)=Γ(Dn(5)), where n≥6. In the paper, as the main result, we show that if n is odd, then G is recognizable by the prime graph and if n is even, then G is quasirecognizable by the prime graph.
Gespeichert in:
| Datum: | 2014 |
|---|---|
| Hauptverfasser: | , |
| Format: | Artikel |
| Sprache: | English |
| Veröffentlicht: |
Інститут математики НАН України
2014
|
| Schriftenreihe: | Український математичний журнал |
| Schlagworte: | |
| Online Zugang: | http://dspace.nbuv.gov.ua/handle/123456789/166067 |
| Tags: |
Tag hinzufügen
Keine Tags, Fügen Sie den ersten Tag hinzu!
|
| Назва журналу: | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| Zitieren: | Groups with the same prime graph as the simple group Dn(5) / A. Babai, B. Khosravi // Український математичний журнал. — 2014. — Т. 66, № 5. — С. 598–608. — Бібліогр.: 33 назв. — англ. |
Institution
Digital Library of Periodicals of National Academy of Sciences of Ukraine| id |
irk-123456789-166067 |
|---|---|
| record_format |
dspace |
| spelling |
irk-123456789-1660672020-02-19T01:26:49Z Groups with the same prime graph as the simple group Dn(5) Babai, A. Khosravi, B. Статті Let G be a finite group. The prime graph of G is denoted by Γ(G). Let G be a finite group such that Γ(G)=Γ(Dn(5)), where n≥6. In the paper, as the main result, we show that if n is odd, then G is recognizable by the prime graph and if n is even, then G is quasirecognizable by the prime graph. Нехай G — скшченна група. Простий граф групи GG позначимо через Γ(G). Нехай G — скінченна група така, що Γ(G)=Γ(Dn(5)), де n≥6. Як основний результат роботи доведено, що для непарних n група G розтзнається простим графом, а для парних n група G є такою, що квазiрозпiзнається простим графом. 2014 Article Groups with the same prime graph as the simple group Dn(5) / A. Babai, B. Khosravi // Український математичний журнал. — 2014. — Т. 66, № 5. — С. 598–608. — Бібліогр.: 33 назв. — англ. 1027-3190 http://dspace.nbuv.gov.ua/handle/123456789/166067 512.5 en Український математичний журнал Інститут математики НАН України |
| institution |
Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| collection |
DSpace DC |
| language |
English |
| topic |
Статті Статті |
| spellingShingle |
Статті Статті Babai, A. Khosravi, B. Groups with the same prime graph as the simple group Dn(5) Український математичний журнал |
| description |
Let G be a finite group. The prime graph of G is denoted by Γ(G). Let G be a finite group such that Γ(G)=Γ(Dn(5)), where n≥6. In the paper, as the main result, we show that if n is odd, then G is recognizable by the prime graph and if n is even, then G is quasirecognizable by the prime graph. |
| format |
Article |
| author |
Babai, A. Khosravi, B. |
| author_facet |
Babai, A. Khosravi, B. |
| author_sort |
Babai, A. |
| title |
Groups with the same prime graph as the simple group Dn(5) |
| title_short |
Groups with the same prime graph as the simple group Dn(5) |
| title_full |
Groups with the same prime graph as the simple group Dn(5) |
| title_fullStr |
Groups with the same prime graph as the simple group Dn(5) |
| title_full_unstemmed |
Groups with the same prime graph as the simple group Dn(5) |
| title_sort |
groups with the same prime graph as the simple group dn(5) |
| publisher |
Інститут математики НАН України |
| publishDate |
2014 |
| topic_facet |
Статті |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/166067 |
| citation_txt |
Groups with the same prime graph as the simple group Dn(5) / A. Babai, B. Khosravi // Український математичний журнал. — 2014. — Т. 66, № 5. — С. 598–608. — Бібліогр.: 33 назв. — англ. |
| series |
Український математичний журнал |
| work_keys_str_mv |
AT babaia groupswiththesameprimegraphasthesimplegroupdn5 AT khosravib groupswiththesameprimegraphasthesimplegroupdn5 |
| first_indexed |
2025-07-14T20:42:03Z |
| last_indexed |
2025-07-14T20:42:03Z |
| _version_ |
1837656413057318912 |
| fulltext |
UDC 512.5
A. Babai, B. Khosravi (Amirkabir Univ. Technology (Tehran Polytechnic), Iran)
GROUPS WITH THE SAME PRIME GRAPH AS THE SIMPLE GROUP Dn(5)
ГРУПИ З ТИМ САМИМ ПРОСТИМ ГРАФОМ, ЩО Й ПРОСТА ГРУПА Dn(5)
Let G be a finite group. The prime graph of G is denoted by Γ(G). Let G be a finite group such that Γ(G) = Γ(Dn(5)),
where n ≥ 6. In the paper, as the main result, we show that if n is odd, then G is recognizable by the prime graph and if
n is even, then G is quasirecognizable by the prime graph.
Нехай G — скiнченна група. Простий граф групи G позначимо через Γ(G). Нехай G — скiнченна група така, що
Γ(G) = Γ(Dn(5)), де n ≥ 6. Як основний результат роботи доведено, що для непарних n група G розпiзнається
простим графом, а для парних n група G є такою, що квазiрозпiзнається простим графом.
1. Introduction. If n is an integer, then we denote by π(n) the set of all prime divisors of n. If G
is a finite group, then π(|G|) is denoted by π(G). The spectrum of a finite group G which is denoted
by ω(G) is the set of its element orders. We construct the prime graph of G which is denoted by
Γ(G) as follows: the vertex set is π(G) and two distinct primes p and q are joined by an edge (we
write p ∼ q) if and only if G contains an element of order pq. Let s(G) be the number of connected
components of Γ(G) and let πi(G), i = 1, . . . , s(G), be the connected components of Γ(G). If
2 ∈ π(G) we always suppose that 2 ∈ π1(G). In graph theory a subset of vertices of a graph is called
an independent set if its vertices are pairwise nonadjacent. Denote by t(G) the maximal number
of primes in π(G) pairwise nonadjacent in Γ(G). In other words, if ρ(G) is some independent set
with the maximal number of vertices in Γ(G), then t(G) = |ρ(G)|. Similarly if p ∈ π(G), then
let ρ(p,G) be some independent set with the maximal number of vertices in Γ(G) containing p and
t(p,G) = |ρ(p,G)|.
A finite group G is called recognizable by prime graph if Γ(H) = Γ(G) implies that H ∼= G.
A non-Abelian simple group P is called quasirecognizable by prime graph if every finite group
whose prime graph is Γ(P ) has a unique non-Abelian composition factor isomorphic to P (see [11]).
Obviously recognition (quasirecognition) by prime graph implies recognition (quasirecognition) by
spectrum, but the converse is not true in general. Also some methods of recognition by spectrum
cannot be used for recognition by prime graph.
Hagie in [8], determined finite groups G satisfying Γ(G) = Γ(S), where S is a sporadic simple
group. It is proved that if q = 32n+1, n > 0, then the simple group 2G2(q) is recognizable by prime
graph [11, 32]. A group G is called a CIT group if G is of even order and the centralizer in G of
any involution is a 2-group. In [14], finite groups with the same prime graph as a CIT simple group
are determined. It is proved that the simple group F4(q), where q = 2n > 2 (see [12]), and 2F4(q)
(see [1]), are quasirecognizable by prime graph. Also in [10], it is proved that if p is a prime number
which is not a Mersenne or Fermat prime and p 6= 11, 13, 19 and Γ(G) = Γ(PGL(2, p)), then G has
a unique non-Abelian composition factor which is isomorphic to PSL(2, p) and if p = 13, then G has
a unique non-Abelian composition factor which is isomorphic to PSL(2, 13) or PSL(2, 27). Then it
is proved that if p and k > 1 are odd and q = pk is a prime power, then PGL(2, q) is recognizable
by prime graph [2].
c© A. BABAI, B. KHOSRAVI, 2014
598 ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 5
GROUPS WITH THE SAME PRIME GRAPH AS THE SIMPLE GROUP Dn(5) 599
In [3], it is proved that if p = 2n + 1 ≥ 5 is a prime number, then 2Dp(3) is quasirecognizable
by prime graph. Then in [4], the authors proved that 2D2m+1(3) is recognizable by prime graph.
Let G be a finite group such that Γ(G) = Γ(Dn(5)), where n ≥ 6. In this paper as the main
result, we show that if n is odd, then G is recognizable by prime graph, and if n is even, then G is
quasirecognizable by prime graph.
In this paper, all groups are finite and by simple groups we mean non-Abelian simple groups.
All further unexplained notations are standard and refer to [5]. Throughout the proof we use the
classification of finite simple groups. In [26] (Tables 2 – 9) independent sets also independent numbers
for all simple groups are listed and we use these results in this paper.
2. Preliminary results.
Lemma 2.1 ([28], Theorem 1). Let G be a finite group with t(G) ≥ 3 and t(2, G) ≥ 2. Then
the following hold:
(1) There exists a finite non-Abelian simple group S such that S ≤ Ḡ = G/K ≤ Aut(S) for the
maximal normal soluble subgroup K of G.
(2) For every independent subset ρ of π(G) with |ρ| ≥ 3 at most one prime in ρ divides the
product |K||Ḡ/S|. In particular, t(S) ≥ t(G)− 1.
(3) One of the following holds:
(a) every prime r ∈ π(G) nonadjacent to 2 in Γ(G) does not divide the product |K||Ḡ/S|; in
particular, t(2, S) ≥ t(2, G);
(b) there exists a prime r ∈ π(K) nonadjacent to 2 in Γ(G); in which case t(G) = 3, t(2, G) =
= 2, and S ∼= Alt7 or L2(q) for some odd q.
Remark 2.1. In Lemma 2.1, for every odd prime p ∈ π(S), we have t(p, S) ≥ t(p,G)− 1.
Lemma 2.2 ([22], Lemma 1). Let N be a normal subgroup of G. Assume that G/N is a Frobe-
nius group with Frobenius kernel F and cyclic Frobenius complement C. If (|N |, |F |) = 1, and F is
not contained in NCG(N)/N, then p|C| ∈ πe(G), where p is a prime divisor of |N |.
Lemma 2.3 [9]. Let G be a finite simple group An−1(q).
(1) If there exists a primitive prime divisor r of qn − 1, then G contains a Frobenius subgroup
with kernel of order r and cyclic complement of order n.
(2) G contains a Frobenius subgroup with kernel of order qn−1 and cyclic complement of order
(qn−1 − 1)/(n, q − 1).
Lemma 2.4 [9]. Let G be a finite simple group.
(1) If G = Cn(q), then G contains a Frobenius subgroup with kernel of order qn and cyclic
complement of order (qn − 1)/(2, q − 1).
(2) If G = 2Dn(q), and there exists a primitive prime divisor r of q2n−2 − 1, then G contains a
Frobenius subgroup with kernel of order q2n−2 and cyclic complement of order r.
(3) If G = Bn(q) or Dn(q), and there exists a primitive prime divisor rm of qm − 1 where
m = n or n − 1 such that m is odd, then G contains a Frobenius subgroup with kernel of order
qm(m−1)/2 and cyclic complement of order rm.
Lemma 2.5 [33]. Let p be a prime and let n be a positive integer. Then one of the following
holds:
(i) there is a primitive prime p′ for pn − 1, that is, p′ | (pn − 1) but p′ - (pm − 1), for every
1 ≤ m < n (usually p′ is denoted by rn);
(ii) p = 2, n = 1 or 6;
(iii) p is a Mersenne prime and n = 2.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 5
600 A. BABAI, B. KHOSRAVI
Remark 2.2 [24]. Let p be a prime number and (q, p) = 1. Let k ≥ 1 be the smallest positive
integer such that qk ≡ 1 (mod p). Then k is called the order of q with respect to p and we denote it
by ordp(q). Obviously by the Fermat’s little theorem it follows that ordp(q)|(p− 1). Also if qn ≡ 1
(mod p), then ordp(q)|n. Similarly if m > 1 is an integer and (q,m) = 1, we can define ordm(q).
If a is odd, then orda(q) is denoted by e(a, q), too. If q is odd, let e(2, q) = 1 if q ≡ 1 (mod 4) and
e(2, q) = 2 if q ≡ −1 (mod 4).
Lemma 2.6 ([27], Proposition 2.5). Let G = Dε
n(q) be a finite simple group of Lie type over a
field of characteristic p. Define
η(m) =
m if m is odd,
m/2 otherwise.
Let r and s be odd primes and r, s ∈ π(G) \ {p}. Put k = e(r, q) and l = e(s, q), and 1 ≤ η(k) ≤
≤ η(l). Then r and s are nonadjacent if and only if 2η(k) + 2η(l) > 2n− (1− ε(−1)k+l), and k, l
satisfy to:
l/k is not an odd natural number,
and if ε = +, then the chain of equalities:
n = l = 2η(l) = 2η(k) = 2k
is not true.
3. Main results.
Lemma 3.1. Let G be a group satisfying the conditions of Lemma 2.1, and let the groups K
and S be as in the claim of Lemma 2.1. Let there exist p ∈ π(K) and p′ ∈ π(S) such that p � p′
in Γ(G), and S contains a Frobenius subgroup with kernel F and cyclic complement C such that
(|F |, |K|) = 1. Then p|C| ∈ ω(G).
Proof. We claim that F � KCG(K)/K. Since KCG(K)/K �G/K, so S ∩KCG(K)/K � S.
Let S ∩KCG(K)/K = S. Then S ≤ KCG(K)/K. So for every t′ ∈ π(S) and t ∈ π(K) we have
t′ ∼ t, which is a contradiction. Consequently S ∩KCG(K)/K = 1, since S is a simple group. So
F � KCG(K)/K, since F ≤ S. Therefore p|C| ∈ ω(G), by Lemma 2.2.
Remark 3.1. Let G = Dn(5), where n ≥ 14. Throughout the paper, we denote a primitive
prime divisor of 5i − 1 by ri. By [30] (Tables 1a – 1c), we have s(G) = 1 and π(G) = π
(
5(5n −
− 1)
∏n−1
i=1
(52i − 1)
)
. Also by [26] (Tables 6, 8) we know that ρ(2, Dn(5)) ⊆ {2, rn, r2(n−1)},
t(Dn(5)) ≥ [(3n + 1)/4] and ρ(Dn(5)) ⊆
{
r2i
∣∣∣∣ [n+ 1
2
]
≤ i < n
}⋃{
ri
∣∣∣∣ [n2
]
≤ i ≤ n, i ≡ 1
(mod 2)
}
.
Therefore if n ≥ 14 and A = {rn, rn−2, r2(n−1), r2(n−2), r2(n−3)}, then A is an independent set
in Γ(Dn(5)).
Corollary 3.1. If G = Dn(5), where n ≥ 14, then t(257, G) ≥ 62, t(193, G) ≥ 40, t(1201, G) ≥
≥ 142, t(14281, G) ≥ 80, t(1129, G) ≥ 65, t(157, G) ≥ 32 and t(19, G) ≥ 11.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 5
GROUPS WITH THE SAME PRIME GRAPH AS THE SIMPLE GROUP Dn(5) 601
Table 1. An upper bound for t(p′, G)
(p, p′) An′(pα) or 2An′(pα) Bn′(pα) or Cn′(pα) Dn′(pα) or 2Dn(pα)
(2, 257) 17 14 15
(3, 193) 17 13 15
(7, 1201) 9 7 9
(13, 14281) 9 7 9
(31, 1129) 9 7 9
(313, 157) – – 6
Proof. We know that e(193, 5) = 192 and so if 193 ∈ π(G), then n ≥ 96. By [26] (Table 8),
B = {r2(n−1), r2(n−2), . . . , r2(n−47)} is an independent set of Γ(G), since (n + 1)/2 ≤ n − 47.
Therefore |B| = 48. If r2i ∈ B, then n− 47 ≤ i ≤ n− 1, therefore 2η(2i) + 2η(192) > 2n. Hence
r2i � 193 in Γ(G) if and only if i/96 and 96/i are not odd natural numbers. Easily we can see that
96/i is an odd number if and only if i = 32 or i = 96. Also 96 divides at most one element of
{n− 47, . . . , n}. Therefore at least 40 elements of B are not adjacent to 193.
Similarly to above since e(257, 5) = 256, e(1201, 5) = 600, e(14281, 5) = 340, e(1129, 5) =
= 282, e(157, 5) = 156 and e(19, 5) = 9 we have t(257, G) ≥ 62, t(1201, G) ≥ 142, t(14281, G) ≥
≥ 80, t(1129, G) ≥ 65, t(157, G) ≥ 32 and t(19, G) ≥ 11.
Corollary 3.1 is proved.
Lemma 3.2. Let G be a finite simple group of Lie type over GF(q), where q = pα. Let p′ be a
prime divisor of |G|. In Table 1, we give some upper bounds for t(p′, G) for some simple groups G
and some prime numbers p′.
Proof. We determine t(257, G) in each case, whenever q = 2α, and the proof of the other cases
are similar. Now we consider each case separately.
Case 1. Let G = An′−1(q), where q = 2α. We know that e(257, q) | 16, since e(257, 2) = 16.
If e(257, q) = 1, then 257 is adjacent to each prime divisor of qi − 1, where i ≤ n′ − 2, by [26]
(Proposition 4.1), so t(257, G) ≤ 3. Otherwise since e(257, q) | 16, then 257 is adjacent to each
prime divisor of qi − 1, where i ≤ n′ − 16, by [26] (Proposition 2.1), so |ρ(257, G) \ {257}| ≤ 16
and so t(257, G) ≤ 17.
Case 2. Let G = 2An′−1(q), where q = 2α. If e(257, q) = 2, then 257 is adjacent to each prime
divisor of qi − 1, where ν(i) ≤ n′ − 2, by [26] (Proposition 4.2), so t(257, G) ≤ 3. Otherwise since
e(257, q) | 16, then 257 is adjacent to each prime divisor of qi − (−1)i, where ν(i) ≤ n′ − 16, by
[26] (Proposition 2.2), so |ρ(257, G) \ {257}| ≤ 16 and so t(257, G) ≤ 17.
Case 3. LetG = Bn′(q), where q = 2α.We have e(257, q) | 16, since e(257, 2) = 16. Therefore
257 is adjacent to each prime divisor of qi − 1, where η(i) ≤ n′ − 8, by [27] (Proposition 2.4), so
|ρ(257, G) \ {257}| ≤ 13 and so t(257, G) ≤ 14.
Case 4. Let G = Dε
n′(q), where q = 2α. Similarly to the above e(257, q) | 16. Therefore 257 is
adjacent to each prime divisor of qi−1, where η(i) ≤ n′−9, by Lemma 2.6, so |ρ(257, G)\{257}| ≤
≤ 14 and so t(257, G) ≤ 15.
Lemma 3.2 is proved.
Theorem 3.1. Let G be a finite group such that Γ(G) = Γ(Dn(5)), where n ≥ 6. If n is odd,
then G is recognizable by prime graph and if n is even, then G is quasirecognizable by prime graph.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 5
602 A. BABAI, B. KHOSRAVI
Proof. If G is a finite group with Γ(G) = Γ(Dn(5)), where n ≥ 6 and K is the maximal
normal soluble subgroup of G, then Lemma 2.1 implies that G has a unique non-Abelian simple
group S such that S ≤ Ḡ = G/K ≤Aut(S). In order to prove the main theorem we must show that
S ∼= Dn(5) and if n is odd, then K = 1 and G ∼= Dn(5). We will prove these statements in the
following lemmas.
Lemma 3.3. The simple group S is not isomorphic to a sporadic group.
Proof. By Lemma 2.1, 11 ≥ t(S) ≥ t(G)− 1, therefore n ≤ 16, by [26].
On the other hand, we know that r2(n−1) ∈ π(S). If n = 16, then r2(n−1) = 7621 divides |S|,
which is a contradiction. Similarly for 6 ≤ n ≤ 15, we get a contradiction.
Lemma 3.3 is proved.
Lemma 3.4. The simple group S is not isomorphic to an alternating group.
Proof. Let S ∼= An′ . We get a contradiction in two steps.
Step 1. Let n ≥ 14. By assumption t(G) ≥ 10 and so t(S) ≥ 9, which implies that n′ ≥ 19. If
x ∈ π(S), such that x � 19 in Γ(S), then n′ − 19 < x ≤ n′, by [26] (Proposition 1.1). Also there
are [20/2] + [20/3]− [20/6] = 13 elements of [n′ − 19, n′] which are divisible by 2 or 3. Therefore
t(19, S) ≤ 8, which is a contradiction by Corollary 3.1.
Step 2. Let 6 ≤ n ≤ 13. If n = 13, then r2(n−1) = 390001 divides |S|. So n′ ≥ 390001,
therefore 37 ∈ π(S), which is a contradiction, since 37 /∈ π(D13(5)). Similarly for 6 ≤ n ≤ 12, we
get a contradiction.
Lemma 3.4 is proved.
Let G be a finite simple group of Lie type over GF(q), where q = pα. In the sequel we denote a
primitive prime divisor of qi − 1 by r′i.
Lemma 3.5. The simple group S is not isomorphic to a finite simple group of Lie type over a
field of characteristic p, where p 6= 5.
Proof. Let S be isomorphic to a finite simple group of Lie type over a field of characteristic p,
where p 6= 5. We get a contradiction in two steps.
Step 1. Let n ≥ 14. By Lemma 2.1, t(S) ≥ t(G) − 1 so t(S) ≥ 9. In the sequel we consider
each possibility for S, by [30] (Tables 1a – 1c).
We denote by A+
n′(q) the simple group An′(q), and by A−n′(q) the simple group 2An′(q).
Case 1. Let S ∼= Aεn′−1(q), where q = pα. By Lemma 2.1, t(S) ≥ t(G)− 1 so
2n′ > 3n− 9. (1)
We know that A = {rn, rn−2, r2(n−1), r2(n−2), r2(n−3)} is an independent set in Γ(G), by Re-
mark 3.1.
If S ∼= An′−1(q), then by [26] (Propositions 3.1, 4.1), each r′i, where i /∈ {n′−1, n′}, is adjacent
to 2 and p in Γ(S).
If S ∼= 2An′−1(q), then by [26] (Propositions 3.1, 4.2), every r′i, where ν(i) /∈ {n′ − 1, n′}, is
adjacent to 2 and p in Γ(S).
On the other hand, by Lemma 2.1, |A∩π(S)| ≥ 4, therefore p is adjacent to at least two elements
of A ∩ π(S) in Γ(S). For example, let r2(n−3) ∼ p ∼ r2(n−2) in Γ(S). Therefore r2(n−3) ∼ p ∼
∼ r2(n−2) in Γ(G). Let a = e(p, 5). Since p ∼ r2(n−2) it follows that 2(n − 2) + 2η(a) ≤ 2n or
2(n− 2)/a is odd, by Lemma 2.6. Similarly since p ∼ r2(n−3) it follows that 2(n− 3) + 2η(a) ≤ 2n
or 2(n− 3)/a is odd. So η(a) ≤ 3, which implies that a ∈ {1, 2, 3, 4, 6} and so p ∈ {2, 3, 7, 13, 31}.
Similarly to the above for every ri, rj ∈ {rn, rn−2, r2(n−1), r2(n−2), r2(n−3)}, if ri ∼ p ∼ rj , then
p ∈ {2, 3, 7, 13, 31}.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 5
GROUPS WITH THE SAME PRIME GRAPH AS THE SIMPLE GROUP Dn(5) 603
Let S ∼= An′−1(q). If p = 2, then since n′ ≥ 16 by (1) and e(257, 2α) | 16 it follows that
257 ∈ π(S). Hence by Lemma 3.2, t(257, S) ≤ 17 and by Corollary 3.1, t(257, G) ≥ 62. Therefore
by Remark 2.1, we get a contradiction. Similarly for every p ∈ {3, 7, 13, 31}, we get a contradiction.
Let S ∼= 2An′−1(q). If p = 3, then since n′ ≥ 16 by (1) and e(193, 3α) | 16 it follows that
193 ∈ π(S). Therefore by Lemma 3.2, t(193, S) ≤ 17 and by Corollary 3.1, t(193, G) ≥ 40.
Therefore by Remark 2.1, we get a contradiction. Similarly for every p ∈ {2, 7, 13, 31}, we get a
contradiction.
Case 2. Let S ∼= Bn′(q) or Cn′(q), where q = pα. By Lemma 2.1, t(S) ≥ t(G)− 1, so
3n′ > 3n− 12. (2)
We know that by [26] (Propositions 3.1, 4.3), every r′i is adjacent to 2 and p in Γ(S), where η(i) 6=
6= n′. On the other hand, by Lemma 2.1, |A ∩ π(S)| ≥ 4. Therefore at least two elements of
A ∩ π(S) are adjacent to p in Γ(S). Denote e(p, 5) by a. Similarly to the above case, we get that
p ∈ {2, 3, 7, 13, 31}.
If p = 7, then since n′ ≥ 11 and e(1201, 7α) | 8 it follows that 1201 ∈ π(S). Hence by
Lemma 3.2, t(1201, S) ≤ 7 and by Corollary 3.1, t(1201, G) ≥ 142. Therefore by Remark 2.1, we
get a contradiction. Similarly for every p ∈ {2, 3, 13, 31}, we get a contradiction.
We denote by D+
n′(q) the simple group Dn′(q), and by D−n′(q) the simple group 2Dn′(q).
Case 3. Let S ∼= Dε
n′(q), where q = pα. By Lemma 2.1, t(S) ≥ t(G)− 1, so
3n′ > 3n− 11. (3)
Let B = A∪{r2(n−4)}. Since n ≥ 14, then by Remark 3.1, B is an independent set in Γ(G).We know
that every r′i, where η(i) /∈ {n′, n′−1} is adjacent to 2 and p in Γ(S), by [26] (Propositions 3.1, 4.4).
On the other hand, by Lemma 2.1, |B ∩ π(S)| ≥ 5, therefore at least two elements of B ∩ π(S) are
adjacent to p in Γ(S). Let a = e(p, 5). Similarly to the above case, we have p ∈ {2, 3, 7, 13, 31, 313}.
If p = 13, then since n′ ≥ 10 and e(14281, 13α) | 8 it follows that 14281 ∈ π(S). Hence by
Lemma 3.2, t(14281, S) ≤ 9 and by Corollary 3.1, t(14281, G) ≥ 80. Therefore by Remark 2.1, we
get a contradiction. Similarly for every p ∈ {2, 3, 7, 31, 313}, we get a contradiction.
Case 4. Let S ∼= E8(q), where q = pα. We know that t(S) = 12. If n ≥ 19, then t(G) ≥ 14,
which is a contradiction, by Lemma 2.1. Therefore n ≤ 18. We know that p ∈ π(S), therefore
p ∈ π(G).
Let n = 18. For every p ∈ π(G)\{5}, easily we can see that π(p30−1) * π(D18(5)) and we get
a contradiction. Similarly for each n and each p ∈ π(G), easily we can see that π(S) * π(Dn(5)).
Step 2. Let 6 ≤ n ≤ 13. If S ∼= E8(q), where q = pα, then we have 19 ∈ π(S), so 9 ≤ n ≤ 13.
Similarly to Case 4 of Step 1, we get a contradiction.
Let S ∼= Bn′(q), Cn′(q) or Dn′(q), where q = pα.
We know that π(S) ⊆ π(G). Therefore p ∈ π(G). On the other hand, since n ≥ 6, by (2) and
(3), it follows that n′ ≥ 3. Therefore
π(p3 − 1) ⊆ π(p3α − 1) = π(q3 − 1) ⊆ π(S) ⊆ π(G).
Let p = 13, then 61 ∈ π(Dn(5)), where 6 ≤ n ≤ 13, which is a contradiction. Similarly for each
p ∈
⋃13
n=6 π(Dn(5)) \ {2, 3, 7, 11, 29, 67} we get a contradiction. We know that r2(n−1) ∈ ρ(2, S).
Hence there is r′i ∈ π(S), such that r2(n−1) = r′i.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 5
604 A. BABAI, B. KHOSRAVI
If n = 6, then r2(n−1) = 521. Let p = 2, we have e(521, 2) = 260. So 260 | αi, and (2260 −
− 1) | (2αi − 1). Consequently π(2260 − 1) ⊆ π(S), which is a contradiction. Let p = 3. Since
e(521, 3) = 520, then π(3520 − 1) ⊆ π(S), which is a contradiction. Similarly for other cases of n
and p, we get a contradiction.
Therefore let S ∼= Aεn′−1(q) or 2Dn′(q) where q = pα. Similarly to the above we have p ∈
∈ π(G). If n′ < 4, then S ∼= 2D3(q) so G = D6(5). We have r10 = 521 ∈ π(S). If p = 521,
then π(5212 − 1) ⊆ π(S) ⊆ π(G), which is a contradiction. Let p = 7, since e(521, 7) = 520,
therefore similarly to the above case we get a contradiction. Similarly for other cases of p, we get a
contradiction.
Therefore n′ ≥ 4 and consequently
π(p4 − 1) ⊆ π(p4α − 1) = π(q4 − 1) ⊆ π(S) ⊆ π(G).
Let p = 13, then 17 ∈ π(G), which is a contradiction. Similarly for each p ∈
⋃13
n=6 π(Dn(5)) \
{2, 3, 7, 41, 67} we get a contradiction. Similarly to the above for p ∈ {2, 3, 7, 41, 67}, we get a
contradiction.
Lemma 3.5 is proved.
Lemma 3.6. If S is isomorphic to a finite simple group of Lie type in characteristic 5, then
S ∼= Dn(5).
Proof. Throughout the proof, since t(S) ≥ 3, using [26] (Tables 8, 9), we consider the following
possibilities. We show that S ∼= Dn(5) in two steps.
Step 1. Let S be isomorphic to a finite simple exceptional group of Lie type.
Case 1. Let S ∼= E8(5
α). By Lemma 2.1, t(S) ≥ t(G) − 1 so n ≤ 18. On the other hand,
we have π(S) ⊆ π(G), therefore 30α ≤ 2(n − 1). Also we know that r2(n−1) ∈ ρ(2, S) =
= {r′15, r′20, r′24, r′30}, so we consider the following cases:
Let r2(n−1) = r′15. Let p0 be a primitive prime divisor of 52(n−1) − 1. Since r2(n−1) = r′15, it
follows that p0 | (515α − 1). Therefore 2(n − 1) ≤ 15α, which is a contradiction. Similarly, when
r2(n−1) = r′20 and r2(n−1) = r′24, we get a contradiction.
Let r2(n−1) = r′30. Similarly to the above 2(n − 1) ≤ 30α. Consequently, n − 1 = 15α and
since 15 | (n − 1) and n ≤ 18 we have n = 16 and α = 1. Therefore S ∼= E8(5). We know that
r13 ∈ π(G) and r13 /∈ π(S). So r13 ∈ π(Ḡ/S) ∪ π(K). Therefore r13 ∈ π(K), since Out(S) = 1.
Using [25], we have D8(5) ≤ E8(5) and D8(5) contains a Frobenius subgroup 521 : r7. Since then
r30 � r13 by Lemma 3.1, we have r13 ∼ r7 in Γ(G), which is a contradiction, by Lemma 2.6.
Case 2. Let S ∼= E7(5
α). By [26], t(S) = 8 and consequently t(G) ≤ 9, by Lemma 2.1.
Therefore n ≤ 12. Since π(S) ⊆ π(G), then 18α ≤ 2(n − 1). On the other hand, we know that
r2(n−1) ∈ ρ(2, S) = {r′14, r′18}. Now we consider the following cases:
Let r2(n−1) = r′14. Let p0 be a primitive prime divisor of 52(n−1) − 1. Similarly to the above
case 2(n− 1) ≤ 14α, which is a contradiction.
Let r2(n−1) = r′18. Similarly to the above 2(n−1) ≤ 18α. Consequently n−1 = 9α. Therefore
n = 10, α = 1 and S ∼= E7(5). We know that r14 ∼ r4 in Γ(G), by Lemma 2.6, but r14 � r4 in
Γ(S), by [26] (Proposition 2.5). Therefore r4 or r14 ∈ π(Ḡ/S)∪π(K). On the other hand, we know
that r16 /∈ π(S). Since {r4, r14, r16} is an independent set, we get a contradiction, by Lemma 2.1.
Similarly to the above discussion it follows that S is not isomorphic to E6(5
α), 2E6(5
α) and F4(5
α).
Step 2. Let S be isomorphic to a finite simple classical group of Lie type.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 5
GROUPS WITH THE SAME PRIME GRAPH AS THE SIMPLE GROUP Dn(5) 605
Case 1. Let S ∼= An′−1(5
α). We know that π(S) ⊆ π(G), therefore n′α ≤ 2(n − 1). Also we
know that r2(n−1) ∈ ρ(2, S) = {r′n′ , r′n′−1}, so we consider the following cases:
Let r2(n−1) = r′n′−1. Let p0 be a primitive prime divisor of 52(n−1)− 1. Since r2(n−1) = r′n′−1,
it follows that p0 | (5(n
′−1)α − 1). Therefore 2(n− 1) ≤ (n′ − 1)α, which is a contradiction.
Let r2(n−1) = r′n′ . Similarly to the above 2(n − 1) ≤ n′α. Consequently 2(n − 1) = n′α. If
α = 1, then r′n′−1 = r2n−3 ∈ π(S) ⊆ π(G), which is a contradiction. Therefore α ≥ 2 so by (1),
we have 2(n − 1) = n′α ≥ 2n′ > 3n − 9, which implies that n = 6. Hence n′ = 5, α = 2 or
n′ = 2, α = 5. If n′ = 2, then we get a contradiction by (1). So S ∼= A4(5
2). Then r5, r10 ∈ π(S)
and r5 and r10 are primitive prime divisors of (52)5 − 1 and so r5 ∼ r10 in Γ(S), but r5 � r10 in
Γ(G), which is a contradiction.
Case 2. Let S ∼= 2An′−1(5
α). We know that π(S) ⊆ π(G). Also we know that r2(n−1) ∈
∈ ρ(2, S), so ν(e(r2(n−1), 5
α)) ∈ {n′, n′ − 1}. Now we consider the following cases:
If n′ is odd, then 2n′α ≤ 2(n − 1), since π(S) ⊆ π(G). By (1), n = 6 hence S ∼= 2A4(5).
We know that r5, r8 /∈ π(S). Therefore r5, r8 ∈ π(Ḡ/S) ∪ π(K). We know that {r5, r8, r10} is an
independent set, which is a contradiction, by Lemma 2.1.
If n′ is even, then 2(n′− 1)α ≤ 2(n− 1). We know that (5α + 1)2 = 2 so by [26] (Table 6), we
have r2(n−1) ∈ {r′2(n′−1), r
′
n′/2}. If r2(n−1) = r′n′/2, then similarly to the above 2(n− 1) ≤ (n′/2)α,
which is a contradiction. Therefore r2(n−1) = r′2(n′−1) so similarly to the above we have 2(n− 1) ≤
≤ 2(n′ − 1)α. Hence by (1), n ≤ 8. Let n = 8, therefore n′ = 8 and S ∼= 2A7(5). We know that
r5, r7 ∈ π(G) and r5, r7 /∈ π(S). So r5, r7 ∈ π(Ḡ/S) ∪ π(K). Since {r5, r7, r14} is an independent
set , we get a contradiction, by Lemma 2.1. Similarly n 6= 6, 7.
Case 3. Let S ∼= Bn′(5α), Cn′(5α) or 2Dn′(5α). We have π(S) ⊆ π(G) so 2n′α ≤ 2(n − 1).
Also we know that r2(n−1) ∈ ρ(2, S) = {r′2n′}. Similarly to the above we have 2(n − 1) ≤ 2n′α,
hence n− 1 = n′α. Now by (2), n′ > n− 4, therefore n′(α− 1) < 3. If α = 2, n′ = 2, then n = 5,
which is a contradiction, since n ≥ 6. Therefore α = 1. Since n ≥ 6, so n′ ≥ 5. We know that if n
is an odd number, then t(2, Dn(5)) = 3, which is a contradiction, since t(2, S) = 2. Therefore n is
even and n′ is odd.
Let S ∼= 2Dn−1(5). We have rn−1 ∈ π(G) and rn−1 /∈ π(S), therefore rn−1 ∈ π(Ḡ/S)∪ π(K).
Since π(Out(S)) = {2}, so rn−1 ∈ π(K). By Lemma 2.4, 2Dn−1(5) contains a Frobenius subgroup
of the form 52(n−2) : r2(n−2). We know that r2(n−1) ∈ π(S) and r2(n−1) � rn−1 in Γ(G). Therefore
by Lemma 3.1, rn−1 ∼ r2(n−2) in Γ(G), which is a contradiction.
Let S ∼= Bn−1(5). We consider two following cases:
1. Let 4 | n. We have r(n−2)/2 � r(n+2)/2 in Γ(S) but r(n−2)/2 ∼ r(n+2)/2 in Γ(G). Therefore
r(n−2)/2 or r(n+2)/2 ∈ π(Ḡ/S)∪π(K). Since π(Out(S)) = {2}, so r(n−2)/2 or r(n+2)/2 ∈ π(K). By
Lemma 2.4, Bn−1(5) contains a Frobenius subgroup of the form 5(n−1)(n−2)/2 : rn−1. We know that
r2(n−1) ∈ π(S) and r2(n−1) � r(n−2)/2, r(n+2)/2 in Γ(G). Therefore by Lemma 3.1, rn−1 ∼ r(n−2)/2
or rn−1 ∼ r(n+2)/2 in Γ(G), which is a contradiction.
2. Let 4 | (n + 2). In this case by r(n−4)/2, r(n+4)/2 similarly to the above case we get a
contradiction.
Similarly for S ∼= Cn−1(5) we get a contradiction.
Case 4. Let S ∼= Dn′(5α). Similarly to the above, we have 2(n′ − 1)α ≤ 2(n − 1), since
π(S) ⊆ π(G). Also we know that r2(n−1) ∈ ρ(2, S) = {r′n′ , r′2(n′−1)}. So we consider the following
cases:
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 5
606 A. BABAI, B. KHOSRAVI
Let r2(n−1) = r′n′ . Let p0 be a primitive prime divisor of 52(n−1) − 1. Since r2(n−1) = r′n′ , it
follows that p0 | (5n
′α − 1). Therefore 2(n− 1) ≤ n′α, which is a contradiction.
Let r2(n−1) = r′2(n′−1). Similarly to the above 2(n − 1) ≤ 2(n′ − 1)α. Consequently n − 1 =
= (n′ − 1)α. By (3), n′ > n − 3. If α ≥ 2, then n − 1 = (n′ − 1)α > (n − 4)α ≥ 2(n − 4). So
n = 6 and n′ = 2, which is a contradiction. Therefore α = 1 and n = n′, so S ∼= Dn(5).
Lemma 3.6 is proved.
Theorem 3.2. If Γ(G) = Γ(Dn(5)), where n ≥ 6, then Dn(5) ≤ G/K ≤ Aut(Dn(5)), where
K = 1 when n is odd and K is a 2-group when n is even.
Proof. By the above lemmas, it follows that Dn(5) ≤ G/K ≤ Aut(Dn(5)), where K is the
maximal normal soluble subgroup of G. We can assume that K is an elementary Abelian p-group by
[16]. Since by [7], we know that Dn(5) acts unisingularly we conclude that p 6= 5.
Let n be odd. We claim that for each element t ∈ π(Dn(5)), we have t � rn or t � r2(n−1). Let
e(t, 5) = a. If t ∼ rn and t ∼ r2(n−1), then by Lemma 2.6, n/a is odd and 2(n−1)+2η(a) ≤ 2n−2,
since a is odd, which is a contradiction.
We know that Dn(5) contains a Frobenius subgroup with kernel of order 5n(n−1)/2 and cyclic
complement of order rn, by Lemma 2.4. Also Dn(5) ≤ G/K, and so G/K contains a Frobenius
subgroup T/K of the form 5n(n−1)/2 : rn. By the above discussion, we know that p � rn or p �
� r2(n−1) in Γ(Dn(5)). Since p 6= 5, it follows that p ∼ rn, by Lemma 3.1. Also we know that
Bn−1(5) ≤ Dn(5), by [25], and so Bn−1(5) ≤ G/K. Similarly G/K contains a Frobenius subgroup
of the form 5(n−2)(n−3)/2 : rn−2, by Lemma 2.4. Since p 6= 5 and p � rn or p � r2(n−1) it follows
that p ∼ rn−2, by Lemma 3.1. Let e(p, 5) = m. Since p ∼ rn it follows that n/m is odd, by
Lemma 2.6. Therefore m is odd. Similarly since p ∼ rn−2 it follows that 2(n − 2) + 2η(m) ≤ 2n
or (n− 2)/m is odd. Consequently, m = 1 and so p = 2. So 2 = p ∼ rn, which is a contradiction.
Therefore K = 1.
Let n be even. We claim that for each element t ∈ π(Dn(5)), we have t � rn−1 or t � r2(n−1).
Let e(t, 5) = a. If t ∼ rn−1 and t ∼ r2(n−1), then by Lemma 2.6, 2(n − 1) + 2η(a) ≤ 2n − (1 −
− (−1)n−1+a) or (n− 1)/a is odd and 2(n− 1) + 2η(a) ≤ 2n− (1− (−1)2(n−1)+a) or 2(n− 1)/a
is odd, which is a contradiction.
Similarly to the above we know that G/K contains a Frobenius subgroup of the form
5(n−1)(n−2)/2 : rn−1, by Lemma 2.4. Now since p 6= 5 and by the above discussion we have
p � r2(n−1) or p � rn−1 so by Lemma 3.1, we conclude that p ∼ rn−1. Also we know that
2Dn−1(5) ≤ Dn(5), by [25]. Similarly G/K contains a Frobenius subgroup of the form 52(n−2) :
r2(n−2), by Lemma 2.4. Similarly p ∼ r2(n−2), by Lemma 3.1. Let e(p, 5) = m. Since p ∼ rn−1, it
follows that 2(n− 1) + 2η(m) ≤ 2n− (1− (−1)m+n−1) or (n− 1)/m is odd, by Lemma 2.6. Simi-
larly since p ∼ r2(n−2) it follows that 2(n− 2) + 2η(m) ≤ 2n− (1− (−1)m+2(n−2)) or 2(n− 2)/m
is odd. Consequently, m = 1, so p = 2. Therefore K is a 2-group.
Theorem 3.2 is proved.
Theorem 3.3. Let n ≥ 6 be odd, if Dn(5) ≤ G ≤ Aut(Dn(5)) and Γ(G) = Γ(Dn(5)), then
G ∼= Dn(5).
Proof. Suppose that G � Dn(5). We know that Out(Dn(5)) = γδ, where γ is the graph
automorphism of order 2 and δ is the diagonal automorphism of order 4. Consequently we consider
the following cases:
1. LetG ∼= Dn(5)〈γ〉. Consider the centralizer CDn(5)(γ), we have π(CDn(5)(γ)) = π(Bn−1(5)),
by [21]. Therefore 2 ∼ r2(n−1), which is a contradiction.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 5
GROUPS WITH THE SAME PRIME GRAPH AS THE SIMPLE GROUP Dn(5) 607
2. Let G ∼= Dn(5)〈δ〉. So if T̂ is a maximal torus of G, then T̂ has order |T ||δ|, where T
is a torus of Dn(5), by [21]. Let |T | = (5n−1 + 1)(5 + 1)/4. Therefore 2 ∼ r2(n−1), which is a
contradiction.
Similarly for G ∼= Dn(5)〈γδ〉 we get a contradiction.
Consequently G ∼= Dn(5) and Dn(5) is recognizable by prime graph.
Theorem 3.3 is proved.
Similarly we can prove the following theorem.
Theorem 3.4. Let n ≥ 6 be even, if Dn(5) ≤ G/O2(G) ≤ Aut(Dn(5)) and Γ(G) = Γ(Dn(5)),
then G ∼= Dn(5)/O2(G).
Corollary 3.2. Let G be a finite group satisfying |G| = |Dn(q)|, where n ≥ 6. If ω(G) =
= ω(Dn(q)), then G ∼= Dn(q).
We note that recently this theorem is proved for each finite simple group (see [31]).
Corollary 3.3. If n ≥ 6 is even, then Dn(q) is quasirecognizable by spectrum, i.e., if G is a finite
group such that ω(G) = ω(Dn(q)), then G has a unique non-Abelian composition factor isomorphic
to Dn(q).
If n ≥ 6 is odd, then Dn(q) is recognizable by spectrum, i.e., if G is a finite group such that
ω(G) = ω(Dn(q)), then G ∼= Dn(q).
1. Akhlaghi Z., Khatami M., Khosravi B. Quasirecognition by prime graph of the simple group 2F4(q) // Acta Math.
hung. – 2009. – 122, № 4. – P. 387 – 397.
2. Akhlaghi Z., Khatami M., Khosravi B. Characterization by prime graph of PGL(2, pk) where p and k > 1 are odd
// Int. J. Algebra and Comput. – 2010. – 20, № 7. – P. 847 – 873.
3. Babai A., Khosravi B., Hasani N. Quasirecognition by prime graph of 2Dp(3) where p = 2n + 1 ≥ 5 is a prime //
Bull. Malays. Math. Sci. Soc. – 2009. – 32, № 3. – P. 343 – 350.
4. Babai A., Khosravi B. Recognition by prime graph of 2D2m+1(3) // Sib. Math. J. – 2011. – 52, № 5. – P. 993 – 1003.
5. Conway J. H., Curtis R. T., Norton S. P., Parker R. A., Wilson R. A. Atlas of finite groups. – Oxford: Oxford Univ.
Press, 1985.
6. Grechkoseeva M. A., Shi W. J., Vasil’ev A. V. Recognition by spectrum of L16(2m) // Alg. Colloq. – 2007. – 14,
№ 3. – P. 462 – 470.
7. Guralnick R. M., Tiep P. H. Finite simple unisingular groups of Lie type // J. Group Theory. – 2003. – 6. – P. 271 – 310.
8. Hagie M. The prime graph of a sporadic simple group // Communs Algebra. – 2003. – 31, № 9. – P. 4405 – 4424.
9. He H., Shi W. Recognition of some finite simple groups of type Dn(q) by spectrum // Int. J. Algebra and Comput. –
2009. – 19, № 5. – P. 681 – 698.
10. Khatami M., Khosravi B., Akhlaghi Z. NCF-distinguishablity by prime graph of PGL(2, p), where p is a prime //
Rocky Mountian J. Math. – 2011. – 41, № 5. – P. 1523 – 1545.
11. Khosravi A., Khosravi B. Quasirecognition by prime graph of the simple group 2G2(q) // Sib. Math. J. – 2007. – 48,
№ 3. – P. 570 – 577.
12. Khosravi B., Babai A. Quasirecognition by prime graph of F4(q) where q = 2n > 2 // Monatsh. Math. – 2011. –
162, № 3. – P. 289 – 296.
13. Khosravi B., Khosravi B., Khosravi B. 2-Recognizability of PSL(2, p2) by the prime graph // Sib. Math. J. – 2008. –
49, № 4. – P. 749 – 757.
14. Khosravi B., Khosravi B., Khosravi B. Groups with the same prime graph as a CIT simple group // Houston J. Math. –
2007. – 33, № 4. – P. 967 – 977.
15. Khosravi B., Khosravi B., Khosravi B. On the prime graph of PSL(2, p) where p > 3 is a prime number // Acta
Math. hung. – 2007. – 116, № 4. – P. 295 – 307.
16. Khosravi B., Khosravi B., Khosravi B. A characterization of the finite simple group L16(2) by its prime graph //
Manuscr. Math. – 2008. – 126. – P. 49 – 58.
17. Khosravi B. Quasirecognition by prime graph of L10(2) // Sib. Math. J. – 2009. – 50, № 2. – P. 355 – 359.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 5
608 A. BABAI, B. KHOSRAVI
18. Khosravi B. Some characterizations of L9(2) related to its prime graph // Publ. Math. Debrecen. – 2009. – 75,
№ 3-4. – P. 375 – 385.
19. Khosravi B. n-Recognition by prime graph of the simple group PSL(2, q) // J. Algebra and Appl. – 2008. – 7,
№ 6. – P. 735 – 748.
20. Khosravi B., Moradi H. Quasirecognition by prime graph of finite simple groups Ln(2) and Un(2) // Acta. Math.
hung. – 2011. – 132, № 12. – P. 140 – 153.
21. Lucido M. S. Prime graph components of finite almost simple groups // Rend. Semin. mat. Univ. Padova. – 1999. –
102. – P. 1 – 14.
22. Mazurov V. D. Characterizations of finite groups by the set of orders of their elements // Algebra and Logic. – 1997. –
36, № 1. – P. 23 – 32.
23. Mazurov V. D., Chen G. Y. Recognizability of finite simple groups L4(2m) and U4(2m) by spectrum // Algebra,
Logika. – 2008. – 47, № 1. – P. 83 – 93.
24. Sierpiński W. Elementary theory of numbers. – Warsaw: Panstwowe Wydawnictwo Naukowe, 1964. – Vol. 42.
25. Stensholt E. Certain embeddings among finite groups of Lie type // J. Algebra. – 1978. – 53. – P. 136 – 187.
26. Vasil’ev A. V., Vdovin E. P. An adjacency criterion in the prime graph of a finite simple group // Algebra and Logic. –
2005. – 44, № 6. – P. 381 – 405.
27. Vasil’ev A. V., Vdovin E. P. Cocliques of maximal size in the prime graph of a finite simple group //
http://arxiv.org/abs/0905.1164v1.
28. Vasil’ev A. V., Gorshkov I. B. On the recognition of finite simple groups with a connected prime graph // Sib. Math.
J. – 2009. – 50, № 2. – P. 233 – 238.
29. Vasil’ev A. V., Grechkoseeva M. A. On recognition by spectrum of finite simple linear groups over fields of
characteristic 2 // Sib. Mat. Zh. – 2005. – 46, № 4. – P. 749 – 758.
30. Vasil’ev A. V., Grechkoseeva M. A. On the recognition of the finite simple orthogonal groups of dimension 2m, 2m +1
and 2m + 2 over a field of characteristic 2 // Sib. Math. J. – 2004. – 45, № 3. – P. 420 – 431.
31. Vasil’ev A. V., Grechkoseeva M. A., Mazurov V. D. Characterization of finite simple groups by spectrum and order //
Algebra and Logic. – 2009. – 48, № 6. – P. 385 – 409.
32. Zavarnitsin A. V. On the recognition of finite groups by the prime graph // Algebra and Logic. – 2006. – 43, № 4. –
P. 220 – 231.
33. Zsigmondy K. Zur theorie der potenzreste // Monatsh. Math. und Phys. – 1892. – 3. – S. 265 – 284.
Received 02.07.12
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 5
|