A New Characterization of PSL(2, q) for Some q
Let G be a finite group and let π e (G) be the set of orders of elements from G. Let k ∈ π e (G) and let m k be the number of elements of order k in G. We set nse (G) := {m k | k ∈ π e (G)}. It is proved that PSL(2, q) are uniquely determined by nse (PSL(2, q)), where q ∈ {5, 7, 8, 9, 11, 13}. As th...
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irk-123456789-1658552020-02-17T01:27:41Z A New Characterization of PSL(2, q) for Some q Asboei, A.K. Amiri, S.S.S. Iranmanesh, A. Статті Let G be a finite group and let π e (G) be the set of orders of elements from G. Let k ∈ π e (G) and let m k be the number of elements of order k in G. We set nse (G) := {m k | k ∈ π e (G)}. It is proved that PSL(2, q) are uniquely determined by nse (PSL(2, q)), where q ∈ {5, 7, 8, 9, 11, 13}. As the main result of the paper, we prove that if G is a group such that nse (G) = nse (PSL(2, q)), where q ∈ {16, 17, 19, 23}, then G ≅ PSL(2, q). Нехай G — скінченна група, а πe(G) — множина порядків елемента з G. Нехай також k∈πe(G), а mk — число елементів порядку k в G. Покладемо nse (G):={mk|k∈πe(G)}. Доведено, що PSL(2,q) однозначно визначаються nse (PSL(2,q)), де q∈{5,7,8,9,11,13}. Основним результатом роботи є доведення того факту, що якщо G є групою, для якої nse (G)=nse(PSL(2,q)), де q∈16,17,19,23, то G≅PSL(2,q). 2015 Article A New Characterization of PSL(2, q) for Some q / A.K. Asboei, S.S.S. Amiri, A. Iranmanesh // Український математичний журнал. — 2015. — Т. 67, № 9. — С. 1155–1162. — Бібліогр.: 14 назв. — англ. 1027-3190 http://dspace.nbuv.gov.ua/handle/123456789/165855 512.5 en Український математичний журнал Інститут математики НАН України |
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Статті Статті Asboei, A.K. Amiri, S.S.S. Iranmanesh, A. A New Characterization of PSL(2, q) for Some q Український математичний журнал |
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Let G be a finite group and let π e (G) be the set of orders of elements from G. Let k ∈ π e (G) and let m k be the number of elements of order k in G. We set nse (G) := {m k | k ∈ π e (G)}. It is proved that PSL(2, q) are uniquely determined by nse (PSL(2, q)), where q ∈ {5, 7, 8, 9, 11, 13}. As the main result of the paper, we prove that if G is a group such that nse (G) = nse (PSL(2, q)), where q ∈ {16, 17, 19, 23}, then G ≅ PSL(2, q). |
| format |
Article |
| author |
Asboei, A.K. Amiri, S.S.S. Iranmanesh, A. |
| author_facet |
Asboei, A.K. Amiri, S.S.S. Iranmanesh, A. |
| author_sort |
Asboei, A.K. |
| title |
A New Characterization of PSL(2, q) for Some q |
| title_short |
A New Characterization of PSL(2, q) for Some q |
| title_full |
A New Characterization of PSL(2, q) for Some q |
| title_fullStr |
A New Characterization of PSL(2, q) for Some q |
| title_full_unstemmed |
A New Characterization of PSL(2, q) for Some q |
| title_sort |
new characterization of psl(2, q) for some q |
| publisher |
Інститут математики НАН України |
| publishDate |
2015 |
| topic_facet |
Статті |
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http://dspace.nbuv.gov.ua/handle/123456789/165855 |
| citation_txt |
A New Characterization of PSL(2, q) for Some q / A.K. Asboei, S.S.S. Amiri, A. Iranmanesh // Український математичний журнал. — 2015. — Т. 67, № 9. — С. 1155–1162. — Бібліогр.: 14 назв. — англ. |
| series |
Український математичний журнал |
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2025-07-14T20:08:54Z |
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UDC 512.5
A. K. Asboei
(Farhangian Univ., Shariati Mazandaran, Iran and Buinzahra Branch, Islamic Azad Univ., Buinzahra, Iran),
S. S. S. Amiri (Babol Branch, Islamic Azad Univ., Babol, Iran),
A. Iranmanesh (Tarbiat Modares Univ., Tehran, Iran)
A NEW CHARACTERIZATION OF PSL(2, q) FOR SOME q *
НОВА ХАРАКТЕРИСТИКА PSL(2, q) ДЛЯ ДЕЯКОГО q
Let G be a finite group and let πe(G) be the set of element orders of G. Let k ∈ πe(G) and let mk be the number of
elements of order k in G. We set nse (G) := {mk|k ∈ πe(G)}. It is proved that PSL(2, q) are uniquely determined by
nse (PSL(2, q)), where q ∈ {5, 7, 8, 9, 11, 13}. As the main result of the paper, we prove that if G is a group such that
nse (G) = nse (PSL(2, q)), where q ∈ {16, 17, 19, 23}, then G ∼= PSL(2, q).
Нехай G — скiнченна група, а πe(G) — множина порядкiв елементiв з G. Нехай також k ∈ πe(G), а mk —
число елементiв порядку k в G. Покладемо nse (G) := {mk|k ∈ πe(G)}. Доведено, що PSL(2, q) однозначно
визначаються nse (PSL(2, q)), де q ∈ {5, 7, 8, 9, 11, 13}. Основним результатом роботи є доведення того факту, що
якщо G є групою, для якої nse (G) = nse (PSL(2, q)), де q ∈ {16, 17, 19, 23}, то G ∼= PSL(2, q).
1. Introduction. If n is an integer, then we denote by π(n) the set of all prime divisors of n. Let
G be a finite group. Denote by π(G) the set of primes p such that G contains an element of order
p. Also the set of element orders of G is denoted by πe(G). A finite group G is called a simple
Kn-group, if G is a simple group with |π(G)| = n. Set mi = mi(G) = |{g ∈ G| the order of g is
i}|. In fact, mi is the number of elements of order i in G, and nse(G) := {mi|i ∈ πe(G)}, the set of
sizes of elements with the same order.
In [11], it is proved that all simple K4-groups can be uniquely determined by nse (G) and |G|.
However, in [13], it is proved that the groups: A4, A5, and A6, and in [9] the groups PSL(2, q)
for q ∈ {7, 8, 11, 13} are uniquely determined by only nse (G). Similar characterizations have been
found for the following groups: A7 and A8 [2], the sporadic simple groups [3], PSL(2, p) [1], the
alternating groups [5], and the symmetric groups Sr, where r is prime [4]. In [9], the authors gave
the following problem:
Let G be a group such that nse (G) = nse (PSL(2, q)), where q is a prime power. Is G isomorphic
to PSL(2, q)?
In this paper, we give a positive answer to this problem and show that the group PSL(2, q) is
characterizable by only nse (G) for q ∈ {16, 17, 19, 23}. In fact, the main theorem of our paper is as
follows:
Main theorem. LetG be a group such that nse (G) = nse (PSL(2, q)),where q ∈ {16, 17, 19, 23}.
Then G ∼= PSL(2, q).
In this paper, we use a new technique for the proof of our main result. Also we apply the
technique of used in [9].
* Partial support by the Center of Excellence of Algebraic Hyper structures and its Applications of Tarbiat Modares
University (CEAHA) is gratefully acknowledge by the third author.
c© A. K. ASBOEI, S. S. S. AMIRI, A. IRANMANESH, 2015
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 9 1155
1156 A. K. ASBOEI, S. S. S. AMIRI, A. IRANMANESH
We note that there are finite groups which are not characterizable by nse (G) and |G|. In 1987,
Thompson gave an example as follows:
Let G1 = (C2 × C2 × C2 × C2) oA7 and G2 = L3(4) o C2 be the maximal subgroups of M23.
Then nse (G1) = nse (G2) and |G1| = |G2|, but G1 6∼= G2. Throughout this paper, we denote by
φ(n) the Euler totient function. If G is a finite group, then we denote by Pq a Sylow q-subgroup
of G and nq(G) is the number of Sylow q-subgroup of G, that is, nq(G) = |Sylq(G)|. All further
unexplained notations are standard, and the reader is referred to [6] if necessary.
2. Preliminary results. In this section, we bring some preliminary lemmas to be used in the
proof of main theorem.
Lemma 2.1 [7] (9.3.1). Let G be a finite solvable group and |G| = mn, where m = pα1
1 . . . pαr
r ,
(m,n) = 1. Let π = {p1, . . . , pr} and hm be the number of π-Hall subgroups of G. Then hm =
= qβ11 . . . qβss satisfies the following conditions for all i ∈ {1, 2, . . . , s}:
1) qβii ≡ 1(mod pj) for some pj ;
2) the order of some chief factor of G is divisible by qβii .
Lemma 2.2 [8]. If G is a simple K3-group, then G is isomorphic to one of the following groups:
A5, A6, PSL(2, 7), PSL(2, 8), PSL(2, 17), PSL(3, 3), PSU(3, 3) or PSU(4, 2).
Lemma 2.3 [14]. Let G be a simple group of order 2a3b5pc, where p 6= 2, 3, 5 is a prime, and
abc 6= 0. Then G is isomorphic to one of the following groups: A7, A8, A9; M11, M12; PSL(2, q),
q = 11, 16, 19, 31, 81; PSL(3, 4), PSL(4, 3), S6(2), PSU(4, 3), or PSU(5, 2). In particular, if
p = 11, then G ∼= M11, M12, PSU(5, 2), or PSL(2, 11); if p = 7, then G ∼= A7, A8, A9, A10,
PSL(2, 49), PSL(3, 4), S4(7), S6(2), PSU(3, 5), PSU(4, 3), J2, or O+
8 (2).
Lemma 2.4 [11]. Let G be a finite group, P ∈ Sylp(G), where p ∈ π(G). Let G have a normal
series K � L�G. If P ≤ L and p - |K|, then the following hold:
(1) NG/K(PK/K) = NG(P )K/K;
(2) |G : NG(P )| = |L : NL(P )|, that is, np(G) = np(L);
(3) |L/K : NL/K(PK/K)|t = |G : NG(P )| = |L : NL(P )|, that is, np(L/K)t = np(G) =
= np(L) for some positive integer t, and |NK(P )|t = |K|.
Lemma 2.5 [7] (9.1.2). Let G be a finite group and m be a positive integer dividing |G|. If
Lm(G) = {g ∈ G|gm = 1}, then m | |Lm(G)|.
Lemma 2.6 [13]. Let G be a group containing more than two elements. Let k ∈ πe(G) and mk
be the number of elements of order k in G. If s = sup {mk|k ∈ πe(G)} is finite, then G is finite and
|G| ≤ s(s2 − 1).
Lemma 2.7 [10]. Let G be a finite group and p ∈ π(G) be odd. Suppose that P is a Sylow
p-subgroup of G and n = psm, where (p,m) = 1. If P is not cyclic and s > 1, then the number of
elements of order n is always a multiple of ps.
Let G be a group such that nse (G) = nse (PSL(2, q)), where q ∈ {16, 17, 19, 23}. By Lemma
2.6, we can assume that G is finite. Let mn be the number of elements of order n. We note that
mn = kφ(n), where k is the number of cyclic subgroups of order n in G. Also we note that if n > 2,
then φ(n) is even. If n ∈ πe(G), then by Lemma 2.5 and the above notation we have
φ(n) | mn,
n |
∑
d|n
md.
(2.1)
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 9
A NEW CHARACTERIZATION OF PSL(2, q) FOR SOME q 1157
In the proof of the main theorem, we often apply (2.1) and the above comments.
3. Proof of the main theorem. In this section, first we prove that the group PSL(2, 16) is
characterizable by nse.
3.1. Characterizability of the group PSL(2, 16) by NSE. Let G be a group such that nse (G) =
= nse (PSL(2, 16)) = {1, 255, 272, 544, 1088, 1920}. First, we prove that π(G) ⊆ {2, 3, 5, 17}.
Since 255 ∈ nse (G), it follows that by (2.1), 2 ∈ π(G) and m2 = 255. Let 2 6= p ∈ π(G), by (2.1),
p | (1 + mp), and (p − 1) | mp, which implies that p ∈ {3, 5, 17}. If 3, 5 and 17 ∈ π(G), then
m3 = 272, m5 = 544 and m17 = 1920, by (2.1). Also we can see easily that G does not contain
any elements of order 9, 25, 51, 85, 289, and 512. Similarly, we can see that if 10 ∈ πe(G), then
m10 = 1920.
Let 3 ∈ π(G). Since 9 6∈ πe(G), exp(P3) = 3. Considering m = |P3| in Lemma 2.5, we have
|P3| | (1 + m3) = 273. Hence |P3| = 3, and n3 = m3/φ(3) = 136 | |G|. Thus if 3 ∈ π(G), then
17 ∈ π(G).
Let 17 ∈ π(G). Since 289 6∈ πe(G), exp(P17) = 17. By Lemma 2.5, |P17| | (1 +m17) = 1921.
Hence |P17| = 17 and n17 = m17/φ(17) = 120 | |G|. Thus if 17 ∈ π(G), then 3 and 5 ∈ π(G).
Now let 5 ∈ π(G). Since 25 6∈ πe(G), exp (P5) = 5. By Lemma 2.5, |P5| | (1 + m5) = 545.
Thus |P5| = 5 and n5 = m5/φ(5) = 136 | |G|. Hence if 5 ∈ π(G), then 17 ∈ π(G). By the above
discussion if 3 or 5, or 17 ∈ π(G), then π(G) = {2, 3, 5, 17}. In follow, we show that π(G) could
not be the set {2}, and hence π(G) must be equal to {2, 3, 5, 17}.
If π(G) = {2}, then πe(G) ⊆ {1, 2, 22, . . . , 28}. Hence |G| = 2m = 4080 + 272k1 + 544k2 +
+1088k3+1920k4,wherem, k1, k2, k3 and k4 are nonnegative integers and 0 ≤ k1+k2+k3+k4 ≤ 3.
It is clear that 4080 ≤ |G| ≤ 9840. Hence m = 12, or 13.
Ifm = 12, then 16 = 272k1+544k2+1088k3+1920k4, which is a contradiction. Hencem = 13,
and we have 4112 = 272k1+544k2+1088k3+1920k4. Using an easy computer calculation, k1 = 1,
k2 = 0, k3 = 0, and k4 = 2. Therefore, πe(G) = {1, 2, 22, . . . , 28}, and |G| = 8192. We prove that
no such group G exists.
By (2.1), clearly m256 = 1920. Since m256 = k φ(256), where k is the number of cyclic
subgroups of order 256 in G, then G have 15 cyclic subgroups of order 256. Set Ω = {Hi|
|Hi| = 256, 1 ≤ i ≤ 15}, where Hi is a cyclic subgroup of order 256 of G and we know that
G acts on Ω by conjugation. If ∆(Hi) is an orbit of this action, then |∆(Hi)| = |G : NG(Hi)|,
and hence by Orbit – Stabilizer theorem, the size of each orbit is a power of 2. On the other
hand, the sizes of these orbits add up to 15, which is odd, so at least one of the orbits must
have size 1. Hence for one of Hi, we have |∆(Hi)| = |G : NG(Hi)| = 1, that is, G = NG(Hi).
Therefore, G has a normal cyclic subgroup of order 256, say N. The index of N in G is |G :
N | = 8192/256 = 32. Let h be a generator of N, then N has only one element of order 2, namely
z = h128. Now we consider any other coset of N in G. If this contains an element of order 2, say
x, then the coset is Nx, and we can proceed as follows. Since x normalizes N and N = 〈h〉, then
x conjugates h to hr for a some odd number r, and since x has order 2, then by relation hx = hr,
we conclude that hrx = hr
2
. Hence (hr)x = hr
2
, then (hx)x = hx
2
= h = hr
2
. Therefore, r2 ≡ 1
(mod 256), then r ≡ 1, 127, 129 or 255 (mod 256). Let L be a subgroup of G generated by N and x
which is a semidirect product of order 2|N | = 512, and is also the union of the two coset N and Nx.
Now we can count the number of elements of order 2 in the coset Nx. If r = 1, then L is isomorphic
to a direct product of C256 by C2 and so it has exactly three elements of order 2 one of them is z,
and the other two are x and zx, lying in Nx. If r = 127, then (hix)2 = hi(xhix) = hi+127i = h128i
and so hix has order 2 when i is even and order 4 when i is odd, hence in this case, Nx contains
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 9
1158 A. K. ASBOEI, S. S. S. AMIRI, A. IRANMANESH
exactly 128 elements of order 2. If r = 129, then (hix)2 = hi(xhix) = hi+129i = h130i and so hix
has order 2 precisely when i is divisible by 128, and so in this case, Nx has exactly two elements
of order 2, x and zx. If r = 255, then (hix)2 = hi(xhix) = hi−i = 1, for all i, and so all 256
elements of Nx has order 2. It follows that every coset Ng of N in G contains 0, 1, 2, 128 or 256
elements of order 2. Finally, let ti be the number of elements of order 2 in the j the coset of N in
G. The number of ti must be 32 and the sum of ti’s is equal to 255, since we have m2 = 255, which
is impossible. Because by choose 32 integers from the set {0, 1, 2, 128, 256}, we don’t obtain the
number 255. Hence no such group G exists.
Therefore, π(G) = {2, 3, 5, 17}. Since 51 6∈ πe(G), the group P17 acts fixed point freely on the
set of elements of order 3, and so |P17| | m3 = 272, which implies that |P17| = 17. Similarly, we
conclude that |P3| = 3. In addition, since 85 6∈ πe(G), the group P5 acts fixed point freely on the set
of elements of order 17, which implies that |P5| = 5.
Now we prove that 10 6∈ πe(G). Suppose that 10 ∈ πe(G). We know that if P and Q are Sylow
5-subgroups of G, then P and Q are conjugate, which implies that CG(P ) and CG(Q) are conjugate
in G. Therefore, m10 = φ(10)n5k, where k is the number of cyclic subgroups of order 2 in CG(P5).
Since n5 = m5/φ(5) = 136, 544 | m10. On the other hand, m10 = 1920, which is a contradiction.
Hence 10 6∈ πe(G).
Since 10 6∈ πe(G), the group P2 acts fixed point freely on the set of elements of order 5. Thus
|P2| | m5 = 544, which implies that |P2| | 25. Since 4080 ≤ |G|, |G| = 25 × 3 × 5 × 17 or
|G| = 24 × 3× 5× 17. If |G| = 25 × 3× 5× 17, then since m5 = 544, n5 = 136.
We claim that G is a nonsolvable group. Suppose G is a solvable group. By Lemma 2.1, 23 ≡ 1
(mod 5), which is a contradiction. Hence G is a nonsolvable group. Since G is a nonsolvable group,
and p ‖ |G| for p ∈ {3, 5, 17}, G has a normal series 1 � N � H � G, such that N is a maximal
solvable normal subgroup of G, and H/N is a nonsolvable minimal normal subgroup of G/N. Then
H/N is a non-Abelian simple K3-group, or K4-group. Let H/N be a non-Abelian simple K3-group.
By Lemma 2.2, H/N ∼= A5. If P5 ∈ Syl5(G), then P5N/N ∈ Syl5(H/N), and n5(H/N)t = n5(G)
for some positive integer t. By Lemma 2.4, 5 - t. Since n5(H/N) = n5(A5) = 6, 136 = 6t,
which is a contradiction. Therefore, H/N is a non-Abelian simple K4-group. By Lemma 2.3,
H/N ≡ PSL(2, 16). Now set H := H/N ∼= PSL(2, 16) and G := G/N. On the other hand, we
have
PSL(2, 16) ∼= H ∼= HCG(H)/CG(H) ≤ G/CG(H) = NG(H)/CG(H) ≤ Aut(H).
Let K = {x ∈ G | xN ∈ CG(H)}, then G/K ∼= G/CG(H). Hence PSL(2, 16) ≤ G/K ≤
≤ Aut(PSL(2, 16)), then G/K ∼= PSL(2, 16), PGO−(4, 4), or PΓL(2, 16). If G/K ∼= PΓL(2, 16),
then since |G| = 2|PSL(2, 16)|, which is a contradiction.
If G/K ∼= PGO−(4, 4), then since |G| = 2|PSL(2, 16)|, |K| = 1, and G ∼= PGO−(4, 4). On
the other hand, nse (G) 6= nse (PGO−(4, 4)), we get a contradiction.
If G/K ∼= PSL(2, 16), then |K| = 2. Since N ≤ K, and N is a maximal solvable normal
subgroup of G, N = K. Now we know that H/N ∼= PSL(2, 16), where |N | = 2, so G has a normal
subgroup N of order 2, generated by a central involution z. Let x be an element of order 5 in G.
Since xz = zx and (o(x), o(z)) = 1, o(xz) = 10. Hence 10 ∈ πe(G), this gives a contradiction.
Therefore, |G| = 24×3×5×17. Now we have |G| = |PSL(2, 16)|, and nse (G) = nse (PSL(2, 16)).
By [11], since PSL(2, 16) is simple K4-group, G ∼= PSL(2, 16).
3.2. Characterizability of the group PSL(2, 17) by NSE. Let G be a group such that nse (G) =
= nse (PSL(2, 17)) = {1, 153, 272, 288, 306, 612, 816}. First, we prove that π(G) ⊆ {2, 3, 17}.
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 9
A NEW CHARACTERIZATION OF PSL(2, q) FOR SOME q 1159
Since 153 ∈ nse (G), it follows that by (2.1), 2 ∈ π(G) and m2 = 153. Let 2 6= p ∈ π(G), by (2.1),
p | (1 +mp), and (p− 1) | mp, which implies that p ∈ {3, 17, 307, 613}.
We will show 307, 613 6∈ π(G). If 307 ∈ π(G), then by (2.1), m307 = 306. On the other hand, by
(2.1), we conclude that if 614 ∈ πe(G), then m614 = 306, or 612, and 614 | (1+m2 +m307 +m614).
Hence 614 | 766, or 614 | 1071, which is a contradiction. Therefore, 614 6∈ πe(G). Since 614 6∈
6∈ πe(G), the group P307 acts fixed point freely on the set of elements of order 2, and |P307| | m2,
which is a contradiction. Hence 307 6∈ π(G). Similar to the above discussion 613 6∈ π(G).
Therefore, π(G) ⊆ {2, 3, 17}. If 3 and 17 ∈ π(G), then m3 = 272, m17 = 288, by (2.1). Also
we can see easily that G does not contain any elements of order 172, 22×17, 3×17 and 34. Similarly,
we can see that if 9 ∈ πe(G), then m9 = 816, and if 27 ∈ πe(G), then m27 ∈ {612, 288}. Let
3 ∈ π(G), since 81 6∈ πe(G), then exp(P3) = 3 or 9, or 27. We will show, in all of these cases if
3 ∈ π(G), then 17 ∈ π(G).
If exp (P3) = 3 , then |P3| | (1 + m3) = 273. Hence |P3| = 3 and n3 = m3/φ(3) = 136 | |G|.
Hence in this case if 3 ∈ π(G), then 17 ∈ π(G).
If exp (P3) = 9, then |P3| | (1 +m3 +m9) = 1089. Hence |P3| = 9 and n3 = m9/φ(9) = 136 |
|G|. Hence if 3 ∈ π(G), then 17 ∈ π(G).
If exp (P3) = 27, then |P3| | (1 + m3 + m9 + m27). Since m27 ∈ {612, 288}, hence |P3| | 81,
or 243. If |P3| = 27, then n3 = m27/φ(27) = 16, or 34. If n3 = 34, then 17 ∈ π(G). If n3 = 16,
then since every element of order 3 lying in the Sylow 3-subgroup and every Sylow 3-subgroup has
at most 2 elements of order 3, m3 ≤ 2× 16 = 32, which is a contradiction. If |P3| ≥ 81 by Lemma
2.7, m27 is a multiple of 27, which is a contradiction. Hence if 3 ∈ π(G), then 17 ∈ π(G).
Now let 17 ∈ π(G). Since 22× 17 /∈ πe(G) and 4 ∈ πe(G), the group P17 acts fixed point freely
on the set of elements of order 4. Hence |P17| | m4 = 306, which implies that |P17| = 17, then
n17 = m17/φ(17) = 18. Therefore, if 17 ∈ π(G), then 3 ∈ π(G). By the above discussion if 3 or
17 ∈ π(G), then π(G) = {2, 3, 5, 17}. In follow, we show that π(G) could not be the set {2}, and
hence π(G) must be equal to {2, 3, 17}.
Let π(G) = {2}. Hence πe(G) ⊆ {1, 2, 22, . . . , 26}. Thus we have |G| = 2m = 2448 + 272k1 +
+ 288k2 + 816k3 + 612k4 + 306k5, where m, k1, k2, k3, k4 and k5 are nonnegative integers and
k1 + k2 + k3 + k4 + k5 = 0, which is a contradiction.
Therefore, π(G) = {2, 3, 17}. Since 51 6∈ πe(G), then P17 acts fixed point freely on the set of
elements of order 3. Hence |P17| | m3 = 272, which implies that |P17| = 17. Similarly, we conclude
that |P3| = 3, or 32. We have |G| = 2m×3n×17 = 2448+272k1+288k2+306k3+612k4+816k5,
wherem, n, k1, k2, k3, k4 and k5 are nonnegative integers 1 ≤ n ≤ 2, and 0 ≤ k1+k2+k3+k4+k5 ≤
≤ 20. Let 34 ∈ πe(G), then m34 = φ(34)n17k, where k is the number of cyclic subgroups of order
2 in CG(P17). Since n17 = m17/φ(17) = 18, 288 | m34, and so by nse (G) we have m34 = 288.
On the other hand, by (2.1), 34 | (1 + m2 + m17 + m34) = 730, which is a contradiction. Hence
34 6∈ πe(G).
Since 34 6∈ πe(G), the group P2 acts fixed point freely on the set of elements of order 17, and
so |P2| | m17 = 288, which implies that 1 ≤ m ≤ 5. Since 2448 ≤ |G|, and 1 ≤ n ≤ 2, m ≥ 4.
Therefore, |G| = 24 × 32 × 17, or |G| = 25 × 32 × 17. If |G| = 25 × 32 × 17, then we show that
G is a nonsolvable group. Suppose that G is a solvable group. Since n17 = 18, then by Lemma
2.1, 32 ≡ 1 (mod 17), which is a contradiction. Hence G is a nonsolvable group. Since G is a
nonsolvable group, and 172 - |G|, G has a normal series: 1�N �H �G, such that N is a maximal
solvable normal subgroup of G and H/N is a nonsolvable minimal normal subgroup of G/N. Then
H/N is a non-Abelian simple K3-group. Hence by Lemma 2.2, H/N = PSL(2, 17). Now set
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H := H/N ∼= PSL(2, 17) and G := G/N. On the other hand, we have
PSL(2, 17) ∼= H ∼= HCG(H)/CG(H) ≤ G/CG(H) = NG(H)/CG(H) ≤ Aut(H).
Let K = {x ∈ G | xN ∈ CG(H)}, then G/K ∼= G/CG(H). Hence PSL(2, 17) ≤ G/K ≤
≤ Aut(PSL(2, 17)), then G/K ∼= PGL(2, 17), or PSL(2, 17). If G/K ∼= PGL(2, 17). Since |G| =
= 2|PSL(2, 17)|, |K| = 1, and G ∼= PGL(2, 17). On the other hand, nse (G) 6= nse (PGL(2, 17)),
we get a contradiction. If G/K ∼= PSL(2, 17), then |K| = 2. Since N ≤ K, and N is a maximal
solvable normal subgroup of G, N = K. Now we know that H/N ∼= PSL(2, 17), where |N | = 2,
so G has a normal subgroup N of order 2, generated by a central involution z. Let x be an element
of order 17 in G. Since xz = zx and (o(x), o(z)) = 1, o(xz) = 34. Hence 34 ∈ πe(G), this gives
a contradiction. If |G| = 24 × 32 × 17, then |G| = |PSL(2, 17)|. Since |G| = | PSL(2, 17)| and
nse (G) = nse (PSL(2, 17)). By [12], G ∼= PSL(2, 17).
3.3. Characterizability of the group PSL(2, 19) by NSE. Let G be a group such that nse (G) =
= nse (PSL(2, 19)) = {1, 171, 360, 380, 684, 1140}. First, we prove that π(G) ⊆ {2, 3, 5, 19}.
Since 171 ∈ nse (G), it follows that by (2.1), 2 ∈ π(G) and m2 = 171. If 2 6= p ∈ π(G), then
p ∈ {3, 5, 7, 19}.
We will show 7 6∈ π(G). Let 7 ∈ π(G), then by (2.1), m7 = 1140. On the other hand, by (2.1),
we conclude that if 14 ∈ πe(G), then m14 ∈ {360, 684, 1140}, and 14 | (1+m2 +m7 +m14), which
is a contradiction. Hence 14 6∈ πe(G). Thus the group P7 acts fixed point freely on the set of elements
of order 2, and |P7| | m2, which is a contradiction. Hence 7 6∈ π(G) and so π(G) ⊆ {2, 3, 5, 19}.
If 3, 5 and 19 ∈ π(G), then m3 = 380, m5 = 684, and m19 = 360. Also we can see easily that
G does not contain any elements of order 27, 32, 95, 125, and 361. Similarly, we can see that if
9 ∈ πe(G), then m9 = 1140, and if 25 ∈ πe(G), then m25 = 1140.
Let 3 ∈ π(G). Since 27 6∈ πe(G), exp(P3) = 3, or 9.
If exp (P3) = 3, then |P3| | (1 + m3) = 381. Thus |P3| = 3 and n3 = m3/φ(3) = 190 | |G|.
Hence in this case if 3 ∈ π(G), then 5 and 19 ∈ π(G).
If exp (P3) = 9, then |P3| | (1 +m3 +m9) = 1521 and so |P3| = 9, and n3 = m9/φ(9) = 190 |
|G|. Hence if 3 ∈ π(G), then 5 and 19 ∈ π(G) .
Now let 5 ∈ π(G). Since 125 6∈ πe(G), exp(P5) = 5, or 25.
If exp (P5) = 5, then |P5| | (1 + m5) = 685, and so |P5| = 5 and n5 = m5/φ(5) = 171 | |G|.
Hence in this case if 5 ∈ π(G), then 3 and 19 ∈ π(G).
If exp (P5) = 25, then |P5| | (1 +m5 +m25) = 1825. Hence |P5| = 25 and n5 = m25/φ(25) =
= 157 | |G|. Therefore, if 5 ∈ π(G), then 3 and 19 ∈ π(G).
In follow, we show that π(G) could not be the sets {2} and {2, 19}, and π(G) must be equal to
{2, 3, 5, 19}.
Let π(G) = {2}, then πe(G) ⊆ {1, 2, 22, 23, 24}. Since nse (G) has six elements, this case is
impossible.
Let π(G) = {2, 19}. Since 361 6∈ πe(G), exp (P19) = 19. Hence |P19| | (1 +m19) = 361 and so
|P19| = 19, or 361.
If |P19| = 19, then n19 = m19/φ(19) = 20| | G|; a contradiction.
If |P19| = 361, then |G| = 2m × 192 = 2736 + 360k1 + 380k2 + 684k3 + 1140k4, where m, k1,
k2, k3, and k4 are nonnegative integers, and 0 ≤ k1 +k2 +k3 +k4 ≤ 2, by |πe(G)| ≤ 8. On the other
hand, 2736 ≤ |G| ≤ 5016, so m = 3 and |G| = 23 × 192. Then 152 = 2736 + 360k1 + 380k2 +
+ 684k3 + 1140k4. It is easy to check this equation has no solution. Hence this case is impossible.
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A NEW CHARACTERIZATION OF PSL(2, q) FOR SOME q 1161
Therefore, π(G) = {2, 3, 5, 19}. Since 95 6∈ πe(G), the group P19 acts fixed point freely on
the set of elements of order 5, and so |P19| | m5 = 684, which implies that |P19| = 19, then
n19 = m19/φ(19) = 20 | |G|. Similarly, we conclude that |P5| = 5. If 15 ∈ πe(G), then m15 =
= φ(15)n5k,where k is the number of cyclic subgroups of order 3 inCG(P5). Since n5 = m5/φ(5) =
= 171, 2× 684 | m15. On the other hand, we know that m15 = 360, which is a contradiction. Hence
15 6∈ πe(G). Similarly, we can see easily 38 6∈ πe(G).
Since 15 6∈ πe(G), the group P3 acts fixed point freely on the set of elements of order 5. So
|P3| | m5, which implies that |P3| | 9. In addition, since 38 6∈ πe(G), |P2| | m19 = 360, and so
|P2| | 24. Because 2736 ≤ |G| and 20 | |G|, |G| = 22 × 32 × 5× 19, or |G| = 23 × 32 × 5× 19. If
|G| = 23 × 32 × 5× 19, then similar to the case PSL(2, 16), or PSL(2, 17), we get a contradiction.
Now nse (G) = nse (PSL(2, 19)), and |G| = |PSL(2, 19)|. By [11], since PSL(2, 19) is a simple
K4-group, G ∼= PSL(2, 19).
3.4. Characterizability of the group PSL(2, 23) by NSE. Let G be a group, such that nse (G) =
= nse (PSL(2, 23)) = {1, 253, 506, 528, 1012, 2760}. First, we prove that π(G) ⊆ {2, 3, 11, 23}.
Since 253 ∈ nse (G), it follows that by (2.1), 2 ∈ π(G) and m2 = 253. Let 2 6= p ∈ π(G), by (2.1),
p | (1 +mp), and (p− 1) | mp, which implies that p ∈ {3, 11, 23, 1013}.
We will show 1013 6∈ π(G). Let 1013 ∈ π(G). Then m1013 = 1012. On the other hand, by (2.1)
we conclude that if 2026 ∈ πe(G), then m2026 = 1012 and 2026 | (1+m2+m1013+m2026) = 2278,
which is a contradiction. Hence 2026 6∈ πe(G). Thus the group P1013 acts fixed point freely on the
set of elements of order 2. Then |P1013| | m2, which is a contradiction. Hence 1013 6∈ π(G) and so
π(G) ⊆ {2, 3, 11, 23}. If 3, 11, and 23 ∈ π(G), then m3 = 506, m11 = 2760, and m23 = 528. Also
we can see easily that G does not contain any elements of order 22, 27, 33, 64, 121, 184, 253, and
529. If 9 ∈ πe(G), then m9 ∈ {528, 2760}.
Let 3 ∈ π(G). Since 27 6∈ πe(G), exp (P3) = 3, or 9. If exp(P3) = 3, then |P3| | (1+m3) = 507.
Hence |P3| = 3 and n3 = m3/φ(3) = 253 | |G|. Hence in this case if 3 ∈ π(G), then 11 and
23 ∈ π(G).
Let exp(P3) = 9. Then |P3| | (1 +m3 +m9) = 1035, or 3267. Hence |P3| = 9, or 27.
If |P3| = 9, then n3 = m9/φ(9) = 88, or 460. If n3 = 460, then since 5 /∈ π(G), we get a
contradiction. Let n3 = 88. Since every element of order 3 lying in the Sylow 3-subgroup and we
have every Sylow 3-subgroup has at most 2 elements of order 3, m3 ≤ 2 × 88 = 176, which is a
contradiction.
If |P3| = 27, by Lemma 2.7, m9 is a multiple of 9, which is a contradiction. Therefore, if
3 ∈ π(G), then 11 and 23 ∈ π(G).
Now let 11 ∈ π(G). We have exp(P11) = 11, hence |P11| | (1+m11) = 2761 and so |P11| = 11.
Then n11 = m11/φ(11) = 276. Since n11 | |G|, 3 ∈ π(G) and so 23 ∈ π(G). In follow, we show
that π(G) could not be the sets {2} and {2, 23}, and π(G) must be equal to {2, 3, 11, 23}.
Let π(G) = {2}, then πe(G) ⊆ {1, 2, 22, 23, 24, 25}. Therefore, |G| = 2m = 5060 + 506k1 +
+528k2+1012k3+2760k4,wherem, k1, k2, k3, k4 are nonnegative integers and k1+k2+k3+k4 = 0,
which is impossible.
Let π(G) = {2, 23}. Since 529 6∈ πe(G), exp(P23) = 23. Then |P23| | (1 +m23) = 529. Hence
|P23| = 23, or 529. If |P23| = 23, then n23 = m23/φ(23) = 24 | |G|; a contradiction. If |P23| = 529,
then |G| = 2m × 232 = 5060 + 506k1 + 528k2 + 1012k3 + 2760k4, where m, k1, k2, k3, and k4 are
nonnegative integers, and 0 ≤ k1 + k2 + k3 + k4 ≤ 3, by |πe(G)| ≤ 9. On the other hand, 5060 ≤
≤ |G| ≤ 13340. Hence m = 4 and |G| = 24×232, and so 3404 = 506k1+528k2+1012k3+2760k4.
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1162 A. K. ASBOEI, S. S. S. AMIRI, A. IRANMANESH
Clearly 23 | k2. Then k2 = 0. Therefore, 74 = 11k1 + 22k3 + 60k4 , where 0 ≤ k1 + k3+ k4 ≤ 3. It
is easy to check this equation has no solution. Hence this case is impossible.
Therefore, π(G) = {2, 3, 11, 23}. Since 23×11 6∈ πe(G), the group P23 acts fixed point freely on
the set of elements of order 11, and so |P23| | m11 = 2760, which implies that |P23| = 23. Similarly,
|P11| = 11. In addition, since 33 6∈ πe(G), |P3| | m11. Then |P3| = 3. As well 22 6∈ πe(G), hence
|P2| | 23. On the other hand, 5060 ≤ |G|, so |G| = 23 × 3 × 11 × 23 = |PSL(2, 23)|. Now we
have nse (G) = nse (PSL(2, 23)), and |G| = |PSL(2, 23)|. By [11], since PSL(2, 23) is a simple
K4-group, G ∼= PSL(2, 23), and the proof is completed.
4. Remark. In this paper, we used from Lemma 2.5 for find the order of group G while the
neither in [9] nor in [13] it did not use. The authors in [9] and [13] for find the order of the group
G being forced to provide proof in several cases and used from Gap program for many cases, but
we did not use from Gap program. By the method [9] and [13] we can not characterized the group
with order more 2000, because they used the GAP program and in the library of GAP, there are only
the groups with order less than 2000. But our method, can work on the groups with order more than
2000. Therefore, this technique can work for the group PSL(2, p), for a given prime number p.
1. Asboei K. A., Amiri S. S., Iranmanesh A. A characterization of the linear groups L2(p) // Czechoslovak Math. J. –
2014. – 64, № 2. – P. 459 – 464.
2. Asboei K. A., Amiri S. S., Iranmanesh A., Tehranian A. A new characterization of A7 and A8 // An. şti. Univ. Ovidius
Constanta. Ser. Mat. – 2013. – 21. – P. 43 – 50.
3. Asboei K. A., Amiri S. S., Iranmanesh A., Tehranian A. A new characterization of sporadic simple groups by nse and
order // J. Algebra and Appl. – 2013. – 12, № 2. – Paper № 1250158.
4. Asboei K. A., Amiri S. S., Iranmanesh A., Tehranian A. A characterization of symmetric group Sr where r is prime
number // Ann. Math. Inform. – 2012. – 40. – P. 13 – 23.
5. Asboei K. A., Amiri S. S., Iranmanesh A. A new characterization of alternating groups // Algebra and Discrete Math.
– 2014. – 18, № 1. – P. 8 – 13.
6. Conway J. H., Curtis R. T., Norton S. P., Parker R. A., Wilson R. A. Atlas of finite groups. – Oxford: Clarendon
Press, 1985.
7. Hall M. The theory of groups. – New York: Macmillan, 1959.
8. Herzog M. On finite simple groups of order divisible by three primes only // J. Algebra. – 1968. – 120, № 10. –
P. 383 – 388.
9. Khatami M., Khosravi B., Akhlaghi Z. A new characterization for some linear groups // Monatsh. Math. – 2009. –
163. – P. 39 – 50.
10. Miller G. Addition to a theorem due to Frobenius // Bull. Amer. Math. Soc. – 1904. – 11. – P. 6 – 7.
11. Shao C. G., Shi W., Jiang Q. Characterization of simple K4-groups // Front Math. China. – 2008. – 3, № 3. –
P. 355 – 370.
12. Shao C. G., Shi W., Jiang Q. A characterization of simple K3-groups // Adv. Math. (China). – 2009. – 38. –
P. 327 – 330.
13. Shen R., Shao C. G., Jiang Q., Shi W., Mazurov V. A new characterization of A5 // Monatsh. Math. – 2010. – 160. –
P. 337 – 341.
14. Shi W. The simple groups of order 2a3b5c7d and Janko’s simple groups // J. South China Normal Univ. Natur. Sci.
Ed. – 1987. – 12, № 4. – P. 1 – 8.
Received 12.04.12,
after revision — 24.06.15
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