Characterization of M₁₁ and L₃(3) by their commuting graphs
For the groups M₁₁ and L₃(3), we show that their commuting graphs are unique
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Інститут математики НАН України
2010
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Цитувати: | Characterization of M₁₁ and L₃(3) by their commuting graphs / M.R. Salarian // Український математичний журнал. — 2010. — Т. 62, № 11. — С. 1583–1584. — Бібліогр.: 11 назв. — англ. |
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irk-123456789-1662902020-02-19T01:27:18Z Characterization of M₁₁ and L₃(3) by their commuting graphs Salarian, M.R. Короткі повідомлення For the groups M₁₁ and L₃(3), we show that their commuting graphs are unique Показано, що комутуючі графи груп M₁₁ та L₃(3) єдині. 2010 Article Characterization of M₁₁ and L₃(3) by their commuting graphs / M.R. Salarian // Український математичний журнал. — 2010. — Т. 62, № 11. — С. 1583–1584. — Бібліогр.: 11 назв. — англ. 1027-3190 http://dspace.nbuv.gov.ua/handle/123456789/166290 en Український математичний журнал Інститут математики НАН України |
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Короткі повідомлення Короткі повідомлення Salarian, M.R. Characterization of M₁₁ and L₃(3) by their commuting graphs Український математичний журнал |
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For the groups M₁₁ and L₃(3), we show that their commuting graphs are unique |
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Article |
author |
Salarian, M.R. |
author_facet |
Salarian, M.R. |
author_sort |
Salarian, M.R. |
title |
Characterization of M₁₁ and L₃(3) by their commuting graphs |
title_short |
Characterization of M₁₁ and L₃(3) by their commuting graphs |
title_full |
Characterization of M₁₁ and L₃(3) by their commuting graphs |
title_fullStr |
Characterization of M₁₁ and L₃(3) by their commuting graphs |
title_full_unstemmed |
Characterization of M₁₁ and L₃(3) by their commuting graphs |
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characterization of m₁₁ and l₃(3) by their commuting graphs |
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Інститут математики НАН України |
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2010 |
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Короткі повідомлення |
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http://dspace.nbuv.gov.ua/handle/123456789/166290 |
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Characterization of M₁₁ and L₃(3) by their commuting graphs / M.R. Salarian // Український математичний журнал. — 2010. — Т. 62, № 11. — С. 1583–1584. — Бібліогр.: 11 назв. — англ. |
series |
Український математичний журнал |
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AT salarianmr characterizationofm11andl33bytheircommutinggraphs |
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2025-07-14T21:07:14Z |
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2025-07-14T21:07:14Z |
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UDC 512.5
M. R. Salarian (Inst. Adv. Stud. Basic Sci., Zanjan, Iran)
CHARACTERIZATION M11 AND L3(3)
BY THEIR COMMUTING GRAPHS
ХАРAКТЕРИЗАЦIЯ M11 TA L3(3)
ЇХ КОМУТУЮЧИМИ ГРАФАМИ
For the groups M11 and L3(3), we show that their commuting graphs are unique.
Показано, що комутуючi графи груп M11 та L3(3) єдинi.
1. Introduction. Throughout this article G is a finite group. One can associate a graph
to G in many different ways (see, for example, [2, 9 – 11]). One of the graphs is the
commuting graph associated to a finite group. For a finite group G, the commuting
graph of G has G − {1} as its vertex set and two distinct vertices x and y are joined
by an edge if [x, y] = 1(x and y commute). In [10, 11] properties of the commuting
graph for finite simple groups were used to prove the Margulis – Platonov conjecture for
arithmetic groups.
Let X and Y be two graphs with vertex sets V (X) and V (Y ), respectively. Then X
is called isomorphic to Y if there is a bijection f : V (X) 7→ V (Y ) such that any two
vertices u and v of V (X) are adjacent in X if and only if f(u) and f(v) are adjacent
in Y . The bijection f is called a graph isomorphism.
In this short note we show that the commuting graphs of the groups M11 and L3(3)
are unique. We note that our proofs do not require the classification of the finite simple
groups.
Theorem 1.1. Let G be a finite group isomorphic to M11 or L3(3) and H be a
finite group such that X(G) ∼= X(H). Then G ∼= H.
Our strategy for identifying the groups M11 and L3(3) is to determine the structures
of the centralizers of involutions. By assumptions and notations in Theorem 1.1 we will
show that H is simple and for each involution h ∈ H we have CH(h) ∼= GL2(3). Then
the main result will follow from the following theorem.
Theorem 1.2 ([3], XII, Theorem 5.2). Let G be a finite simple group and t ∈ G be
an involution in the center of a Sylow 2-subgroup of G. If CG(t) ∼= GL2(3), then either
G ∼=M11 or G ∼= L3(3).
For a finite group G, O(G) is the largest normal subgroup of G of odd order and
M(G) is the schur multiplier of G. We have used the Atlas [4] notations for simple
groups. The other notations follow [1] and [8].
For a vertex x ∈ V (X), d(x) is the number of vertices adjointed with x. In this
paper all graphs are simple and without loop. We have used [5] for other notations in
graph theory.
2. Preliminaries. In this section we recall some known theorems in finite groups.
Theorem 2.1 [6]. Let G be a finite group which contains a self-centralizing sub-
group of order three. Then one of the following statements is true:
i) G contains a nilpotent normal subgroup N such that G/N is isomorphic to either
Z3 or S3;
c© M. R. SALARIAN, 2010
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 11 1583
1584 M. R. SALARIAN
ii) G contains a normal 2-subgroup N and G/N ∼= A5;
iii) G ∼= L3(2) ∼= L2(7).
Theorem 2.2 (Glauberman Z?-theorem [7]). Let G be a finite group and 1 6= t ∈ G
be an involution. If G 6= O(G)CG(t), then t is conjugate in G to an involution in
CG(t) \ 〈t〉 .
Lemma 2.1. Let G be a finite group. If a Sylow 2-subgroup of G is cyclic or is
isomorphic to Z2 ×D8 or Q8, then G is not a non-Abelian simple group.
Proof. Let G be a finite group and T ∈ Syl2(G). Let T be isomorphic to Q8 or T
be cyclic. Then there is a unique involution t ∈ T . So by Theorem 2.2, we get that G
is not a non-Abelian simple group. Now let T be isomorphic to Z2 ×D8. Let 〈t〉 = T ′
and Z(T ) = 〈s, t〉. Then s is not conjugate to t in G and so Aut(T ) is a 2-group. Since
T ∈ Syl2(G), we have NG(T ) = T and then by [8] (Theorem 7.1.1) we get that no
two involutions in Z(T ) are conjugate in G. Let N be a subgroup of T isomorphic
to D8, then t ∈ Z(N) and N is a maximal subgroup of T . If t is conjugate to an
involution of N \ {t} in G, then s is not conjugate to an involution of N in G. Then by
Thompson transfer lemma ([3], XII.8.2) we get that G is not simple. If s is conjugate
to an involution of N in G, then t is not conjugate to an involution of T \ {t} in G.
Then by Theorem 2.2 we get that G is not simple. If neither s nor t is conjugate to an
involution in N \ {t}, then by Theorem 2.2 or by Thompson transfer lemma we get that
G is not simple and the lemma is proved.
3. Proof of Theorem 1.1. In this section we shall prove Theorem 1.1. We recall
that for a finite group G, X(G) is the commuting graph of G.
Notations. In this section G is a finite group isomorphic to M11 or L3(3) and H is
a finite group such that X(G) ∼= X(H). Let φ : X(G) → X(H) be an isomorphism
and g ∈ V (X(G)) be an involution. Set h = φ(g).
Lemma 3.1. For each involution x ∈ H we have that CH(x) is isomorphic to
either GL2(3) or Z2 × S4.
Proof. Since X(G) ∼= X(H), we have |G| = |H| and X(CG(g)) ∼= X(CH(h)). By
[4, p. 13, 18] we have that CG(g) ∼= GL2(3). As X(CG(g)) ∼= X(CH(h)) and there is a
vertex of degree 4 in X(CG(g)), there is a vertex of degree 4 in X(CH(h)) say r. Since
d(r) = 4, we get that |CH(r, h)| = 6 and so CH(r, h) = 〈r〉 × 〈h〉. Hence r is of order
3 and h is of order 2. Therefore a Sylow 3-subgroup S of CH(h) is of order three and
CCH(h)(S) = S×〈h〉. It gives us that a Sylow 3-subgroup S of CH(h)/ 〈h〉 is of order
three and S is self-centralizing. Now by Theorem 2.1 and as |CH(h)| = |CH(g)| = 24 ·3
we get that CH(h)/ 〈h〉 ∼= S4. Therefore CH(h) ∼= GL2(3) or CH(h) is isomorphic to
Z2 × S4 and the lemma is proved.
Proof of Theorem 1.1. By Lemma 3.1, CH(h) is isomorphic to GL2(3) or Z2×S4.
Assume that H is a simple group. Then by Lemma 2.1 we get that CH(h) ∼= Z2 × S4
does not happen and therefore CH(h) ∼= GL2(3). This and Theorem 1.2 give us that
H ∼=M11 or L3(3). Since |H| = |G|, we get that H ∼= G and theorem is proved. Hence
it is enough to show that H is simple. We assume that H is not simple and N is a
minimal normal subgroup of H . Let 〈h, f〉 be an elementary Abelian 2-group of order
4 in H . Then by coprime action we have
O(H) =
〈
CH(x) ∩O(H);x ∈ 〈h, f〉]
〉
.
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 11
Since for each involution x ∈ H , either CH(x) ∼= GL2(3) or CH(x) ∼= Z2×S4, we
deduce that O(H) = 1. This gives us that 2 divides the order of N . First assume that
CH(x) ∼= GL2(3) for each involution x ∈ H . Then by Lemma 2.1 and as CH(h)∩N is
normal in CH(h), we get that CH(h) ≤ N . Now by Theorem 1.2, N ∼= M11 or L3(3).
By [4, p. 13, 18], Out(M11) = 1 and |Out(L3(2))| = 2. As |N |2 = |H|2, we get that
N = H and the theorem is proved in this case.
Now assume that CH(x) ∼= Z2×S4 for each involution x ∈ H . Then by Lemma 2.1
and as CH(h) ∩ N is normal in CH(h), we get that a Sylow 2-subgroup of N is
elementary Abelian of order 8, N is simple and for each involution t ∈ N we have
that CN (t) is elementary Abelian of order 8. Now by ([8], Theorem 16.6.1) we get that
N ∼= J1 or a group of Ree type. In a group of Ree type the centralizer of an involution
is isomorphic to Z2×L2(q) for q ≥ 3. Therefore N is not a group of Ree type. Assume
that N ∼= J1, then by [4, p. 36] we get that the centralizer of each involution in N is
isomorphic to Z2 × A5, therefore N is not isomorphic to J1 and hence this case does
not happen. Now the theorem is proved.
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Received 25.01.10
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