A New Characterization of Sporadic Groups
aa r X i v : . [ m a t h . G R ] S e p A New Characterization of Sporadic Groups ∗ Zhongbi Wang, Heng Lv, Yanxiong Yan , Guiyun Chen † School of Mathematics and Statistics, Southwest University,Beibei, Chongqing, 400715, ChinaE-mail: [email protected]; [email protected]; [email protected]; [email protected]
Abstract: Let G be a finite group, n a positive integer. π ( n ) denotes the set of all primedivisors of n and π ( G ) = π ( | G | ) . The prime graph Γ( G ) of G , defined by Grenberg and Kegel,is a graph whose vertex set is π ( G ) , two vertices p, q in π ( G ) joined by an edge if and only if G contains an element of order pq . In this article, a new characterization of sporadic simplegroups is obtained, that is, if G is a finite group and S a sporadic simple group. Then G ∼ = S ifand only if | G | = | S | and Γ( G ) is disconnected. This characterization unifies the several charac-terizations that can conclude the group has non-connected prime graphs, hence several knowncharacterizations of sporadic simple groups become the corollaries of this new characterization.Keywords : sporadic groups; order; prime graph; characterizationMathematics Subject Classification (2020): 20D08 In the past three decades, as a very interesting topic, quantitative characterization of a group, especiallya simple group, has been being an active topic in the theory of finite simple group since classification of finitesimple groups completed in the early of 1980s. When Wujie Shi began to investigate the topic whether a finitesimple group can be uniquely determined by its order and the set of its element orders, he proposed a famousconjecture in 1987, which was recorded as Problem 12.39 in [1].
Shi’s Conjecture.
Let G be a finite group, S a finite simple group, then G ∼ = S if and only if | G | = | S | and π e ( G ) = π e ( S ), where π e ( G ) denotes the set of element orders in G .Research on Shi’s conjecture began an era of quantitative characterization of finite simple groups. At last,this conjecture was completely proved in 2009. In the series of papers to prove Shi’s conjecture, an importantconcept the prime graph of a finite group was frequently used for dealing with those simple groups with non-connected prime graph, which was defined by Gruenberg and Kegel in [2] as following:Let G be a finite group, n a positive integer. π ( n ) denotes the set of all prime divisors of n and π ( G ) = ∗ This work was supported by the National Natural Science Foundation of China (Grant No.11671324, 11971391), FundamentalResearch Funds for the Central Universities (No. XDJK2019B030). † corresponding author ( | G | ). The prime graph Γ( G ) of G , defined by Grenberg and Kegel, is a graph whose vertex set is π ( G ), twovertices p, q in π ( G ) joined by an edge if and only if G contains an element of order pq . We denote the numberof connected components of Γ( G ) as t ( G ), the connected components of Γ( G ) as { π i , i = 1 , · · · , t ( G ) } , and wealways assume 2 is in π ( G ) if 2 (cid:12)(cid:12) | G | . The components of prime graphs of all finite simple groups were given byJ. S. William, A. S. Kondrat´ e v, M. Suzuki, N. Iiyora and H. Yamaki etc. (see [2-5]).The prime graph once be used by the second author to study the famous Thompson’s conjecture: Thompson’s Conjecture.
Let G be a finite group with Z ( G ) = 1, N ( G ) = { n ∈ N | G has a conjugacyclass of length n } . If M is a finite simple group such that N ( G ) = N ( M ), then G ∼ = M. During the second author studying Thompson’s conjecture, he proved that | G | = | M | if G and M satisfyconditions of Thompson’s Conjecture and the prime graph Γ( M ) is non-connected. For a finite group withnon-connected prime graph, he divided its order into co-prime divisors, each of them exactly corresponding toeach of components of the prime graph, and called these divisors the order components of the finite group andfound that many finite simple groups can be uniquely determined by their order components. Actually, manysimple groups with non-connected prime graphs have been proved to be uniquely determined by their ordercomponents, for example, in a series of articles , for example [6-11], etc, it is shown many simple groups withnon-connected prime graphs are characterized by order components of their prime graphs. There are some othertopics on characterization of a finite simple group by its order and some other quantitative properties relatedto non-connected prime graph, for example, in [12-16], characterization of a finite simple group by its orderand maximal element order (the largest element order or the second largest element order, or both of them),characterization of a finite simple group by its order and the set of orders of maximal abelian subgroups, etc.In these topics, the discussed finite group are usually ascribed to a finite group having non-connected primegraph, whose order is the same as some finite simple group. Therefore, it is a meaningful topic to study thefinite group having its prime graph non-connected and its order being equal to a finite simple group. In thisarticle, we shall discuss this topic and specially focus on a finite group having its prime graph non-connectedand its order being equal to a sporadic group. we shall prove the following theorem: Main Theorem . Let G be a finite group and S a sporadic simple group. Then G ∼ = S if and only if | G | = | S | and the prime graph of G is disconnected. By above theorem, the following known characterizations of sporadic simple groups including Shi’ Con-jecture and Thompson Conjecture, become its corollaries since under the hypothesis the prime graphs of thegroups are non-connected, for example: 2 orollary 1.1
Let G be a finite group and S a sporadic simple group. Then G ∼ = S if and only if | G | = | S | and π e ( G ) = π e ( S ) . Corollary 1.2
Let G be a finite group and S a sporadic simple group. Then G ∼ = S if and only if Z ( G ) = 1 and N ( G ) = N ( S ) . Corollary 1.3
Let G be a finite group and S a sporadic simple group. Then G ∼ = S if and only if G and S have the same order components. Corollary 1.4
Let G be a finite group and S a sporadic simple group. Then G ∼ = S if and only if | G | = | S | and the sets of orders of maximal abelian subgroups of G and S are equal. Corollary 1.5
Let G be a finite group and S a sporadic simple group. Then G ∼ = S if and only if | G | = | S | and the largest element orders of G and S are the same. Throughout the paper, the actions unspecified always means conjugate action. Let G be a finite group, fora prime p ∈ π ( G ), G p denotes the Sylow p − subgroup of G . In addition, G = U ⋉ V denotes G is the semidirectproduct of U and V , especially, V E G . In order to prove the Main Theorem, we first present some lemmas which are useful in our proof.
Lemma 2.1 ( [2], Theorem A ) If G is a finite group whose prime graph has more than one components, thenone of the following holds:(1) G is a F robenius group;(2) G is a − F robenius group;(3) G has a normal series E H E K E G such that H and G/K are π groups, K/H is a non-abeliansimple group, H is a nilpotent group, where ∈ π . And | G/K | (cid:12)(cid:12) | Out(
K/H ) | . Lemma 2.2 ( [17], Lemma 2.6(1)) Let G be a F robenius group with
F robenius kernel K and F robenius complement H . Then K is nilpotent and | H | (cid:12)(cid:12) | K | − . Moreover, t ( G ) = 2 , and Γ( G ) = { π ( H ) , π ( K ) } . Lemma 2.3 ( [17], Lemma 2.6(2)) Let G be a − F robenius group, then G = DEF, where D and DE arenormal subgroups in G , DE and EF are F robenius groups with kernels D and E , respectively. Moreover t ( G ) = 2 , π ( G ) = π ( D ) ∪ π ( F ) and π ( G ) = π ( E ) . emma 2.4 ( [18], 8.2.3) Let G and A be two groups. Let p be a prime divisor of | G | . Suppose that the actionof A on G is coprime . Then there exists an A − invariant Sylow p − subgroup of G. Lemma 2.5 ( [17], Lemma 2.4) Let G be a p − group of order p n , and G act on a q − group H of order q m ,where p, q are two distinct primes. If | G | ∤ | GL(m , q) | , then pq ∈ π e ( G ⋉ H ) . By the above Lemma, we have the following corollary.
Corollary 2.6
Let G be a p − group of order p n , and G act on a q − group H of order q m . If p r ∤ Q mi =1 ( q i − ,where ≤ r ≤ n , then pq ∈ π e ( G ⋉ H ) . For convenience, we call a section of a group G that is a F robenius group as a section of
F robenius type.
Lemma 2.7
Let G be a finite group, p, q ∈ π ( G ) . If | G q | = q n , and p ∤ | GL(n , q) | , then G has no section of F robenius type H ⋉ K such that p (cid:12)(cid:12) | H | and | K q | = q r , where r ≤ n . Furthermore, if G does have a section of F robenius type H ⋉ K such that pq (cid:12)(cid:12) | HK | and p (cid:12)(cid:12) | H | , then ( q, | K | ) = 1 and pq (cid:12)(cid:12) | H | . Proof. If G has a section of F robenius type H ⋉ K such that p (cid:12)(cid:12) | H | and | K q | = q r , then H p ⋉ K q is a F robenius group and H p ⋉ K q has no element of order pq . But under the assumption p ∤ | GL(n , q) | , we have | H p | ∤ | GL(n , q) | , since | GL(r , q) | (cid:12)(cid:12) | GL(n , q) | , it follows that | H p | ∤ | GL(r , q) | , thus, it follows by Lemma 2.5 that H p ⋉ K q has an element of order pq , a contradiction. The first part of the lemma follows. And the second partfollows straightforward from the first part. Remark.
Let G be a finite group satisfying the hypothesis of the Main Theorem, then by Lemma 2.1, G may be a F robenius group or a 2 − F robenius group. In order to prove the Main Theorem, we first show that G cannot be a F robenius group or a 2 − F robenius group, the proof will be separated into six lemmas. In theproofs of next six lemmas, when we mention a
F robenius group or a 2 − F robenius group, we are referringgroups and notations in Lemma 2.2 and Lemma 2.3 without explanations. Moreover, we shall frequently useLemma 2.7 to get information of π ( H ) or come to a contradiction. G is neither a F robenius group nor a − F robenius group
Now we start to show that G is neither a F robenius group nor a 2 − F robenius group.
Lemma 2.8
Let G be a group and S a Mathieu group. If | G | = | S | , then G is neither a F robenius group nora − F robenius group. roof. From [19], we get the orders of Mathieu groups are the following: | M | = 2 · · · | M | = 2 · · · | M | = 2 · · · · | M | = 2 · · · · · | M | = 2 · · · · · U ⋉ V be a Hall subgroup of G with 11 (cid:12)(cid:12) | U V | or 23 (cid:12)(cid:12) | U V | , which is a F robenius group with V the F robenius kernel and U the F robenius complement. Let S be one of M , M and M . Since 11 ∤ | GL(7 , | ,11 ∤ | GL(3 , | and 5 , <
11, we have by Lemma 2.7 that 11 ∈ π ( V ). Thus, | V | = 11. Since V U is a F robenius group, we deduce that | U | (cid:12)(cid:12)
10. Noticing 2 (cid:12)(cid:12) | G | , we conclude that | U | = 5. If G is a 2 − F robenius group, then 11 ∈ π ( D ) for both E and F are two F robenius complements. Surely, | D | = 11. Thus, | E | = 5,and hence, | F | (cid:12)(cid:12)
4. Therefore, 3 (cid:12)(cid:12) | D | , it follows that 15 ∈ π e ( DE ), a contradiction. If G is a F robenius group,then G = U V and 3 (cid:12)(cid:12) | V | , but 5 ∤ | GL(2 , | , so G has an element of order 15 by Lemma 2.5, a contradiction.Let S be one of M or M . We consider the prime 23. Since 23 ∤ | GL(10 , | , 23 ∤ | GL(3 , | , and5 , , <
23, in the same reason we have that 23 ∈ π ( V ). Now | V | = 23, it follows that | U | = 11 by Corollary2.6. Suppose that G is a 2 − F robenius group, then 23 ∈ π ( D ), thus, | E | = 11, therefore, | F | (cid:12)(cid:12) | | . Consequently,7 ∈ π ( D ) and | D | = 7, which is impossible since D E cannot be a F robenius group. If G is a F robenius group,then G = U V . Hence,
U V is a F robenius group, contradicting to Lemma 2.7. This concludes the lemma.
Lemma 2.9
Let G be a group and S a Janko group. If | G | = | S | , then G is neither a F robenius group nor a − F robenius group.
Proof. Step 1
To show the lemma follows while | G | = | J | or | J | .Noticing | J | = 2 · · · · · | J | = 2 · · · ·
19, 19 ∤ | GL (7 , | and 19 ∤ | GL (5 , | , we get byLemma 2.7 that G has no subgroups which is a F robenius group with complement of order divided by 19 if | G | = | J | or | J | . Hence, if G is a 2 − F robenius group, then 19 ∈ π ( D ), | E | (cid:12)(cid:12) | E | is odd, further, | F | (cid:12)(cid:12) | F | (cid:12)(cid:12) ( | E | − | D | = 5, which means that ED is a F robenius group of order 15 or 45, itis impossible. If G is a F robenius group, then | H | (cid:12)(cid:12) | K | = 5. Thus, H K is a F robenius group, which isimpossible.
Step 2
To show the lemma follows while | G | = | J | .(1) Let U ⋉ V be a section of F robenius type of G . Notice | G | = | J | = 2 · · ·
7. If 7 ∈ π ( V ), then U V is F robenius group, thus, | U | (cid:12)(cid:12)
6. If 7 ∈ π ( U ), by 7 ∤ GL(3 , | and 7 ∤ | GL(2 , | , we get by Lemma 2.7 that V is 2 − group.(2) Assume that G is a F robenius group. If 7 ∈ π ( K ), then | H | (cid:12)(cid:12) π ( K ) = π ( G ) and t ( G ) = 1,5 contradiction. If 7 ∈ π ( H ), then | K | = 2 , | H | = 3 · ·
7, contradicting | H | (cid:12)(cid:12) ( | K | − G is a 2 − F robenius group. If 7 (cid:12)(cid:12) | D | , then | E | (cid:12)(cid:12) | F | (cid:12)(cid:12)
2, thus, π ( D ) = π ( G ) and t ( G ) = 1, a contradiction. If 7 ∈ π ( E ), then D is a 2 − group by (1), so | F | (cid:12)(cid:12)
6. Since 3 (cid:12)(cid:12) | G | , we deduce that | E | = 3 · ·
7, contradicting | E | (cid:12)(cid:12) ( | D | − ∈ π ( F ), since EF is a F robenius group, (1) impliesthat E is a 2 − group, a contradiction. Step 3
To show the lemma follows while | G | = | J | .(1) By | G | = | J | = 2 · · · · · · · · ·
43, we get that G has no nilpotent normal subgroupof order divided by 5, otherwise, t ( G ) = 1. Moreover, by 43 ∤ GL(3 , | and 43 ∤ GL(3 , | , we get that if U ⋉ V is a section of F robenius type with 43 (cid:12)(cid:12) | U | , then V is a 2 − group.(2) Assume that G is a F robenius group. Then 5 ∈ π ( H ). Moreover, if 43 ∈ π ( K ) , then Corollary2.6 indicates that 43 · ∈ π e ( G ), this is impossible. Hence, 43 ∈ π ( H ), so | K | = 2 by (1), thus, | H | =3 · · · · · · · ·
43, contradicting | H | (cid:12)(cid:12) ( | K | − G is a 2 − F robenius group, then 5 / ∈ π ( D ) by (1). Moreover, if 43 ∈ π ( D ), then π ( E ) ⊆ { , } , and thus, 5 ∈ π ( F ). But by Lemma 2.7, EF cannot be a F robenius group, a contradiction. If43 ∈ π ( E ), then D can only be a 2 − group by (1). At this moment, F is a { , }− group for E F is a F robenius group. Hence, 5 ∈ π ( E ) and F E is a F robenius group, which is impossible for π ( F ) ⊆ { , } , a contradiction.Now 43 ∈ π ( F ), (1) indicates that E is a 2 − group, a contradiction.The lemma follows from Steps 1-3. Lemma 2.10
Let G be a group and S a Conway group. If | G | = | S | , then G is neither a F robenius group nora − F robenius group.
Proof.
From [19], we get the orders of Conway groups are the following: | Co | = 2 · · · · · · , | Co | = 2 · · · · · , | Co | = 2 · · · · · . (1) Now 2 · · divides | G | . Noticing | GL(2 , | = 2 · ·
7, we have that if G has a nilpotent normalsubgroup of order divided by 7 or 11, then t ( G ) = 1 by Corollary 2.6. Hence, G and G are not normalin G . Let U ⋉ V be a section of F robenius type, then, by 23 ∤ | GL(9 , | , 23 ∤ | GL(4 , | , 23 ∤ | GL(2 , | and11 , <
23, we get that if 23 (cid:12)(cid:12) | U | then V is a 2 − group.(2) If G is a F robenius group. Suppose that 23 ∈ π ( K ), then | H | = 11, so that 7 ∈ π ( K ), contradicting(1). Consequently, 23 ∈ π ( H ) and | K | = | G | by (1), a contradiction to | H | (cid:12)(cid:12) ( | K | − G be a 2 − F robenius | E | is odd, we have 23 ∤ | F | by (1). If 23 ∈ π ( E ), then D is a 2 − group and | F | = 11. Therefore,7 ∈ π ( E ), which yields | F | is a { , }− group, a contradiction to | F | = 11. Hence, 23 ∈ π ( D ), we deduce that | E | = 11 and | F | (cid:12)(cid:12)
10. Thus, 7 ∈ π ( D ), contradicting (1). This concludes the lemma. Lemma 2.11
Let G be a group and S a Fischer group. If | G | = | S | , then G is neither a F robenius group nora − F robenius group.
Proof.
From [19], we get the orders of Fischer groups are the following: | F i | = 2 · · · · · | F i | = 2 · · · · · · · | F i ′ | = 2 · · · · · · · · | GL(3 , | = 2 · · ·
19, we can show that G p (if there is) is not normal in G , where p = 5 , , , ,
17. Let U ⋉ V be a section of F robenius type of G ,then by π (GL(2 , { , , } and π (GL(3 , { , , , } and Corollary 2.6, we get that if 13 (cid:12)(cid:12) | U | , then V is a { , }− group.(2) Assume that G is a F robenius group. If | G | = | F i | , we have 5 , , , ∈ π ( H ), and K is a 2 − group,contradicting | H | (cid:12)(cid:12) ( | K | − | G | = | F i | or | F i ′ | , then we come to a contradiction by (1) and 13 does notdivide 2 −
1, 2 −
1, 3 − − G is a 2 − F robenius group. Then either 13 ∈ π ( E ) or 13 ∈ π ( F ) by (1). Supposethat 13 ∈ π ( E ) , since E F is a F robenius group, it follows that | F | (cid:12)(cid:12) ·
3, so that 11 ∈ π ( E ) by (1), thus, | F | = 2. Hence, | D | = | G | /
2. Note that D is a { , }− group, we have | E | = | G | = 5 . Furthermore, by5 ∤ t − , t = 9 , ,
16, we deduce that 3 ∤ | D | . Therefore, D is a 2 − group. And since DE is a F robenius group, we immediately get a contradiction by 3 ∤ k −
1, where k = 16 , ,
20. As a result, 13 ∈ π ( F ), then E can only be a 3 − group. Which implies that D is a 2 − group. By calculating 2 m − m ≤
21, we get that3 ∤ m −
1, a contradiction. This completes the proof of the lemma.
Lemma 2.12
Let G be a group and S the Monster group or the Baby group. If | G | = | S | , then G is neither a F robenius group nor a − F robenius group.
Proof.
From [19], the orders of the Monster group and the Baby group are the following: | M | = 2 · · · · · · · · · · · · · · | B | = 2 · · · · · · · · · · t ( G ) = 1 if G p is normal in G , where p = 17 , , , ,
41 or 71, hence, G has no normal subgroup with order p . Moreover, by calculating, it follows that 47 does not divide | GL(20 , | ,7 GL(9 , | , | GL(6 , | , | GL(2 , | and | GL(3 , | . Hence, if G has a section of F robenius type U ⋉ V with47 (cid:12)(cid:12) | U | , then V can only be a 2 − group.(2) Assume that G is a F robenius group. If 47 ∈ π ( K ), then | K | = 47 and K E G . So that | H | = 23,which yields 17 ∈ π ( K ), a contradiction. Therefore, 47 ∈ π ( H ), which implies | K | = | G | by (1), contradicting | H | (cid:12)(cid:12) ( | K | − G is a 2 − F robenius group. Then 17 π ( D ) by (1). If 47 ∈ π ( D ), then | E | = 23, hence, | F | (cid:12)(cid:12) ·
11, from which it follows that 17 ∈ π ( D ), a contradiction to (1). If 47 ∈ π ( E ), then | F | divides 2 · D is a 2 − group by (1). Consequently, for any p ∈ π ( G ) \ { , } , all p − elements are contained in E , so | E | ≥ · · · · · · · · > , contradicting | E | (cid:12)(cid:12) ( | D | − ∈ π ( F ), then E canonly be a 2 − group by (1) since EF is a F robenius group, a contradiction to | E | is odd. The Lemma follows. Lemma 2.13
Let G be a group and S one of Suz, HS, M c L, He, HN, T h, O ′ N, Ly and Ru . If | G | = | S | ,then G is neither a F robenius group nor a − F robenius group.
Proof.
We write the proof in five steps.
Step 1
To show the lemma follows while | G | equals one of | Suz | , | HS | and | M c L | .From [19], we get that: | Suz | = 2 · · · · · | HS | = 2 · · · · | M c L | = 2 · · · · G E G , then by | G | = 11 we can show that t ( G ) = 1, a contradiction. Hence, G G . Similarly,we get that G G . Let U ⋉ V be a section of F robenius type of G such that 11 (cid:12)(cid:12) | U | . Since 11 ∤ | GL(3 , | ,then V can only be a { , }− group.(2) Assume that G is a F robenius group. Then 11 ∈ π ( H ) by (1), so that K can only be a { , }− group.Note that | H | (cid:12)(cid:12) ( | K p | −
1) for p = 2 or 3, it is impossible by trivial calculating.(3) Assume that G is a 2 − F robenius group. Then 7 , π ( D ) by (1). Suppose that 11 ∈ π ( E ), then | F | (cid:12)(cid:12) · D is a { , }− group by (1). This indicates that 7 ∈ π ( E ). Hence, | F | = 2, thereby, 5 ∈ π ( E ).If 3 ∈ π ( D ), then, in view of the facts that 5 k | GL (7 , | and 5 (cid:12)(cid:12) | G | , there exists an element of order 5 in E commutes an element of order 3, a contradiction. So 3 ∈ π ( E ). Therefore, D is a 2 − subgroup, and thus, all the2 ′ − elements are in E , it is impossible for | E | (cid:12)(cid:12) ( | D | −
1) and | D | = | G | /
2. Now 11 ∈ π ( F ), then E is a 3 − groupby (1) and | E | odd. In this case E = G , but by checking we know that 11 ∤ | G | −
1, a contradiction to EF is a F robenius group.
Step 2
To show the lemma follows while | G | = | He | .(1) By | G | = | He | = 2 · · · ·
17, one has that | G | = 17, hence, G has no nilpotent normal subgroup8ith order divided by 17 for t ( G ) >
1. Moreover, by 17 does not divide the orders of GL (3 , GL (2 ,
5) and GL (3 , G has a section of F robenius type U ⋉ V with 17 (cid:12)(cid:12) | U | then V is a 2 − group.(2) Assume that G is a F robenius group. Then by (1) it follows that 17 ∈ π ( H ), and | K | = 2 . Hence, | H | = 3 · · ·
17, contradicting | H | (cid:12)(cid:12) ( | K | − G is a 2 − F robenius group. Then we get that17 ∈ π ( E ) or 17 ∈ π ( F ) by (1). Suppose that 17 ∈ π ( E ), then D and F are two 2 − groups by (1) and | F | (cid:12)(cid:12) | .As a result, π ( E ) = { , , , } . Contradicting | E | (cid:12)(cid:12) ( | D | − ∈ π ( F ), yielding that E is a 2 − groupby (1), a contradiction to | E | is odd. Step 3
To show the lemma follows while | G | equals one of | HN | , | T h | and | O ′ N | .(1) By [19], | HN | = 2 · · · · · | T h | = 2 · · · · · · | O ′ N | = 2 · · · · · · | G | = 19, and G is not normal in G , otherwise, t ( G ) = 1. Moreover, since 19 does not divide | GL(15 , | , | GL(10 , | , | GL(6 , | and | GL(2 , | , we get that if G has a section of F robenius type U ⋉ V with19 (cid:12)(cid:12) | U | , then | V | = 7 .(2) Assume that G is a F robenius group, then by (1) it follows that 19 (cid:12)(cid:12) | H | , K = G and | K | (cid:12)(cid:12) < | H | ,contradicting | H | (cid:12)(cid:12) ( | K | − G is a 2 − F robenius group. Then 19 ∈ π ( E ) or 19 ∈ π ( F ) by (1). If 19 ∈ π ( E ), since DE is a F robenius group, we have D is a 7 − group, further, by 19 (cid:12)(cid:12) | D | −
1, we have | D | = 7 . Therefore, | G | = | O ′ N | and | F | = 2 for | F | (cid:12)(cid:12) | E | −
1. As a result, | E | > | D | −
1, a contradiction. If 19 ∈ π ( F ), then | E | = 7 by (1) and previous argument, so | G | = | O ′ N | . Noticing that 7 − · · . Therefore, | D | = 11,which is impossible for ED is a F robenius group.
Step 4
To show the lemma follows while | G | = | Ly | .(1) In such case, | G | = | Ly | = 2 · · · · · · ·
67. Since 67 does not divide | GL(8 , | , | GL(7 , | and | GL(6 , | , we come to that G has no section of F robenius type U ⋉ V with 67 (cid:12)(cid:12) | U | .(2) Assume that G is a F robenius group. Then 67 ∈ π ( K ) by (1), hence, | H | (cid:12)(cid:12)
66. Therefore, 7 ∈ π ( K ) and | K | = 7, we get a contradiction for K H is a F robenius group.(3) Assume that G is a 2 − F robenius group. Then 67 ∈ π ( D ) by (1) and E , F are two F robenius complements. Hence, | E | (cid:12)(cid:12)
33, which concludes | F | (cid:12)(cid:12) ·
5. Thereby, 7 ∈ π ( D ), since | D | = 7, we get a contradictionfor D E is a F robenius group.
Step 5
To show the lemma follows while | G | = | Ru | .(1) In this case, | G | = | Ru | = 2 · · · · ·
29. Since 29 does not divide | GL(14 , | , | GL(3 , | and | GL(3 , | , we come to that G has no section of F robenius type U ⋉ V with 29 (cid:12)(cid:12) | U | . Noticing that | G | = 13,9e deduce that G has no nilpotent normal subgroup with order divided by 13.(2) Assume that G is a F robenius group. Then, 29 ∈ π ( K ) by (1), thus, | H | (cid:12)(cid:12)
28. Hence, 13 ∈ π ( K ), acontradiction to (1). Assume that G is a 2 − F robenius group. Then, 29 ∈ π ( D ). And thus, | E | = 7 for | E | odd, it follows that | F | (cid:12)(cid:12) ·
3. Thereby, 13 ∈ π ( D ), a contradiction to (1).This Lemma follows from Step 1-5. Corollary 2.14
Let G be a finite group and S one of 26 sporadic groups. If | G | = | S | and the prime graph of G is disconnected, then G has a normal series E H E K E G such that the following property (*):(*) H and G/K are π groups, K/H is a non-abelian simple group, H nilpotent, and | G/K | (cid:12)(cid:12) | Out(
K/H ) | . Proof.
It follows straightforward from Lemma 2.1 and Lemma 2.8 − G ∼ = S The necessity of the Main Theorem is obvious. Now we start to prove the sufficiency of The Main Theorem. Inthe following discussion, K and H always means subgroups of G in Corollary 2.14. We shall frequently considerthe possibilities of K/H by | K/H | dividing | S | by using the list of simple groups in [19]. But [19] lists onlyall the non-abelian simple groups of order less than 10 , except that L ( q ) , L ( q ) , U ( q ), L ( q ) , U ( q ) , S ( q )and G ( q ), which are stopped at orders 10 ( q ≤ , ( q ≤ , ( q ≤ ( q ≤ , ( q ≤ , ( q ≤ , ( q ≤ K/H not appeared in the listof [19] when | S | ≥ . Noticing the fact that orders of sporadic simple groups except B and M are all oforder < , | B | (cid:12)(cid:12) | M | and 10 < | M | < . Hence, it is necessary to determine two kinds of simple groups T not contained in the list of [19], one is T such that | T | (cid:12)(cid:12) | M | and | T | < ; another one is contained in the seriesof L ( q )( q > , L ( q )( q > , U ( q )( q > L ( q )( q > , U ( q )( q > , S ( q )( q >
41) and G ( q )( q > | T | dividing the order of some sporadic simple group except B and M . We have the following lemma. Lemma 2.15
Let S be a sporadic group, and T a non-abelian simple group, then(I) If S = M , | T | (cid:12)(cid:12) | S | , and T is not contained in the list of [19], then T is one of A n (26 ≤ n ≤ , L (2 ) and L (13 ) .(II) If S = M , | T | (cid:12)(cid:12) | S | , then T is contained in the list of [19]. Proof.
For a simple group T satisfying the hypothesis of (I), since 10 < | M | < , we have that | T | ≤ ,hence we have to check if there exists T of order ≤ among the series of L ( q )( q > , L ( q )( q > , U ( q )( q > L ( q )( q > , U ( q )( q > , S ( q )( q >
41) and G ( q )( q > of order greater than 10 and smaller than 10 among all non-abelian simple groups except sporadic simplegroups. In order to make the proof easy to read, we check all series of non-abelian simple groups except sporadicsimple groups. By Classification Theorem of Simple Groups, we have the following steps.(1) Assume that T ∼ = A n , an alternating group. Note that | A | < < | A | , we have n ≥
26. Onthe other hand, in view of the fact that 11 k| M | , we come to that 26 ≤ n ≤
32. By calculating the order of A n (26 ≤ n ≤ T ∼ = A n (26 ≤ n ≤ T ∼ = A n ( q ). Then | T | = q n ( n +1)2 n Q i =1 ( q i +1 − / ( n + 1 , q − | T | (cid:12)(cid:12) | M | andthe largest exponent of prime power of | M | is 46 that n ≤
9. Therefore, T can only be one of the Lie typesimple groups: A ( q )( q > A ( q )( q > A ( q )( q > A ( q )( q > A ( q )( q > A ( q )( q > A ( q )( q > A ( q )( q > A ( q )( q < T = A ( q )( q > | T | divides | M | , we deduce that one of the cases holds: q = 2 t (7 ≤ t ≤ q = 3 t (5 ≤ t ≤ q = 5 t (4 ≤ t ≤ q = 7 t (3 ≤ t ≤ q = 13 t (2 ≤ t ≤ q is either 2 or 13 .If T = A ( q )( q > q divides one of 2 , 3 , 5 and 7 . Hence, q is one of: 2 t (5 ≤ t ≤ q = 3 t (4 ≤ t ≤ q = 5 and q = 7 . Now calculating the order of these groups, for example, by using Maple,we have | T | ∤ | M | , a contradiction.By the same approach above, we can get contradictions if T = A n ( q ), where 3 ≤ n ≤ T = B n ( q )( n ≥
2) or C n ( q )( n ≥ | T | = | B n ( q ) | = | C n ( q ) | = q n n Q i =1 ( q i − / (2 , q − | T | (cid:12)(cid:12) | M | that n ≤
6. Hence, T may be one of B ( q )( q > B ( q )( q > B ( q )( q > B ( q )( q >
2) and B ( q )( q > T ∼ = B ( q )( q >
41) as an example, and the other cases can be studied by the same way. In view of the factsthat | T | (cid:12)(cid:12) | M | , q >
41 and the exponent of q is 4, by comparing the powers in | M | , we conclude that q divides2 or 3 . Hence, q = 2 t , ≤ t ≤
11 or q = 3 t , ≤ t ≤
5. Now calculating the order of each group, we getthat | M | 6≡ | T | ), a contradiction.(4) To show T = D n ( q )( n ≥ | T | = q n ( n − ( q n − n − Q i =1 ( q i − / (4 , q n − | T | (cid:12)(cid:12) | M | that n ≤
7. Therefore, T may be one of D ( q )( q > D ( q )( q > D ( q )( q > D ( q )( q > T ∼ = D ( q )( q > | T | (cid:12)(cid:12) | M | and n ≥
4, we deduce that q (cid:12)(cid:12) | T | , thus, q is a power of 2 or 3. But q >
19, hence, q = 2 r or 3 s , where r ≥ s ≥
3, which implies 2 or 3 divides | T | , a contradiction.(5) To show T = G ( q )( q > F ( q )( q > E ( q )( q > q and q divides the orders of11 ( q )( q >
3) and E ( q )( q > T = F ( q )( q > E ( q )( q > T ∼ = G ( q )( q > | T | = q ( q − q − | M | and noticing q >
25, we get that q equals one of 2 t , ≤ t ≤
7, 3 . Now calculating the orders of these groups, we get that | M | 6≡ | T | ), a contradiction.(6) To Show T = E ( q ) and E ( q ). Since q (cid:12)(cid:12) | E ( q ) | and q (cid:12)(cid:12) | E ( q ) | , we have | T | can not divide | M | .(7) To show T = A n ( q )( n ≥ T ∼ = A n ( q )( n ≥ | T | = q n ( n +1)2 n Q i =1 ( q i +1 − ( − i +1 ) / ( n +1 , q + 1). It follows from | T | (cid:12)(cid:12) | M | that n ≤
9. Therefore, T may be one of A ( q )( q > A ( q )( q > A ( q )( q > A ( q )( q > A ( q )( q > A i ( q )( i = 7 , , q >
2) and A ( q )( q < T ∼ = A ( q )( q >
32) as an example, and the other cases may be studied similarly. In view of the facts that | T | (cid:12)(cid:12) | M | and q >
32, we deduce that q may be a power of 2, 3, 5 or 7 and divides 2 , 3 , 5 or 7 . Hence, q equals one of 2 t , ≤ t ≤
15, 3 t , ≤ t ≤
6, 5 and 7 . By calculation, we get that | M | 6≡ | T | ), acontradiction.(8) To show T = B ( q ) , q = 2 m +1 ( m > T ∼ = B ( q ) , q = 2 m +1 ( m > | T | = q ( q + 1)( q − | M | , we get that q equals one of 2 t , t = 2 m + 1 , ≤ m ≤ | M | 6≡ | T | ), a contradiction.(9) To show T = D n ( q )( n ≥ T ∼ = D n ( q )( n ≥ | T | = q n ( n − ( q n + 1) n − Q i =1 ( q i − / (4 , q n + 1), hence, n ≤
7. Therefore, T may be one of D ( q )( q > D ( q )( q > D ( q )( q >
3) and D ( q )( q > T ∼ = D ( q )( q >
17) as an example, and the other cases may be dealt with similarly. If T ∼ = D ( q )( q > q (cid:12)(cid:12) | T | , so that q divides 2 or 3 by observing the powers of primes dividing | M | .Hence, q = 2 r or 3 s since q >
17, where r ≥ s ≥
3. Therefore 2 or 3 dividing | T | , a contradiction.(10) To show T = D ( q )( q > T ∼ = D ( q )( q > | T | = q ( q + q +1)( q − q − q (cid:12)(cid:12) | M | . Since q >
7, comparing the prime power of | M | , it follows that q = 2 , and T = D (8), we get that | M | 6≡ | T | ), a contradiction.(11) To show T = G ( q )( q = 3 m +1 , m > T ∼ = G ( q )( q = 3 m +1 , m > | T | = q ( q + 1)( q −
1) and q (cid:12)(cid:12) , hence, q (cid:12)(cid:12) , contradicting q ≥ .(12) To show that T = F ( q )( q = 2 m +1 ), E ( q ). Assume T ∼ = F ( q )( q = 2 m +1 ), E ( q ). Then | T | = q ( q + 1)( q − q + 1)( q − q divides 2 , thus, q = 2 , but | M | 6≡ | F (8) | ), acontradiction. Using the similar arguments as above, we can easily get a contradiction if T ∼ = E ( q ).By (1)-(12), we come to that (I) follows.Now we start to prove (II). Let S be a sporadic group, then | S | < if S = B and M ; and | S | (cid:12)(cid:12) | M | if12 = Ly and J . Hence if S = Ly and J , then T is among simple groups discribed in (I), i.e., L (2 ), L (13 )and A n (26 ≤ n ≤ | L (2 ) | , | L (13 ) | and | A n | (26 ≤ n ≤
32) can di-vide the order of any sporadic group except M , which implies that (II) follows for S = M , Ly and J . If S ∼ = Ly or J , T is one of L ( q )( q > , L ( q )( q > , U ( q )( q > L ( q )( q > , U ( q )( q > , S ( q )( q >
41) and G ( q )( q > < among these series,we can show that | T | does not divide | S | . So (II) and then Lemma 2.15 follows.Now we begin to prove G ∼ = S , we divide the whole proof into six lemmas. Lemma 2.16
Let G be a group and S one of groups: M , M , M , M , M and J . If | G | = | S | and theprime graph of G is disconnected, then G ∼ = S . Proof.
By assumptions, | G | is one of | M | , | M | , | M | , | M | , | M | and | J | , also, Γ( G ) is disconnected.Applying Corollary 2.14, G has a unique simple section K/H such that K and H satisfying (*). By Lemma2.15, K/H can only be simple groups in the list of [19]. By comparing the order of G with the orders of thesimple groups in [19], we can obtain all possibilities of K/H . We discuss step by step following what S is. Step 1 If S = M , then G ∼ = M .By | G | = | M | = 2 · · ·
11, checking simple groups in the list of [19], we get that
K/H is isomorphicto one of groups in Table 1.Table 1: The non-abelian simple groups
K/H with orders dividing | M | K/H | K/H | |
Out(
K/H ) | A · · A · · L (11) 2 · · ·
11 2 M · · ·
11 1If
K/H ∼ = A or A , then 11 ∈ π ( H ), and | H | = 11. Since H is nilpotent, we know H E G . Notethat | G/C G ( H ) | (cid:12)(cid:12) | Aut( H ) | , one has that { , , } ⊆ π ( C G ( H )). Since Γ( G ) is disconnected, we have5 / ∈ π ( C G ( H )). Consider the action on C G ( H ) by an element g of order 5 . Applying Lemma 2.4, C G ( H )has a
2, a contradiction.Now we have
K/H ∼ = M , so G ∼ = S ∼ = M by | G | = | M | , as desired.13 tep 2 If S = M , then G ∼ = M .Since | G | = | M | = 2 · · ·
11, we get by [19] that
K/H is isomorphic to one of the groups in Table 2.Table 2: The non-abelian simple groups
K/H with orders dividing | M | K/H | K/H | |
Out(
K/H ) | A · · A · · L (11) 2 · · ·
11 2 M · · ·
11 1 M · · ·
11 2If
K/H is isomorphic to one of the groups in Table 2 except M , then | ( K/H ) | (cid:12)(cid:12) and 3 ∤ | Out(K / H) | ,hence, 3 ∈ π ( H ). Clearly, | H | = 3 or 3 , also H E G . Note that | G/C G ( H ) | (cid:12)(cid:12) | Aut( H ) | , and | Aut( H ) | (cid:12)(cid:12) · , we have π ( G ) = π ( C G ( H )) , thus, Γ( G ) is connected, a contradiction.Now we have K/H ∼ = M , so that G ∼ = S ∼ = M , as desired. Step 3 If S = M , then G ∼ = M .Since | G | = | M | = 2 · · · ·
11. by the same reason, we get that
K/H is isomorphic to one of thegroups in Table 3. Table 3: The non-abelian simple groups
K/H with orders dividing | M | K/H | K/H | |
Out(
K/H ) | A · · L (2) 2 · · A · · L (8) 2 · · L (11) 2 · · ·
11 2 A · · · M · · ·
11 1 A · · · L (4) 2 · · · · M · · · ·
11 2From Table 3, we conclude that 11 ∈ π ( H ) while K/H is isomorphic to one of the groups: A , L (2), A , L (8), A , A and L (4). By using the same arguments as the second paragraph in Step 1, we also getcontradictions.If K/H ∼ = L (11) or M , then we see that 7 ∈ π ( H ) . Surely, | H | = 7 and H E G . So π ( G ) = π ( C G ( H ))for | G/C G ( H ) | (cid:12)(cid:12) | Aut( H ) | . Consequently, Γ( G ) is connected, a contradiction.Now we have K/H ∼ = M , then G ∼ = S ∼ = M , as desired. Step 4 If S = M , then G ∼ = M . 14n this case, | G | = | M | = 2 · · · · ·
23. By [19], we get that
K/H is isomorphic to one of the groupsin Table 4. Table 4: The non-abelian simple groups
K/H with orders dividing | M | K/H | K/H | |
Out(
K/H ) | A · · L (2) 2 · · A · · L (8) 2 · · L (11) 2 · · ·
11 2 A · · · L (23) 2 · · ·
23 2 M · · ·
11 1 A · · · L (4) 2 · · · · M · · · ·
11 2 M · · · · ·
23 1If
K/H is isomorphic to one of the groups in Table 4 except L (23) and M , then 23 ∈ π ( H ). Clearly, | H | = 23 and H E G. Since | G/C G ( H ) | (cid:12)(cid:12) | Aut( H ) | , it follows that { , , , , } ⊆ π ( C G ( H )) . Taking into account the fact that Γ( G ) is not connected, we have 11 / ∈ π ( C G ( H )). Now consider the actionof an element g of order 11 on C G ( H ) and applying Lemma 2.4, we see that C G ( H ) has a h g i − invariant Sylow − subgroup, which implies that 77 ∈ π e ( G ), therefore, Γ( G ) is connected, a contradiction.Suppose that K/H ∼ = L (23), we deduce that 7 ∈ π ( H ). Surely, | H | = 7 and H E G . Furthermore, weconclude that π ( G ) = π ( C G ( H )) for | G/C G ( H ) | (cid:12)(cid:12) | Aut( H ) | . Consequently, Γ( G ) is connected, a contradiction.Now we have K/H ∼ = M , then G ∼ = S ∼ = M , as desired. Step 5 If S = M or J , then G ∼ = S .In these cases, either | G | = | M | = 2 · · · · ·
23 or | G | = | J | = 2 · · ·
7. By Lemma 2.15, wecheck the list of simple groups in [19] and come to that
K/H is a simple group in Table 5 and 6, respectively.Suppose that
K/H is isomorphic to one of the groups in Table 5 and Table 6 except L (8), L (4), U (3)and S , then | ( K/H ) | (cid:12)(cid:12) and 3 ∤ | Out(K / H) | , hence, 3 ∈ π ( H ). Moreover, | H | = 3 or 3 . By using the samearguments as the second paragraph in Step 2, we can get contradictions.Suppose that
K/H ∼ = L (8) or U (3), then 5 ∈ π ( H ) and | H | = 5 or 5 . Clearly, H E G. Thus, we get that t ( G ) = 1 for | G/C G ( H ) | (cid:12)(cid:12) | Aut( H ) | and | GL(2 , | = 2 · ·
5. Consequently, Γ( G ) is connected, a contradiction.Suppose that K/H ∼ = L (4), then 11 ∈ π ( H ). Note that 5 ∤ | GL(3 , | , using the same argument as thesecond paragraph in Step 1, we also get a contradiction.Now we have K/H ∼ = S , then G ∼ = S by | G | = | K/H | , as desired.15able 5: The non-abelian simple groups K/H with orders dividing | M | K/H | K/H | |
Out(
K/H ) | A · · L (2) 2 · · A · · L (8) 2 · · L (11) 2 · · ·
11 2 A · · · U (3) 2 · · L (23) 2 · · ·
23 2 M · · ·
11 1 A · · · L (4) 2 · · · · M · · · ·
11 2 M · · · · ·
23 1 M · · · · ·
23 1Table 6: The non-abelian simple groups
K/H with orders dividing | J | K/H | K/H | |
Out(
K/H ) | A · · L (7) 2 · · A · · L (8) 2 · · A · · · U (3) 2 · · A · · · L (4) 2 · · · · J · · · Lemma 2.17
Let G be a group and S one of groups: J , J and J . If | G | = | S | and the prime graph of G isdisconnected, then G ∼ = S . Proof.
Similar to Lemma 2.16, we need consider the possibilities of the unique simple section
K/H for each S , and have the following steps. Step 1 If S = J , then G ∼ = J .If S = J , then | G | = | J | = 2 · · · · ·
19. Applying Lemma 2.15, we get by [19] that
K/H isisomorphic to one of the groups in Table 7.Table 7: The non-abelian simple groups
K/H with orders dividing | J | K/H | K/H | |
Out(
K/H ) | A · · L (2) 2 · · L (11) 2 · · ·
11 2 J · · · · ·
19 1If
K/H is isomorphic to one of A , L (2) and L (11), then 19 ∈ π ( H ) and | H | = 19. Since H E G ,16e have | G/C G ( H ) | (cid:12)(cid:12) · , thus, { , , , , } ⊆ π ( C G ( H )). The fact that Γ( G ) is disconnected yields3 π ( C G ( H )). Let a 3 − element g act on C G ( H ), then, by Lemma 2.4, C G ( H ) has a h g i − invariantSylow 5 − subgroup. Since 5 k | G | , which implies that 15 ∈ π e ( G ), therefore, Γ( G ) is connected, a contradiction.Now we have K/H ∼ = J , so G ∼ = S ∼ = J by | G | = | J | , as desired. Step 2 If S = J , then G ∼ = J .In this case, | G | = | J | = 2 · · · ·
19. Applying Lemma 2.15, we get by [19] that
K/H is isomorphicto one of the groups in Table 8.Table 8: The non-abelian simple groups with orders dividing | J | K/H | K/H | |
Out(
K/H ) | A · · A · · L (17) 2 · ·
17 2 L (19) 2 · · ·
19 2 L (16) 2 · · ·
17 4 U (2) 2 · · J · · · ·
19 2While
K/H is isomorphic to one of the groups in Table 8 except L (19) and J , we conclude that 19 ∈ π ( H ).Then we deduce that t ( G ) = 1 since | G/C G ( H ) | (cid:12)(cid:12) · , a contradiction.Suppose that K/H ∼ = L (19), then 17 ∈ π ( H ), also we get a contradiction by t ( G ) = 1 as | G/C G ( H ) | (cid:12)(cid:12) .Therefore, K/H ∼ = J , since | G | = | K/H | = | J | , then G ∼ = S ∼ = J . Step 3 If S = J , then G ∼ = J .If S = J , then | G | = | J | = 2 · · · · · · · · ·
43. Applying Lemma 2.15, we get by [19]that
K/H is isomorphic to one of the groups in Table 9.Table 9: The non-abelian simple groups
K/H of orders dividing | J | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · A · · · L (2) 2 · · L (4) 2 · · · · A · · L (32) 2 · · ·
31 5 L (8) 2 · · L (43) 2 · · · ·
43 2 L (11) 2 · · ·
11 2 M · · ·
11 2 A · · · M · · · ·
11 2 U (3) 2 · · L (2) 2 · · · · ·
31 2 L (23) 2 · · ·
23 2 M · · · · ·
23 1 M · · ·
11 1 U (11) 2 · · · ·
37 2 · L (29) 2 · · · ·
29 2 M · · · · ·
23 1 L (31) 2 · · ·
31 2 J · · · · · · · · ·
43 1If
K/H is isomorphic to one of the groups in Table 9 except L (43) and J , then 43 ∈ π ( H ) and | H | = 43.17lso H E G. Note that | G/C G ( H ) | (cid:12)(cid:12) | Aut( H ) | , we have π ( G ) r { } ⊆ π ( C G ( H )) . Since Γ( G ) is notconnected, we see that 7 / ∈ π ( C G ( H )). Considering the action on C G ( H ) by an element g of order 7. Lemma2.4 indicates that C G ( H ) has a h g i − invariant Sylow − subgroup, which implies 23 · ∈ π e ( G ), therefore,Γ( G ) is connected, a contradiction.If K/H ∼ = L (43), then 5 ∈ π ( H ) and | H | = 5. We immediately get a contradiction by t ( G ) = 1 since | G/C G ( H ) | (cid:12)(cid:12) .At last, K/H ∼ = J , then G ∼ = S ∼ = J , as desired. Lemma 2.18
Let G be a group and S one of Co , Co , Co , F i . If | G | = | S | and the prime graph of G isdisconnected, then G ∼ = S . Proof.
We write the proof upon what S is step by step. Step 1 If S = Co , then G ∼ = Co .By assumption, | G | = | Co | = 2 · · · · · ·
23. Applying Lemma 2.15 and checking [19], we getthat
K/H is isomorphic to one of the groups in Table 10.Suppose that
K/H is isomorphic to one of the groups in Table 10 except Co , Co , Co , M , M and L (23), then 23 ∈ π ( H ) and | H | = 23. Also H E G . Since | G/C G ( H ) | (cid:12)(cid:12)
22, it follows that π ( G ) \ { } ⊆ π ( C G ( H )). As Γ( G ) is disconnected, 11 π ( C G ( H )). By Lemma 2.4 and using the fact that 11 ∤ | GL (2 , | ,the action of an 11 − element g on C G ( H ) implies that C G ( H ) has a h g i − invariant Sylow − subgroup, sothat 11 · ∈ π e ( G ), which implies that t ( G ) = 1, a contradiction.Suppose that K/H is isomorphic to one of Co , Co , M , M and L (23), we deduce that 7 ∈ π ( H ) and | H | = 7 or 7 , H E G . By | G/C G ( H ) | (cid:12)(cid:12) | GL (2 , | and | GL (2 , | = 2 · ·
7, we have π ( G ) = π ( C G ( H )),hence, t ( G ) = 1, a contradiction.If K/H ∼ = Co , then G ∼ = S by | G | = | K/H | . Step 2 If S is one of Co , Co and F i , then G ∼ = S .Since | G | equals one of | Co | = 2 · · · · · | Co | = 2 · · · · ·
23 and | F i | =2 · · · · · · ·
23, by the same reason as above we get that
K/H is one of simple groups in Table11-13 respectively. 18able 10: The non-abelian simple groups
K/H of orders dividing | Co | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · G (3) 2 · · ·
13 2 L (7) 2 · · S (5) 2 · · ·
13 2 A · · L (3) 2 · · ·
13 2 L (8) 2 · · M · · · · ·
23 1 L (11) 2 · · ·
11 2 U (2) 2 · · ·
11 2 L (13) 2 · · ·
13 2 F (2) ′ · · ·
13 2 A · · · A · · · ·
11 2 L (3) 2 · ·
13 2 L (9) 2 · · · ·
13 2 U (3) 2 · · HS · · · ·
11 2 L (23) 2 · · ·
23 2 O +8 (2) 2 · · · · L (25) 2 · · ·
13 2 D · · ·
13 3 M · · ·
11 1 A · · · ·
11 2 L (27) 2 · · ·
13 2 · M · · · · ·
23 1 A · · · G (4) 2 · · · ·
13 2 M c L · · · ·
11 2 S (8) 2 · · · ·
13 2 · L (4) 2 · · · · A · · · · ·
13 2 U (2) 2 · · S (3) 2 · · · ·
13 2 Sz (8) 2 · · ·
13 3 O (3) 2 · · · ·
13 2 L (49) 2 · · · U (2) 2 · · · ·
11 2 · U (4) 2 · · ·
13 2 U (5) 2 · · · ·
13 2 M · · ·
11 2 A · · · · ·
13 2 U (5) 2 · · · · Suz · · · · ·
13 2 A · · · Co · · · · ·
23 1 L (64) 2 · · · ·
13 2 · A · · · · ·
13 2 M · · · ·
11 2 A · · · · ·
13 2 J · · · Co · · · · ·
23 1 S (2) 2 · · · F i · · · · ·
13 2 A · · · Co · · · · · ·
23 1 U (3) 2 · · · Table 11: The non-abelian simple groups
K/H of orders dividing | Co | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · A · · · L (7) 2 · · M · · · ·
11 2 A · · J · · · L (8) 2 · · S (2) 2 · · · L (11) 2 · · ·
11 2 A · · · A · · · U (3) 2 · · · U (3) 2 · · M · · · · ·
23 1 L (23) 2 · · ·
23 2 U (2) 2 · · ·
11 2 M · · ·
11 1 A · · · ·
11 2 A · · · HS · · · ·
11 2 L (4) 2 · · · · O +8 (2) 2 · · · · U (2) 2 · · A · · · ·
11 2 M · · ·
11 2 M · · · · ·
23 1 U (5) 2 · · · · M c L · · · ·
11 2 U (2) 2 · · · ·
11 2 · Co · · · · ·
23 119able 12: The non-abelian simple groups
K/H of orders dividing | Co | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · A · · · L (7) 2 · · M · · · ·
11 2 A · · J · · · L (8) 2 · · S (2) 2 · · · L (11) 2 · · ·
11 2 A · · · A · · · U (3) 2 · · · U (3) 2 · · M · · · · ·
23 1 L (23) 2 · · ·
23 2 U (2) 2 · · ·
11 2 M · · ·
11 1 A · · · ·
11 2 A · · · HS · · · ·
11 2 L (4) 2 · · · · A · · · ·
11 2 U (2) 2 · · M · · · · ·
23 1 M · · ·
11 2 M c L · · · ·
11 2 U (5) 2 · · · · Co · · · · ·
23 1Table 13: The non-abelian simple groups
K/H of orders dividing | F i | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · A · · · L (7) 2 · · U (3) 2 · · · A · · G (3) 2 · · ·
13 2 L (8) 2 · · L (3) 2 · · ·
13 2 L (11) 2 · · ·
11 2 M · · · · ·
23 1 L
13 2 · · ·
13 2 U (2) 2 · · ·
11 2 L (17) 2 · ·
17 2 F (2) ′ · · ·
13 2 A · · · A · · · ·
11 2 L (16) 2 · · ·
17 2 L (9) 2 · · · ·
13 2 L (3) 2 · ·
13 2 O +8 (2) 2 · · · · U (3) 2 · · O − (2) 2 · · · ·
17 2 L (23) 2 · · ·
23 2 S (2) 2 · · · L (25) 2 · · ·
13 2 A · · · ·
11 2 M · · ·
11 1 M · · · · ·
23 1 L (27) 2 · · ·
13 2 · G (4) 2 · · · ·
13 2 A · · · L (4) 2 · · · ·
17 2 L (4) 2 · · · · L (16) 2 · · · · ·
17 2 · U (2) 2 · · A · · · · ·
13 2 Sz (8) 2 · · ·
13 3 O (3) 2 · · · ·
13 2 M · · ·
11 2 S (3) 2 · · · ·
13 2 U (4) 2 · · ·
13 2 U (2) 2 · · · ·
11 2 · A · · · S (2) 2 · · · ·
17 1 L (64) 2 · · · ·
13 2 · Suz · · · · ·
13 2 M · · · ·
11 2 O +8 (3) 2 · · · ·
13 2 · J · · · F i · · · · ·
13 2 S (4) 2 · · ·
17 2 F i · · · · · · ·
23 1From Table 11-13, we have(1) If 23 ∤ | K/H | , then 23 ∤ | Out(K / H) | .(2) If 23 (cid:12)(cid:12) | K/H | , then K/H is isomorphic to one of: S , L (23), M and M . Moreover, 5 ∤ | Out(K / H) | .20or K/H satisfying (1), with the same arguments as in
Step 1 , we can get contradictions. For
K/H satisfying (2), if
K/H ∼ = S , then G ∼ = S , we are done. For the remain cases, observing that | L (23) | = 2 · · · | M | = 2 · · · · · | M | = 2 · · · · ·
23, and by (2) , we deduce that 5 ∈ π ( H ), | H | (cid:12)(cid:12) and H E G . It forces that π ( G ) = π ( C G ( H )) since | G/C G ( H ) | (cid:12)(cid:12) | GL (3 , | and | GL (3 , | = 2 · · ·
31, hence, t ( G ) = 1, a contradiction. Lemma 2.19
Let G be a group and S = F i or F i ′ . If | G | = | S | and the prime graph of G is disconnected,then G ∼ = S . Proof.
The proof is divided into following steps.
Step 1 If S = F i , then G ∼ = F i .In this case, | G | = | F i | = 2 · · · · ·
13. By the same reason, checking the list in [19], we get that
K/H is isomorphic to one of the groups in Table 14.Table 14: The non-abelian simple groups with orders dividing | F i | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · A · · · L (7) 2 · · U (3) 2 · · · A · · G (3) 2 · · ·
13 2 L (8) 2 · · L (3) 2 · · ·
13 2 L (11) 2 · · ·
11 2 F (2) ′ · · ·
13 2 L (13) 2 · · ·
13 2 U (2) 2 · · ·
11 2 A · · · A · · · ·
11 2 L (9) 2 · · · ·
13 2 O +8 (2) 2 · · · · L (3) 2 · ·
13 2 S (2) 2 · · · U (3) 2 · · A · · · ·
11 2 L (25) 2 · · ·
13 2 A · · · M · · ·
11 1 G (4) 2 · · · ·
13 2 L (27) 2 · · ·
13 2 · A · · · L (4) 2 · · · · A · · · · ·
13 2 U (2) 2 · · O (3) 2 · · · ·
13 2 Sz (8) 2 · · ·
13 3 S (3) 2 · · · ·
13 2 M · · ·
11 2 U (2) 2 · · · ·
11 2 · U (4) 2 · · ·
13 2 Suz · · · · ·
13 2 L (64) 2 · · · ·
13 2 · J · · · M · · · ·
11 2
F i · · · · ·
13 2Suppose that
K/H is one of the groups in Table 14 except L (25), U (4), J , A , F (2) ′ , O +8 (2), G (4), A , A , A , Suz and
F i , then 5 ∈ π ( H ) and | H | = 5 or 5 . Clearly H E G . Further, we deducecontradictions to t ( G ) = 1 since | G/C G ( H ) | (cid:12)(cid:12) | GL (2 , | and | GL (2 , | = 2 · · K/H is isomorphic to one of L (25), U (4), J , A , F (2) ′ , O +8 (2) and G (4), then 11 ∈ π ( H )and | H | = 11, also H E G . Thereby, we deduce contradictions to t ( G ) = 1 since | G/C G ( H ) | (cid:12)(cid:12) · K/H is isomorphic to A or A , then 13 ∈ π ( H ) and | H | = 13, also H E G . Similarly,we get contradictions to t ( G ) = 1 since | G/C G ( H ) | (cid:12)(cid:12) · K/H is isomorphic to A or Suz , then 3 ∈ π ( H ) and | H | = 3 or 3 , also H E G . Since | G/C G ( H ) | (cid:12)(cid:12) | GL (4 , | and | GL (4 , | = 2 · · ·
13, it follows that π ( G ) r { } ⊆ π ( C G ( H )). In view ofthe fact that Γ( G ) is disconnected, we have 13 π ( C G ( H )). Now looking at the action of a 13 − element g on C G ( H ) and applying Lemma 2.4, we get that C G ( H ) has a h g i − invariant Sylow 7 − subgroup. Therefore,13 · ∈ π e ( G ), which implies that Γ( G ) is connected, a contradiction.Now we have K/H ∼ = F i , then G ∼ = S ∼ = F i as desired. Step 2 If S = F i ′ then G ∼ = F i ′ .In this case, | G | = | F i ′ | = 2 · · · · · · · ·
29. By [19] and Lemma 2.15, we get that
K/H may be one of the groups in Table 15.Table 15: the non-abelian simple groups
K/H of orders dividing | F i ′ | K/H | K/H | | Out(
K/H ) | K/H | K/H | | Out(
K/H ) | A · · A · · · L (7) 2 · · U (3) 2 · · · A · · G (3) 2 · · ·
13 2 L (8) 2 · · L (3) 2 · · ·
13 2 L (11) 2 · · ·
11 2 M · · · · ·
23 1 L
13 2 · · ·
13 2 U (2) 2 · · ·
11 2 L (17) 2 · ·
17 2 F (2) ′ · · ·
13 2 A · · · A · · · ·
11 2 L (16) 2 · · ·
17 2 L (9) 2 · · · ·
13 2 L (3) 2 · ·
13 2 O +8 (2) 2 · · · · U (3) 2 · · O − (2) 2 · · · ·
17 2 L (23) 2 · · ·
23 2 D · · ·
13 3 L (25) 2 · · ·
13 2 A · · · ·
11 2 M · · ·
11 1 M · · · · ·
23 1 L (27) 2 · · ·
13 2 · G (4) 2 · · · ·
13 2 L (29) 2 · · · ·
29 2 L (4) 2 · · · ·
17 2 A · · · S (8) 2 · · · ·
13 2 · L (4) 2 · · · · L (16) 2 · · · · ·
17 2 · U (2) 2 · · A · · · · ·
13 2 Sz (8) 2 · · ·
13 3 O (3) 2 · · · ·
13 2 L (49) 2 · · · S (3) 2 · · · ·
13 2 U (4) 2 · · ·
13 2 U (2) 2 · · · ·
11 2 · M · · ·
11 2 A · · · · ·
13 2 A · · · S (2) 2 · · · ·
17 1 L (64) 2 · · · ·
13 2 · Suz · · · · ·
13 2 M · · · ·
11 2 O +8 (3) 2 · · · ·
13 2 · J · · · F i · · · · ·
13 2 S (4) 2 · · ·
17 2 F i · · · · · · ·
23 1 S (2) 2 · · · F i ′ · · · · · · · ·
29 222uppose that
K/H is isomorphic to one of the groups in Table 15 except
F i ′ or L (29), then 29 ∈ π ( H ).Of course, | H | = 29 and H E G . It follows from | G/C G ( H ) | (cid:12)(cid:12) · t ( G ) = 1, a contradiction.Suppose that K/H ∼ = F i ′ , then G ∼ = S ∼ = F i ′ , we are done. Now suppose that K/H ∼ = L (29). Observethat | L (29) | = 2 · · · ·
29, and | Out(L (29)) | = 2, we deduce that 11 ∈ π ( H ), | H | = 11 and H E G .Then by | G/C G ( H ) | (cid:12)(cid:12) ·
5, we get t ( G ) = 1, a contradiction. Lemma 2.20
Let G be a group and S a Monster group or Baby Monster group. If | G | = | S | and the primegraph of G is disconnected, then G ∼ = S . Proof.
We have the following two steps.
Step 1 If S = M , then G ∼ = M .Here, | G | = | M | = 2 · · · · · · · · · · · · · ·
71. By Lemma 2.15,
K/H maybe a simple group in the list of [19] and in Lemma 2.15, we list all of them in Table 16.Table 16: the non-abelian simple groups
K/H of orders dividing | M | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · Sz (32) 210 · · ·
41 5 L · · HS · · · ·
11 2 A · · J · · · ·
19 2 L · · S · · ·
74 2 L · · ·
11 2 O +8 (2) 212 · · · · L · · ·
13 2 O − · · · ·
17 2 L · ·
17 2 3 D · · ·
13 3 A · · · A
12 29 · · · ·
11 2 L · · ·
19 2 M
24 210 · · · · ·
23 1 L · · ·
17 22 G · · · ·
13 2 L · ·
13 2
McL · · · ·
11 2 U · · L · · · ·
13 22 L · · ·
23 2 L · · · ·
17 22 L · · ·
13 22 U · · · ·
17 22 M
11 24 · · ·
11 1 S · · · ·
13 2 · L · · ·
13 2 · L · · · · ·
17 23 · L · · · ·
29 2 S · · ·
41 22 L · · ·
31 2 A
13 29 · · · · ·
13 2 A · · · He · · · ·
17 2 L · · · · S · · · · ·
13 2 U · · O · · · ·
13 2 Sz (8) 26 · · ·
13 3 G · · · ·
31 1 L · · ·
31 5 L · · · ·
31 8 L · · · ·
41 2 U · · · ·
11 2 · L · · ·
47 2 U · · · ·
13 22 L · · ·
72 22 L · · · ·
31 2 U · · ·
13 22 A
14 210 · · · · ·
13 2 M
12 26 · · ·
11 2 S · · · ·
17 1 L · · · ·
59 2 L · · · · ·
31 12 U · · · Ru · · · · ·
29 1 J · · · · ·
19 1 L · · · ·
13 2 L · · · ·
71 2
Suz · · · · ·
13 2 A · · · O ′ N · · · · · ·
31 2 L · · · ·
13 2 · Co · · · · ·
23 1 L · · ·
41 23 A
15 210 · · · · ·
13 4 L · · ·
31 2 L · · · · ·
13 2 M
22 27 · · · ·
11 2 S · · · · ·
17 2 J · · · O +8 (3) 212 · · · ·
13 23 · L · · · ·
31 2 · O − · · · · ·
41 22 S · · ·
17 22 A
16 214 · · · · ·
13 2 S · · · O +10(2) 220 · · · · ·
31 2 A
10 27 · · · O − · · · · ·
17 2 L · · ·
19 2 · U · · · · ·
19 6 U · · · Co · · · · ·
23 1 G · · ·
13 2 L · · · · ·
41 20 S · · ·
13 2 U · · · · ·
19 6 U · · ·
19 2 Fi
22 217 · · · · ·
13 2 L · · ·
13 22 A
17 214 · · · · · ·
17 2 L · · · ·
31 2 S · · · · ·
31 2 M
23 27 · · · · ·
23 1 O · · · · ·
31 2 U · · ·
11 2 L · · · · · ·
31 22 able 16: (continued) K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | F ′ · · ·
13 2 HN · · · · ·
19 2 A
11 27 · · · ·
11 2 A
18 215 · · · · · ·
17 2 F · · · · ·
17 2 L · · · · · · ·
31 12 L · · · · ·
132 22 A
22 218 · · · · · · ·
19 2 S · · · · · ·
31 1 U · · · · · · ·
41 4 A
19 215 · · · · · · ·
19 2 O +10(3) 215 · · · · · ·
41 22 S · · · · ·
41 2 A
23 218 · · · · · · · ·
23 2 O · · · · ·
41 2 2 E · · · · · · ·
19 6 Th · · · · · ·
41 1 S · · · · · · ·
31 1 A
20 217 · · · · · · ·
19 2 A
24 221 · · · · · · · ·
23 2 Fi
23 218 · · · · · · ·
23 1 Fi ′
24 221 · · · · · · · ·
29 2 Co · · · · · ·
23 1 A
25 221 · · · · · · · ·
23 2 A
21 217 · · · · · · ·
19 2 L · · · · ·
17 22 O +12(2) 230 · · · · · ·
31 2 L · · · · ·
41 2 · O − · · · · · · ·
31 2 A
26 222 · · · · · · · ·
23 2 A
27 222 · · · · · · · ·
23 2 A
28 224 · · · · · · · ·
23 2 A
29 224 · · · · · · · · ·
29 2 A
30 225 · · · · · · · · ·
29 2 A
31 225 · · · · · · · · · ·
31 2 A
32 230 · · · · · · · · ·
29 2 M · · · · · · · · · · · · · ·
71 1 B · · · · · · · · · ·
47 1
Since all groups in Table 16 except L (71) and M have their orders and their automorphism group ordersnot divided by 71, we have 71 ∈ π ( H ). Hence, | H | = 71 and H E G . Thereby, we get a contradiction to t ( G ) = 1 since | G/C G ( H ) | (cid:12)(cid:12) · · K/H ∼ = M , then G ∼ = S ∼ = M , we are done. Assume that K/H ∼ = L (71), since | L (71) | =2 · · · ·
71 and | Out(L (71)) | = 2, we deduce that 19 ∈ π ( H ). Similarly, we get a contradiction to t ( G ) = 1for | G/C G ( H ) | (cid:12)(cid:12) · . Step 2 If S = B , then G ∼ = B .In this case, | G | = | B | = 2 · · · · · · · · · ·
47. By Lemma 2.15,
K/H is a group inthe list of [19], which are groups in Table 17.Table 17: The non-abelian simple groups
K/H of orders dividing | B | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · U · · ·
11 2 L · · HS · · · ·
11 2 A · · J · · · ·
19 2 L · · L · · · · · ·
31 22 L · · ·
11 2 O +8 (2) 212 · · · · L · · ·
13 2 O − · · · ·
17 2 L · ·
17 2 3 D · · ·
13 3 A · · · A
12 29 · · · ·
11 2 L · · ·
19 2 M
24 210 · · · · ·
23 1 L · · ·
17 22 G · · · ·
13 2 L · ·
13 2
McL · · · ·
11 2 U · · L · · · ·
13 22 L · · ·
23 2 L · · · ·
17 22 L · · ·
13 22 U · · · ·
17 22 M
11 24 · · ·
11 1 S · · · ·
13 2 · L · · ·
13 2 · L · · · · ·
17 23 · L · · · ·
29 2 A
13 29 · · · · ·
13 2 L · · ·
31 2 S · · · · ·
13 2 A · · · O · · · ·
13 2 L · · · · G · · · ·
31 1 U · · L · · · ·
31 8 Sz (8) 26 · · ·
13 3 U · · · ·
11 2 · L · · ·
31 5 L · · · ·
31 2 L · · ·
47 2 A
14 210 · · · · ·
13 2 U · · · ·
13 22 S · · · ·
17 1 L · · ·
72 22 U · · ·
13 22 M
12 26 · · ·
11 2 L · · · · ·
31 12 U · · · J · · · · ·
19 1
Suz · · · · ·
13 2 A · · · L · · · ·
13 2 · Co · · · · ·
23 1 A
15 210 · · · · ·
13 4 L · · · · ·
13 2 L · · ·
31 2 S · · · · ·
17 2 M
22 27 · · · ·
11 2 O +8 (3) 212 · · · ·
13 23 · J · · · L · · · ·
31 2 · S · · ·
17 22 A
16 214 · · · · ·
13 2 able 17: (continued) K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | S · · · O +10(2) 220 · · · · ·
31 2 A
10 27 · · · O − · · · · ·
17 2 L · · ·
19 2 · G · · ·
13 2 U · · · Co · · · · ·
23 1 S · · ·
13 2 L · · · ·
31 2 U · · ·
19 2 Fi
22 217 · · · · ·
13 2 L · · ·
13 22 A
17 214 · · · · · ·
17 2 M
23 27 · · · · ·
23 1 S · · · · · ·
31 12 F ′ · · ·
13 2 HN · · · · ·
19 2 A
11 27 · · · ·
11 2 A
18 215 · · · · · ·
17 2 F · · · · ·
17 2 L · · · · · · ·
31 12 A
19 215 · · · · · · ·
19 2 2 E · · · · · · ·
19 6 S · · · · · · ·
31 1 A
20 217 · · · · · · ·
19 2 Fi
23 218 · · · · · · ·
23 1 Co · · · · · ·
23 1 O − · · · · · · ·
31 2 O +12(2) 230 · · · · · ·
31 2 B · · · · · · · · · ·
47 1
Suppose that
K/H is isomorphic to one of the groups in Table 17 except L (47) and B , then 47 ∈ π ( H ).Clearly, | H | = 47 and H E G . It follows from | G/C G ( H ) | (cid:12)(cid:12) ·
23 that π ( G ) r { } ⊆ π ( C G ( H )). Andsince Γ( G ) is disconnected, we have 23 π ( C G ( H )). Considering the action of a 23 − element g on C G ( H ),because 13 ∈ π ( C G ( H )), in view of Lemma 2.4 and 2.5, we deduce that 23 · ∈ π e ( G ). This implies thatΓ( G ) is connected, a contradiction.Suppose that K/H ∼ = B , then G ∼ = S ∼ = B , we are done. If K/H ∼ = L (47). Since | L (47) | = 2 · · ·
47 and | Out(L (47)) | = 2, we see that 13 ∈ π ( H ). Similarly, we get a contradiction to t ( G ) = 1 by | G/C G ( H ) | (cid:12)(cid:12) · Lemma 2.21
Let G be a group and S one of the groups: Ly , O ′ N , M c L , T h , HN , He , Ru , HS , and Suz .If | G | = | S | and the prime graph of G is disconnected, then G ∼ = S . Proof.
We prove the lemma upon what S is one by one. Step 1 If S = Ly , then G ∼ = Ly .In this case, | G | = | Ly | = 2 · · · · · · · K/H is one of the groups in Table 18 by [19] andLemma 2.15. Table 18: The non-abelian simple groups
K/H of orders dividing | Ly | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · J · · · L (7) 2 · · L (125) 2 · · · ·
31 6 A · · A · · · L (8) 2 · · U (3) 2 · · · L (11) 2 · · ·
11 2 A · · · ·
11 2 A · · · M c L · · · ·
11 2 U (3) 2 · · G (5) 2 · · · ·
31 1 M · · ·
11 2 L (31) 2 · · ·
31 2 A · · · L (4) 2 · · · U (2) 2 · · L (32) 2 · · ·
31 5 M · · ·
11 2 U (5) 2 · · · A · · · L (5) 2 · · ·
31 2 M · · · ·
11 2 Ly · · · · · · ·
67 125uppose that
K/H is isomorphic to one of the groups in Table 18 except Ly , then 67 ∈ π ( H ). Surely, | H | = 67 and H E G . It follows from | G/C G ( H ) | (cid:12)(cid:12) · ·
11 that π ( G ) r { } ⊆ π ( C G ( H )). Since Γ( G ) isdisconnected , we see that 11 π ( C G ( H )). Considering the action of a 11 − element g on C G ( H ), because31 ∈ π ( C G ( H )) and using Lemma 2.4 and 2.5, we deduce that 11 · ∈ π e ( G ), which implies that Γ( G ) isconnected, a contradiction. Hence, K/H ∼ = Ly . Therefore, G ∼ = S ∼ = Ly , as desired. Step 2 If S = O ′ N , then G ∼ = O ′ N .In this case, | G | = | O ′ N | = 2 · · · · · ·
31. Then
K/H is one of the groups in Table 19 by [19]and Lemma 2.15. Table 19: The non-abelian simple groups K/H of orders dividing | O ′ N | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · A · · · L (7) 2 · · L (4) 2 · · · · A · · U (2) 2 · · L (8) 2 · · L (32) 2 · · ·
31 5 L (11) 2 · · ·
11 2 M · · ·
11 2 A · · · A · · · L (19) 2 · · ·
19 2 M · · · ·
11 2 U (3) 2 · · S (2) 2 · · · M · · ·
11 1 U (8) 2 · · ·
19 18 L (31) 2 · · ·
31 2 O ′ N · · · · · ·
31 2Suppose that
K/H is isomorphic to one of the groups in Table 19 except O ′ N , L (31) and L (32), then31 ∈ π ( H ). Of course, | H | = 31, H E G . It follows from | G/C G ( H ) | (cid:12)(cid:12) · · π ( G ) r { } ⊆ π ( C G ( H )).Further, since Γ( G ) is disconnected , we have 5 π ( C G ( H )). Now considering the action of a 5 − element g on C G ( H ), and using the fact that 19 ∈ π ( C G ( H )). Lemma 2.4 and 2.5 yield that 5 · ∈ π e ( G ), which impliesthat Γ( G ) is connected, a contradiction.Suppose that K/H ∼ = L (31) or L (32), then 19 ∈ π ( H ) and | H | = 19, also H E G . Hence, we getcontradictions to t ( G ) = 1 as | G/C G ( H ) | (cid:12)(cid:12) · .Now we have K/H ∼ = O ′ N , then G ∼ = S ∼ = O ′ N , as desired. Step 3 If S = M c L , then G ∼ = M c L .By | G | = | M c L | = 2 · · · ·
11 and Lemma 2.15, we get that
K/H is a simple group in the list of [19],which are groups in Table 20. 26able 20: The non-abelian simple groups
K/H orders dividing | M c L | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · U (2) 2 · · L (7) 2 · · M · · ·
11 2 A · · U (5) 2 · · · · L (8) 2 · · A · · · L (11) 2 · · ·
11 2 M · · · ·
11 2 A · · · J · · · U (3) 2 · · A · · · M · · ·
11 1 U (3) 2 · · · A · · · A · · · ·
11 2 L (4) 2 · · · · M c L · · · ·
11 2Suppose that
K/H is isomorphic to one of the groups in in Table 19 except M c L , A , M , M , M and L (11). We conclude that 11 ∈ π ( H ). Surely, | H | = 11 and H E G . Hence, we come to contradictions to t ( G ) = 1 for | G/C G ( H ) | (cid:12)(cid:12) · K/H is isomorphic to one of M c L , A , M , M , M and L (11). If K/H ∼ = M c L ,then G ∼ = S ∼ = M c L , as desired. For the remaining cases, it always follows that 5 ∈ π ( H ). Here | H | = 5or 5 and H E G . Consequently, we can deduce contradictions to t ( G ) = 1 by | G/C G ( H ) | (cid:12)(cid:12) | GL(2 , | and | GL(2 , | = 2 · · Step 4 If S = T h , then G ∼ = T h .By | G | = | T h | = 2 · · · · · ·
31 and Lemma 2.15, we get that
K/H is in the list of [19], whichare in Table 21. Table 21: The non-abelian simple groups
K/H of orders dividing | T h | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · L (5) 2 · · ·
31 2 L (7) 2 · · J · · · A · · L (125) 2 · · · ·
31 6 L (8) 2 · · A · · · L (13) 2 · · ·
13 2 U (3) 2 · · · A · · · G (3) 2 · · ·
13 2 L (3) 2 · ·
13 2 U (8) 2 · · ·
19 18 U (3) 2 · · ·
13 2 L (3) 2 · · ·
13 2 L (25) 2 · · ·
13 2 L (2) 2 · · · ·
31 2 L (27) 2 · · ·
13 6 F (2) ′ · · ·
13 2 A · · · L (9) 2 · · · ·
13 2 L (4) 2 · · · · O +8 (2) 2 · · · U (2) 2 · · D (2) 2 · · ·
13 3 Sz (8) 2 · · ·
13 3 G (4) 2 · · · ·
13 2 L (32) 2 · · ·
31 5 S (8) 2 · · · ·
13 6 L (49) 2 · · · S (3) 2 · · · ·
13 2 U (4) 2 · · ·
13 4 O (3) 2 · · · ·
13 2 U (5) 2 · · · · L (2) 2 · · · ·
31 227able 21: (continued)
K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · · T h · · · · · ·
31 1 L (64) 2 · · · ·
13 6 L (19) 2 · · ·
19 2Suppose that
K/H is isomorphic to one of the groups in Table 21 except
T h , L (19) and U (8), then19 ∈ π ( H ). Surely, | H | = 19 and H E G . Suppose that K/H is one of
T h , L (19) and U (8). If K/H ∼ = T h ,then G ∼ = S ∼ = T h , as desired. For the remain cases, we have 13 ∈ π ( H ), | H | = 13 and H E G . Therefore,we can get contradictions to t ( G ) = 1 by | G/C G ( H ) | (cid:12)(cid:12) · | G/C G ( H ) | (cid:12)(cid:12) · , respectively. Step 5 If S = HN , then G ∼ = HN .By | G | = | HN | = 2 · · · · ·
19, Lemma 2.15 and [19], we have,
K/H is one of groups in Table 22.Table 22: The non-abelian simple groups
K/H of orders dividing | HN | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · A · · · L (7) 2 · · M · · · ·
11 2 A · · J · · · L (8) 2 · · S (2) 2 · · · L (11) 2 · · ·
11 2 A · · · A · · · U (3) 2 · · · L (19) 2 · · ·
19 2 U (8) 2 · · ·
19 18 U (3) 2 · · U (2) 2 · · ·
11 2 M · · ·
11 1 A · · · ·
11 2 A · · · HS · · · ·
11 2 L (4) 2 · · · · M c L · · · ·
11 2 U (2) 2 · · O +8 (2) 2 · · · M · · ·
11 2 A · · · ·
11 2 U (5) 2 · · · · M c L · · · ·
11 2 J · · · · ·
19 1 HN · · · · ·
19 2Suppose that
K/H is isomorphic to one of the groups in Table 22 except HN , L (19), J and U (8),then 19 ∈ π ( H ). Surely, | H | = 19 and H E G . Further, we deduce contradictions to t ( G ) = 1 since | G/C G ( H ) | (cid:12)(cid:12) · .Suppose that K/H is isomorphic to one of HN , L (19), J and U (8). If K/H ∼ = HN , then G ∼ = S ∼ = HN ,as desired. For the remain cases, we can argue as follows.Assume that K/H ∼ = L (19) or U (8). Since | L (19) | = 2 · · · | Out(L (19)) | = 2; | U (8) | = 2 · · · | Out(U (8)) | = 2 · .We conclude that 11 ∈ π ( H ) and | H | = 11, also H E G . Further, we deduce contradictions to t ( G ) = 1 as | G/C G ( H ) | (cid:12)(cid:12) ·
5. Now assume that
K/H ∼ = J , notice that | J | = 2 · · · · · | Out(J ) | = 2. We see28hat 5 ∈ π ( H ) and | H | = 5 , H E G . Since | G/C G ( H ) | (cid:12)(cid:12) | GL (5 , | and | GL (5 , | = 2 · · · · · · { , , , } ⊆ π ( C G ( H )). In view of the fact that Γ( G ) is disconnected, we see that 11 π ( C G ( H )). Considering the action of a 11 − element g on C G ( H ), we obtain that C G ( H ) has a h g i − invariant Sylow − subgroup, so 11 · ∈ π e ( G ), which implies that t ( G ) = 1, a contradiction. Step 6 If S = He , then G ∼ = He .In this case, | G | = | He | = 2 · · · ·
17. By [19] and Lemma 2.15, we get that
K/H is a group inTable 23. Table 23: The non-abelian simple groups
K/H of orders dividing | He | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · U (3) 2 · · L (7) 2 · · A · · · A · · L (4) 2 · · · · L (8) 2 · · L (49) 2 · · · L (17) 2 · ·
17 2 J · · · A · · · S (4) 2 · · ·
17 2 L (16) 2 · ·
17 4 He · · · ·
17 2Suppose that
K/H is isomorphic to one of the groups in Table 23 except He , L (17), L (16), S (4), then17 ∈ π ( H ). Clearly, | H | = 17 and H E G . Hence, we get contradictions to t ( G ) = 1 by | G/C G ( H ) | (cid:12)(cid:12) .Suppose that K/H ∼ = He , L (17), L (16), S (4). If K/H ∼ = He , then G ∼ = S ∼ = He , as desired. For theremaining cases, we can argue as follows.Observe that | L (17) | = 2 · · | Out(L (17)) = 2; | L (16) | = 2 · · · | Out(L (16)) | = 4; | S (4) | = 2 · · · | Out(S (4)) = 4. We deduce that 7 ∈ π ( H ). Clearly | H | = 7 also H E G . Since | G/C G ( H ) | (cid:12)(cid:12) | GL (3 , | and | GL (3 , | = 2 · · ·
19, we conclude that { , , } ⊆ π ( C G ( H )). Because t ( G ) >
1, we have 3 π ( C G ( H )). Now considering the action of an element g of order 3 on C G ( H ) and usingLemma 2.4 and 2.5, we deduce that C G ( H ) has a h g i − invariant Sylow − subgroup, therefore, 3 · ∈ π e ( G ),which implies that t ( G ) = 1, a contradiction. Step 7 If S = Ru , then S ∼ = Ru .By | G | = | Ru | = 2 · · · · ·
29, Lemma 2.15 and [19], we get that
K/H is a group in Table 24.29able 24: The non-abelian simple groups
K/H of orders dividing | Ru | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · L (29) 2 · · · ·
29 2 L (7) 2 · · A · · · A · · L (4) 2 · · · · L (8) 2 · · S z (8) 2 · · ·
13 3 L (13) 2 · · ·
13 2 U (4) 2 · · ·
13 4 A · · · U (5) 2 · · · L (3) 2 · ·
13 2 L (64) 2 · · · ·
13 6 U (3) 2 · · J · · · L (25) 2 · · ·
13 4 F (2) ′ · · ·
13 2 L (27) 2 · · ·
13 6 G (4) 2 · · · ·
13 2 Ru · · · · ·
29 1Suppose that
K/H is isomorphic to one of the groups in Table 24 except Ru and L (29), then 29 ∈ π ( H ), | H | = 29 and H E G . Since | G/C G ( H ) | (cid:12)(cid:12) ·
7, it follows that { , , , } ⊆ π ( C G ( H )). Because t ( G ) > π ( C G ( H )). Viewing the action of an element g of order 7 on C G ( H ) and using Lemma 2.4 and2.5, we deduce that C G ( H ) has a h g i − invariant Sylow − subgroup. Hence, 7 · ∈ π e ( G ), so that t ( G ) = 1,a contradiction.Suppose that K/H ∼ = Ru or L (29). If K/H ∼ = Ru , then G ∼ = S ∼ = Ru , we are done. Assume that K/H ∼ = L (29). Because | L (29) | = 2 · · · · | Out(L (29)) | = 2, we have that 13 ∈ π ( H ) and | H | = 13,and H E G . Consequently, we deduce a contradiction to t ( G ) = 1 from | G/C G ( H ) | (cid:12)(cid:12) · Step 8 If S = HS , then G ∼ = HS .In this case, | G | = | HS | = 2 · · · ·
11. Applying Lemma 2.15, we get from [19] that
K/H is one ofthe groups in Table 25.Table 25: the non-abelian simple groups
K/H of orders dividing | HS | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · M · · ·
11 1 L (7) 2 · · A · · · A · · L (4) 2 · · · · L (8) 2 · · U (5) 2 · · · · L (11) 2 · · ·
11 2 M · · · ·
11 2 A · · · HS · · · ·
11 2Suppose that
K/H is isomorphic to one of the groups in Table 25 except HS , M , M and L (11),then 11 ∈ π ( H ) and | H | = 11. Since H E G , we have | G/C G ( H ) | (cid:12)(cid:12) ·
5, which implies that t ( G ) = 1, acontradiction.Suppose that K/H is isomorphic to one of M , M and L (11), then 5 ∈ π ( H ) and | H | = 5 . Since30 E G , we have | G/C G ( H ) | (cid:12)(cid:12) | GL (2 , | . Notice | GL (2 , | = 2 · ·
5, we come to π ( C G ( H )) = π ( G ), acontradiction to t ( G ) = 1.Therefore K/H ∼ = HS , then G ∼ = S ∼ = HS , as desired. Step 9 If S = Suz , then G ∼ = Suz .In this case, | G | = | Suz | = 2 · · · · ·
13. By the same reason, we get that
K/H is isomorphic to agroup in Table 26. Table 26: The non-abelian simple groups
K/H of orders dividing | Suz | K/H | K/H | |
Out(
K/H ) | K/H | K/H | |
Out(
K/H ) | A · · M · · · ·
11 2 A · · U (2) 2 · · ·
11 2 U (2) 2 · · A · · · ·
11 2 L (7) 2 · · A · · · ·
11 2 L (8) 2 · · L (13) 2 · · ·
13 2 A · · · L (3) 2 · ·
13 2 U (3) 2 · · L (25) 2 · · ·
13 2 A · · · L (27) 2 · · ·
13 2 · L (4) 2 · · · · Sz (8) 2 · · ·
13 3 A · · · U (4) 2 · · ·
13 2 J · · · L (64) 2 · · · ·
13 2 · S (2) 2 · · · G (3) 2 · · ·
13 2 A · · · L (3) 2 · · ·
13 2 U (3) 2 · · · F (2) ′ · · ·
13 2 O +8 (2) 2 · · · · L (9) 2 · · · ·
13 2 L (11) 2 · · ·
11 2 G (4) 2 · · · ·
13 2 M · · ·
11 1 A · · · · ·
13 2 M · · ·
11 2
Suz · · · · ·
13 2Suppose that
K/H is isomorphic to one of the groups in Table 26 except L (13), L (3), L (25), L (27), Sz (8), U (4), L (64), G (3), L (3), F (2) ′ , L (9), G (4), A and Suz , then 13 ∈ π ( H ) , | H | = 13 and H E G . Consequently, we get contradictions to t ( G ) = 1 by | G/C G ( H ) | (cid:12)(cid:12) · K/H is isomorphic to one of L (13), L (3), L (25), L (27), Sz (8), U (4), L (64), G (3), L (3), F (2) ′ , L (9), G (4), then 11 ∈ π ( H ), | H | = 11, and H E G . Hence, | G/C G ( H ) | (cid:12)(cid:12) ·
5, which implies t ( G ) = 1, a contradictions.Suppose that K/H ∼ = A , then 3 ∈ π ( H ) and | H | = 3 . Since H E G , similarly we get contradictions to t ( G ) = 1 from | G/C G ( H ) | (cid:12)(cid:12) · K/H ∼ = Suz , then G ∼ = S ∼ = Suz , as desired. The lemma holds by Step 1-9.
Proof.
The Main Theorem follows from Lemma 2.8-2.13 and 2.16 − eferences [1] Mazurov, V. D., Khukhro, E. I., Unsolved problems in group theory, Russian Academy of Sciences, Instituteof Mathematics, Novosibirsk. 17 (2010), no. 18, 171-180.[2] Williams, J. S., Prime graph components of finite groups, J. Algebra. 69 (1981), no. 2, 487-513.[3] Kondrat’ev, A. S., On prime graph components of finite simple groups, Mat. Sb. 180 (1989), no. 6, 787-797.[4] Iiyori, N., Yamaki, H., Prime graph components of the simple groups of Lie type over the field of evencharacteristic, J. Algebra. 155 (1993), no. 2, 335-343.[5] Suzuki, M., On the prime graph of a finite simple group-an application of the method of Feit-Thompson-Bender-Glauberman. Groups and combinatorics-in memory of Michio Suzuki, Adv. Stud. Pure Math. 32(2001), 41-207.[6] Khosravi, A., Khosravi, B., A new characterization of PSL( p, q ), Comm. Algebra. 32 (2004), no. 6, 2325-2339.[7] Chen, G. Y., Characterization of D ( q ), Southeast Asian Bull. Math. 25 (2001), no. 3, 389-401.[8] Darafsheh, M. R., Characterizability of the group D p (3) by its order components, where p ≥ m + 1, Acta Math. Sin. 24 (2008), no. 7, 1117-1126.[9] Khosravi, B., Khosravi, B., Khosravi, B., Characterizability of PSL( p + 1 , q ) by its order component(s),Houston J. Math. 32 (2006), no. 3, 683-700.[10] Iranmanesh, A., Khosravi, B., A characterization of C ( q ) where q >
5, Comment. Math. Univ. Carolin.43 (2002), no. 1, 9-21.[11] Nosratpour, P., Darafsheh, M. R., Characterization of G ( q ), where 2 < q ≡ L ( q ) by the largest element orders, Math. Rep.(Bucur.) 17 (2015), no. 4, 353-358.[13] He, L.G., Chen, G.Y., A new characterization of simple K -groups, Comm. Algebra. 40 (2012), 3903-3911.3214] Jiang, Q. H., Shao, C. G., Shi , W. J., Zhang, Q. L., A new characterization of L ( q ) by largest elementorders, Bull. Iranian Math. Soc. 43 (2017), No. 5, 1143-1151.[15] Chen, G. Y., Characterization of alternating groups by the set of orders of maximal abelian subgroups,Sib. Math. J. 47 (2006), no. 3, 594-596.[16] Li L., Chen G. Y., A new characterization of the simple group A ( p n ), Sib. Math. J. 53 (2012), no. 2,243-247.[17] Wang, Z. B., He, L. G and Chen, G. Y, An ONC-Characterization of A and A15