summary:Let $G$ be a finite group, and let $N(G)$ be the set of conjugacy class sizes of $G$. By Thompson's conjecture, if $L$ is a finite non-abelian simple group, $G$ is a finite group with a trivial center, and $N(G)=N(L)$, then $L $ and $G$ are isomorphic. Recently, Chen et al.\ contributed interestingly to Thompson's conjecture under a weak condition. They only used the group order and one or two special conjugacy class sizes of simple groups and characterized successfully sporadic simple groups (see Li's PhD dissertation). In this article, we investigate validity of Thompson's conjecture under a weak condition for the alternating groups of degrees $p+1$ and $p+2$, where $p$ is a prime number. This work implies that Thompson's conjecture holds for the alternating groups of degree $p+1$ and $p+2$

Khalili Asboei, Alireza

Mohammadyari, Reza

Institute of Mathematics AS CR, v. v. i.

Czechoslovak Mathematical JournalAlireza Khalili Asboei; Reza MohammadyariCharacterization of the alternating groups by their order and one conjugacy class lengthCzechoslovak Mathematical Journal, Vol. 66 (2016), No. 1, 63–70Persistent URL: http://dml.cz/dmlcz/144875Terms of use:© Institute of Mathematics AS CR, 2016Institute of Mathematics of the Czech Academy of Sciences provides access to digitizeddocuments strictly for personal use. Each copy of any part of this document must contain theseTerms of use.This document has been digitized, optimized for electronic delivery andstamped with digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://dml.czCzechoslovak Mathematical Journal, 66 (141) (2016), 63–70CHARACTERIZATION OF THE ALTERNATING GROUPSBY THEIR ORDER AND ONE CONJUGACY CLASS LENGTHAlireza Khalili Asboei, Sari, Buin Zahra,Reza Mohammadyari, Buin Zahra(Received December 9, 2014)Abstract. Let G be a finite group, and let N(G) be the set of conjugacy class sizes of G.By Thompson’s conjecture, if L is a finite non-abelian simple group, G is a finite groupwith a trivial center, and N(G) = N(L), then L and G are isomorphic. Recently, Chenet al. contributed interestingly to Thompson’s conjecture under a weak condition. Theyonly used the group order and one or two special conjugacy class sizes of simple groupsand characterized successfully sporadic simple groups (see Li’s PhD dissertation). In thisarticle, we investigate validity of Thompson’s conjecture under a weak condition for thealternating groups of degrees p+1 and p+2, where p is a prime number. This work impliesthat Thompson’s conjecture holds for the alternating groups of degree p+ 1 and p+ 2.Keywords: finite simple group; conjugacy class size; prime graph; Thompson’s conjectureMSC 2010 : 20D08, 20D60, 20D061. IntroductionIf n is an integer, then we denote by π(n) the set of all prime divisors of n. LetG be a finite group. Denote by π(G) the set of primes p such that G contains anelement of order p. The set of conjugacy class sizes of G is denoted by N(G).We construct the prime graph of G, denoted by Γ(G), as follows: the vertex setis π(G) and two distinct vertices p and p′ are joined by an edge if and only if G hasan element of order pp′. Let t(G) be the number of connected components of Γ(G)and let π1, π2, . . . , πt(G) be the connected components of Γ(G). If 2 ∈ π(G), then wealways suppose 2 ∈ π1. Some of the characterizations by prime graph can be foundin [8], [16].We can express |G| as a product of integers m1,m2, . . . ,mt(G), where π(mi) = πifor each i. The numbers mi are called the order components of G. In particular,63if mi is odd, then we call it an odd order component of G. Write OC(G) for theset {m1,m2, . . . ,mt(G)} of order components of G and T (G) for the set of connectedcomponents of G. For related results, in [10] the authors show that L3(q), is deter-mined up to isomorphism by order components. Similar characterizations have beenfound in [12], [11]. According to the classification theorem of finite simple groupsand [13], [17], [9], we can list the order components of finite simple groups withdisconnected prime graphs as in Tables 1–3 in [10].In 1980, Thompson posed the following conjecture (see [15], Question 12.38).Thompson’s conjecture: If L is a finite non-abelian simple group, G is a finitegroup with a trivial center, and N(G) = N(L), then L and G are isomorphic.In 1994, Chen proved in his PhD dissertation [4] that Thompson’s conjectureholds for all simple groups with a non-connected prime graph (also see [3], [2]).Recently, Chen and Li contributed interestingly to Thompson’s conjecture undera weak condition. They only used the group order and one or two special conjugacyclass sizes of simple groups and characterized successfully sporadic simple groups(see Li’s PhD dissertation [14]) and simple K3-groups (a finite simple group is calleda simpleKn-group if its order is divisible by exactly n distinct primes), by which theyverified Thompson’s conjecture for sporadic simple groups and simple K3-groups.Also Chen et al. in [5] verified Thompson’s conjecture for the projective special lineargroup L2(p) by its order and one special conjugacy class size, where p is a prime.In [1], we have verified Thompson’s conjecture for the alternating groups of degree pby its order and one special conjugacy class size, where p is a prime. Hence, it is aninteresting topic to characterize simple groups with their orders and few conjugacyclass sizes. In this paper, we characterize the alternating groups of degrees p+1 andp+ 2 by their order and one special conjugacy class length, where p is a prime. OurMain theorem is as follows.Main theorem. Let G be a group. Then(a) G ∼= Altp+1 if and only if |G| = (p+ 1)!/2 and G has a conjugacy class of length(p+ 1)!/2p, where 7 6= p > 5 is a prime number. For the case p = 7, G ∼= Alt8or L3(4) if and only if |G| = 8!/2 and G has a conjugacy class of length 2 880.(b) G ∼= Altp+2 if and only if |G| = (p+ 2)!/2 and G has a conjugacy class of length(p+ 2)!/2p, where p > 5 is a prime number.We write pk ‖ m if pk | m and pk+1 ∤ m. The other notation and terminology inthis paper are standard, and the reader is referred to [6] if necessary.642. Preliminary resultsFor the proof of the Main theorem we need the following lemmas.Lemma 2.1 ([7], Theorem 10.3.1). Let G be a Frobenius group with Frobeniuskernel H and Frobenius complement K. Then |K| ≡ 1 (mod |H |).Lemma 2.2 ([2], Lemma 8). Let G be a finite group with t(G) > 2, and Na normal subgroup of G. If N is a πi-group for some prime graph component of G,and µ1, µ2, . . . , µr are orders of some components of G but not a πi-number thenµ1µ2 . . . µr is a divisor of |N | − 1.Now we state the following lemma which is proved in [13], Lemma 6, with somedifferences and classify the simple groups of Lie type with prime odd order componentby the θ function which will be introduced later.Lemma 2.3. If L is a simple group of Lie type and has prime odd order componentp > 17 and π(L) has at most θ(L) prime numbers t, where p+ 1/2 < t < p, thenθ(L) 6 3.Throughout the proof of the above lemma, we can divide simple groups of Lietype, L, with prime odd order component p > 17, into the following cases.(1) θ(L) = 0 if L is isomorphic to Ap′−1(q), Ap′(q),where q − 1 | p′ + 1, A2(2),2Ap′−1(q),2Ap′(q),where q + 1 | p′ + 1, 2A3(2), Bn(q),where n = 2m′and q is odd, Bp′(3), Cn(q),where n = 2m′or (n, q) = (p′, 3), Dp′+1(3), Dp′(q) for q = 3, 5,2Dn(q) for (n, q) = (2m′ , q), (p′, 3),where 5 6 p′ 6= 2m′+ 1 (or 2m′+ 1, 3),where 5 6 p′ 6= 2m′+ 1, G2(q),where q ≡ ε (mod 3) for ε = ±1, 3D4(q), E6(q) or2E6(q);(2) θ(L) = 1 if L is isomorphic to one of the simple groups A1(q),where 2 | q, A2(4),2A5(2), Cp′(2), Dn(2),where n = p′ or p′ + 1, 2Dn(2),where (n, q) = (2m′+ 1, 2) (or p′ = 2m′+ 1, 3),where m′ > 2, E7(2), E7(3), F4(q),2F4(q),where q = 22n+1 > 2, or G2(q),where 3 | q;(3) θ(L) = 2 if L is isomorphic to the simple groups A1(q),where q ≡ ε (mod 4) for ε = ±1, 2B2(q),65where q = 22m′+1 > 2, or 2G2(q),where q = 32m′+1 > 3;(4) θ(L) = 3 if L is isomorphic to the simple groups E8(q) or2E6(2).Lemma 2.4 ([13], Lemma 1). If n > 6 is a natural number, then there are atleast s(n) prime numbers pi such that n+ 1/2 < pi < n. Heres(n) = 6 for n > 48;s(n) = 5 for 42 6 n 6 47;s(n) = 4 for 38 6 n 6 41;s(n) = 3 for 18 6 n 6 37;s(n) = 2 for 14 6 n 6 17;s(n) = 1 for 6 6 n 6 13.In particular, for every natural number n > 6 there exists a prime p such thatn+ 1/2 < p < n, and for every natural number n > 3 there exists an odd primenumber p such that n− p < p < n.3. Proof of the Main theoremSince the necessity of the theorem can be checked easily, we only need to provethe sufficiency.P r o o f of Main theorem (sufficiency). By hypothesis, there exists an element xof order p in G such that CG(x) = 〈x〉 and CG(x) is a Sylow p-subgroup of G. By theSylow theorem, we have that CG(y) = 〈y〉 for any element y in G of order p. So, {p}is a prime graph component of G and t(G) > 2. In addition, p is the maximal primedivisor of |G| and an odd order component of G. The proof of the main theoremfollows from the following lemmas.Lemma 3.1. G has a normal series 1 E H E K E G such that H and G/K areπ1-groups, K/H is a non-abelian simple group and H is a nilpotent group.P r o o f. Let x ∈ G be an element of order p, then CG(x) = 〈x〉. Set H = Op′(G)(the largest normal p′-subgroup of G). Then H is a nilpotent group since x acts onH fixed point freely. Let K be a normal subgroup of G such that K/H is a minimalnormal subgroup of G/H . Then K/H is a direct product of copies of some simplegroup. Since p | |K/H | and p2 ∤ |K/H |, K/H is a simple group. Since 〈x〉 is a Sylow66p-subgroup of K, G = NG(〈x〉)K by the Frattini argument and so |G/K| dividesp− 1.If |K/H | = p, then by Lemma 2.4 and |G| = (p+ 1)!/2 or (p+ 2)!/2, we deducethat there exists r ∈ π(G) such that p− 1/2 < r < p. So, r ∤ p − 1 and hence,|G/K| | (p− 1) implies that r ∈ π(H).Let R be an r-Sylow subgroup of H . Thus |G| = (p+ 1)!/2 or (p+ 2)!/2 showsthat R is a cyclic subgroup of order r. On the other hand, the same reasoning asbefore shows that R⋊P is a Frobenius group, where P is a p-Sylow subgroup of K,so Lemma 2.1 forces p | r − 1 and hence, p+ 1 < r, which is a contradiction. If K/H has an element of order rq where r and q are primes, then G has also suchan element. Hence by definition of order components, an odd order component of Gmust be an odd order component of K/H . Note that t(K/H) > 2.Lemma 3.2. (a) If t ∈ π(H), then t 6 p+ 1/2;(b) A group K/H cannot be isomorphic to a sporadic simple group. Moreover,if K/H is isomorphic to an alternating simple group, then we must have G ∼= Altn,where n = p+ 1 or p+ 2.P r o o f. (a) If t divides |H | where p+ 1/2 < t < p, then since H is a nilpotentsubgroup of G and the order of T , which is the Sylow t-subgroup of H , is equal to t.By Lemma 2.2, we must have p | t− 1, which is impossible. Thus |H | is not divisibleby the primes t with p+ 1/2 < t < p.(b) We note that if H 6= 1, by nilpotency of H we may assume that H is a t-groupfor t ∈ π1(G).If K/H ∼= J4, then p = 43. Since 19 ∈ π(G) − π(Aut(J4)), hence 19 ∈ π(H). ByLemma 2.2, 43 | 19i − 1 for i = 1 or 2, which is impossible.If K/H ∼= M22, then p = 11. Since 52 | |G| and 5 ‖ |Aut(M22)|, 5 ∈ π(H). So byLemma 2.2, we get a contradiction.If K/H ∼= J2, then p = 7, but 24 ‖ |G| and 27 ‖ |K/H |, which is impossible. IfK/H is isomorphic to other sporadic simple groups we can obtain a contradictionsimilarly.Now let K/H be isomorphic to an alternating group. By Tables 1 and 2 in [10],K/H must be isomorphic to Altn, where n = p, p+ 1 or p+ 2.For the case p+ 1, if K/H ∼= Altp, then G/H ∼= Altp or Symp. Let G/H∼= Altp,then |H | = p+ 1.If there is an odd number t ∈ π(H), then t | (p + 1)/2, by nilpotency of H , wemay assume H is a t-group, and by Lemma 2.5, we get a contradiction.67If p + 1 = 2a, then due to nilpotency of H we may assume H is an elementaryabelian 2-group. So CK(H)/H E K/H . Since K/H is a simple group, H = CK(H)or K = CK(H).If K = CK(H), then 2 and p are linked in the prime graph of G, which is a con-tradiction. Hence H = CK(H). So K/H = NK(H)/CK(H) 6 Aut(H) and then|K/H | divides f(a) = Πa−1i=0 (2a−2i). Since K/H ∼= Altp, p−2 = 2a−3 | f(a), whichis impossible.If G/H ∼= Symp, then |H | = (p+ 1)/2 and this contradicts Lemma 2.2.If K/H ∼= Altp+1, then H = 1 and G ∼= Altp+1.If K/H ∼= Altp+2, then |K/H | > |G|, which is impossible.For the case p + 2, if K/H ∼= Altp, since G/H ∼= Altp or Symp, we must have|H | = (p+ 1)(p+ 2) or (p+ 1)(p+ 2)/2, respectively. Let t be an odd prime divisorof p + 2. By nilpotency of H , we may assume H is a t-group. Lemma 2.2 yieldsp | ta − 1. Thus ta = p+ 1, which contradicts gcd(p+ 1, p+ 2) = 1.If K/H ∼= Altp+1, by the same method we must have |H | = p + 2 when G/H ∼=Altp+1, and thus we obtain a contradiction by Lemma 2.2. Therefore,K/H ∼= Altp+2and G ∼= Altp+2. Lemma 3.3. If t ∈ π(G/H) and p+ 1/2 < t < p, then t ∈ π(K/H).P r o o f. It follows from Lemma 3.2 (a), and the proof of Lemma 6 (d) in [13]. Lemma 3.4. A group K/H cannot be isomorphic to a simple group of Lie type.P r o o f. By Lemmas 3.3 and 2.4, we must have 17 6 p 6 37 and θ(K/H) > 2.Therefore K/H is isomorphic to one of the following simple groups:(1) L2(q), where q ≡ ε (mod 4) for ε = ±1;(2) 2B2(q), where q = 22m′+1 > 2;(3) 2G2(q), where q = 32m′+1 > 3;(4) E8(q) or2E6(2).Since one of the odd order components of K/H is equal to p, by Tables 2 and 3in [10], we must have:(1) K/H ∼= L2(17) for p = 17;(2) K/H ∼= L2(19),2G2(27) or2E6(2) for p = 19;(3) K/H ∼= L2(q), where p = q is equal to 23, 29 or 31;(4) K/H ∼= L2(37),2G2(27) or2E6(2) for p = 37.Let p = 17. Then K/H ∼= L2(17). Since 13 /∈ π(L2(17)), we get a contradictionby Lemma 3.3.68Let p = 19. If K/H ∼= L2(19), then since 17 /∈ π(L2(19)), we get a contradictionby Lemma 3.3. Since 37 | |2G2(27)|, K/H ≇2G2(27). For the case K/H ∼=2E6(2),we obtain a contradiction by 236 | |2E6(2)|.Let K/H ∼= L2(q), where p = q is equal to 23, 29, 31, or 37. If p = 37, thensince 23 /∈ π(L2(37)), we get a contradiction by Lemma 3.3. Similarly, we obtaina contradiction when p = 23, 29, or 31.Let p = 37. If K/H ∼= L2(37), then since 23 /∈ π(L2(37)), we get a contradiction.Since 23 /∈ π(2G2(27)), K/H ≇2G2(27). If K/H ∼=2E6(2), a contradiction followsby 236 | |2E6(2)|. Lemma 3.5. If n = 8, then G ∼= L3(4), or Alt8.P r o o f. We know that G has a normal series 1 ⊳ H ⊳ K ⊳ G such that H andG/K are π1-groups, K/H is a non-abelian simple group, and H is a nilpotent group.By Lemma 3.2 (b), K/H cannot be isomorphic to a sporadic simple group, and ifK/H is isomorphic to an alternating simple group, then we must have G ∼= Alt8.Let K/H be isomorphic to the simple group of Lie type G(q) where q = sm and sis a prime number. We know that 7 is one of the odd order components of K/H . Sos = 2, 3, or 7. Then the order of all Sylow t-subgroups of G are less than or equalto 64 or 81. Therefore, K/H ∼= L2(7), L2(8), L3(4), L4(2) ∼= Alt8 or U3(3).If K/H ∼= L2(7), L2(8) or U3(3), then since 5 ∈ π(G) − π(Aut(K/H)), we have5 ∈ π(H), which contradicts Lemma 3.2 (a).Therefore, K/H ∼= L3(4), or L4(2) ∼= Alt8. Since |G| = |L3(4)| = |Alt8|, G ∼=L3(4), or Alt8. Now Lemmas 3.2 (b), 3.4, and 3.5 imply Main theorem.Corollary 3.6. Thompson’s conjecture holds for the alternating groups of degreesp+ 1 and p+ 2, where p is a prime number.P r o o f. Let G be a group with trivial central and N(G) = N(Altn), wheren = p + 1 or p+ 2. Then it is proved in [3], Lemma 1.4, that |G| = |Altn|. Hence,the corollary follows from Main theorem. Acknowledgment. The authors are thankful to the referee for carefully readingthe paper and for his suggestions and remarks.69References[1] A.K.Asboei, R.Mohammadyari: Recognizing alternating groups by their order and oneconjugacy class length. J. Algebra. Appl. 15 (2016), 7 pages.doi: 10.1142/S0219498816500213.[2] G.Chen: Further reflections on Thompson’s conjecture. J. Algebra 218 (1999), 276–285.[3] G.Chen: On Thompson’s conjecture. J. Algebra 185 (1996), 184–193.[4] G.Y.Chen: On Thompson’s Conjecture. PhD thesis, Sichuan University, Chengdu, 1994.[5] Y.Chen, G. Chen: Recognizing L2(p) by its order and one special conjugacy class size.J. Inequal. Appl. 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Algebra. 69 (1981), 487–513.Authors’ addresses: A l i r e z a K h a l i l i A s b o e i (corresponding author), Depart-ment of Mathematics, Farhangian University, Shariati Mazandaran, Sari, Iran, and De-partment of Mathematics, Buinzahra Branch, Islamic Azad University, Buin Zahra, Iran,e-mail: khaliliasbo@yahoo.com; R e z a Mo h ammad y a r i, Department of Mathematics,Buinzahra Branch, Islamic Azad University, Buin Zahra, Iran, e-mail: rezamohammadyari@gmail.com.70

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Characterization of the alternating groups by their order and one conjugacy class length

Alireza Khalili Asboei

Reza Mohammadyari

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Characterization of the alternating groups by their order and one conjugacy class length

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