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Publicly Available Published by De Gruyter February 14, 2017

On Thompson’s conjecture for alternating groups of large degree

  • Ilya B. Gorshkov EMAIL logo
From the journal Journal of Group Theory

Abstract

For a finite group G, let N(G) denote the set of conjugacy class sizes of G. We show that if G is a finite group with trivial center such that N(G)=N(Altn), where n>1361 and n or n-1 is a sum of two primes, then GAltn.

1 Introduction

Let G be a finite group. Put N(G)={|gG|:gG}. In 1987 Thompson posed the following conjecture concerning N(G).

Thompson’s Conjecture (see [10, Question 12.38]).

If L is a finite simple non-abelian group, G is a finite group with trivial center, and N(G)=N(L), then GL.

It has been proved that Thompson’s conjecture is valid for many finite simple groups of Lie type (see [1]). Denote the alternating group of degree n by Altn, and the symmetric group of degree n by Symn. Alavi and Daneshkhah [2] proved that the groups Altn, where at least one the of numbers n, n-1 or n-2 is a prime, are characterized by N(G). This conjecture has also been confirmed for Alt10,Alt16, Alt22 and Alt26 in [12], [5], [14] and [9], respectively. In [7], the author showed that if N(G)=N(Altn) for n5, then G is non-solvable. In [6], we obtained some important information about the composition factors of a group G with N(G)=N(Altn), where n>1361.

The aim of this paper is to prove the following theorem.

Main Theorem.

Let G be a finite group with trivial center such that N(G) is equal to N(Altn), n27, G has a composition factor SAltn-ε, where ε is a non-negative integer such that the set {n-ε,,n} does not contain primes and at least one of the numbers n or n-1 is a sum of two primes. Then GAltn.

This theorem and the main result of [6] (see Lemma 2.13 below) imply immediately the following corollary.

Corollary.

Thompson’s conjecture is true for the alternating group of degree n if n>1361 and at least one of the numbers n or n-1 is a sum of two primes.

At this moment it is not known if there are even positive integers that cannot be decomposed into a sum of two primes.

2 Notation and preliminary results

All notation is standard and can be found in [4].

Lemma 2.1 ([12, Lemma 4]).

Suppose that G is a finite group with trivial center and p is a prime in π(G) such that p2 does not divide |xG| for all x in G. Then a Sylow p-subgroup of G is elementary abelian.

Lemma 2.2 ([3, Theorem 1]).

Let G be a finite group acting transitively on a set Ω with |Ω|>1. Then there exist a prime r and an r-element gG such that g acts without fixed points on Ω.

Lemma 2.3.

Let G be a finite group, gG, |π(|g|)|=1 and |gG|>1. Then there exists gG such that π(|g|)π(|(g)G|) and |π(|g|)|=1.

Proof.

The statement follows from Lemma 2.2. ∎

Lemma 2.4.

Let G=Tg, (|T|,|g|)=1, |π(|g|)|=1 and |gG|>1. Then there exist normal subgroups K and R in G such that K<R, R/K is a minimal normal subgroup of G/K, |gR|>1 and [g,T]R.

Proof.

This statement follows from Lemma 2.3, the Frattini argument, and [4, Lemma 5.3.2]. ∎

Lemma 2.5 ([4, Theorem 5.2.3]).

Let A be a π(G)-group of automorphisms of an abelian group G. Then G=CG(A)×[G,A].

Lemma 2.6 ([6, Lemma 1.6]).

Let S be a non-abelian finite simple group. If pπ(S), then there exist aN(S) and gS such that |a|p=|S|p, |gS|=a and |π(|g|)|=1.

Lemma 2.7 ([11, Lemma 14]).

Let S be a non-abelian finite simple group. Any odd element from π(Out(S)) either belongs to π(P) or does not exceed m2, where m is the largest element of π(S).

Lemma 2.8 ([6, Lemma 1.4]).

Let G be a finite group, KG, G¯=G/K, xG and x¯=xKG/K. The following assertions hold:

  1. |xK| and |x¯G¯| divide |xG|.

  2. If L and M are neighboring members of a composition series of G, L<M, S=M/L, xM and x~=xL, then |x~S| divides |xG|.

  3. If yG,xy=yx, and (|x|,|y|)=1, then CG(xy)=CG(x)CG(y).

  4. If (|x|,|K|)=1, then CG¯(x¯)=CG(x)K/K.

Lemma 2.9 ([6, Lemma 2.3]).

Let n>26, let t be a prime such that n2<tn, let αN(Altn) and suppose that t does not divide α. Then α is equal to |Altn|/t|C| or |Altn|/|Altt+i||B|, where C=CAltn-t(g) for some gAltn-t, t+in and B=CAltn-t-i(h) for some hAltn-t-i.

Lemma 2.10 ([8]).

The product of k consecutive integers n(n-1)(n+k-1), which are greater than k, contains a prime divisor greater than 3k2 with the exceptions 34,89 and 678910.

Lemma 2.11 ([13, Lemma 3.6]).

Let s and p be distinct primes, H be a semidirect product of a normal p-subgroup T and a cyclic subgroup g of order s, and [T,g]1. Suppose that H acts faithfully on a vector space V over a field of positive characteristic t not equal to p. If the minimal polynomial of g on V does not equal to xs-1, then the following assertions hold:

  1. CT(g)1,

  2. T is non-abelian,

  3. p=2 and s=22δ+1 is a Fermat prime.

Lemma 2.12.

We have N(Symn)N(Altn)

Proof.

The proof is obvious. ∎

Lemma 2.13 ([6]).

Let G be a finite group with N(G)=N(Altn), where n>1361. Then G has a composition factor isomorphic to an alternating group Altm, where mn and the half-interval (m,n] contains no primes.

3 Proof of the Main Theorem

Suppose that 26<n, where n, n-1 and n-2 are not primes, at least one of the numbers n or n-1 is decomposed into a sum of two primes, G is a finite group with Z(G)=1, N(G)=N(Altn), and G has a composition factor SAltn-ε, where ε is a non-negative integer such that the set {n-ε,,n} does not contain primes. Let K be a maximal normal subgroup of G such that G/K contains a subgroup isomorphic to S. Set Ω={t:t is a prime, n2<tn}. It is clear that SG/KAut(S).

Lemma 3.1.

We have that |Altn| divides |G|.

Proof.

If n is odd, there exist α,γN(G) such that α=n!/2n, γ=n!/4(n-1). If n is even, there exist α,γN(G) such that α=n!/4n, γ=n!/2(n-1). The statement follows from Lemma 2.8. ∎

Lemma 3.2.

We have Ωπ(K)=.

Proof.

Assume that Ωπ(K). Let T be a normal subgroup of G lying in K such that π(K/T)Ω and π((K/T)/R)Ω= for a minimal normal subgroup R of K/T. Let G~=G/T, K~=K/T, tπ(K~)Ω and S~G~ be a pre-image of S of minimal order. Since R is a minimal normal subgroup of G~, we have R=R1××Rk, where R1,,Rk are isomorphic simple groups. Assume that R is non-solvable and k>1. It follows from Lemma 2.6 that there exist y1 and y2 in R1 and R2, respectively, such that |π(|y1|)|=|π(|y2|)|=1, tπ(|y1R1|), and tπ(|y2R2|). We have that t2 divides |(y1y2)R1R2| and there exists an element gG such that t2 divides |gG|; a contradiction. Then k=1. We have S~<NG~(R). Since Out(A) is solvable for any finite simple group A and (NG~(R)/CG~(R))/R<Aut(R), we get S~<CG~(R). It follows from Lemma 2.6 that there exists gR such that tπ(|gR|). Using Lemma 2.8, we obtain that there exists an element sS~ such that |s|Ωπ(g) and tπ(|sS~|). Therefore, t2 divides |(sg)G~|; a contradiction.

It follows that R is an abelian t-group for some prime t. Let gS~, |g|Ω{t}. It is clear that g centralizes R. Therefore, S~ centralizes R. Since t|K~/R|, we see that CK~(R)=R×L. Since CK~(R)G~, we see that LG~. Hence, L=1. Therefore, S~S and G~(RK¯)×S~, where K¯=K~/R. If K¯1, then K¯ acts faithfully on R. It follows from Lemma 2.3 that there exists hK~ such that |π(|h|)|=1 and t divides |hK~|. Therefore, t2 divides |(hg)G~|; a contradiction. Thus, K¯=1. We have GT.(R×S). Similarly we can obtain that GW.(F×S), where π(W)Ω= and π(F)Ω. It is easy to prove that F is an abelian group.

Since Z(G)=1, we see that |hG|>1 for any element hF. It follows from Lemma 2.4 that there exist subgroups L and H such that LG, L<W, H is an unique normal subgroup of G^=G/L, |xH|>1 for some xF and |h(G^)/H|=1 for any hF. In particular, we have (W.S)/L.H<CG^/H(F) and |hG^|=|hH| for any hF.

Let xF,|x|=t, rΩ and tr2j+1. Since H is a minimal normal subgroup of G^, we have HH1××Hl, where H1,,Hl are isomorphic finite simple groups. Assume that H is an elementary abelian k-group. It is clear that CH(x)G^. By the definition of H we obtain that CH(x)=1. Therefore, x acts on H fixed-point-freely. There exists a Frobenius group A=yz in S such that |y|=r,|z|π(r-1){k} (since r-1 is not a power of prime, π(r-1){k} is not empty). Hence, tπ(|yS|)π(|zS|). Let A^<G^ be a pre-image of A of minimal order. Since r|W^|, using the Frattini argument we obtain that y~A^, where y^A^ is a pre-image of y of minimal order. It follows from Lemma 2.11 that one of the groups CR(y^) or CR(z^) is not trivial. Let v{y^,z^} be such that CH(v)>1 and fCH(v). Then the subgroup CG^(vf) intersects xG^ and any Sylow t-subgroup of S^ trivially. It follows that t2 divides |(vf)G¯|; a contradiction.

We have that H is non-solvable and H1 is a non-abelian simple group. Since t>s for any sπ(H), it follows from Lemma 2.7 that HixHi for 1il. Let bH, b1, b=b1b2bv, where if bi,bjHc for any 1cl, then i=j. Put length(b)=v. Then t divides length(b) for any bCH(x). If a1H1 with a=a1a1yatyr, then length(a)=1 or length(a)=r. Hence, we have aCG^(y^) and CG^(a)xG^=. We obtain that t|CG^(ay^)|. It follows that t2 divides |(ay^)G^|; a contradiction. ∎

By the hypotheses of the Main Theorem, n or n-1 is equal to p+q, where p and q are primes. We can assume that pq. Further our proof is divided into three propositions.

Proposition 3.3.

If p=q, then the assertion of the Main Theorem is true.

Proposition 3.4.

If q=3, then the assertion of the Main Theorem is true.

Proposition 3.5.

We have GAltn.

4 Proof of Proposition 3.3

Assume that n or n-1 is equal to 2p.

Lemma 4.1.

We have δ:=n!/2p2N(G)

Proof.

Consider the element g=(1,2,,p)(p+1,,2p) of Altn. It is obvious that |CAltn(g)|=p2. ∎

Lemma 4.2.

If pα for some αN(G), then α=δ or α=n!/(p(n-p)!).

Proof.

The proof is obvious. ∎

Lemma 4.3.

We have p|K|

Proof.

Assume that p divides |K|. Let PSylp(K). By the Frattini argument, NG(P)/NK(P)G/K. Let N=NG(P)/Op(NK(P)), let S¯<N be a subgroup of minimal order having a composition factor isomorphic to S, and let M be a minimal normal subgroup of N. Let us take an element gNG(P) such that |g|=p and g¯=gOp(K)/Op(K)M. It is obvious that M<Z(Q) for any QSylp(N). In particular, any p-element of N centralizes M. Since S¯ is generated by all its p-elements, we see that S¯<CN(M). In particular, tπ(CN(g¯)) for any tΩ. Using Lemma 2.8, we get that p|gG|. Therefore, |gG|=δ. Combining Lemmas 3.2 and 2.8, we obtain that π(|gG|)Ω=; a contradiction. ∎

Lemma 4.4.

We have GAltn

Proof.

It follows from Lemmas 3.1 and 4.3 that p2 divides |S| and, consequently, ε1. Assume that K1. Let gG, |g|=p, |gG|=δ and g¯=gK. Then |g¯S|=δ or |g¯S|=δ/n. Suppose that |g¯S|=δ/n. Then g acts on K non-trivially. Using Lemma 2.8, we get |gK|=n. It is easy to prove that p divides |n|t-1 for any tπ(n). But this is impossible, since n-1=2p; a contradiction. Therefore, |g¯S|=δ, ε=0 and K<CG(g). Since Z(G)=1, there exists an element hK with |hG|>1. Hence, δ divides |(hg)G| and δ|(hgG)|. However, δ is maximal with respect to divisibility in N(G); a contradiction. Thus, K=1 and SAltn. Now the statement of the lemma follows from Lemma 2.12. ∎

The proposition is proved.

5 Proof of Proposition 3.4

Assume that q=3. Let gG, |g|=p and g¯=gKS. By Lemma 2.8 we have that |g¯S| divides |gG| and |gG| divides n!/p. Suppose that ε>0. Then |gK| divides n-ε+1ni. It is easy to prove that p divides |n-ε+1ni|t-1 for any tπ(n-ε+1ni); a contradiction. Therefore, ε=0. Now the statement can be proved similarly to Lemma 4.4.

6 Proof of Proposition 3.5

Let p>q>3,gG,|g|=p and g¯=gKS.

Lemma 6.1.

We have α=|gG|=n!/(p(n-p)!).

Proof.

We have Ω{p}π(g¯S). Using Lemma 2.8, we get Ω{p}π(α). It follows from Lemma 2.9 that α=n!/(p|C|), where C=CAltn-p(h) for some element hAltn-p.

Assume that |C|<(n-p)!. Let hG be such that |hG|=n!/(pq). Since |G|p=p, we can assume that hCG(g). If pπ(|h|), then α divides |hG| and, consequently, q divides |C|. Therefore, α=n!/(pq). Let x be an element of CG(g) such that |xG|=n!/(pa), where 1<aq. Since α|xG|, we obtain using Lemma 2.8 that pπ(|x|). Hence, |xG| and α divide |(xg)G|, but α is maximal with respect to divisibility in N(G); a contradiction. Therefore, pπ(|h|) and we can assume that α|hG|. Since |hG| is maximal with respect to divisibility in N(G), we obtain that |(gh)G|=|hG|. Therefore, α divides |(gh)G|. Thus, α=n!/(p(n-p)!). ∎

Let C=CG(g), α=|gG|, γ=n!/(pq), Θ={n!/(pb):b(n-p)!}N(G) and QSylq(C).

Lemma 6.2.

The following statements hold:

  1. QZ(C).

  2. If vQZ(C), then |(gv)G|=γ.

  3. |Q|/|QZ(C)|=q,

  4. Suppose that vQZ(C) and f is a {p,q}-element of CG(gv)Z(C). Then |(gf)G|=γ.

Proof.

(i) It follows from Proposition 3.4 that n-p>3. Therefore, there exists an element βΘ such that βq>αq. Let hC and |hG|=β. Since CG(gh)C and |CG(gh)|q<|C|q, we obtain using Lemma 2.8 that there exists rQZ(C).

(ii) We have |(gr)G|α. Assume that for some lG such that |(lg)G|=γ, where lCG(g) and π(l)={t}, we have tq. Let bC and |bG|=γ. Since b centralizes some Sylow q-subgroup of C, we can assume that rCG(b). Therefore, γ divides |(grb)G| and |(gr)G| divides |(grb)G|; a contradiction. Thus, there exists rQZ(Q) such that |(gr)G|=γ.

Suppose that there exists r′′QZ(C) such that |(r′′g)G|=δγ. Then there exists a {p,q}-element aCG(gr′′). It is obvious that |(ga)G|γ. Therefore, rCG(ga) and |(gar′′)G|q>|(gr′′)G|q; a contradiction.

(iii) Let vC and |vG|=β. Since |(gb)G|=γ for any bQZ(C), we see that CG(gv)Q=Z(C)Q, and, consequently, |Q|/|Z(C)Q|=q.

(iv) We have that |(gf)G|γ. Therefore |(gf)G|{α,γ}. Since fZ(C), we obtain |(gf)G|α and consequently |(gf)G|=γ. ∎

Lemma 6.3.

We have n-p-ε<5.

Proof.

Assume that n-p-ε5. Therefore QK. It follows from the Frattini argument and Lemma 6.2 (i) that there exists a subgroup H in NC(Q) such that H/KHAlt5. Let N¯=NC(Q)/Z(Q)Z(C) and Q¯=Q/Z(Q)Z(C). Note that by Lemma 6.2 (iii) we have |Q¯|=q. Therefore, N¯/CN¯(Q¯) is isomorphic to a cyclic subgroup of order dividing q-1 in Aut(Q¯). Hence, CN¯(Q¯)H. Let h1,h2HZ(C),|h1|=2 and |h2|=3. Since |(gh1)G|=|(gh2)G|=γ, we see that CG(gh1)=CG(gh2); a contradiction. ∎

Lemma 6.4.

If n-p-ε=4, then 3 divides q-1.

Proof.

Assume that n-p-ε=4. Therefore QK. It follows from the Frattini argument and Lemma 6.2 (i) that there exists a subgroup H<NC(Q) such that H/KHAlt4. Let N¯=NC(Q)/Z(Q)Z(C) and Q¯=Q/Z(Q)Z(C). Note that by Lemma 6.2, |Q¯|=q. Therefore, N¯/CN¯(Q¯) is isomorphic to a cyclic subgroup of order dividing q-1 in Aut(Q¯). If CN¯(Q¯)H, then similarly to the proof of Lemma 6.3 we obtain a contradiction. Thus, |H/CN¯(Q¯)H|=3. ∎

Lemma 6.5.

The number Π=n-ε+1ni is not divisible by primes larger than q.

Proof.

Suppose that there exists a prime t dividing Π such that t>q. Then t divides only one number from the set {n-ε+1,,n}. In particular, |Π|t-1<p. It follows from Lemma 6.1 that |gG|=α and |g¯G/K|=(n-ε)!/(n-ε-p)!. Since K is not divisible by p, we obtain |gK|=Π/(n-p-ε+1n-pi). Therefore, t divides |gK|. Let H be a maximal normal subgroup in K such that t divides |g~K~|, where K~=K/H and g~=gHG/H. Let N be a minimal normal subgroup of K~. By the maximality of H, t divides |g~N|. We have NN1××Nk, where N1,,Nk are isomorphic simple groups. It is easy to prove that |g~N|tp+1; a contradiction. ∎

Lemma 6.6.

If ε>0, then ε4,q7; if n-ε-p=2, then n-p5.

Proof.

The statement follows from Lemmas 2.10, 6.3, 6.4 and 6.5. ∎

Lemma 6.7.

We have ε=0.

Proof.

Assume that ε>0. Then |gK|=α/|g¯G/K|. It is clear that if ε>1, then there exists an odd prime divisor t of |gK|. We can show that p divides |gK|t-1. If t divides only one number from the set Λ={n-ε+1,,n}, then we can obtain a contradiction as in the proof of Lemma 6.5. It follows from Lemma 6.6 that ε4. Since t divides two numbers from the set Λ, we have t=3. Therefore, |gK|ttl+1, where tl divides only one number from the set Λ. Analyzing the residues of the division of tl by p, it is easy to get a contradiction. ∎

Lemma 6.8.

We have GAltn.

Proof.

Since |gS¯|=|gS|, we have K<CG(g). Therefore, K.S is a central extension. Since Z(G)=1, it follows that Z(K)=1. Assume that K1. Let hK be an element of prime order t. Then t is coprime to a number θ{n-1,n-2}. There exists xG such that |x|{θ,θ/2}, and |xS|=n!/θ. It is obvious that |xS| is maximal with respect to divisibility in N(G). By Lemma 2.8 we obtain |(hx)G|>|xS|, where xG is a pre-image of x in G; a contradiction. Thus, K=1. By Lemma 2.12 we obtain the statement of the lemma. ∎

Proposition 3.5 and the Main Theorem are proved.

The Corollary follows from Lemma 2.13 and the Main Theorem.


Communicated by Evgenii I. Khukhro


Award Identifier / Grant number: 15-11-10025

Funding statement: The work is supported by the Russian Science Foundation (project no. 15-11-10025).

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Received: 2016-8-22
Published Online: 2017-2-14
Published in Print: 2017-7-1

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