The Characterization of Finite Simple Groups with no Elements of Order Six by Their Element Orders

of Order Six by Their Element Orders

Jianbei An

Department of Mathematics, University of Auckland Auckland, New Zealand. Email: an@math.auckland.ac.nz

and Wujie Shi*

Institute of Mathematics, Southwest-China Normal University Chongqing, P. R. China. Email: wjshi@swnu.edu.cn

ABSTRACT. We show that all butA⁶ non-abelian nite simple groups with no elements of order6are characterized by their element orders.

INTRODUCTION

Let G be a nite group ande(G) the set of orders of all elements in G. Then e(G) is a subset of positive integers ^Z+. Let be a subset of^Z+ and h() the number of non- isomorphic nite groups K such that e(K) = . Then h(e(G)) 1 for any nite group G. Following 16] we say that a nite groupG isnon-distinguishableif h(e(G)) =¹, and distinguishable if h(_e(G)) < ¹. Moreover, G is k-distinguishable if h(_e(G)) = k and characterizible if h(e(G)) = 1. From 4], 14, Theorem 1.1] and 5], we have the following propositions.

Proposition 1

. Suppose a nite group Ghas an elementary abelian normal subgroup.

Then Gis non-distinguishable. In particular, a solvable nite group is non-distinguishable.

Proposition 2

. The following non-abelian nite simple groups are characterizible.

1991 Mathematics Subject Classication. Primary 20D05, 20D06

* Project supported by the National Natural Science Foundation of China and the State Education Ministry Foundation of China

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(a) Alternating groups A_n for n^2f5789111213^g. (b) All sporadic simple groups except J².

(c) Simple groups L²(q) with q ⁶= 9, Suzuki-Ree groups, L³(4), L³(7), L³(8), L⁴(3), S⁴(7), U³(4), U⁴(3), U⁶(2), G²(3), O^;8(2) and O^10;(2).

In this paper, we proved that, except forA⁶, all non-abelian nite simple groups with no elements of order 6 are characterizible, that is we proved the following Theorem.

Theorem

. Let G be a nite group and H a nite non-abelian simple group with no elements of order 6 such that H ^6'A⁶. Then G^'H if and only if _e(G) =_e(H).

The alternating groupA⁶ is not characterizible, in fact, innitely many groups share the same element orders with A⁶ (see 2, Lemma 2]).

In the following, we suppose a group is always a nite group and a simple group is always non-abelian. We denote by (G) the set of prime divisors of the order ^jG^j of G. The other notation are standard following from 3] and 13].

1. PRELIMINARY RESULTS AND LEMMAS

The results of the paper depends on the classication of nite simple groups and the following lemmas.

Lemma 1.1

. Let G be a simple group with no elements of order 6. Then G is iso- morphic to one of the following groups:

(a) L²(q) with q ⁶1(mod12) (b) Sz(2²ⁿ⁺¹) with n 1 (c) L³(2ⁿ) with n⁶0(mod6)

(d) U³(2ⁿ) with n 2 and n⁶3(mod6). Proof. See 8] and 9].

Remark

. From paper 9], a unitary groupU³(2ⁿ) has no elements of order 6 whenever n 2 and n⁶35(mod6). But, in fact, U³(2ⁿ) with n 5(mod6) also has no elements of order 6. Indeed, suppose n = 6k +y for some integers k 0 and 0 y 5. Then 2ⁿ+ 12^y + 10(mod9) if and only if y = 3, that isn= 6k+ 3.

From paper 6], SU³(2ⁿ) has a single conjugacy class of involutions each with a cen- tralizer of order 2ⁿ(2ⁿ+1). IfU³(2ⁿ) withn5(mod6) has elements of order 6, it follows

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that 3 divides 2ⁿ + 1, so then 9 must divides 2ⁿ + 1 (note that Z(SU³(2ⁿ)) is cyclic of order gcd(32ⁿ+ 1)), which is impossible.

The prime graph ;(G) of a group G is the graph whose vertex set is the set (G) and two vertices pr ² (G) are adjacent if and only if G contains an element of order pr. Denote by t(G) the number of connected components of ;(G), and by i =i(G) for i = 12:::t(G) the connected components of ;(G). If ^jG^j is even, then suppose 2 is a vertex of ¹. We shall also use the following unpublished result of K.W. Gruenberg and O.H. Kegel, (see 19, Theorem A]).

Lemma 1.2

. If Gis a nite group whose prime graph has more than one component, then G has one of the following structures: (a) Frobenius or 2-Frobenius (b) simple (c) an extension of a ¹-group by a simple group (d) simple by¹ or (e) ¹ by simple by¹. A group G is called 2-Frobenius if there exists a normal series 1 H K G of G such that H is the Frobenius kernel ofK and K=H is the Frobenius kernel ofG=H.

Suppose = + or ^; and let L³(q) = L³(q) or U³(q) according as = + or ^;. For simplicity, we always identify q^; with q^;1.

Lemma 1.3

. Suppose q= 2ⁿ and = = ^{gcd (31}_q^;). Then

e(L³(q)) =^f124s:s^j(q^;)s^j2(q^;)s^j(q^2;1) ors^j(q²+q+ 1)^g:

In addition, if (q) ^{2 f}(2+)(4+)^g, then ¹(L³(q)) = ^f2^g, ²(L³(q)) = ^f3^g for q ²

f24^g, ³(L³(q)) = ^f7^g or ^f5^g according as q = 2 or 4 and ⁴(L³(4)) = ^f7^g if (q) ⁶²

f(2+)(4+)^g, then ¹(L³(q)) = ^fr prime : r^j2(q^;) or r^j(q +)^g and ²(L³(q)) =

frprime :r^j(q²+q+ 1)^g. In particular, t(L³(q)) 2.

Proof. The sete(L³(q)) is determined by calculations from conjugacy SL³(2ⁿ)-classes given, say by 6, Tables VII-1 and IX-1], and _i(L³(q)) for all i are given by 12].

2. THE PROOF OF THE THEOREM

From Lemma 1.1, if H is a simple group with no elements of order 6, then H is isomorphic toL²(q), Sz(2²ⁿ⁺¹) or L³(2ⁿ) for suitableq orn. By 1] and 17],L²(q) (q⁶= 9 and L²(9) ^' A⁶) and Sz(2²ⁿ⁺¹) (n 1) are characterizible. It suces to show that H =L³(2ⁿ) is characterizible.

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Theorem 2.1

. Let G be a nite group and H = L³(2ⁿ) with n ⁶ 0 or 3(mod6) according as = + or ^;. Then G^'H if and only if e(G) =e(H).

Proof. It is clear that if G^'H, thene(G) =e(H).

Let q = 2ⁿ with n ⁶ 0 or 3(mod6) according as = + or ^;, and suppose e(G) = _e(H). By Lemma 1.3, _e(L³(q)) =^f124s:s^j(q^;)s^j2(q^;)s^j(q^2;1) ors^j(q²+ q+ 1)^g where = ^{gcd (31}_q^;). Thus G has no elements of order 6 and t(G) 2.

(1). G is non-solvable.

SupposeGis solvable andK is its^f2pr^g-Hall subgroup, where pr²(G) such that p^jq+ andr^j(q²+q+1). Since Ghas no elements of order 2p, 2r andpr, it follows that K is a solvable group all of whose elements are of prime power orders. By 11, Theorem 1], ^j(K)^j2, which is impossible.

(2). G is neither Frobenius nor 2-Frobenius.

SupposeGis a Frobenius group with complementC. By the structure of non-solvable Frobenius complements (see 15, Theorem 18.6]), we have that C has a normal subgroup C of index2 such thatC ^'SL²(5)Z, where every Sylow subgroup ofZ is cyclic and (Z)^\f235^g = . Since _e(SL²(5)) = ^f12:::610^g, it contradicts with 6 ⁶² _e(G).

imilarly, G is not 2-Frobenius.

By Lemma 1.2, G has a normal series 1 N < G¹ G such that (a) N and G=G¹ are ¹-groups (b) G=G¹=N is simple with 6⁶²e( G). We claim that

(3). G^'H.

By Lemma 1.1, G is isomorphic to one of the groups, L²(q⁰) with q^{0 6} 1(mod12), Sz(2^2m+1) with m 1,L³(2^m) with m⁶0 or 3(mod6) according as = + or ^;.

Case (i). Suppose G ^' L²(q⁰) and q⁰ = 2^m with m 2. Since the vertexes set of _i is a subset of the set (G) and dierent vertexes sets are disjoint, each vertex of _i is coprime to any vertex of j for i⁶=j and both G=G¹ andN are ¹-groups, it follows that (q²+q+1)²e( G). Note that(q²+q+ 1) is the largest or second largest number in the set

f124(q^;)2(q^;)(q^2;1)(q²+q+ 1)^g

according as = + or ^;, and in addition, G and hence G have no element x such that (q² + q + 1) is a proper divisor of ^jx^j. It follows that (q² + q + 1) = q⁰ + 1 and (q^2;q+ 1)^2fq⁰+ 1q^0;1^g, which is impossible as both q and q⁰ are powers of 2.

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Suppose q (mod4) and q = p^m, where p is an odd prime and = + or ^;. A similar proof to above shows that(q²+q+1)^2fp⁰¹²(q⁰+)^g, since(q²+q+1) is odd and

1

2(q^0;) is even. If(q²+q+1) =p⁰, thenm= 1, since otherwise ¹²(q⁰+1)> ¹²(q^0;1)> p⁰. Thus ¹²(q⁰+ 1) = ¹²(p⁰+ 1) = ¹²((q²+q+ 1) + 1)²_e(G).

If ¹²((q²+q+ 1) + 1) = 2, then (q² +q+ 1) = 3, which is impossible as 3 is not a divisor of (q²+q + 1). If ¹²((q² +q + 1) + 1) = 4, then (q² +q + 1) = 7 = p⁰, so that =^;, and (q) = (2+) or (4+). In the former case, L²(p⁰) = L²(7), which is isomorphic toH =L³(2). In the later case,H =L³(4), G=L²(7) (note that e(L²(7)) =

f12347^ge(L³(4)) =^f123457^g), and by Proposition 2,L³(4) is characterizible, so that e(G)⁶=e(H), which is impossible.

Ifr 5 is a common prime factor of ¹²((q²+q+1)+1) andq^;, thenq (modr) and ¹²((q²+q+1)+1) ³²⁺¹(modr). But ^{32+1 6}0(modr), soris not a common factor of ¹²((q²+q+ 1) + 1) and q^; (respectively, and 2(q^;)). Similarly, a prime r 5 is not a common factor of ¹²((q²+q+1)+1) and q+. Thus ¹²((q²+q+1)+1) is a power of 3. Moreover, ifq >2 and = 1, then ¹²((q²+q+1)+1) = ¹²(q²+q+2) ¹²(^;1)^{n 6} 0(mod3), which is impossible. If = ¹³, then ¹²((q²+q+ 1) + 1) = ¹⁶(q²+q+ 4) = 3^t has no solution for n and t, which is a contradiction. If q = 2, then = + and = 1, so (q²+q+ 1) = 7, which is discussed above.

Suppose (q² +q + 1) = ¹²(q⁰+). Then ¹²(q^0;) = (q² +q + 1)^{; 2} e(G), which is impossible.

It follows that G ^6' L²(q⁰) except when q⁰ = 7 and H = L³(2), in which case G = L²(7)^'H.

Case (ii). Suppose G ^' Sz(2^2m+1) with m 1. A similar proof to that of Case (i) shows that(q²+q+1) = 2^2m+1+2^m+1+1 and(q²+q+1)^2f2^2m+1;12^2m+1+2^m+1+1^g, which are impossible. Thus G^6'Sz(2^2m+1).

Case (iii). Suppose G^'L^;3(2^m), where m 2 when = +. A similar proof to that of Case (i) shows that (q²+q+ 1) =(2^2m;1) and

(q^2;q+ 1)^2f(2^2m+ 2^m+ 1)(2^2m;1)^g

where = 1=gcd(32^m+). If (q²+q + 1) = (2^2m;1) and (2^m+) ⁶= 1, then an odd prime divisor r of (2^m +) divides both (q² +q + 1) and 2(2^m +) ² e( G).

In particular, 2r ² e( G) and so 2r ² e(H). A contradiction, since by Lemma 1.3, 2r⁶²e(H) for r^j(q²+q+1). If(2^m+) = 1, then(q²+q+1) = (2^2m;1) = 2^m;

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has no solution. Similarly,(q^2;q+1) =(2^2m+2^m+1) can also induce a contradiction.

It follows that G^6'L^;3(2^m).

Case (iv). Suppose G ^' L³(2^m). A similar proof to that of Case (i) shows that (q²+q+ 1) =(2^2m+ 2^m+ 1) and

(q^2;q+ 1)^2f(2^2m;1)(2^2m;2^m+ 1)^g

where= 1=gcd(32^m;). A similar proof to that of Case (iii) shows that(q^2;q+1)⁶= (2^2m;1), so (q²+q+ 1) =(2^2m+2^m+ 1) and m=n. Thus G^'L³(2ⁿ)^'H.

(4). N = 1.

Suppose N ⁶= 1. Since (N)¹(G), N is a normal ¹-subgroup of G. By t(G) 2, a subgroup R=N of order d ^{2 2}(G=N) acts xed-point freely on N, so by a result of Thompson, 10, p.337],N is nilpotent. ReplacingN by a factor group, we may suppose N is an elementary abelianr-subgroup, where r is a prime such thatr= 2, r^jq+ orr^jq^;. Case (i). Supposer^jq+. Since 2r ⁶²_e(G), it follows that a Sylow 2-subgroupP² of G acts xed-point freely on N, so thatP² is cyclic or generalized quaternion, a contradiction since G¹ is non-solvable.

Case (ii). Suppose r= 2. LetQ be a cyclic subgroup of order(q²+q+1) in G¹=N, and N(Q) = NG¹=N(Q). By 7, Theorem 3.1], N(Q) = Q:^Z3 and suppose N](Q:^Z3) is its preimage inG¹. It follows by 10, Chapter 3, Theorem 4.4] thatG¹ contains an element of order 6, which is impossible.

Case (iii). Suppose r^jq^;. If r= 3, then repeat the proof of Case (ii) above, so that by 1, Lemma 7],G¹ has an element of order 9 and 9²e(H). Since 3^jq^;, it follows that 9^jq^;, n0 or 3(mod6) according as = + or^;, and by Lemma 1.3, 6 ²_e(H), which is also impossible. If r ⁶= 3, we denote S ² Syl²(G¹). Let G = G¹=N ^' L³(q) (The case of U³(q) is similar since the centralizer of every involution has a normal Sylow 2-subgroup in the two classes simple groups 18]). We have the following observations about G: The case of q = 4 is discussed, see Proposition 2.(c). Z(S) is an elementary abelian group of orderq, and for any 1⁶=x²Z(S), C_G(x)^'SK, where K is of order (q^;1)=gcd(3q^;1) and acts on S as the set of diagonal matrices. And all elements of Z(S) are conjugate in G¹.

By co-prime action (10, Chapter 5, Theorem 3.16]), N =^hCN(x)^j1⁶=x ²Z(S)ⁱ and 6

C_N(x)⁶= 1. Without lose any generality, we may assume that x=

0

@

1 0 1 0 1 0 0 0 1

1

A

put U = C_N(x), C = C_G(x), so C acts on U. Assume C_N(f) = 1 for each element f of order 4 in S. Then f inverts each element ofU, so fC]C_C(U). PutD =^hf ²S^jf² = xⁱ. As q > 4, S=Z(S) is the direct sum of two non-isomorphic K-submodules of order q. As D is K-invariant, D = S. Thus SC] C_C(U). But S = SC], so SU] = 1 which is a contradiction.

Therefore N = 1 and G^{1 '} L³(q). Now G¹ is normal in G and by t(G) 2, CG(G¹) = 1, so

G¹ GAut(G¹) since NG(G¹)=CG(G¹)Aut(G¹). Finally, we show that

(5). G=G¹.

Suppose G = G¹:T and q = 2ⁿ, where T Out(G¹). In the notation of Kleidman and Liebeck, 13, Propositions 2.2.3 and 2.3.5] and the fact that the automorphism group of L³(q) splits over L³(q),

A= Out(G¹) =

(

h ^i'Z(3_q^;1):^Z_n:^Z2 if = +,

h ^i'Z(3q⁺¹⁾:^Z2_n if =^;,

where = diag^f11^g2GL³(q), ((aij)) = (a²_ij) for each matrix (aij)²GL³(q) and is the inverse-transpose of GL³(q). Let

L= GL²(q) =

det(X)^;1 0

0 X

:X ²GL²(q)

K = SL³(q):

Then L^'Z_q^;L²(q) and LC_K(), where C_K() is the xed-point set of in K. So C_L( A) =C^Zq;( A)C_L²⁽_q⁾( A)

and C_L²⁽_q⁾() = L²(2) = ^hi' S³ G¹, where =

0 1 1 1

and =

0 1 1 0

. Thus

j^j= 3, ^jj= 2 and ^;1 = =^;1. If =^;, then T ^{h i}.

Suppose = + and w = <sup>k` 2</sup> T is a 2-element for some integers k and `. Then ^w = and 6 is a divisor of ^jw^j, which is impossible. Thus T ^{h i}, since w ²

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hw^{2i h i} when w is odd. So L²(2) C_G¹(^{h i}) C_G¹(T) and neither 2 nor 3 is a factor of ^jT^j.

Since ^jj= 1 or 3, it follows that T ^\hi= 1 and T^hi=^hi'T. But ^hi=^hi'hi is cyclic, so is T. If T =^hwⁱ, thenw ^2fkk;1kg for some k.

Ifw= ^k withkodd, then by 13, Propositions 2.2.3 and 2.3.5], w² = ^k;k2k = ^2k. If w = ^k with k even, then w³ = ^3k. Since T = ^hwⁱ = 3D^hw²ⁱ = ^hw³ⁱ, we may suppose T = ^hwⁱ with w = <sup>`</sup> for some `. Similarly, if w = ^;1k, then we may also suppose T =^hwⁱ withw = <sup>`</sup> for some `.

If G¹ = L³(q), then C_G¹() = L³(2) C_G¹(^{`</sup>) = C<sub>G</sub><sup>1</sup>(T). If G<sup>1</sup> = U<sup>3</sup>(q), then T = <sup>h</sup>w<sup>2i</sup> = <sup>h2`i} and CG¹(²) = U³(2) CG¹(^2`) = CG¹(T). Since L³(2) has an element of order 4, it follows that 4^jw^{j 2} e(G) = e(H). But 4t ⁶² e(H) for any t 1, so T = 1 and G=G¹. This proves Theorem 2.1.

Since up to now, only nitely many non-abelian nite simple groups are found to be non-distinguishable, we raise the following conjecture, which is opposite to Proposition 1.

Conjecture

. Almost all non-abelian nite simple groups are characterizible.

Acknowledgments

The results of the paper were obtained during the rst author stay at Southwest-China Normal University. He would like to thank Mr. Yuanhu Li and Mathematics Department for hospitality. The authors would also like to thank the referee for very helpful suggestions and comments.

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