NSE characterization of projective special linear group <span class="mathjax-formula">${L}_{5}(2)$</span>

NSE characterization of projective special linear group L

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SHITIANLIU(*)

ABSTRACT - LetGbe a group andv(G) be the set of element orders of G. Let k2v(G) and sk be the number of elements of order k in G. Let nse(G)ˆ fskk2v(G)g. In Khatami et al. and Liu,L3(2) andL3(4) are uniquely determined by nse(G). In this paper, we prove that ifGis a group such that nse(G)ˆnse(L5(2)), thenGL5(2).

MATHEMATICSSUBJECTCLASSIFICATION(2010). 20D05, 20D06, 20D20.

KEYWORDS. Element order, projective special linear group, Thompson's problem, number of elements of the same order.

1. Introduction

Here we introduce some notations which will be used. Ifnis an integer, then we denote byp(n) the set of all prime divisors ofn. LetGbe a group.

The set of element orders ofG is denoted byv(G). Letk2v(G) and sk

be the number of elements of order kin G. Let nse(G)ˆ fs_kk2v(G)g.

Let p(G)ˆp(jGj). Let n_p or n_p(G) denote the number of the Sylow p subgroupsP_pofG. Other notations are standard (see [1]).

A finite groupGis called a simpleK_n-group, ifGis a simple group with jp(G)j ˆn. In 1987, J. G. Thompson posed a very interesting problem re- lated to algebraic number fields as follows (see [2]).

(*) Indirizzo dell'A.: School of Science, Sichuan University of Science and Engineering, 643000, Zigong Sichuan, P. R. China.

E-mail: [email protected]

The author is supported by the Department of Education of Sichuan Province (Grant No: 12ZB085, 12ZB291 and 13ZA0119) and by the Opening Project of Sichuan Province University Key Laboratory of Bridge Non-destruction Detecting and Engineering Computing (Grant No: 2013QYJ02) and by the Scientific Research Project of Sichuan University of Science and Engineering (Grant No:

2014RC02). The author is very grateful for the helpful suggestions of the referee.

Thompson's Problem.LetT(G)ˆ f(n;sn)n2v(G) andsn 2nse(G)g, where sn is the number of elements with order n. Suppose that T(G)ˆT(H) for some (finite?) groupH. IfGis a finite solvable group, is it true thatHis also necessarily solvable?

It was proved that: LetGbe a group and M someK_i-group,iˆ3;4, thenGMif and only ifjGj ˆ jMjand nse(G)ˆnse(M) (see [3, 4]). And also the groupsA12,A13andL5(2) are characterizable by order and nse (see [5, 6, 7]). Recently, all sporadic simple groups are characterizable by nse and order (see [8]).

Comparing the sizes of elements of same order but disregarding the ac- tual orders of elements inT(G) of theThompson's Problem, in other words, it remains only nse(G), whether can it characterize finite simple groups? Up to now, some groups especially for L2(q), where qˆ7;8;9;11;13, can be characterized by only the set nse(G) (see [9, 10]). The author has proved that the groupsL₃(4) andL₂(16) are characterizable by nse (see [11, 12] respec- tively).

In this paper, it is shown that the groupL5(2), which the number of the set of the same order is 13, also can be characterized by nse(L5(2)).

2. Some Lemmas

In this section we will give some lemmas used in the proof of the main theorem.

LEMMA 2.1 [14]. Let G be a finite group and m be a positive integer dividingjGj. If Lm(G)ˆ fg2Gg^mˆ1g, then mjLm(G)j.

LEMMA2.2 [13]. Let G be a finite group and p2p(G)be odd. Suppose that P is a Sylow p-subgroup of G and nˆp^sm with(p;m)ˆ1. If P is not cyclic, then the number of elements of order n is always a multiple of p^s.

LEMMA2.3 [10]. Let G be a group containing more than two elements. If the maximal number s of elements of the same order in G is finite, then G is finite andjGj s(s² 1).

LEMMA 2.4 [15, Theorem 9.3.1]. Let G be a finite solvable group and jGj ˆmn, where mˆp^a₁¹ p^a_r^r,(m;n)ˆ1. Letpˆ fp1; ;prgand hmbe the number of Hall p-subgroups of G. Then hm ˆq^b₁¹ q^bs^s satisfies the following conditions for all i2 f1;2; ;sg:

(1) q^b_iⁱ1(mod pj) for some pj.

(2) The order of some chief factor of G is divided byq^b_iⁱ.

To prove GL5(2), we need the structure of simpleKn-groups with nˆ4;5.

LEMMA2.5 [16]. Let G be a simple K₄-group. Then G is isomoorphic to one of the following groups:

(1) A₇, A₈, A₉or A₁₀. (2) M₁₁, M₁₂or J₂. (3) One of the following:

(a) L₂(r), where r is a prime and r² 1ˆ2^a3^bv^c with a1, b1, c1, and v is a prime greater than3.

(b) L2(2^m), where2^m 1ˆu,2^m‡1ˆ3t^bwith m2, u, t are primes, t>3, b1.

(c) L₂(3^m), where 3^m‡1ˆ4t, 3^m 1ˆ2u^c or 3^m‡1ˆ4t^b, 3^m 1ˆ2u, with m2, u, t are odd primes, b1, c1.

(d) One of the following28simple groups: L₂(16), L₂(25), L₂(49), L₂(81), L3(4), L3(5), L3(7), L3(8), L3(17), L4(3), S4(4), S4(5), S4(7), S4(9), S6(2), O^‡₈(2), G2(3), U3(4), U3(5), U3(7), U3(8), U3(9), U4(3), U5(2), Sz(8), Sz(32),²D₄(2)or²F₄(2).

LEMMA2.6 [17]. Each simple K5-group is isomorphic to one of the fol- lowing simple groups:

(1) L2(q)withjp(q² 1)j ˆ4.

(2) L3(q)withjp((q² 1)(q³ 1))j ˆ4.

(3) U3(q)with q satisfiesjp((q² 1)(q³‡1))j ˆ4.

(4) O₅(q)withjp(q⁴ 1)j ˆ4.

(5) Sz(2^2m‡1)withjp((2^2m‡1 1)(2^4m‡2‡1))j ˆ4.

(6) R(q) where q is an odd power of 3 and jp(q² 1)j ˆ3 and jp(q² q‡1)j ˆ1.

(7) The following 30 simple groups: A11, A12, M22, J3, HS, He, McL, L₄(4), L₄(5), L₄(7), L₅(2), L₅(3), L₆(2), O₇(3), O₉(2), PSp₆(3), PSp₈(2), U₄(4), U₄(5), U₄(7), U₄(9), U₅(3), U₆(2), O^‡₈(3), O₈(2),³D₄(3), G₂(4), G₂(5), G₂(7)or G₂(9).

LEMMA 2.7. Let G be a simple Kn-group with nˆ4;5 and 31j jGj j2¹⁰3²5731. Then GL₂(31)or L₅(2).

PROOF. We prove the Lemma by the following two steps.

Step a.Gis a simpleK4-group.

Order consideration rules out the cases (1) and (2) of Lemma 2.5.

So we consider Lemma 2.5(3). We will deal with this with the following cases.

Case 1.GL₂(r), wherer2 f5;7;31g.

- Let rˆ5 or 7, then jp(q² 1)j ˆ2, which contradicts jp(q² 1)j ˆ3.

- Letrˆ31, thenjp(q² 1)j ˆ3. So we haveGL2(31).

Case 2.GL2(2^m), whereu2 f3;5;7;31g.

- Let uˆ3, then mˆ2 and so 5ˆ3t^b. But the equation has no solution inN, a contradiction.

- Let uˆ5 then the equation 2^m 1ˆ5 has no solution in N, a contradiction.

- Letuˆ7, thenmˆ3, and 2³‡1ˆ3t^b. Thustˆ3 andbˆ1. But t>3, a contradiction.

- Letuˆ31, thenmˆ5, 2⁵‡1ˆ311. Thus we haveGL₂(2⁵).

But 11j jL2(2⁵)j, a contradiction.

Case 3.GL2(3^m)

We will consider the case by the following two subcases.

- Subcase 3.1. 3^m‡1ˆ4tand 3^m 1ˆ2u^c. We can suppose thatt2 f3;5;7;31g

Lettˆ3;5 or 31, the equation 3^m‡1ˆ4thas no solution. So we rule out the case.

Lettˆ7, thenmˆ3 and so 3³ 1ˆ211¹, which means 11j jGj, a contradiction.

- Subcase 3.2. 3^m‡1ˆ4t^b and 3^m 1ˆ2u.

We can suppose thatu2 f3;5;7;31g

Letuˆ3;7 or 31, then the equation 3^m 1ˆ2uhas no solution in N, a contradiction.

Letuˆ5, then 3^mˆ11, a contradiction.

For the remaining case, considering the order ofGwe can rule out this case.

Step b.Gis a simpleK₅-group.

From Lemma 2.6, we will consider the following cases.

LetGL2(q), we can suppose thatqˆ2^m;3²;5;7;31.

Ifqˆ2^m, then we havemˆ6;8;9;10.

Ifmˆ6, then 13 divides the order of G.

Ifmˆ8, then17j jGj. Ifmˆ9, then 19 and 73 belong top(G).

Ifmˆ10, then 11 and 41 belong top(G).

Thus we rule out this case.

Similarly we can rule out the other groups except forL₅(2).

From Steps a and b, we have thatGL₂(31) orL₅(2).

This completes the proof of the Lemma. p

3. Main result and its proof

LetGbe a group such that nse(G)ˆnse(L₅(2)), ands_nbe the number of elements of ordernofG. By Lemma 2.3 we haveGis finite. We note that snˆkf(n), wherekis the number of cyclic subgroups of ordern. Also we note that ifn>2, thenf(n) is even. Ifm2v(G), then by Lemma 2.1 and the above discussion, we have

f(m)js_m mj P

djms_d (

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THEOREM3.1. Let G be a group withnse(G)ˆnse(L₅(2))ˆ f1, 6975, 75392, 416640, 476160, 624960, 666624, 833280, 952320, 1249920, 1333248, 1428480, 1935360g, where L5(2) is the projective special linear group of degree 5 over the field of order 2. Then GL5(2).

PROOF. We prove the theorem by first proving that p(G)ˆ f2;3;5;7;31g, second showing thatjGj ˆ jL₅(2)j, and soGL₅(2).

B y (1),p(G)ˆ{2, 3, 5, 7, 31, 41, 373, 624961, 833281, 952321, 1249921, 1333249}. Ifm>2, thenf(m) is even, hences2ˆ6975, 22p(G).

In the following, we prove that 4162p(G). If 412p(G), then by (1), s₄₁ˆ1428480. If 2412v(G), then s₈₂62nse (G). Therefore 8262v(G).

Now we consider Sylow 41-subgroupP₄₁acts fixed point freely on the set of elements of order 2, thenjP₄₁j js₂(ˆ46975), a contradiction. Similarly we can prove that the primes 373, 624961, 833281, 952321, 1249921, and 1333249 do not belong top(G). Hence we havep(G)ˆ f2;3;5;7;31g. Fur- thermore, by (1)s₃ˆ75392,s₅ˆ666624,s₇ˆ476160 ands₃₁ˆ1935360.

If 7132v(G), then by (1) 91j1‡s₇‡s₁₃‡s₉₁andf(91)js₉₁, and so we haves₉₁62nse(G), a contradiction. Hence 71362v(G).

If 2^a12v(G), thenf(2^a1)ˆ2^{a1 1}js2^a1 and so 0a112.

If 5^a22v(G), then 0a22. If 5²2v(G), thens2562nse(G), a con- tradiction. Hence 0a21.

If 7^a32v(G), then 0a₃2. If 7²2v(G), thens₄₉62nse(G), a con- tradiction. Therefore 0a₃1.

If 31^a4 2v(G), then 0a₄2. Sinces₃₁262nse(G), then 0a₄1.

If 3^a52v(G), then 0a54.

In the following, we consider p(G) as a subset of f2;3;5;7;31g which contains the prime 2.

Case a.p(G)ˆ f2g.

In this case, v(G) f1;2;2²; ;2¹²g. But the number of the set of nse(G) is 13, it is easy to have a contradiction since the equation 9999360‡ 75392k₁ ‡ 416640k₂ ‡ 476160k₃ ‡ 624960k₄ ‡ 666624k₅ ‡ 833280k₆ ‡ 952320k7‡1249920k8‡1333248k9‡1428480k10‡1935360k11ˆ2^m, where k1;k2; ;k11;mare nonnegative integers and 0k1‡k2‡k3‡k4‡:::‡ k₁₁0. Then the equation has no solution inN.

Case b.p(G)ˆ f2;3g.

If 2^a3^b2v(G), then by Lemma 2.1, we have 0a11 and 0b4.

We have exp(P3)ˆ3, 9, 27 or 81.

Suppose that exp(P3)ˆ3, then by Lemma 2.1, we have thatjP3j j1‡ s₃(ˆ79393) and sojP₃j j9.

Let jP3j ˆ3. Then n3ˆs3=f(3)ˆ75392=2ˆ2⁶1931, which means19;312p(G), a contradiction.

Let jP3j ˆ9. Then 9999360 ‡ 75392k1 ‡ 416640k2 ‡ 476160k3 ‡ 624960k₄‡666624k₅‡833280k₆‡952320k₇‡1249920k₈‡1333248k₉

‡1428480k₁₀ ‡ 1935360k₁₁ˆ2^m3², where k₁;k₂; ;k₁₁;m;n are nonnegative integers and0k₁‡k₂ ‡k₃‡k₄ ‡:::‡k₁₁36. So 156240‡1178k1‡6510k2‡7440k3‡9765k4‡10416k5‡13020k6‡ 14880k7‡19530k8 ‡20832k9‡22320k10‡30240k11ˆ2^{m 6}3²and since31(5040‡38k₁‡210k₂‡240k₃‡315k₄‡336k₅‡420k₆‡480k₇

‡630k₈‡672k₉‡720k₁₀)‡30240k₁₁ˆ2^{m 6}3², the equation has no solution inN.

Suppose that exp(P₃)ˆ9, then by Lemma 2.1,jP₃j j1‡s₃‡s₉and so jP₃j j27.

Let jP₃j ˆ9, then s₉ ˆ1249920;624960;1428480;1935360 and so n₃ˆs₉=f(9). It follows that52p(G), a contradiction.

Let jP₃j ˆ27, then by Lemma 2.2, we have s₉ˆ1935360, and so n3ˆs9=f(9). We also have that52p(G), a contradiction.

Suppose that exp(P3)ˆ27, then if jP3j ˆ27 and by Lemma 2.1, s27ˆ1244920;1935360 and so n3ˆs27=f(27). It follows that 52p(G), a contradiction. If jP3j 3⁴, then similarly as the Case ``exp(P3)ˆ3 and jP₃j ˆ9'', we also can rule out these cases.

Suppose that exp(P₃)ˆ81, thens₈₁ˆ1935360 and son₃ˆs₈₁=f(81)ˆ 7680, it follows that 52p(G). IfjP₃j 243, then by Lemma 2.2,s₈₁ˆ81t for some integert, but the equation has no solution inN.

Case c.p(G)ˆ f2;5g

Since 5²62v(G), then by Lemma 2.1,jP5j j1‡s5(ˆ5333125) and so jP5j j125.

Let jP₅j ˆ5, then n₅ˆs₅=f(5)ˆ2⁸3731. It follows that 3;7;312p(G), a contradiction.

LetjP5j5². Then by Lemma 2.2,s5ˆ25tfor some integert, but the equation has no solution inN.

Case d.p(G)ˆ f2;7g.

Since 7²62v(G), then jP₇j j1 ‡ s₇(ˆ12481) and so jP₇j j7. So n₇ˆs₇=f(7)ˆ79360ˆ2⁹331, which means 3;312p(G), a contra- diction.

Case e.p(G)ˆ f2;31g

Since 31²62v(G), then exp(P₃₁)ˆ31. Then jP₃₁j j1 ‡ s₃₁ and so jP31j ˆ31. Sincen31ˆs31=f(31)ˆ2¹⁰3²7, 3;72p(G), a contradiction.

Case f. p(G) isequal tof2;3;5g,f2;5;7g,f2;5;31g,f2;3;5;7g,f2;3;5;31g orf2;5;7;31g.

Similarly as the Case c, we also get a contradiction.

Case g.p(G)ˆ f2;3;7g, orp(G)ˆ f2;7;31g

Similarly as the Case d or Case e, we get a contradiction.

Case h.p(G)ˆ f2;3;5;7;31g.

We have known, as for Cases c and d, thatjP7j ˆ7 andjP31j ˆ31.

Step 1. jGj ˆ2¹⁰3²5731, jGj ˆ2¹⁰3³5²731, jGj ˆ2¹¹3² 5731, andjGj ˆ2¹¹3³5²731.

We show that 5762v(G).

If 572v(G), setPandQare Sylow 7-subgroups ofG, thenPandQ are conjugate in G and so C_G(P) and C_G(Q) are also conjugate in G.

Therefore we have s₃₅ ˆf(35)n₇k, where k is the number of cyclic subgroups of order 5 in C_G(P7). As n7ˆs7=f(7)ˆ476160=6ˆ2080, 7936024js15. We haves3562nse(G). We conclude that 3562v(G). It fol-

lows that the Sylow 5-groupP5 ofGacts fixed point freely on the set of elements of order 7,jP5j js7, and sojP5j ˆ5.

We show that 23162v(G).

If 2312v(G), setPandQare Sylow 31-subgroups ofG, thenPandQ are conjugate in G and so C_G(P) and C_G(Q) are also conjugate in G.

Therefore we have s₆₂ˆf(62)n₃₁k, where k is the number of cyclic subgroups of order 2 inC_G(P31). Asn31ˆs31=f(31)ˆ1935360=30ˆ64512, 64512js15 and so s62ˆs31. On the other hand, 62j1 ‡ s2 ‡ s31 ‡ s₆₂(ˆ3870721), we also get a contradiction. We conclude that 23162v(G).

It follows that the Sylow 2-group ofGacts fixed point freely on the set of elements of order 31, and sojP₂j js₃₁(ˆ1935360). ThusjP₂j j2¹¹.

We show that 53162v(G).

If 5312v(G), then s155ˆf(155)n31k, where k is the number of cyclic subgroups of order 5 inC_G(P₃₁), and so 7849440js₁₅₅. So we have s₁₅₅ 62nse(G), a contradiction. Hence 53162v(G). Similarly we can prove that 33162v(G) and sojP₃j js₃₁. ThusjP₃j j3³.

Therefore jGj ˆ2^m3ⁿ5731. But P

sk2nse(G)skˆ9999360ˆ2¹⁰3² 57312^m3ⁿ5731. Hence we have that jGj ˆ2¹⁰3²5731, jGj ˆ2¹⁰3³5731,jGj ˆ2¹¹3²5731, andjGj ˆ2¹¹3³5731.

Step 2.GL₅(2).

In this step, we first prove that there are no groups such that jGj ˆ 2¹⁰3³5731, jGj ˆ2¹¹3²5731, and jGj ˆ2¹¹3³5731 with nse(G)ˆnse(L5(2)), thenjGj ˆ2¹⁰3²5731 and nse(G)ˆnse(L5(2)), it follows, from [6] thatGL₅(2).

LetjGj ˆ2¹⁰3³5731 with nse(G)ˆnse(L₅(2)).

IfGis soluble, then by Lemma 2.4, for every primep,p^pi 1(modq) for some q. Thus we have 51(mod 13), a contradiction. SoG is insoluble.

Therefore we can suppose thatGhas a normal series 1/K/L/Gsuch that L=Kis isomorphic to a simpleK_i-group withiˆ3;4;5 as 5²;7²and 31²do not divide the order ofG.

IfL=Kis isomorphic to a simpleK₃-group, then by [18],L=K is iso- morphic toA₅,A₆,L₂(7),L₂(8),U₃(3) orU₄(2).

- If L=KA5, then by [1], n3(L=K)ˆn3(A5)ˆ10, and so n3(G)ˆ10t for some integer t with 3j⁶t. Hence the number of elements of order 3 inGis:s3ˆ20tˆ75392. But the equation has no solution inN, a contradiction.

- Similarly, for the groupsA₆,L₂(7),L₂(8),U₃(3) andU₄(2), we also can rule out these cases.

If L=K is isomorphic to a simple Kn-group with nˆ4;5, then by Lemma 2.7,L=Kis isomorphic toL2(31) orL5(2).

- IfL=KL2(31), then from [1],n31(L=K)ˆn3(L2(31))ˆ32, and so n₃₁(G)ˆ32tfor some integer t with 31j⁶t. Hence the number of elements of order 31 inGiss₃₁ˆ32t30ˆ960tand sotˆ2016. So 2⁵3²7j jKj j2⁵3²7, HencejKj ˆ2⁵3²7, andjN_K(P₃₁)j ˆ1.

So K\N_G(P31)ˆK\C_G(P31), which means that K^jP31 is a Frobenius group and sojP31jjAut(K)j, a contradiction.

- IfL=K L₅(2), setGˆG=K andLˆL=K. Then

L₃(4)LLC_G(L)=C_G(L)G=C_G(L)ˆN_G(L)=C_G(L)Aut(L) Set Mˆ fxKjxK2C_G(L)g, then G=MG=C_G(L) and so L5(2)G=MAut(L5(2)). ThereforeG=Mis isomorphic toL5(2) or 2:L5(2)

IfG=ML₅(2), thenjMj ˆ3 and soMˆZ(G). Hence Ghas an element of order 331, a contradiction.

IfG=M2:L₅(2), order consideration rules out this case.

Similarly as the arguments of ``jGj ˆ2¹⁰3³5731 with nse(G)ˆ nse(L5(2))'', we can rule out these cases: jGj ˆ2¹¹3²5731, and jGj ˆ2¹¹3³5731 with nse(G)ˆnse(L₅(2)).

Therefore we havejGj ˆ2¹⁰3²5731ˆ jL₅(2)jand by assumption nse(G)ˆnse(L₅(2)), then by [6], we haveGL₅(2).

This completes the proof. p

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Manoscritto pervenuto in redazione il 5 Gennaio 2013.