On the Dirichlet Polynomial of Finite Groups of Lie Type.
ERIKADAMIAN(*) - ANDREALUCCHINI(*)
Dedicated to Guido Zappa on his 90th birthday
ABSTRACT- For a given finite groupGthere exists a uniquely determined Dirichlet polynomialPG(s) with the property that fort2Nthe numberPG(t) coincides with the probability of generatingGbytrandomly chosen elements. We discuss whether the isomorphism type of a simple groupGcan be determined by the knowledge ofPG(s):
1. Introduction.
For a given finite group G one may define a sequence of integers fan(G)gn2N as follows:
an(G) X
jG:Hjn
mG(H):
HeremGdenotes the MoÈbius function defined on the subgroup lattice ofG.
In particular, one hasmG(G)1 andmG(H) P
H<KmG(K) for anyH<G.
Let
PG(s)X
n2N
an(G) ns
be the Dirichlet generating function associated with the sequence fan(G)gn2N. The Dirichlet polynomial PG(s) gives a great amount of in- formation aboutG:In [13] P.Hall observed that for anyt2N the series PG(t) gives the probability thattrandomly chosen elements ofGgenerate G. Moreover (see for example [3]), the complex function PG(s), and in (*) Indirizzo degli AA.: UniversitaÁ di Brescia, Dipartimento di Matematica, Via Valotti, 25133 Brescia, Italy.
E-mail: [email protected] E-mail: [email protected]
particular its value at -1, can be used to investigate the topological prop- erties of the poset of proper cosets ofG:
A problem that has been tackled from various points of view ([10], [9], [6]) concerns the study of which properties ofPG(s), as an element of the ring of Dirichlet polynomials with integer coefficients, reflect on properties of the groupG. Recently [8] we proved that the knowledge ofPG(s) allows us to decide whether the factor groupG=Gis a simple group or not: ifG1is simple andPG2(s)PG1(s);thenG2=(G2) is simple. We conjecture that a stronger result is true: if G1 is simple and PG2(s)PG1(s) then G2=G2G1. This has been proved for alternating simple groups [7], so the aim of this paper is to investigate the case of simple groups of Lie type. The main result we will obtain is that if we know thatGis a simple group of Lie type over a field of characteristicp;then with the help of the functionPG(s) we can determine the order of a Sylowp-subgroup ofG. This result will be employed to prove that a sporadic simple group can be identified from its Dirichlet polynomial, and also to show that ifG1andG2are non isomorphic simple groups of Lie type defined in the same characteristic, then PG2(s)6PG1(s):
2. Known results and open questions.
LetGbe a finite simple group. We want to discuss about the properties ofGthat can be deduced from the knowledge of the Dirichlet polynomial PG(s):The simplest thing to do is to consider the set
y(G) fn2Njan(G)60g:
Ifn2y(G) thenGmust have a subgroup of indexn;moreover the smallest n2y(G) with n>1 coincides withm(G); the smallest index of a proper subgroup ofG:The main obstacle when we work withy(G) is that even when we know that G has a subgroup H with index n we are not sure than n2y(G);in fact, in order to decide whetheran(G)60 we should focus on the set of subgroupsHGwith indexnand such thatmG(H)60;but here we meet another problem as the suman(G) P
jG:HjnmG(H) can be zero even if the terms are different from zero. A case in which it is easy to deduce that an(G)60 is when we know thatGhas a maximal subgroup of indexnand there is no maximal subgroup with index a proper divisor ofn(in that case an(G) is precisely the number of maximal subgroups of indexn). One can expect that the sety(G) is large enough to give good hints concerning the
order of the group G:Let us formulate two conjectures in this direction.
Define the probabilistic order ofGas follows:
po(G)l:c:m:fnjn2y(G)g
Clearly po(G) dividesjGj;our first conjecture is that po(G) jGj;a weaker conjecture is that we can deduce fromy(G) which are the prime divisors of jGj, namely:p(po(G))p(G): Both these conjectures are open questions and it seems a difficult but intriguing task to prove something in this di- rection without a heavy use of the classification of finite simple groups and of their maximal subgroups.
In [9] and [7] it is proved that an analysis of y(G) allows to decide whetherGis an alternating group; more precisely we have:
THEOREM1. Let G be a finite nonabelian simple group. Set nm(G) and let p be the minimal prime number which divides n;
(1) if np then: GAlt (n)if and only if(n 2)!2y(G);
(2) if n>p and n2 f6,24g= then: GAlt (n) if and only if an(G) n and n
p is the smallest integer iny(G) which is different from 1 and not divisible by n;
(3) if n6then: GAlt (6)if and only if a6(G) 12;
(4) if n24then: GAlt (24)if and only if253, 7592=y(G):
Hence for the rest of the paper we shall restrict our attention to sporadic simple groups and simple groups of Lie type.
3. Dealing with groups of Lie type.
The problem of recognizing a simple group Gfrom its Dirichlet poly- nomialPG(s) is still open. In this section we shall prove a result which turns out to be useful for the analysis of this problem for groups of Lie type.
Indeed we shall show that if we know thatGis a simple group of Lie type over a field of characteristicp; then we may use the polynomialPG(s) in order to determine the order of a Sylowp-subgroup ofG:
We need to define some other Dirichlet polynomials associated withG and its subgroups.
LetRbe the ring of Dirichlet polynomials with integer coefficients; for any finite set of prime numberspwe may define a ring endomorphism ofR as follows:
hp: R ! R f(s)X1
n1
an
ns 7! f(p)(s)X1
n1
bn
ns where
bn an if nis ap0-number 0 otherwise:
(
We are mainly interested inP(q)G(s), withqa prime number.
Moreover, for any subgroup K of G, we may define a Dirichlet poly- nomial as follows:
PG(K;s)X
n2N
an(G;K)
ns where an(G;K) X
jG:Hjn;
KHG
mG(H):
LEMMA2. Let P be a Sylow p-subgroup of a finite group G, p a prime number; suppose that each maximal subgroup of G which contains P contains also NG(P):Then
P(p)G (s)PG(P;s 1)PG(NG(P);s 1):
PROOF. First we claim that mG(H)jNH(P)j mG(H)jNG(P)j for each subgroup PHG: Indeed, either mG(H)0 or H can be written as intersection of maximal subgroups, say M1;. . .;Mt (see [14]); as PHMi; by hypothesis we get that NG(P)Mi, which implies NG(P)M1\ \MtHand NG(P)NH(P): Now set Vp
fHGj jHjp jGjpg, then P(p)G (s) X
H2Vp
mG(H) jG:Hjs
X
Q2Sylp(G)
X
QH
mG(H)
jG:Hjs 1 jH:NH(Q)j
1 jNG(P)j
X
PH
mG(H)jNH(P)j jG:Hjs 1
1 jNG(P)j
X
PH
mG(H)jNG(P)j jG:Hjs 1
X
PH
mG(H)
jG:Hjs 1PG(P;s 1):
This proves the first equality in our statement. The other one, PG(P;s 1)PG(NG(P);s 1); is again an immediate consequence of the previous remark that if mG(H)60 andPH, then NG(P)H:
p THEOREM3. Suppose that G is a finite group of Lie type defined over a field of characteristic p and let U2Sylp(G):ThenjP(p)G(0)j jUj.
PROOF. A finite groupGof Lie type over the fieldFq,qpf;can be constructed starting from a connected reductive algebraic groupXdefined over an algebraically closed field of characteristic p and considering the subgroupGXFof fixed points under a Frobenius mapF:LetBbe anF- stable Borel subgroup ofX:The unipotent radicalU ofBF is a Sylowp- subgroup ofGFandNG(U)BF. As it is well known, a maximal subgroup ofGFwhich containsUshould containBF, hence it is a maximal parabolic subgroup ofGF, so we can apply the previous lemma in order to deduce that
P(p)G(0)PG(BF; 1) X
BFH
mG(H)jG:Hj:
To the mapFa symmetryron the Dynkin diagram ofXis associated (ris trivial in the untwisted case). LetI: fO1;. . .;Okgbe the set of ther-or- bits on the nodes of the Dynkin diagram. IfJI, thenJ[
j2J
Ojis ar- stable subset of the set of nodes of the Dynkin diagram and one may as- sociate anF-stable parabolic subgroupPJofXwithJ. Moreover, the map J7!PFJis an isomorphism between the latticeP(I) of subsets ofIordered by inclusion, and the lattice of subgroups ofGcontainingBF. In particular, mG(PJ)mP(I)(J)( 1)k jJj(see [20], 3.8.3). As described in [4], Ch. 9, to any subset Jof I, a parabolic subgroup WJ of the Weyl groupWF and a polynomialPWJ(x) are associated with the property thatPWJ(q) jPFJj:So one has
P(p)G(0) X
BFH
mG(H)jG:Hj X
JI
mG(PFJ)jG:PFJj
X
JI
( 1)k jJjjG:PFJj ( 1)kX
JI
( 1)jJj PW(q) PWJ(q)
By a theorem of Solomon (see 9.4.5. and Ch. 14 in [4]) X
JI
( 1)jJj PW(q) PWJ(q)
jUj
so we conclude thatP(p)G(0)( 1)kjUj: p
COROLLARY4. LetSbe a simple group of Lie type defined over a field K; if we know that the characteristic ofK isp;then we may determine from the Dirichlet polynomialPS(s)the order of a Sylowp-subgroup ofS:
PROOF. Let S f2F4(2)0;G2(2)0;2G2(3)0;B2(2)0g: If S2 S= then there exists a finite group of Lie typeGwithSG=Z(G); moreoverpdoes not dividejZ(G)jand, by Theorem 3,P(p)G(0)P(p)S (0) is the order of a Sylow p-subgroup of S: On the other hand a direct computation shows that jP(p)S (0)j pjUjwhenS2 SandU is a Sylowp-subgroup ofS:Note that m(B2(2)0)6; m(G2(2)0)28; m(2G2(3)0)9;m(2F4(2)0)1600: Now let Sbe a simple group of Lie type defined over a field of characteristic p:
From the series PS(s) we recover the value of m(G): If (p;m(S))2=
=
2f(2, 6);(2, 28);(2, 1600); (3, 9)g, thenS62 SandjP(p)S (0)jcoincides with the order of the Sylow p-subgroup of S. Ifp2 andm(S)28 then either SB3(2) and jP(2)G(0)j 29 or SG2(2)0 and jP(2)G(0)j 26: If p2 and m(S)1600 then G2F4(2)0; if p2 and m(S)6 then GB2(2)0; if
p3 andm(S)9 thenG2G2(3)0: p
4. Connection between Theorem 3and the Solomon-Tits Theorem.
In the previous section we proved Theorem 3 by computing directly PG(BF; 1), which is possible as we can compute the index and the MoÈbius function of any parabolic subgroup ofG. We would like now to present a different proof for the same result; this is less immediate and direct, but it makes evident the connection between the study of the Dirichlet polynomial PG(s) and some topological properties of the poset of proper cosets inG.
We first revise and generalize a well-known result of K. S. Brown [3].
LetKbe a proper subgroup of a finite groupG:We define two posets:
the first one, C C(G;K); consists of the proper cosets Hx (KH<
<G;x2G) ordered by inclusion. The secondC C(G;K) is the subposet ofCconsisting of the cosetsHxwhereHsatisfies the additional property of being intersection of maximal subgroups ofG:As it is well known, we can apply topological concepts to a posetPby using the simplicial complexD(P) (the order complex ofP) whose simplices are the finite chains inP:
LEMMA5. The complexesD(C)andD(C)are homotopy equivalent.
PROOF. We recall the following criterion due to Quillen ([18] Proposi- tion 1.6): if there exists an order preserving mapf :X!Y between two
posets, with the property that for any y2Y the lower fiber f=y
fx2Xjf(x)ygis contractible, then the complexesD(X) andD(Y) are equivalent. In our case we may define an order preserving map f :C ! C by sending Ux to Ux; being U the intersection of the maximal sub- groups of G containing U: For any Vx2 C the lower fiber f=Vx
fUyjy2VxandKUVg is contractible as Vx is a least upper
bound for f=Vx: p
IfG is ann-dimensional complex, the Euler-PoincareÁ characteristic of G is the integer x(G) P
0qn( 1)qaq;where aq is the number ofq-sim- plices of G: The reduced Euler-PoincareÁ characteristic is defined as:
~
x(G)x(G) 1:
Generalizing a result due to Brown in the particular case K1; we prove that the Euler-PoincareÁ characteristic of the complexes D(C) and D(C) can be computed with the help of the functionPG(K;s):Indeed one has:
PROPOSITION6. PG(K; 1) ~x(D(C(G; K))) ~x(D(C(G; K))):
PROOF. SinceD(C(G;K)) andD(C(G;K)) are equivalent, it suffices to prove the first equality. LetCbe the poset obtained by adding toC(G;K) a greatest element (which we may well take asG) and a least element (that we denote by 0). Recall that the MoÈbius functionmPassociated to a posetPhas the property that if a<b then mP(a;b) is the number of chains of even length in the interval (a;b) fc2 P ja<c<bg minus the number of chains of odd length (here we include the empty chain, which we agree to consider of length -1); this implies that mP(a;b)~x(D(a;b)):In our parti- cular case we obtain mC(0;G)~x(D(C(G;K)): Now let Hx2 C(G;K); the interval [H;G] in the subgroup lattice of Gis isomorphic to the interval [Hx;G] inCvia the mapU!Ux;thusmC(Hx;G)mG(H) and
PG(K; 1) X
KHG
mG(H)jG:Hj mG(G) X
Hx2C(G;K)
mC(Hx;G)
mC(G;G) X
Hx2C(G;K)
mC(Hx;G) mC(0;G) ~x(D(C(G;K)):
This concludes our proof. p
We shall assume for the remaining part of this section thatGis a finite group of Lie type of rankn; as it is well knownGadmits a (B;N)-pair with Ba Borel subgroup ofG:LetfM1;. . .;Mngbe the set of maximal parabolic
subgroups of G containing B: The Tits building T(G;B;N) of G is the simplicial complex whose vertices are the cosets Mix with x2G and 1in and whose simplices are collections of vertices with nonempty intersection. If p is the characteristic of the underlying field of G, then UOp(B) is a Sylowp-subgroup ofG;and the following holds.
PROPOSITION7. The complexes D(C(G; U))andT(G;B; N)are homo- topy equivalent.
PROOF. Recall that ifBis a complex, then one can consider the faced posetP(B), consisting of the simplices ofBordered by inclusion; the order complex D(P(B)) (the first barycentric subdivision of B) is homotopy equivalent toB:Moreover ifPis a poset, thenD(P)D(Pop), beingPopthe dual poset ofP:These remarks together with Lemma 5, implies that our statement is proved if we can show thatD(P(T(G;B;N))) andD(C(G;U)op) are equivalent. By definition an element of P(T(G;B;N)) is a set fMi1x;. . .;Mirxg with 1i1< <irn; on the other hand the ele- ments ofC(G;U) are cosetsHxwhereHis intersection of maximal sub- groups containingU; since such anHcan be written in a unique way as intersection of maximal parabolic subgroups in fM1;. . .;Mng we obtain that the mapfMi1x;. . .;Mirxg 7!(Mi1\ \Mir)xin an order preserving bijection between the posetsP(T(G;B;N)) andC(G;U)op. p
COROLLARY8. PG(U; 1) ~x(T(G;B; N)):
So we may computePG(U; 1) with the help of the celebrated Borel- Tits Theorem [19], which asserts that the Tits buildingT(G;B;N) has the homotopy type of a wedge ofjUjspheres, each one of dimension (n 1).
Since the reduced Euler-PoincareÁ characteristic of a wedge oft r-dimen- sional spheres is ( 1)rtwe get:
COROLLARY9. PG(p)(0)PG(U; 1)( 1)njUj.
5. Consequences of Theorem 3.
PROPOSITION10. IfGis a finite simple group of Lie type over a field of characteristicp;thenp2p(po(G)):
PROOF. By Theorem 3, we have thatP(p)G(0)60;while, by the definition of the MoÈbius function, PG(0) P
HGmG(H)0: This implies PG(s)6
6P(p)G (s);which is possible only when there exists a positive integerndi-
visible bypwithan(G)60: p
It is useful to define the following set:
~y(G) is the set of n2N with the following property: G contains a maximal subgroup M of index n but it does not contain any proper subgroupHsuch thatjG:Hjis a proper divisor ofn:
In this section we will make a large use of the following fact (already recalled in section 2):~y(G)y(G):
THEOREM11. Let G be a sporadic simple group; if H is a finite simple group with PG(s)PH(s), then GH:
PROOF. By Theorem 1,Hcannot be an alternating group. Suppose that HLn(q) is a group of Lie type of ranknover a fieldFqof characteristicp;
we know thatm(G)m(H);and the possible values form(G) andm(H) are listed in Table 1; for any familyLof groups of Lie type, this table provides a function f(L;n;q) such that m(Ln(q))f(L;n;q): So we have to check whether there are some choices ofL;n;qfor whichf(L;n;q)m(G):The key tool here is that thep-adic expansion off(L;n;q) is of a very particular and recognizable shape (for example the first and last digits are equal to 1 and the second is 0 or 1); moreover by Proposition 10,p2p(po(H))p(po(G));
hencepmust be a prime divisor ofjGj;so what we have to do is to write thep- adic expansion ofm(G) for any prime divisorpofjGjand check whether for some choice ofL;n;q, withqap-power, thep-adic expansions ofm(G) and of f(L;n;q) are the same. This is almost never the case; in fact there are only two possibilities: (G;H)2 f(M11;A1(11));(M24;A1(23))g. In the first case one can check thatmM11(1)60;sojM11j 79202y(M11);while 79202=y(A1(11)) (as 7920 does not divide jA1(11)j 660): this is enough to conclude that PM11(s)6PA1(11)(s):In the second case, one can notice that 17712~y(M24) and does not dividejA1(23)j 6072:To conclude our proof it remains to consider the case whenGandHare both sporadic simple groups. From Table 1, we have that if G and H are non isomorphic sporadic simple groups with m(G)m(H), thenfG;Hg fJ2;HSg:But we can conclude thatPJ2(s)6
6PHS(s) as 1762~y(HS) and 176 does not dividejJ2j: p For the remaining part of this paper our attention will be restricted to finite simple groups of Lie type. It is well known, see [1], [16], that ifGand Hare non isomorphic simple groups of the same order, thenGandHei-
TABLE1. Minimal index for sporadic and Lie groups.
G m(G) G m(G)
M11 11 M12 12
M22 22 M23 23
M24 323 J4 11229313743
J2 2252 J3 223419
J1 2719 Co1 23335713
Co2 225223 Co3 22323
Fi22 233513 Fi23 341723
Fi024 23337229 Ly 2234113767
McL 5211 He 22373
Ru 225729 O'N 233251131
Suz 23411 B 233454233147
HS 2252 Th 2335531931
HN 2535419 M 371445565229415971
G2(3) 3313 A1(7) 7
G2(4) 2513 A1(11) 11
A3(2) 8 2F4(2)0 2652
B2(3) 27 2A2(5) 50
An(q) qn1 1
q 1 Bn(3); (n3) 1
23n 1(3n 1) Bn(q) q2n 1
q 1 Cn(q) q2n 1
q 1 Bn(2) 2n 1(2n 1) 2An(2);(6jn1) 2n(2n1 1)
3
2A2(q) q31 2A3(q) (q1)(q31)
Dn(q) (qn1)(qn 1 1)
q 1 2An(q);(n4) [qn1 ( 1)n1][qn ( 1)n] q2 1
2Dn(q) (qn1)(qn 1 1)
q 1 3D4(q) (q1)(q8q41) G2(q) q6 1
q 1 E6(q) (q41)(q9 1)(q12 1) (q3 1)(q 1)
2B2(q) q21 2E6(q) (q41)(q91)(q12 1)
(q31)(q 1) F4(q) (q41)(q12 1)
q 1 2F4(q) (q1)(q31)(q61)
2G2(q) q31 E7(q) (q14 1)(q8 1)(q12 1)
(q6 1)(q4 1)(q 1) Dn(2) 2n 1(2n 1) E8(q) (q101)(q16 1)(q24 1)
(q6 1)(q 1)
ther are A2(4) and A3(2) or are Bn(q) and Cn(q) for somen3 and some odd q: So our task of recognizing a simple group G from its Dirichlet polynomial would be nearly completed if we could determine jGj from PG(s);however, as we mentioned in Section 2, this is not an easy problem, and it can be tackled only with an intensive use of results on the maximal subgroups of Lie groups. However the following remark is useful and easy to prove.
LEMMA 12. Let G be a simple group of Lie type and let B be a Borel subgroup. IfjG:Bj u, then u2y(G):
PROOF. As we noticed in the proof of Theorem 3,mG(B)( 1)lwherel is the Lie rank in the untwisted case and the number ofr-orbits on the nodes of the Dynkin diagram in the twisted case. Now suppose jHj jBj and mG(H)60;asjG:Hj uis coprime withp;the subgroupHcontains a Sylow p-subgroupPofG; the normalizerB NG(P) is again a Borel subgroup and is contained in each maximal subgroup of Gwhich containsP; asHis an intersection of maximal subgroups,B H, butjG: Bj jG:Hj u;hence HB; as the Borel subgroups inGare all conjugated and self-normalizing, we conclude thatau(G) jG:NG(B)jmG(B)( 1)lu: p The previous lemma tells us that the p0-part of po(G) cannot be too different fromjGjp0;indeedjGjp0 jG:BkHj;beingHa Cartan subgroup of the Borel subgroup B, hence (jGj=po(G))p0 divides jHj: What is more difficult to understand is how much smaller can (po(G))p be compared to jGjp:However, if we already know thatp is the characteristic of the Lie group G;then, by Corollary 4, with the help of the Dirichlet polynomial PG(s) we may compute the number po(G)l:c:m:(jGjp;po(G)), and use this number as a good approximation forjGj:For some groups of low rank (A1(q);2B2(q);2A2(q)); we will need a more accurate estimate of po(G):
Before starting the analysis of these particular cases, let us recall some definitions and results concerning Zsigmondy primes. A prime numberuis called aprimitive prime divisorofab 1 if it dividesab 1 but it does not divide ae 1 for any integer 1eb 1. It was proved by Zsigmondy [22] that ifaandbare integers greater than 1 and (a;b)6(2;6);then there exists a primitive prime divisor of ab 1 except when a2 and b is a Mersenne prime.
LEMMA13. The following hold:
(1) If GA1(q)then q(q 1)=2dividespo(G);
(2) if G2B2(q)then q 1dividespo(G);
(3) if G2A2(q)with q3rand r>1;then(q1)2(q 1)divides po(G):
PROOF. (1) IfGA1(q);then looking at the list of its maximal sub- groups (see for example [15], Satz 8.27, pag. 213) one can notice the fol- lowing: if q67;9;11; then GA1(q) has a maximal subgroup D iso- morphic to the dihedral group of order 2(q1)=(2;q 1) andjMj jDjfor eachMmaximal subgroup ofGwith index dividingjG:Dj(more precisely eitherMis conjugate toDorq59 andMAlt (5));so ifq67;9; 11, then jG:Dj q(q 1)=22~y(G). Finally a direct computation shows that po(G) jGjwhenG2A1(7);A1(9);A1(11):
(2) IfG2B2(q) thenq2ewithe3 odd;jGj q2(q21)(q 1) and the maximal subgroups ofGare the following [21]:
(a) B(the parabolic subgroup) withjBj q2(q 1);
(b) XawithjXaj 4(qa
p2q
1) anda2 f 1;1g;
(c) D'D2(q 1) withjDj 2(q 1);
(d) 2B2(q0) whereqqa0andais a prime divisor ofe.
Notice that (q
p2q
1)(q
p2q
1)q21 so there exists a2 f 1;1gsuch thatjXajis divisible by a primitive prime divisoruof 24e 1.
A maximal subgroupMhas order divisible byuonly whenMXa, hence jG:Xaj q2(q a
p2q
1)(q 1)=42~y(G);andq 1 divides po(G):
(3) Forq3r;r61;G2A2(q)PSU(3;q) has orderq3(q31)(q2 1) and contains the following maximal subgroups [2] (hereZndenotes the cyclic group of ordern):
(a) B[q3]:Zq2 1; (b) GU(2;q);
(c) (Zq1)2:Sym(3);
(d) Zq2 q1:3;
(e) PSU(3;q0) withqqa0 andaprime;
(f) SO(3;q):
Let now ube a primitive prime divisor of 36r 1: IfM is a maximal subgroup ofGwith order divisible byuthenMis of the kind described in (d), hencejGj=(3(q2 q1))q3(q1)2(q 1)=32~y(G): p
Now we are ready to start with the proof of our main result.
THEOREM14. Let G1and G2be two simple groups of Lie type defined over fields with the same characteristic; if PG1(s)PG2(s), then G1G2:
The proof is quite long and will be splitted in consecutive steps. The main ingredient is an analysis performed in [16], which explains how one can recognize a finite simple group of Lie type from its order. For the convenience of the reader we recall some definitions introduced in [16]. Let n>1 be a natural number, thecontributionof a primeptonis the highest power ofpdividingn; the following invariants can be defined:
p(n): is called thedominant primein nand it is the prime number whose contribution tonis maximal.
l(n): is the exponent ofp(n) so thatp(n)l(n)is the largest power ofp(n) dividingn:
v(n): is the largest order of p(n) modulo a prime divisor p1 of n=p(n)l(n); such a prime number is called aprominent prime.
c(n): is the largest order ofp(n) modulo a non-prominent prime; one putsc(n)0 in casenhas no non-prominent prime divisor other thanp(n).
If G is a group of order n than the numbers p(G)p(n); l(G)
l(n); v(G)v(n) and c(G)c(n) are called the Artin invariants of jGj: In [16] the authors prove that apart from few exceptions, there is a unique simple group of Lie type with given values of the Artin invariants.
Our first task will be to prove that we can obtain the Artin invariants by looking at the Dirichlet polynomial of the group. From now on,Gwill be a finite group of Lie type over a field of characteristic p: In generalp(G) coincides with the characteristicpof our group of Lie type; indeed we have the following (see [16, Theorem 3.3]):
STEP 1. Assume that G is a group of Lie type over a field of char- acteristic p with G62A2(3);2A3(2);then p(G)6p if and only if m(G)is a prime-power with m(G)>jGjp:Moreover one of the following occurs:
(1) m(G)is a 2-power, in which case pm(G) 1is a Mersenne prime and GA1(p);
(2) m(G) is a Fermat prime, in which case p2 and G
A1(m(G) 1);
(3) m(G)9;in which case p2and GA1(8):
This has the following consequence:
STEP2. Let G be a simple group of Lie type defined over a field F; if we know that the characteristic of F is p;then we may determine from the Dirichlet polynomial PG(s)whether p(G)p or not, and when p(G)6p we may determine G up to isomorphism.
PROOF. First notice thatG2A2(3) is the unique simple group of Lie type withm(G)27;on the other handP(2A3(2))28 andG2A3(2) is the unique group of Lie type withm(G)28 and characteristic 2. First notice thatG2A3(2)'B2(3) is the unique group of Lie type withm(G)27 (see for example the list of primitive groups of small degree in [11]) so we may recognizeGfrom its Dirichlet polynomial. Next let us considerG2A2(3), here we getm(G)28 andp(G)263p; ifHis a group of Lie type with m(H)28, thenH2 f2A2(3);A1(27);B3(2)g. It can be checked (see Lemma 13 and [5]) that 13272~y(A1(27)) and 5242~y(B3(2)) but neither of these numbers dividesjGjso ifH6'G, thenPG(s)6PH(s).
When G62 f2A3(2);2A2(3)g we proceed as follows: we determinem(G) andjGjp with the help of the Dirichlet polynomialPG(s);ifm(G) is not a prime-power orm(G) jGjp);thenpp(G);otherwise we are in one of the three cases described in step 1 and in each of them the knowledge ofm(G)
suffices to determineGup to isomorphism. p
So our theorem is proved whenp(G)6p;and for the rest of the proof our attention will be restricted on groups of Lie type satisfyingp(G)p:
Clearly, in this case we can obtain alsol(G) from the knowledge ofPG(s) as pl(G) jGjp:Now we start to discuss the other two Artin invariants,v(G) andc(G). Let us first recall other results from [16]. The order ofGL(q) has a standard factorization:
jL(q)j 1 dqhP(q);
whered,handP(q) are given in [16, Table L1]. In particular (see [16]) this order has thecyclotomic factorization in terms of p:
jL(q)j 1 dplY
m
Fm(p)em;
where Fm(x) is the m-th cyclotomic polynomial. Let a(G) be the largest value ofmfor whichem 60 and defineb(G) as the next largest value ofm for whichem60;as it is explained in [16], a consequence of Zsigmondy's Theorem is that, apart from the few exceptional cases listed in [16] Lemma 4.6,a(G) andb(G) coincides withv(G) andc(G) (and the precise values are reported in [16] Table A.1).
STEP3. If p(G)p;then(v(G);c(G))(v(po(G));c(po(G))):
PROOF. As we noticed after the proof of Lemma 12,jGj=po(G) divides the order of a Cartan subgroupHofG:For most cases, in order to prove our
statement it will suffice to check that the prime numbers dividingjHjare not relevant when one wants to computev(G) andc(G), which is equivalent to verify the following condition:
() there exist at least two primes u1and u2 which dividejGj=jHj such that the following holds: for any prime u dividingjHj;the order of p modulo u is not grater than the order of p modulo u1
and u2.
So before starting with a case by case analysis we recall the value ofjHj for the different classes of groups of Lie type. IfGLn(q) is of untwisted type, thenjHj (q 1)n=d(see Section 8.6 in [4]); ifGiLn(q) is a twisted Lie group and all the roots have the same length, thenjHj Q
J (qjJj 1)=d;
whereJruns on the set of ther-orbits on the nodes of the Dynkin diagram (see Section 14.1 in [4]); ifG2 f2B2(q);2G2(q);2F4(q)g(i.e. whenGhas roots of different length), thenjHj (q 1)e;withe2 f1;2g. Now we start with the analysis of the different possibilities; according to [16] Lemma 4.6, there are three cases:
a) (v(G);c(G))(a(G);b(G)):
In this case the values of (v(G);c(G)) are given in [16] Table A.1.
Assume that G is a Lie group over the field Fq with qpr; if G is of untwisted type, then any primeuwhich divides jHj, also dividesq 1, hencephas order at mostrmodulou; this means that condition () is certainly verified when v(G)>c(G)>r and, by [16] Table A.1, this is true with the only exception ofGA1(q);in this last case we can easily conclude with the help of Lemma 13. Now assume that Gis a twisted Lie group. First consider the cases when all the roots have the same length. If uis a prime divisor ofjHj, then the order of pmoduloudi- vides rs, with s3 when G3D4(q); and s2 otherwise. If G62A2(q), then v(G)>c(G)>rs and () is satisfied. If G2A2(q);
then (v(G);c(G))(6r;2r);this information may be recovered from the index of the Borel subgroup B as jG:Bj q31 is divisible by F6r(p)F2r(p): If G2 f2B2(q);2G2(q);2F2(q)g, then p has order at most r modulo any prime divisor of jHj q 1 and condition () is certainly satisfied except forG2B2(2r) with r 1 mod 6:In this last case one can apply Lemma 13 (2).
b)p2;a(G)6 orb(G)6:
The pairs (v(G);c(G)) and (a(G);b(G)) does not coincide when p2 and either a(G)6 or b(G)6; there are precisely 18 groups of Lie type in this situation, and the values (v(G);c(G)) for them are given in
[16, Table A.2(a)]; for most of them a direct and easy computation shows that the exist at least two primes u1 and u2 dividing the index of the Borel subgroup and such that p has order v(G) modulo u1 and c(G) modulou2; only two of these groups require more attention. The first of them isGA2(8): in this casejGj 29327273;v(G)9 is the order of 2 modulo 73 and c(G)3 is the order of 2 modulo 7; but the Borel subgroup B has index 3273, not divisible by 7, so we need to check more carefully whether 7 divides po(G):this can be deduced by looking at the indices of the maximal subgroups, indeed it turns out that 293722~y(G):A similar situation occurs whenG2A2(8): in this case jGj 2934719;and v(G)18;c(G)3 are the orders of 2 modulo 19 and 7; 7 does not divide the index of the Borel subgroup but 287192~y(G):
c) p is Mersenne, b(G)2; c(G)1 andGA1(p2);A2(p); B2(p) or
2A2(p):
This case does not give us particular problems; same argument which has been applied in case (a) tells us that v(G)a(G) is the order of p modulo a prime divisor of the index of the Borel subgroupB(see [16, Table A.3]). MoreoverjG:Bjis an even number andc(G)1 is the order ofp
modulo 2. p
In [16] the authors prove that the are only few pairs of non-isomorphic simple groups of Lie type with the same Artin invariants (the possibilities are listed in [16], Table 5.2). In particular we have:
STEP 4. Suppose that G1 and G2 are two non isomorphic simple groups of Lie type with the same Artin invariants; if m(G1)m(G2)then one of the following occurs:
(1) G12A2(q);G22G2(q);
(2) G12A2(q2);G22B2(q3);
(3) G1A1(26); G22A2(4);
(4) G1Bn(q);G2Cn(q)with q odd and n3:
STEP5. P2A2(q)(s)6P2G2(q)(q):
PROOF. Assume, by contradiction, that P2A2(q)(s)P2G2(q)(s): This would imply po(2A2(q))po(2G2(q));since, by Lemma 13, (q1)2 divides po(2A2(q)); we would get that (q1)2 divides j2G2(q)j q3(q31)(q 1);
which is false. p
STEP6. P2A2(q2)(s)6P2B2(q3)(s):
PROOF. Assume, by contradiction, that P2A2(q2)(s)P2B2(q3)(s): This would imply po(2B2(q3))po(2A2(q2));since, by Lemma 13, q3 1 divides po(2B2(q3)); we would get thatq3 1 divides j2A2(q2)j q6(q61)(q4 1);
which is false. p
STEP7. PA1(26)(s)6P2A2(4)(s).
PROOF. Assume by contradiction that PA1(26)(s)P2A2(4)(s). Thus po(A1(26))po(2A2(4)). By Lemma 13 we get that 72po(A1(26)), but j2A2(4)j 2635213 is not divisible by 7. p
STEP8. If q is odd and n3:then PBn(q)(s)6PCn(q)(s):
PROOF. Recall that jBn(q)j jCn(q)j qn2 Q
1in(q2i 1)
=2: As-
sume that qpr with p an odd prime, and let p be the union of the sets p1 and p2, where p1 is the set of the primitive prime divisors of p2nr 1 and p2 the set of the primitive prime divisors of p(2n 2)r 1:
Our aim is to prove that P(p)Bn(q)(s)6P(p)Cn(q)(s): We will apply a theorem of Feit ([12, Theorem A]) which asserts that if (2n;q)2 f(6;= 3);(6;5)g;
then there exists u2p1 such that either u>2nr1 or u2 divides p2nr 1q2n 1: By definition, the non trivial contributions to the computation of the Dirichlet polynomial P(p)G(s) come only from the subgroups containing a Sylow u-subgroup of G for each prime u2p:
The maximal subgroups of Cn(q)PSp (2n;q) whose order is divisible by at least a prime u12p1 and a prime u22p2 are described in [13]
and [17, Table 2.5]. In particular we deduce that if Cn(q)PSp (2n;q) contains a maximal subgroup M with this property, then (2n;q)2 f(6;= 3);(6;5)g, r1; u12n1 and jMj is not divisible byu21; as a consequence we get that uu1 and u2 divides jCn(q)j whereas u does not dividejMj. This implies that there is no maximal subgroup of Cn(q) containing a Sylow u-subgroup for each u2p; hence P(p)Cn(q)(s)1: The situation is different for Bn(q)V2n1(q): Indeed if W is a non-singular 1-dimensional subspace of the orthogonal space V F2n1q with the property that W? has typeO2n; then the stabilizer MStabV2n1(q)(W) is a maximal subgroup of V2n1(q) with jV2n1(q):Mj qn(qn 1)=2, a p0-number. This implies P(p)Bn(q)(s)61
P(p)Cn(q)(s): p
6. An example.
The following is the Dirichlet polynomial associated with a finite simple group G; we want to use the results established in the paper in order to identifyGup to isomorphism.
PG(s)(S)1 62
31s 186
(2331)s 775 (5231)s
3100 (225231)s 3875
(5331)s
4000 (2553)s
4650 (235231)s
11625
(35331)s 15500
(225331)s 18600 (2335231)s
31000 (235331)s
186000
(2335331)s 124000 (255331)s First we notice thatm(G)31;asa31(G) 62, from Theorem 1 we de- duce that Gis not of alternating type; moreover by Table 1, there is no sporadic simple groupHwithm(H)31;henceGis a simple group of Lie type. Now we computeP(p)G (0) for any primep2p(po(G)) f2;3;5;31g;we get the following values:
p 2 3 5 31
P pG 0 24232 3311723 53 33143
Only forp5 the numberjP(p)G(0)jis ap-power, hence by Theorem 3 the Lie group G is defined over a field of characteristic 5, and 53 is the order of a Sylow 5-subgroup ofG:Moreoverm(G)31 jGj553so by the argument in Step 1 of the proof of Theorem 14,p(G)5 andl(G)3:
The orders of 5 modulo 2,3,31 are, respectively, 1,2,3 so v(G)3 and c(G)2: G has Artin invarians (p(G);(G);v(G);c(G))(5;3;3;2); this allows us to concludeGA2(5)(3;5):
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Manoscritto pervenuto in redazione il 7 settembre 2005.