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Recognizing the Non-Frattini Abelian Chief Factors of aFinite Group from Its Probabilistic Zeta FunctionMassimiliano Patassini aa Dipartimento di Matematica, Università di Padova, Padova, Italy

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To cite this article: Massimiliano Patassini (2012): Recognizing the Non-Frattini Abelian Chief Factors of a Finite Group fromIts Probabilistic Zeta Function, Communications in Algebra, 40:12, 4494-4508

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Communications in Algebra®, 40: 4494–4508, 2012Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927872.2011.610075

RECOGNIZING THE NON-FRATTINI ABELIAN CHIEFFACTORS OF A FINITE GROUP FROM ITS PROBABILISTICZETA FUNCTION

Massimiliano PatassiniDipartimento di Matematica, Università di Padova, Padova, Italy

Given a finite group G, the Dirichlet polynomial of G is

PG�s� =∑H≤G

�G�H�

�G � H�s �

The multiplicative counterpart of this polynomial is called the Probabilistic Zetafunction of the group G. We prove that if H is a finite group such that PG�s� = PH�s�,then G and H have the same non-Frattini abelian chief factors.

Key Words: Chief factors; Dirichlet polynomial; Probabilistic Zeta function.

2010 Mathematics Subject Classification: Primary 20D30; Secondary 11M41, 20D06, 20E28.

Let G be a finite group and N be a normal subgroup of G. The Dirichlet polynomialof G given N is defined as follows:

PG�N �s� =∑n≥1

an�G�N�

ns� where an�G�N� = ∑

H≤G��G�H�=n�HN=G

�G�H��

Here �G is the Möbius function of the subgroup lattice of G, defined by �G�G� =1� �G�H� = −∑H<K≤G �G�K� for H < G. In particular, if G = N we write PG�s� =PG�G�s� and we call PG�s� the Dirichlet polynomial of G. The multiplicative inverseof PG�s� is known as the Probabilistic Zeta function of G (see [1, 19] for references).

A problem that arises naturally is to determine which properties of the groupG are encoded by the polynomial PG�s�. It is known that PG/Frat�G��s� = PG�s� (seeLemma 2), so from the Dirichlet polynomial of G we can only hope to read offproperties of G/Frat�G�. Further, it was noted in [11] that PG�s� does not uniquelydetermine the isomorphism class of G/Frat�G�.

Nevertheless, certain group theoretic properties are given by the Dirichletpolynomial. For instance, if G and H are groups such that PG�s� = PH�s� and G is

Received June 24, 2011. Communicated by P. Tiep.Address correspondence to Massimiliano Patassini, Dipartimento di Matematica, Università di

Padova, Via Trieste, Padova 63-35121, Italy; E-mail: frapmass@gmail.com

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CHIEF FACTORS AND DIRICHLET POLYNOMIAL OF GROUPS 4495

soluble (or p-soluble, or perfect), then H has the same property (see [3, 7]). If G issimple and PG�s� = PH�s�, then H/Frat�H� � G (see [4, 21]).

The aim of this article is prove the following theorem.

Theorem 1. Let G and H be two finite groups and assume that PG�s� = PH�s�. ThenG and H have the same non-Frattini abelian chief factors.

We introduce some notations that we shall use throughout the article.

• The ring of the Dirichlet finite series (or Dirichlet polynomials) is

� ={∑

n≥1

an

ns� an ∈ �� ��n � an �= 0�� < �

}�

Note that PG�N �s� and PG�s� are elements of �.• Let be a set of prime and let ′ be the set of prime numbers that does not lie in

. Denote by � the subring of � given by{∑m

am

ms∈ � � am �= 0 ⇒ m is a ′ − number

}�

Note that both � and � are factorial domains (for instance, see [5]).• We define an homomorphism of rings between � and � in the following way

� � → �

f�s� = ∑n∈�

an

ns�→ f ���s� = ∑

n∈�

bnns

where

bn ={an if n is a ′ number

0 otherwise.�

It was already known to Gaschütz (see [11]) that PG�s� = PG/N �s�PG�N �s�. So,if 1 = G0 � G1 � � � � Gk = G is a chief series of G, then we have:

PG�s� =k−1∏i=0

PG/Gi�Gi+1/Gi�s��

For a prime p, let Ip = �i ∈ �0� � � � � k− 1� � Gi+1/Gi is a p-group�, and define

QG�p�s� =∏i∈Ip

PG/Gi�Gi+1/Gi�s��

So the Dirichlet polynomial

QG�s� =∏p

QG�p�s�

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4496 PATASSINI

collects the contribution of the abelian chief factors of G to PG�s�. It is easy tosee that if i ∈ Ip, then P

�p′�G/Gi�Gi+1/Gi

�s� = PG/Gi�Gi+1/Gi�s� (see Section 2), hence QG�p�s�

divides �PG�s�� P�p′�G �s��. We would be happy to say that QG�p�s� = �PG�s�� P

�p′�G �s��,

but it is not always the case. For example, it fails when G = PSL2�7� and p = 2, aswe will see. So, we shall study what happens when Gi+1/Gi is non abelian in orderto gather enough informations to find QG�p�s� from the knowledge of the Dirichletpolynomial PG�s�.

In Section 1, we introduce some notation and lemmas we shall use throughoutthe article. In Section 2, we give some information on the knowledge about thefactorization of the Dirichlet polynomial PG�s�, with respect to the chief factors ofG. In particular, we see that the factors of PG�s� are Dirichlet polynomials of typeP̃L�N �s� = PL�N �s�+ ��N �1−s for some � ∈ � and some monolithic primitive group L

with socle N . In order to prove Theorem 20, we study the common factors of P̃�p′�L�N �s�

and P̃L�N �s�. We are interested in the case when P̃�p′�L�N �s� �= 1 (these cases are listed in

[14, Theorem 1]). Section 3 is devoted to the proof of �P̃�p′�L�N �s�� P̃L�N �s�� = 1 when N

is a direct product of some copies of an alternating group of prime power degree.Moreover, we give an explicit formula of the Dirichlet polynomial P�r�

X�S�s�, when X

is an almost simple group with socle S � Altk and r is a prime number such thatk/2 < r < k− 2. In the last section we complete the proof of Theorem 1, which is aconsequence of Theorem 20.

1. SOME DEFINITIONS AND LEMMAS

We introduce some notations that we shall use throughout the article.

• Let

�′ ={∑

m≥1

am

ms� am ∈ m�� ��m � am �= 0�� < �

}�

which is a subring of �. Let �′ = �′ ∩�. We define the map

� � �′ → � X′ �

given by ��p1−s� = xp for each p ∈ ′, where X′ is the set of commutingindeterminates �xr � r ∈ ′�. Clearly, � is a ring isomorphism; hence, �′

and �′

are factorial domains.• Let k be a positive integer. We let �k� be the set of prime numbers that divides

k. If H is a finite group, we let �H� = ��H��.• Let k be a positive integer and let r be a prime number. We denote by �k�r thegreatest power of r that divides k, i.e., �k�r = ri, where ri divides k but ri+1 doesnot divide k.

• Let f�s� =∑k≥1

akks

be an element of � and let r be a prime number. Welet ak�f�s�� = ak. Moreover, �f�s��r = max��k�r � ak�f�s�� �= 0�. Finally, �f�s�� =�l.c.m�k � ak�f�s�� �= 0�.

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CHIEF FACTORS AND DIRICHLET POLYNOMIAL OF GROUPS 4497

We need some remarks on the Möbius function of a poset � with a topelement 1̂. The Möbius function of the poset � is defined as:

���x� =

1 if x = 1̂�

− ∑y∈��y≥x

���y� if x < 1̂��

In particular, if � is the subgroup lattice of a group G, the top element is the groupG and the Möbius function is denoted by �G. We will use repeatedly, often withoutmention, the following results.

Lemma 2 (see [15]). Let G be a finite group and H a subgroup of G. If �G�H� �= 0,then H is intersection of maximal subgroups of G.

Lemma 3. Let � be a poset with top element 1̂ and let �′ be a subposet such that

• 1̂ ∈ �′,• if x ∈ �′� y ∈ � and y ≥ x, then y ∈ �′.

Then ��′�x� = ���x� for each x ∈ �′.

Finally, we report a technical lemma on the reducibility of a polynomial.

Lemma 4 (see [18, Chap. VI, Theorem 9.1]). Let D = � x1� � � � � xk� be a ring ofpolynomials with integer coefficients. Suppose that a� b ∈ D and let f�x� = a+ bxm ∈D x�. If f�x� is reducible in D x�, then

• �a� b� �= 1 or• a� b ∈ Du for some prime divisor u of m.

2. FACTORIZATION OF THE DIRICHLET POLYNOMIAL PG�s�

As we have seen in the introduction, if 1 = G0 � G1 � � � � Gk = G is a chiefseries of G, then we have:

PG�s� =k−1∏i=0

PG/Gi�Gi+1/Gi�s�� (∗)

The factorization �∗� is well understood, thanks to the work of Gaschütz (see [11,12]), Detomi and Lucchini (see [6]) on the crowns. In order to describe it, we needsome definitions. Let N be a minimal normal subgroup of a group G, and let

LN ={G/CG�N� if N is not abelian�

G/CG�N�� N otherwise,

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4498 PATASSINI

be the monolithic primitive group associated with N . Clearly N is isomorphic to thesocle of LN . Define:

P̃LN �1�s� = PLN �N

�s�� P̃LN �i�s� = PLN �N

�s�− �1+ qN + · · · + qi−2N ��N

�N �s for i > 1�

where �N = �CAut�N��LN/N�� and qN = �EndLN�N�� if N is abelian, qN = 1 otherwise.

In particular, note that if N is abelian, then N is an elementary abelian group ofprime power order and

PLN �N�s� = 1− �Der�LN/N�N��

�N �s �

where �Der�LN/N�N�� is the number of complement of N in LN .Finally, let A1 and A2 be two chief factors of G. We say that A1 is G-equivalent

to A2 if A1 and A2 are G-isomorphic to the minimal normal subgroups (possiblyto the same) of a primitive epimorphic image of G (recall that this epimorphicimage has one or two minimal normal subgroups and in the latter case they are notG-isomorphic). For a non-Frattini chief factor A of G, let �G�A� be the number offactors of G, which are G-equivalent to A (it is independent on the choice of thechief series). Now we can state the main result of [6].

Theorem 5 ([6, Theorems 17 and 18]). Let G be a finite group. Then

PG�s� =∏A∈�

( ∏1≤i≤�G�A�

P̃LA�i�s�

)�

where � is a set of representatives of the G-equivalence classes of the non-Frattini chieffactors of G, and LA is the monolithic primitive group associated with A. In particular,if A = H/K is a non-Frattini chief factor, then

P̃LA��G/K�A��s� = PG/K�H/K�s��

We can say some more words on the Dirichlet polynomial of a monolithicprimitive group L with non-abelian socle N . Assume that S is a simple componentof L, define X = NL�S�/CL�S� and n = �L � NL�S��. We have that N � Sn. Since S �soc�X�, assume that S ≤ X. The following result shows a connection between theDirichlet polynomials PL�N �s� and PX�S�s�.

Theorem 6 (see [22, Theorem 5]).

P�r�L�N �s� = P

�r�X�S�ns − n+ 1��

for each prime divisor r of the order of S.

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CHIEF FACTORS AND DIRICHLET POLYNOMIAL OF GROUPS 4499

In particular, if A = H/K is a non-Frattini chief factor of G and L = LA isthe primitive monolithic group associated with A, by Theorems 5 and 6, and by thedefinition of P̃L�i�s� we have that

P�r�G/K�H/K�s� = P̃

�r�

L��G/K�A��s� = P

�r�L�N �s� = P

�r�X�S�ns − n+ 1�

for each prime divisor r of the order of S.Let p be a prime number. In this article, we want to study the Dirichlet

polynomials f�s�, which divide both PG�s� and P�p′�G �s�. Thanks to the description of

the factors of PG�s� given by Theorem 5, we can focus our attention to the Dirichletpolynomials P̃LA�i

�s�. If A is abelian and �A� = pa for some positive integer a, thenP̃

�p′�LA�i

�s� = P̃LA�i�s�. If A = H/K is not abelian, then P̃

�p′�LA�i

�s� does not depend on i,and P̃

�p′�LA�i

�s� �= 1 if and only if there exists a subgroup H of X such that HS = X and�X � H� is a power of p (under the notation of Theorem 6). This happens just in afew cases, as the following Proposition shows.

Proposition 7 (see [14, Theorem 1]). Let X be an almost simple group with non-abelian socle S and let p be a prime number. There exists a subgroup H of X such thatHS = X and �X � H� = pa for some positive integer a if and only if one of the followingholds:

• X has socle S = Altpa with pa ≥ 5, H = NX�Altpa−1�, where Altpa−1 is a subgroupof S.

• X has socle S = PSLn�q�, where pa = qn−1q−1 , the group X does not contain non-

trivial graph automorphisms and H = NX�P�, where P is the stabilizer of a line orhyperplane in S.

• X = S = PSL2�11�, H � Alt5, and pa = 11.• X = S = M11, H � M10, and pa = 11.• X = S = M23, H � M22, and pa = 23.• X has socle S = PSU4�2�, H = NX�P�, where P is a parabolic subgroup of indexpa = 27 of S.

Notation. From now on, let L be a primitive monolithic group with non-abeliansocle N � Sn for some non-abelian simple group S and some positive integer n.Moreover, let X = NL�S�/CL�S� and assume that S ≤ X. Let � be any integernumber and let

P̃L�N �s� = PL�N �s�+ ��N �1−s�

By Theorem 6, we have that

P̃��L�N �s� = P

��L�N �s� = P

��X�S�n�s − 1�+ 1��

where ⊆ �S�.

In the following section we study the case when S is an alternating group ofprime power degree.

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4500 PATASSINI

3. THE ALTERNATING CASE

Let p be a prime number and let a be a positive integer such that pa ≥ 5.Assume that S � Altpa , so X = Altpa or Sympa . We want to prove the following.

Theorem 8. The greatest common multiple of P̃L�N �s� and P̃�p′�L�N �s� is 1.

3.1. On the Intransitive Subgroups of Altk and Symk

In this subsection, let k be a positive integer, k ≥ 8. Let X be either Altk orSymk.

Now, we prove a Lemma on the subgroups of Altk containing a Sylowr-subgroup where r is a prime such that k

2 < r < k− 2 (which exists by Bertrandpostulate for k ≥ 8).

Lemma 9. Let X be an almost simple group with socle Altk. Let r be a prime numbersuch that k/2 < r < k− 2. If K is a proper subgroup of X containing a Sylow r-subgroup of X and KS = X, then K is an intransitive subgroup of X.

Proof. Assume that K is a maximal primitive subgroup of X. Note that K mustcontain a cycle of length r. By a result of Jordan (see [17, Note C]), we have thatK is at least k− r + 1-transitive. By [16, Théoréme I], since k− r + 1 > 2, we havethat K must contain Altk−r , hence, K is an intransitive maximal subgroup of X.

It is easy to see that if K is an imprimitive maximal subgroup of X, then r

divides �X � K�. �

The above theorem implies that the Dirichlet polynomial P�p�X�S�s� is completely

determined by the knowledge of the intersection of the maximal intransitivesubgroups of X. So, in the sequel we want to describe the intransitive subgroupsof X that are intersection of maximal subgroups of X. Let � = �1� � � � � k�, so X �Alt��� or Sym���. Clearly X acts on � by permutations. Let ���� be the set ofpartitions of �. Let �X be the set consisting of the intersection of the intransitivemaximal subgroups of X and the group X. It turns out that

�X = �StabX�P� � P ∈ ���

where StabX�P� = ∩�∈PStabX���.When X = Symk, the situation was studied in [8, Section 2]. In particular it

was proved in the following.

Proposition 10 (see [8, Proposition 2.1]). We have that

∑H∈�Symk

��Symk�H�

�Symk � H�s = ∑�∈�k

����

��������s−1�

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CHIEF FACTORS AND DIRICHLET POLYNOMIAL OF GROUPS 4501

where �k is the set of partitions of k, i.e., non-decreasing sequences of positive integerswhose sum is k, we have � = �k1� � � � � kl� ∈ �k for some l ≤ k� k1� � � � � kl ∈ �− �0�such that k1 + · · · + kl = k, and we define

���� = �−1�l−1�l− 1�!� ���� = k!∏li=1 ki!

� ���� =k∏

i=1

�i!�

where �i = ��i � ki = i��.

There is a deep connection between the posets �Symkand �Altk

, as the followingLemma shows.

Lemma 11. Let H ∈ �Symkand assume that H �= 1. Then H � Altk.

Moreover, the map � � �Symk− �StabSymk

�P� �∏

�∈P ��� = 2� → �Altkgiven by

��H� = H ∩Altk is an isomorphism of posets.

Proof. Let H ∈ �Symkand assume that H �= 1. We have that H = StabSymk

�P� forsome partition P of �, such that there exists i� j ∈ � ∈ P with i �= j. Hence, �ij� ∈ H ,thus, H � Altk.

We want to prove the second part of the statement. Given a subgroup H ofSymk we can define an equivalence relation on �: if i� j ∈ �, then i ∼H j if and onlyif there exists h ∈ H such that i�h = j. So we say that i�H = �i�h � h ∈ H� is the H-orbit of i. Let X be Symk or Altk. Let P be a partition of � such that

∏�∈P ��� �= 2.

We have that P = � i�StabX�P� � i ∈ ��. So, it is easy to verify that the inverse map of �is � � �Altk

→ �Symk− �StabSymk

�P� �∏

�∈P ��� = 2�, given by ��H� = StabSymk�� i�H �

i ∈ ���. �

Finally, we are ready to prove the result we need.

Proposition 12. Let k ≥ 8. Let X be either Altk or Symk, and let r be a primenumber such that k

2 < r < k− 2. Then P�r�X�S�s� = R

�r�Symk

�s�, where

RSymk�s� = ∑

H∈�Symk

��Symk�H�

�Symk � H�s = ∑�∈�k

����

��������s−1�

Proof. By Proposition 9, if M is a maximal subgroup of X such that M contains aSylow r-subgroup, then M is intransitive. So, let � r

X be the poset of the subgroupsH of X such that HS = X and the subgroup H contains a Sylow r-subgroup.Moreover, let �r

Symkbe the subposet of �Symk

consisting of the elements H ∈ �Symk

such that H contains a Sylow r-subgroup.Now, thanks to Lemma 11, we have that � r

X and �rSymk

are isomorphic asposets. Thus, using Proposition 10 and Lemma 3, we conclude that:

P�r�X�S�s� =

∑H∈�r

X

�X�H�

�X � H�s = ∑H∈�r

X

��rX�H�

�X � H�s = ∑H∈�r

Symk

��rSymk

�H�

�X � H�s = R�r�Symk

�s��

This completes the proof. �

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4502 PATASSINI

3.2. The Proof of Theorem 8

We recall a number of theoretical results on the prime factors of the binomialcoefficients.

Theorem 13 ([9]). Let m and k be two natural numbers such that m ≥ 2k. Let u andv be two natural numbers such that

(m

k

) = uv, where the prime factors of u are less thank and the prime factors of v are greater than or equal to k. If m ≥ 57, then v > u.

Proposition 14. Let m ≥ 6 and 3 ≤ k ≤ m/2 be two natural numbers such that�m� k� �∈ ��8� 3�� �12� 5��. Then there exists a prime number r such that r divides

(m

k

)and r does not divide

(m

2

).

Proof. Assume that k = 3. For a contradiction, suppose that (m

2

) ⊇ (m

3

). This

implies that �m− 2� ⊆ �2� 3�. Assume that 4 divides m− 2. In this case 2 �∈ (m

2

) =(m

3

), thus, 4 cannot divide m− 2. Similarly, 9 does not divide m− 2. Hence, m− 2

divides 6 and the claim follows by direct inspection.Assume that k = 4. For a contradiction, suppose that

(m

2

) ⊇ (m

4

). This

implies that ��m− 2��m− 3�� ⊆ �2� 3�. A direct inspection shows that this cannotoccur.

Assume that k ≥ 5. For a contradiction, assume that (m

2

) ⊇ (m

k

). Let v be

as in Theorem 13. Since G.C.D.�m�m− 1�� �m− 2� � � � �m− k+ 1�� divides �k− 1�!and

(m

2

) ⊇ (m

k

), we get v ≤ (

m

2

). By Theorem 13, if m ≥ 57, then v2 >

(m

k

), hence,(

m

2

)2>(m

k

). However, it is easy to see that

(m

2

)2<(m

5

) ≤ (m

k

)for m ≥ 57 and k ≥ 5,

hence, we have a contradiction. A direct computation shows that the result holdsfor the remaining cases. �

Arguing in the same way, we obtain the following.

Proposition 15. Let m ≥ 8 and 4 ≤ k ≤ m/2 be two natural numbers such that�m� k� �= �9� 4�. Then there exists a prime number r such that r divides

(m

k

)and r does

not divide(m

3

).

Proposition 16. Let p > 2. Let be the set of prime divisors of(pa

2

). We have that

P�′�X�S �s� = 1− pa�1−s� −

(pa�pa − 1�

2

)�1−s�

+ ��pa�pa − 1���1−s��

where

� ={1 if 2 ∈ �

0 otherwise.

Proof. If pa = 5� 7, then the result holds (use [10, GAP]). As in Lemma 9, let r bea prime number such that pa/2 < r < pa − 2. Note that r ∈ ′.

Let H be a subgroup of X such that HS = X and �X � S� is a -number. ThenH ∩ X is contained in a maximal subgroup M of S such that �S � M� is a -number.In particular, we have that M contains a Sylow r-subgroup, hence, by Lemma 9,

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CHIEF FACTORS AND DIRICHLET POLYNOMIAL OF GROUPS 4503

we have that M is a maximal intransitive subgroup. Thus, �S � M� = (pa

k

)for some

1 ≤ k < pa/2. By Proposition 14, we have that k ≤ 2. By Proposition 12 we get theclaim. �

Proposition 17. Let a ≥ 4. Let be the set of prime divisors of(2a3

). We have that

P�′�X�S �s� = 1− 2a�1−s� −

(2a�2a − 1��2a − 2�

6

)�1−s�

+ �1(−�2a−1�2a − 1���1−s� + �2a�2a − 1���1−s�

)+ �1�2

(2�2a−1�2a − 1��2a − 2���1−s� − �2a�2a − 1��2a − 2���1−s�

)�

where

�1 ={1 if �2a − 1�3 �= 3�

0 otherwise,

�2 ={1 if �2a − 2�3 �= 3�

0 otherwise.

Proof. The argument is similar to the proof of Proposition 16, so we omit it. Notethat instead of Proposition 14 we apply Proposition 15. �

Now, we are ready to prove Theorem 8.

Proof of Theorem 8. Let

={(pa

2

)if p �= 2 or pa = 8

(2a3

)otherwise.

Since ⊆ �S�, by Theorem 6, we have that P̃��L�N �s� = P

��X�S�n�s − 1�+ 1�. We claim

that P̃��L�N �s� is irreducible.By [10, GAP], if pa = 8, then we have that

P�′�L�N �s� = 1− 8n�1−s� − 28n�1−s��

which is irreducible.Assume that pa �= 8. Let q be the greatest prime divisor of pa − 1. Let

xp = ��p1−s� and xq = ��q1−s� (recall the notation explained in Section 1). ByPropositions 16 and 17, we have that

�(P̃

�′�L�N �s�

) = 1− xnap + ��xp�xbq�

where b is a positive integer and ��xp� ∈ E xp�, with E = � X−�p�q��. For acontradiction suppose that P̃

�′�L�N �s� is reducible. By Lemma 4, we have that �1−

xnap � ��xp�� �= 1 or 1− xnap � ��xp� ∈ �E xp��u for some prime number u that divides b.

Suppose that there exists a common irreducible factor f�xp� ∈ E xp� of 1− xnap and

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4504 PATASSINI

��xp�. Then f�xp� divides g�xp� = ��P̃�′�L�N �s�� ∈ �E xq�� xp�. Since f�xp� divides 1−

xnap , we have that xp − � divides f�xp� in � X�, where � is a na-th root of 1. Thus,xp − � divides g�xp� in � X�. By Propositions 16 and 17, we have that g��� �= 0 forany � ∈ �, hence, we get a contradiction. Finally, it is clear that 1− xnap �∈ E xp�

u foreach prime u. Thus, we conclude that P̃�′�

L�N �s� is irreducible.Now we can complete the proof. Note that if f�s� and g�s� are Dirichlet

polynomial, � is a set of prime numbers and h�s� = �f�s�� g�s��, then h����s� =�f ����s�� g����s��. Let h�s� = �P̃

�p′�L�N �s�� P̃L�N �s��. By definition, h�s� divides P̃

�p′�L�N �s� =

1− pna�1−s�. Since p ∈ , then h�′��s� = h�s�. But h�′��s� = �P̃�p′�L�N �s�� P̃

�′�L�N �s�� = 1,

hence, h�s� = 1. �

4. THE MAIN THEOREM

Let L�N� S, and X be as in the end of Section 2.

Proposition 18. Let p be a prime number. If p = 2, then assume that S �� PSL2�q�

with q a Mersenne prime. We have that �P̃L�N �s�� P̃�p′�L�N �s�� = 1. Moreover, if f�s� is an

irreducible factor of P̃L�N �s� such that a1�f�s�� = 1, then f �p′��s� = P̃�p′�L�N �s� = P

�p′�L�N �s�

or f �p′��s� = 1.

Proof. The result is obvious if P̃�p′�L�N �s� = 1, so assume that P̃�p′�

L�N �s� �= 1. We havethat X and S are as in Proposition 7.

Assume that S � Altpa . By Theorem 8, we have that �PL�N �s�� P�p′�L�N �s�� = 1.

Assume that S � PSLk�q� for k ≥ 3 and X does not contain non-trivial graphautomorphisms. By [20, Theorem 2], if �k� q� �= �3� 2�, then the polynomial P̃L�N �s�

is irreducible, so �P̃L�N �s�� P̃�p′�L�N �s�� = 1. If S � PSL3�2� then p = 7. Using [10, GAP],

we have that

P̃�2�L�N �s� = 1− 2 · 7n�1−s� + 21n�1−s��

which is irreducible. Hence, �P̃L�N �s�� P̃�p′�L�N �s�� = 1.

Assume that S � PSL2�q� and pa = q + 1. We have that �pa� q� ∈ ��2b� 2b −1�� �2b + 1� 2b�� for some b ∈ �. Since we assumed that S �� PSL2�q� with q aMersenne prime, we have that q = 2b� b ≥ 3. It turns out that if = �2� 2b + 1�, then

P̃�′�L�N �s� = 1− �2b + 1�n�1−s� − �2b−1�2b + 1��n�1−s� + ��2b�2b + 1��n�1−s�

for some � ∈ � (see [13, Theorem 1.1]). Clearly P̃�′�L�N �s� is irreducible, hence,

�P̃L�N �s�� P̃�p′�L�N �s�� = 1.

Assume that X = S � PSL2�11�. Using [10, GAP], we have that

P̃�2�L�N �s� = 1− 2 · 11n�1−s� + 165n�1−s��

which is irreducible. Hence, �P̃L�N �s�� P̃�p′�L�N �s�� = 1.

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CHIEF FACTORS AND DIRICHLET POLYNOMIAL OF GROUPS 4505

Assume that X = S � M23. Using [10, GAP], we have that

P̃��2�3�5��L�N �s� = 1− 23n�1−s� − 2 · 253n�1−s��

which is irreducible. Hence, �P̃L�N �s�� P̃�p′�L�N �s�� = 1.

Assume that X = S � M11. Using [10, GAP], we have that

P̃��2�3��L�N �s� = 1− 11n�1−s� − 55n�1−s��

which is irreducible. Hence, �P̃L�N �s�� P̃�p′�L�N �s�� = 1.

Assume that S � PSU4�2�. Using [10, GAP], we have that

P̃�2�L�N �s� = 1− 27n�1−s� − 36n�1−s� + 2 · 216n�1−s��

which is irreducible. Hence �P̃L�N �s�� P̃�p′�L�N �s�� = 1.

Let f�s� be an irreducible factor of P̃L�N �s� such that a1�f�s�� = 1. As we haveseen above and in Propositions 16 and 17 for S � Altpa , there exists a set of primenumbers such that P̃�′�

L�N �s� is irreducible and p ∈ . Since f �′��s� divides P̃�′�L�N �s�,

we have that f �′��s� = P̃�′�L�N �s� or f

�′��s� = 1. Thus, f �p′��s� = P̃�p′�L�N �s� or f

�p′��s� = 1.�

Proposition 19. Assume that S � PSL2�q� with q = 2b − 1 a Mersenne prime. Then�P̃L�N �s�� P̃

�2′�L�N �s�� divides 1− 2n�1−s�, hence, P̃

�2′�L�N �s� = 1− 2bn�1−s� does not divide

P̃L�N �s�. Moreover if P̃L�N �s� is reducible, then there exists an irreducible factor f�s� ofP̃L�N �s� such that �f�s��q = �P̃L�N �s��q, 1 �= f �2′��s� �= 1− 2k�1−s� for every k ∈ �− �0�

and P̃L�N �s�

f�s�divides 1− 2n�1−s�.

Proof. Assume that q = 7. By [10, GAP], if X = S, then

P̃�3�L�N �s� = 1− 2 · 7n�1−s� − 8n�1−s� + 28n�1−s� + 56n�1−s�

= �1− 2n�1−s���1+ 2n�1−s� + 22n�1−s� − 2 · 7n�1−s� − 2 · 14n�1−s� − 28n�1−s���

and the second factor is irreducible. By [10, GAP], if X > S, then,

P̃�3�L�N �s� = 1− 8n�1−s� − 28n�1−s� + 56n�1−s�

= �1− 2n�1−s���1+ 2n�1−s� + 22n�1−s� − 28n�1−s���

and the second factor is irreducible. Hence, in both cases, �P̃L�N �s�� P̃�2′�L�N �s�� divides

1− 2n�1−s�.Let q = 2b − 1 for some b ≥ 5. Let = �2� 2b − 1�. By [2, Section 3], we have that

P̃�′�L�N �s� = 1− �2b�n�1−s� − �2b−1�2b − 1��n�1−s� + �2b�2b − 1��n�1−s�

= �1− 2n�1−s���1+ 2n�1−s� + · · · + 2�b−1�n�1−s� − �2b−1�2b − 1��n�1−s���

and the second factor is irreducible (use Lemma 4).

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4506 PATASSINI

Assume that P̃L�N �s� is reducible. By the above argument,P̃�′�L�N �s�

1−2n�1−s� is irreducible.

Hence, there exists an irreducible factor f�s� of P̃L�N �s� such thatP̃�′�L�N �s�

1−2n�1−s� dividesf �′��s�. Let P̃L�N �s� = f�s�g�s� with g�s� �= 1. Since �f�s��q = qn we have that�g�s��q = 1. By [2, Section 3], we get

f �2��s�g�2��s� = P̃�2�L�N �s� =

{1− 2 · 7n�1−s� + 21n�1−s� if n = 7�

1− ��2b − 1��2b−1 − 1��n�1−s� otherwise,

hence, f �2��s� = P̃�2�L�N �s�, otherwise �g�s��q > 1. So we have that �f�s��r = �PL�N �s��r

for each r ∈ �S�− �2�. It follows that �g�s��r = 1 for each r ∈ �S�− �2�, so 1 �=g�s� = g�2

′��s� divides 1− 2n�1−s�.Finally, note that f �2′��s� �= 1− 2k�1−s� for every k ∈ �− �0�. Indeed, since 1 �=

g�s� = g�2′��s� divides 1− 2n�1−s� and �b�2

n�1−s�� dividesP̃�2′�L�N �s�

1−2n�1−s� , we may assume that

f �2′��s� = �b�2n�1−s��

∏d∈D

�d�2�1−s���

where D is a proper subset of the set of divisors of n and �d�x� is the d-thcyclotomic polynomial. If

f �2′��s� = �b�2n�1−s��

∏d∈D

�d�2�1−s�� = 1− 2k�1−s�

for some k ∈ �− �0�, then bn divides k, hence, 1− 2n�1−s� divides∏

d∈D �d�2�1−s��,

against the assumption that D is a proper subset of the set of divisors of n. �

The following theorem implies Theorem 1.

Theorem 20. Let G be a finite group and let p be a prime number.

(1) If p is odd, then QG�p�s� = �PG�s�� P�p′�G �s��.

(2) Let

PG�s� =l∏

j=1

fj�s�

be the factorization into irreducible factors of PG�s�, where a1�fj�s�� = 1 for 1 ≤j ≤ l. Let J be the set of j ∈ �1� � � � � l� such that fj�s� �= f

�2′�j �s� �= 1 and f

�2′�j �s� �=

1− 2k�1−s� for each k ∈ �. Then

�PG�s�� P�2′�G �s�� = QG�2�s�

∏j∈J

1− 2bjnj�1−s�

f�2′�j �s�

�

where 2bj − 1 = max �fj�s�� and nj = log2bj−1 �fj�s��2bj−1 for j ∈ J .

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CHIEF FACTORS AND DIRICHLET POLYNOMIAL OF GROUPS 4507

(3) Suppose that

QG�p�s� =∏

1≤i≤k1

(1− ci

pnis

)= ∏

1≤j≤k2

(1− dj

pmjs

)�

where c1� � � � � ck1� d1� � � � � dk2are positive integers. Then k1 = k2 and there exists

� ∈ Sym�k1� such that mi� = ni and di� = ci for 1 ≤ i ≤ k1. So the non-Frattinichief factor of G, which are p-groups, are elementary abelian groups of sizepn1� � � � � pnk1 .

Proof. Assertion (1) follows from Proposition 18.Let us prove (2). Let I be the set of index i ∈ �0� � � � � k− 1� such that

PG/Gi�Gi+1/Gi�s� is reducible andGi+1/Gi � PSL2�q

′i�

n′i where q′i = 2b

′i − 1 is a Mersenne

prime and n′i is a positive integer. By Section 2, for i ∈ I , there exists a monolithic

primitive group Li with non-abelian socle Ni � PSL2�q′i�

n′i and � ∈ � such thatPG/Gi�Gi+1/Gi

�s� = PLi�Ni�s�+ ��Ni�1−s. By Proposition 19, PG/Gi�Gi+1/Gi

�s� = hi�s�gi�s�,where

• gi�s� is the irreducible factor of PG/Gi�Gi+1/Gi�s� such that �gi�s��q′i = �PG/Gi�Gi+1/Gi

�s��q′iand 1 �= g�2

′��s� �= 1− 2t�1−s� for any t ∈ �− �0�;• hi�s� divides 1− 2n

′i�1−s�.

In particular, hi�s� =P�2′�G/Gi�Gi+1/Gi

�s�

g�2′�i �s�

= 1−2b′in′i �1−s�

g�2′�i �s�

� Hence, we have that

�PG�s�� P�2′�G �s�� = QG�2�s�

∏i∈I

hi�s� = QG�2�s�∏i∈I

1− 2b′in

′i�1−s�

g�2′�i �s�

�

By Propositions 18 and 19, it is clear that there exists a bijection � � I → J such thatf��i��s� = gi, hence, b��i� = b′i� q��i� = q′

i and n��i� = n′i.

Finally, (3) follows by [3, Lemma 16]. �

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