A finiteness condition on the coefficients of the probabilistic zeta function

0.13 MB·English

by  Duong Hoang Dung

A FINITENESS CONDITION ON THE COEFFICIENTS
OF THE PROBABILISTIC ZETA FUNCTION
DUONGHOANGDUNGAND ANDREALUCCHINI
3
1
Abstract. We discuss whether finiteness properties of a profinite group G can be deduced from the
0
2 coefficients of the probabilistic zeta function PG(s). In particular we prove that if PG(s) is rational

n and all but finitely many non abelian composition factors of G are isomorphic to PSL(2,p) for some
a prime p,then G contains only finitely many maximal subgroups.
J

6
1

] 1. Introduction
R
G
LetGbeafinitelygenerated profinitegroup. AsGhasonlyfinitelymanyopensubgroupsof agiven
.
h index,foranyn ∈Nwemaydefinetheinteger a (G)asa (G) = µ (H),wherethesumisover all
t n n H G
a opensubgroupsH ofGwith|G : H|= n.Hereµ (H)denotesthePM¨obiusfunctionoftheposetofopen
m G
subgroups of G, which is defined by recursion as follows: µ (G) = 1 and µ (H) = − µ (K) if
[ G G H<K G

H < G. Then we associate to G a formal Dirichlet series P (s), defined as P
1 G
v
a (G)
8 P (s)= n .
3 G ns
6 nX∈N
3
Notice that if H is an open subgroup of G and µ (H) 6= 0, then H is an intersection of maximal
. G
1
subgroups of G. Therefore the formal Dirichlet series P (s) encodes information about the lattice
0 G
3
generated by the maximal subgroups of G, just as the Riemann zeta function encodes information
1
: about the primes, and combinatorial properties of the probabilistic sequence {a (G)} reflect on the
v n
i structure of G.
X
r If G contains only finitely many maximal subgroups (i.e. if the Frattini subgroup FratG of G has
a
finite index in G), then there are only finitely many open subgroups H of G with µ (H) 6= 0 and
G
consequently a (G) = 0 for all but finitely many n ∈ N (i.e. P (s) is a finite Dirichlet series). A
n G
natural question is whether the converse is true.
Let {Gn}n∈N be a countable descending series of open normal subgroups with the properties that
G = G, G = 1 and G /G is a chief factor of G for each n ∈ N. The factor group G/G is
1 n∈N n n n+1 n
finite, so Tthe Dirichlet series P (s) is also finite and belongs to the ring D of Dirichlet polynomials
G/Gn
with integer coefficients. Actually, P (s) is a divisor of P (s) in the ringD, i.e. there exists a
G/Gn G/Gn+1
Dirichlet polynomial P (s) such that P (s) = P (s)P (s). As explained in [3], the Dirichlet
n G/Gn+1 G/Gn n
MSC(2010): Primary: 20E18;Secondary: 20D06,20P05,11M41.
Keywords: Probabilisticzetafunction,speciallineargroups.
1
2 DUONGHOANGDUNGANDANDREALUCCHINI
series P (s) can be written as an infinite formal product
G
P (s) = P (s),
G n
nY∈N
and if we change the series {Gn}n∈N, the factorization remains the same up to reordering the fac-
tors. Moreover it turns out that P (s) = 1 if G /G is a Frattini chief factor (i.e. G /G ≤
n n n+1 n n+1
Frat(G/G ). Notice that G has finitely many open maximal subgroupsif and only if the chief series
n+1
{Gn}n∈N contains onlyfinitelymanynon-Frattinifactors. Thiscouldsuggestawrongargument: ifthe
product P (s) = P (s) is a Dirichlet polynomial, then P (s)= 1 for all but finitely many n∈ N
G n∈N n n
and consequentlyQthe series {Gn}n∈N contains only finitely many non-Frattini factors. The problem is
that it is possible that a Dirichlet polynomial can be written as a formal product of infinitely many
non trivial elements of D. To give an idea of what can occur, let us recall a related question, with an
unexpected solution: if G is prosolvable, then we can consider the p-local factor
apm
P (s) = .
G,p pms
mX∈N
It turns out that P (s) = P (s) where Ω is the set of indices n such that G /G has
G,p n∈Ωp n p n n+1
p-power order. It is not diffiQcult to prove that a finitely generated prosolvable group G contains
only finitely many maximal subgroups whose index is a power of p if and only if a chief series of
G contains only finitely many non-Frattini factors whose order is a p-power. Therefore the previous
tempting wrong argument would suggest the following conjecture: if the p-factor P (s) is a Dirichlet
G,p
polynomial, then G has only finitely many maximal subgroups of p-power index. However this is
false; in [4] it is constructed a 2-generated prosolvable group G such that, for any prime p, G contains
infinitelymanymaximalsubgroupsofp-powerindex,whileP (s)isafiniteDirichletseries. Knowing
G,p
that P (s) can be a polynomial even when Ω is infinite, could lead to believe in the existence of a
G,p p
counterexample to the conjecture that P (s) ∈ D implies G/Frat(G) finite. However, using results
G
from number theory, in [4] it was prove that if G is prosolvable and P (s) is a polynomial, then
G,p
either Ω is finite or, for every prime q, there exists n ∈ Ω such that the dimension of G /G as
p p n n+1
F G-module is divisible by q; using standard arguments of modular representation theory one deduce
p
thatthisispossibleonlyifinfinitelymanyprimesappearsbetweenthedivisorsoftheorderofthefinite
images of G; but then P (s) 6= 1 for infinitely many primes r and P (s) cannot be a polynomial.
G,r G
So P (s) can be a polynomial only if Ω is finite for every prime and empty for all but finitely many,
G p
and therefore only if G/Frat(G) is finite. Really a stronger result holds: if G is a finitely generated
prosolvable group, then P (s) is rational (i.e. P (s) = A(s)/B(s) with A(s) and B(s) finite Dirichlet
G G
series) if and only if G/FratG is a finite group. Partial generalization has been obtained in [5] and in
[6]. All these results can be summarized in the following statement:
Theorem 1.1. Let G be a finitely generated profinite group. Assume that there exist a prime p and
a normal open subgroup N of G such that the set S of nonabelian composition factors of N satisfies
one of the following properties:
• all the groups in S are alternating groups;
RATIONAL PROBABILISTIC ZETA FUNCTION 3
• all the groups in S are of Lie type over fields of characteristic p, where p is a fixed prime;
• all the groups in S are sporadic simple groups.
Then P (s) is rational if and only if G/Frat(G) is a finite group.
G
Themain ingredient in the proof of the previous results is the following result, proved with the help
of the Skolem-Mahler-Lech Theorem and where π(G) is the set of the primes q with the properties
that G contains at least an open subgroup H whose index is divisible by q.
Proposition 1.2. Let G be a finitely generated profinite group, assume that π(G) is finite and let r
i
be the sequence of the composition lengths of the non-Frattini factors in a chief series of G. Assume
that there exists a positive integer q and a sequence c of nonnegative integers such that the formal
i
product
c
i
1−
(cid:18) (qri)s(cid:19)
Yi
is rational. Then c = 0 for all but finitely many indices i.
i
In our applications, c /(qri)s is one of the summands of the Dirichlet polynomial P (s) associated
i i
to the chief factor G /G and must be chosen so that the polynomials P∗(s) = 1−c /(qri)s satisfy
i i+1 i i
two conditions:
(1) if P (s)6= 1 for infinitely many i ∈N, then also P∗(s) 6= 1 for infinitely many i∈ N;
i i
(2) if the infinite product P (s) is rational, then also P∗(s) is rational.
i i i i
Q Q
Thechoice ofthe“approximation” P∗(s)ofP (s)isnoteasyandwearenotabletousethisstrategy
i i
in the general case; roughly speaking, it requires an order on the set of the primes numbers with the
property that “small” simple groups in S have order not divisible by “large” primes. When S is the
set of the alternating groups, we just consider the natural order, but when S consists of simple groups
of Lie type in characteristic p, we say that a prime q is smaller than a prime q if the multiplicative
1 2
order of p mod q is smaller than its order modulo q . This order depends on the choice of p and
1 2
our arguments do not work if S contains groups of different kinds, for example groups of Lie type in
different characteristics. However we expect that our statement remains true in the general case. In
the present paper we prove a new result in this direction:
Theorem 1.3. Let G be a finitely generated profinite group. Assume that there exists a normal open
subgroup N of G such that any nonabelian composition factor of N is isomorphic to PSL(2,p) for
some prime p. Then P (s) is rational if and only if G/Frat(G) is a finite group.
G
2. Preliminaries and notations
LetRbetheringofformalDirichletserieswithintegercoefficients. WesaythatF(s) = a /ns
n∈N n
∈ R is a Dirichlet polynomial if a = 0 for all but finitely many n ∈ N. The set D of tPhe Dirichlet
n
polynomials is a subringof R. We will say that F(s) ∈ R is rational if there exist A(s),B(s) ∈ D with
F(s) = A(s)/B(s).
4 DUONGHOANGDUNGANDANDREALUCCHINI
For every set π of prime number, we consider the ring endomorphism of R defined by:
a a∗
F(s) = n 7→ Fπ(s) = n
ns ns
nX∈N nX∈N
where a∗ = 0 if n is divisible by some prime p ∈ π, a∗ = a otherwise. We will use the following
n n n
remark:
Remark 2.1. For every set π of prime numbers, if F(s) is rational then Fπ(s) is rational.
Thefollowing result is a consequence of the Skolem-Mahler-Lech Theorem (see [4] for more details):
Proposition 2.2. Let I ⊆ N and let q,r ,c be positive integers for each i∈ I. Assume that
i i
(i) for every n∈ N, the set {i ∈ I | r divides n} is finite;
i
(ii) there exists a prime t such that t does not divide r for any i ∈ I.
i
If the product
c
i
F(s)= 1−
(cid:18) (qri)s(cid:19)
Yi∈I
is rational, then I is finite.
Proposition 2.3. [5, Corollary 5.2] Let G be a finitely generated profinite group and assume that
π(G) is finite. For each n, there are only finitely many non-Frattini factors in a chief series whose
composition length is at most n. Moreover there exists a prime t such that no non-Frattini chief factor
of G has composition length divisible by t.
Proof of Proposition 1.2. It follows immediately from Propositions 2.2 and 2.3. (cid:3)
Finally let us recall the following result.
Proposition 2.4. [5, Proposition 4.3] Let F(s) be a product of finite Dirichlet series:
b
i,n
F(s) = F (s), where F (s) =
i i ns
Yi∈I nX∈N
Let q be a prime and Λ the set of positive integers divisible by q. Assume that there exists a set {r }
i i∈I
of positive integers such that if n ∈ Λ and b 6= 0 then n is an r -th power of some integer and
i,n i
v (n) = r (where v (n) is the q-adic valuation of n). Define
q i q
w = min{x ∈ N vq(x) = 1 and bi,xri 6=0 for some i ∈I}
(cid:12)
(cid:12)
If F(s) is rational, then the product
(2.1) F∗(s) = 1+ bi,wri
(cid:18) (wri)s(cid:19)
Yi∈I
is also rational.
RATIONAL PROBABILISTIC ZETA FUNCTION 5
Now let G be a finitely generated profinite group and let {Gi}i∈N be a fixed countable descending
series of open normal subgroups with the property that G = G, G = 1 and G /G is a chief
0 i∈N i i i+1
factor of G/G for each i ∈ N. In particular, for each i ∈ N,Tthere exist a simple group S and
i+1 i
a positive integer r such that G /G ∼= Sri. Moreover, as described in [3], for each i ∈ N a finite
i i i+1 i
Dirichlet series
b
i,n
(2.2) P (s)=
i ns
nX∈N
is associated with the chief factor G /G and P (s) can be written as an infinite formal product of
i i+1 G
the finite Dirichlet series P (s):
i
(2.3) P (s)= P (s).
G i
iY∈N
Moreover, this factorization is independenton thechoice of chief series(see [2,3])andP (s)= 1unless
i
G /G is a non-Frattini chief factor of G.
i i+1
We recall some properties of the series P (s). If S is cyclic of order p , then P (s) = 1−c /(pri)s,
i i i i i i
where c is the number of complements of G /G in G/G . It is more difficult to compute the
i i i+1 i+1
series P (s) when S is a non-abelian simple group. In that case a relevant role is played by the group
i i
L = G/C (G /G ). This is a monolithic primitive group and its unique minimal normal subgroup
i G i i+1
is isomorphic to G /G ∼= Sri. If n 6= |S |ri, then the coefficient b in (2.2) depends only on the
i i+1 i i i,n
knowledge of L ; more precisely we have
i
b = µ (H).
i,n Li
X
|Li:H|=n
Li=Hsoc(Li)
Some help in computing the coefficients b comes from the knowledge of the subgroup X of AutS
i,n i i
induced by the conjugation action of the normalizer in L of a composition factor of the socle Sr (note
i i
that X is an almost simple group with socle isomorphic to S ). More precisely, given an almost simple
i i
group X with socle S, we can consider the following Dirichlet polynomial:
c (X)
n
(2.4) P (s) = , where c (X) = µ (H).
X,S ns n X
Xn |XX:H|=n
X=SN
The following can be deduced from [7]:
Lemma 2.5. If S is nonabelian and π is a set of primes containing at least one divisor of |S | then
i i
Pπ(s)= Pπ (r s−r +1).
i Xi,Si i i
In particular, if n is not divisible by some prime in π, then there exists m ∈ N with n = mri and
b = c (X )·mri−1.
i,n m i
For an almost simple group X, let Ω(X) be the set of the odd integers m ∈ N such that
• X contains at least one subgroup Y such that X = YsocX and |X :Y| = m;
• if X = YsocX and |X :Y| = m, then Y is a maximal subgroup if X.
6 DUONGHOANGDUNGANDANDREALUCCHINI
Note that if m ∈ Ω(X), X = YsocX and |X : Y| = m, then µ (Y) = −1: in particular c (X) < 0.
X m
Combined with Lemma 2.5, this implies:
Remark 2.6. If m ∈ Ω(Xi), then bi,mri < 0.
Lemma 2.7. Let X be an almost simple group with soc(X) = PSL(2,p), where p ≥ 5 is an odd prime
and let w(X) = min{m | m ∈ Ω(X)}. Then we have:
q(q−1)/2 if q ≡ 3 mod 4 and q ∈/ {5,7,11,19,29},

q(q+1)/2 if q ≡ 1 mod 4 and q ∈/ {5,7,11,19,29},



 5 if q = 5,



 7 if q = 7 and X =PSL(2,7),



 3·7 if q = 7 and X =PGL(2,7),



w(X) =  11 if q = 11 and X = PSL(2,11),



11·5 if q = 11 and X = PGL(2,11),

 19·3 if q = 19 and X = PSL(2,19),


 19·32 if q = 19 and X = PGL(2,19),



 29·7 if q = 29 and X = PSL(2,29),



 29·3·5 if q = 29 and X = PGL(2,29).





Proof. Assume that S ∼= PSL(2,q) where q ≥ 5 be an odd prime: Aut(S) = PGL(2,q) and X =
PSL(2,q) or X = PGL(2,q). In both the cases the conclusion follows easily form the list of maximal
subgroups of X given in [1]. (cid:3)
3. Proof of Theorem1.3
We start now the proof of our main result. We assume that G is a finitely generated profinite
group G with the properties that P (s) = a /ns is rational. As described in Section 2, P (s) can
G n n G
be written as a formal infinite product of PDirichlet polynomials P (s) = b /ns corresponding
i n∈N i,n
to the factors G /G of a chief series of G. Let J be the set of indicePs i such that G /G is a
i i+1 i i+1
non-Frattini chief factor. Since P (s) = 1 if i ∈/ J, we have
i
P (s) = P (s).
G j
jY∈J
For C(s) = c /ns ∈ R, we define π(C(s)) to be the set of the primes q for which there
n∈N n
exists at least oPne multiple n of q with c 6= 0. Notice that if C(s) = A(s)/B(s) is rational then
n
π(C(s)) ⊆ π(A(s)) π(B(s)) is finite. Let S be the set of the finite simple groups that are isomorphic
to a composition faSctor of some non-Frattini chief factor of G. The first step in the proof of Theorem
1.3 is to show that S is finite. The proof of this claim requires the following result.
Lemma 3.1 ([5, Lemma 3.1]). Let G be a finitely generated profinite group and let q be a prime with
q ∈/ π(P (s)). If q divides the order of a non-Frattini chief factor of G, then this factor is not a
G
q-group.
RATIONAL PROBABILISTIC ZETA FUNCTION 7
Lemma 3.2. If G satisfies the hypothesis of Theorem 1.3, then the sets S and π(G) are finite.
Proof. Since P (s) is rational, we have that π(P (s)) is finite. Therefore, it follows from Lemma 3.1
G G
that S contains only finitely many abelian groups. Assume by contradiction that S is infinite. This
is possible only if the subset S∗ of the simple groups in S that are isomorphic to PSL(2,p) for some
prime p is infinite. Let
I := {j ∈ J |S ∈ S∗}, A(s) := P (s) and B(s):= P (s).
j i i
Yi∈I Yi∈/I
Notice that π(B(s)) ⊆ π(S) is a finite set. Since P (s) = A(s)B(s) and π(P (s)) is finite,
S∈S\S∗ G G
if follows that the set πS(A(s)) is finite. In particular, there exists a prime number q ≥ 5 such that
q ∈/ π(A(s)) but PSL(2,q) ∈S∗. Let Λ be the set of the odd integers n divisible by q but not divisible
by any prime strictly greater than q and set
r := min{r |S = PSL(2,q)},
i i
I∗ := {i ∈ I | S = PSL(2,q) and r = r},
i i
w := min{w(X )| S = PSL(2,q) and r =r},
i i i
β := min{n > 1| n ∈ Λ,v (n) = r and b 6= 0 for some i∈ I}.
q i,n
Assume i ∈ I, n ∈ Λ and b 6= 0. We have that S ∼= PSL(2,q ) for a suitable prime q and, by
i,n i i i
Lemma 2.5, n= xri and X contains a subgroup whose index divides x ; using the classification of the
i i i
maximal subgroups of X in [1], one can notice that q divides the indices of all the subgroups of X
i i i
with odd index. Hence q divides x and consequently n. Since q is the largest prime divisor of |S |
i i i i
and q is the largest prime divisor of n, we deduce that q = q . It follows that β = wr and b 6= 0 if
i i,β
and only if i ∈ I∗ and w(X ) = w; moreover in this last case b < 0. Hence the coefficient c of 1/βs
i i,β β
in A(s) is
c = b < 0.
β i,β
i∈I∗,wX(Xi)=w
This implies that q ∈ π(A(s)) which is a contradiction. So we have proved that S is finite. By [5,
Lemma 3.2], if follows that π(G) is also finite. (cid:3)
Proof of Theorem 1.3. Let T bethe set of the almost simplegroups X such that there exists infinitely
many i ∈ J with X ∼= X and let I = {i ∈ J | X ∈ T}. By Lemma 3.2, J \I is finite. We have to
i i
prove that J is finite; this is equivalent to show that I = ∅. Let i ∈ I : by the hypothesis of Theorem
1.3, there exists a prime q such that S ∈ {C ,PSL(2,q )}. Set q = max{q |i ∈I} and let Λ be the
i i qi i i
set of odd integers n divisible by q. Assume n ∈ Λ and b 6= 0 for some i ∈ I. If S is cyclic, then
i,n i
P (s) = 1−c /ns where n = |G /G | = qri and c is the number of complements of G /G in
i n i i+1 i n i i+1
G/G : this implies q = q . Otherwise S = PSL(2,q ) with q ≤ q and, by Lemma 2.5, n = xri and
i+1 i i i i i
X contains a subgroup whose index divides x ; since x divides |SL(2,q )|, q divides x and q ≤ q, we
i i i i i i
must have q = q . In both the cases, we have n = xri where x a positive integer with v (x ) =1. Let
i i i i q i
w = min{x ∈ Λ | vq(x) = 1 and bi,xri 6= 0 for some i∈ I}.
8 DUONGHOANGDUNGANDANDREALUCCHINI
Since J \I is finite and P (s) = P (s) is rational, also P (s) is rational. In particular, the
G i∈J i i∈I i
following series is rational: Q Q
{2}
Q(s) = P (s).
i
Yi∈I
Let I∗ = {i ∈ I | bi,wri 6= 0}. By the above considerations and Remark 2.6, i∈ I∗ if and only if either
S ∼= C and w = q or soc(X ) = PSL(2,q) and w(X ) = w. In particular if i ∈ I∗ then there exist
i q i i
infinitely many j ∈ I with X ∼= X and all of them are in I∗, hence I∗ is an infinite set. Moreover
i j
bi,wri < 0 for every i ∈ I∗, and therefore applying Proposition 2.4to the Dirichlet series Q(s), we
deduce that the product
bi,wri bi,wri
H(s) = 1+ = 1+
(cid:18) wris (cid:19) (cid:18) wris (cid:19)
Yi∈I iY∈I∗
is rational. By Corollary 1.2 the set I∗ must be finite, a contradiction. (cid:3)
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First Author
Mathematisch Instituut,Leiden Universiteit, Niels Bohrweg 1, 2333 CA Leiden,The Netherlands
Dipartimento di Matematica, Universit`a degli studidi Padova, Via Trieste 63, 35121 Padova, Italy
Email: [email protected]
Second Author
Dipartimento di Matematica, Universit`a degli studidi Padova, Via Trieste 63, 35121 Padova, Italy
Email: [email protected]