RECOGNIZING SOLUBLE GROUPS FROM THEIR PROBABILISTIC ZETA FUNCTIONS.
ELOISA DETOMI AND ANDREA LUCCHINI
Abstract. We answer a question by Bouc, proving that if PG(s) has an Euler product expansion with all factors of the form 1−ci/qiswhere each qiis a prime power, then G is soluble.
1. Introduction Given a finite group G we define the complex function
PG(s) = X
H≤G
µ(H, G) [G : H]s,
where µ is the M¨obius function of the subgroup lattice of G. This function is the multiplicative inverse of the probabilistic zeta function of G, as described by Boston [1] and Mann [8]. The function PG(s) belongs to the ring R of the the finite Dirichlet series; R is a unique factorization domain (since it is a polynomial ring overC in 1/2s, 1/3s, 1/5s, . . . ) and Boston suggested that the study of the factors of PG(s) could be useful to understand what properties are satisfied by the probabilistic zeta function associated to the finite group G.
If N is a normal subgroup of G define PG,N(s) = X
HN =G
µ(H, G) [G : H]s.
We have that PG,N(s) is a divisor of PG(s) in R; more precisely, as it is noticed in [3],
PG(s) = PG/N(s)PG,N(s). (1.1) When G is a soluble group, by repeated use of (1.1), Gasch¨utz [5] obtains an Euler product expansion for the Dirichlet series PG(s) (i.e. a factorization in which the factors are naturally associated to prime numbers). Very briefly, if
1 = N0≤ N1≤ · · · ≤ Nk= G
is a chief series, then PG(s) is the product of factors of the form 1 − ci/qsi, and there is a factor for each index i = 1, . . . , k such that Ni/Ni−1 has a complement in G/Ni−1. Here ci is the number of complements and qi = |Ni : Ni−1| is a prime power. Bouc [2] has asked whether there is a converse to this result: if G is a finite group such that PG(s) has an Euler product expansion with factors of the form1 − ci/qsi where eachqi is a prime power, isG soluble?
1991 Mathematics Subject Classification. 20P05, 20F05, 11M41.
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In this paper we give an affirmative answer to this question. More precisely, for each prime p set
Rp= (
X
r∈N
apr
(pr)s ap
r= 0 for almost all r )
; we prove:
Theorem 1. LetG be a finite group. The following are equivalent:
(i) G is soluble;
(ii) PG(s) is a product of factors of the form 1 − ci/qsi where eachqi is a prime power;
(iii) PG(s) = f1(s) · · · ft(s) with fi(s) ∈ Rpi.
An easy calculation shows that PG(s) = P an/ns satisfies condition (iii) in Theorem 1 if and only if the sequence n → an is multiplicative, that is anm = anamwhenever n and m are coprime. This implies that we can deduce whether a finite group G is soluble from the knowledge of the function PG(s). Note that the analogous result for nilpotent groups does not hold. Indeed, as it was shown by Gasch¨utz, if G1=Z2×Z3×Z3 and G2= Sym(3) ×Z3, then
PG1(s) = PG2(s) = 1 − 1
2s
1 − 1 3s
1 − 3 3s
.
On the other hand we can deduce whether a finite group G is supersoluble from the function PG(s); namely we have:
Theorem 2. A finite group G is supersoluble if and only if PG(s) is a product of factors of the form 1 − ci/psi where each pi is a prime and each ci is a positive integer.
The proof of Theorem 1 (and consequently of Theorem 2) depends on the classi- fication of finite simple groups. Indeed we use a result due to Guralnick [6], which states that PSL(2, 7) is the only finite simple group with subgroups of two different prime power indices. When PG(s) ∈ Rp the solubility of G can be easily deduced without employing the classification; in that case we have the additional result:
Theorem 3. A finite group G is a p-group if and only if PG(s) ∈ Rp. 2. Proofs
Recall that a primitive monolithic finite group is a finite group L such that soc(L) is a minimal normal subgroup and there exists a maximal subgroup of L with trivial core.
Lemma 4. LetL be a primitive monolithic finite group with nonabelian socle N ∼= Sr whereS PSL(2, 7) is a nonabelian simple group. There is at most one prime p satisfying: there exists H < L such that L = HN and |L : H| = pa.
Proof. Let H be a supplement of N in L and let |L : H| = pa where p is a prime and 0 6= a ∈ N. Then |N : H ∩ N| = |L : H| = pa. Let N = S1× · · · × Sr, with Si ∼= S for i = 1, . . . , r. Since H acts transitively on the set {S1, . . . , Sr}, we get that either H ∩ N is isomorphic to a subgroup of Kr, where K < S, or H ∩ N ∼= Su with u < r (see for example the proof of O’Nan-Scott theorem in [7]). Clearly, since
|N : H ∩ N | = pa, only the first case can occur. In particular, H ∩ N is isomorphic
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to a subgroup of Kr, where K is a proper subgroup of S with index a power of p.
The uniqueness of p is now a consequence of the already quoted result by Guralnick [6] which states that PSL(2, 7) is the only finite simple group with subgroups of
two different prime power indices.
Lemma 5. LetL be a primitive monolithic finite group with nonabelian socle N ∼= S1× · · · × Sr whereSi∼= PSL(2, 7) for i = 1, . . . , r. If NL(S1) 6= S1CL(S1) and H is a proper subgroup of L such that L = HN and |L : H| is a prime power, then
|L : H| is a power of 2.
Proof. Let |L : H| = |N : H ∩ N | = pa for a prime p and an integer a > 0.
Moreover, let π1 be the projection of N onto the first component S1of N and set K1= π1(H ∩N ); clearly K1is a proper subgroup of S1(otherwise H ∩N ∼= Suwith u < r) and it has p-power index in S1. Note also that NH(S1) ≤ NH(K1). Since L = HN and N ≤ NL(S1), then NL(S1) = N H ∩ NL(S1) = N (H ∩ NL(S1)) = N NH(S1). If NH(S1) ≤ S1CH(S1), then NL(S1) ≤ N S1CH(S1) ≤ S1CL(S1), against our assumption. Thus NH(S1) S1CH(S1), and so there is an element that normalizes K1 and induces an outer automorphism on S1. By the structure of Aut(PSL(2, 7)) ∼= PGL(2, 7), we conclude that |S1: K1| is a power of 2, that is p = 2. Indeed, we can produce an element x ∈ PGL(2, 7) \ PSL(2, 7) with 2-power order and normalizing a subgroup K of p-power index in S = PSL(2, 7). Thus Khxi has p-power index in PGL(2, 7), and we conclude by observing that PGL(2, 7) has
no maximal subgroup of p-power index for p 6= 2.
Proof of Theorem 1. Clearly, it is sufficient to prove that if G is a finite group and PG(s) = f1(s) · · · ft(s) with fi(s) ∈ Rpi and pi 6= pj for i 6= j, then G is soluble. Assume, by contradiction, that G has a nonabelian chief factor H/K and set C = CG(H/K). The quotient group L = G/C is a monolithic primitive group with nonabelian socle N = HC/C. Since
PG(s) = PG/C(s)PG,C(s) = PG/HC(s)PG/C,HC/C(s)PG,C(s), the series PL,N(s) = PG/C,HC/C(s) divides PG(s) and we can write
PL,N(s) = g1(s) · · · gt(s) where gi(s) = GCD(PL,N(s), fi(s)) ∈ Rpi. Observe that the constant term of PL,N(s) =P
HN =Lµ(H, L)/|L : H|sis precisely 1 and, since N is a non-Frattini subgroup of L, PL,N(s) has a non constant term (re- lated to a maximal supplement of N in L). Thus we can assume that g1(s), . . . , gu(s) are non constant series, for 1 ≤ u ≤ t, and gu+1(s) = . . . = gt(s) = 1. In particular, there exist some positive integers l1, . . . , lusuch that
gi(s) = ai+ bi
(pili)s +X
j>li
ci,j
(pij)s where bi6= 0, for i = 1, . . . , u, and
u
Y
i=1
a1= 1;
and so we can write
PL,N(s) = g1(s) · · · gu(s) = 1 + d1
(p1l1)s + · · · + du
(pulu)s +X
n∈N
en
ns, (2.1)
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where di6= 0 for i = 1, . . . , u and en= 0 if n is divisible by a prime q /∈ {p1, . . . , pu} or if n = pmi i and mi≤ li. Since
PL,N(s) = X
HN =L
µ(H, L)
[L : H]s =Xhn
ns where hn= X
HN =L
|L:H|=n
µ(H, L)
and di6= 1, there exist proper subgroups Hi of L such that
HiN = L and |L : Hi| = pili, i = 1, . . . u. (2.2) Note that, by the minimality of li in (2.1), each Hi is a maximal subgroup of L.
Let S be the nonabelian simple group such that N is a direct product of some copies of S.
a) If S 6= PSL(2, 7), by Lemma 4 there is at most one prime p such that a supplement H of N in L has index a power of p. Therefore, recalling that pi6= pj for i 6= j, by (2.2) we get u = 1, p1= p and hence PL,N(s) ∈ Rp.
Now, as |N : N ∩ H1| = |L : H1| = pl1 6= 1, p divides |N | and N has a non trivial Sylow p-subgroup P . Since L = N NG(P ) and NG(P ) 6= L contains a Sylow p-subgroup of L, there is a maximal subgroup M in L such that M N = L and
|L : M | is prime to p. Set m = min{c ∈ N | (c, p) = 1 and there exists K ≤ L s.t. KN = L, |L : K| = c}. By minimality of m, every subgroup K with index m and satisfying L = KN is maximal in L, and so µ(K) = −1. Therefore PL,N(s) contains a non trivial term c/ms∈ R/ p and so PL,N(s) /∈ Rp, a contradiction.
b) Now assume S = PSL(2, 7) and let N = S1× · · · × Sr, Si∼= S.
If NL(S1) 6= CL(S1)S1, then, by Lemma 5 there is at most one prime p such that a supplement H of N in L has index a power of p. Thus, by the same arguments used in the previous case, we get a contradiction.
So, let NL(S1) = CL(S1)S1. This implies that L = N X where X acts faithfully as permutations on {S1, . . . , Sr} (we can see L as a wreath product of S1 by a transitive subgroup of Sym(r)).
If r = 1, then L = PSL(2, 7) and we have (see for example [1])
PPSL(2,7)(s) = 1 − 8/8s− 14/7s+ 21/21s+ 28/28s+ 56/56s− 84/84s; this gives a contradiction, because by (2.1) we see that u = 2 and {p1, p2} = {2, 7}
(hence PPSL(2,7)(s) ∈ R2R7), but 21/21s∈ R/ 2R7.
If r 6= 1, let D be the diagonal subgroup of N ; then DX is a subgroup of L with index | PSL(2, 7)|r−1 = 168r−1. Let m be the minimal positive integer such that L has a subgroup H with |L : H| = 168m and L = HN . Clearly, m ≤ r − 1.
Note that if M is a maximal subgroups of L and L = M N , then either M ∩ N is isomorphic to Kr, where K is a subgroup of S, or M ∩ N is isomorphic to Sv with v < r. In particular, since the maximal subgroups of PSL(2, 7) have index 7 and 8, we get that 7ror 8rdivides |L : M | = |N : M ∩ N | in the first case, |L : M | = 168v in the second one. Thus, by (2.2), we obtain u ≤ 2 and PL,N(s) ∈ R2R7. Moreover, since qrdoes not divide 168m= (8 · 7 · 3)m, for q = 7, 8, and m < r, by minimality of m it follows that each subgroup H satisfying L = HN and |L : H| = 168m, is a maximal subgroup of L, and so µ(H) = −1. Therefore PL,N(s) contains a non trivial term c/(168m)s∈ R/ 2R7, and thus PL,N(s) /∈ R2R7, a contradiction. 2 Remark 6. The argument used in the proof of Theorem 1 can be adapted to prove a more general result: if H/K is a chief factor of a finite group G and PG/K,H/K(s) = g1(s) · · · gt(s) with gi∈ Rpi, thenH/K is abelian.
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Indeed, let H/K be a nonabelian chief factor; consider the monolithic primitive group L = G/CG(H/K) and let N = soc(L) ∼= H/K. In [4] it is proved that PG/K,H/K(s) − PL,N(s) = c/|N |s for some c; on the other hand, to deduce from PL,N(s) = g1(s) · · · gt(s) that N is abelian, we argue only on terms an/nsof PL,N(s) with “small” n (in particular n 6= |N |).
Proof of Theorem 2. If G is soluble, then, as described in the introduction, Gasch¨utz obtains a product expansion for the Dirichlet series PG(s) indexed by the non- Frattini factors in a chief series of G; the factors are all of the kind 1 − ci/qis, where qi is the order of the corresponding chief factor. If the group G is supersoluble, qi is a prime. Conversely, suppose that PG(s) = Q(1 − ci/psi), where, for each i, pi is a prime and ci is a positive integer. Since a group is G supersoluble if and only if G/ Frat(G) is supersoluble, it suffices to prove than any non-Frattini chief factor of G has prime order. Suppose, by contradiction, that H/K is a non- Frattini chief factor of G of order q, which is not a prime. Note that, by Theorem 1, G is soluble and so H/K is abelian. Thus, by a result of Gasch¨utz [5], we have PG/K,H/K(s) = 1 − c/qs, where c 6= 0 is the number of complements of H/K in G/K. Now, PG/K,H/K(s) divides PG(s). Since the factors 1 − ci/psi are irreducible in R, 1 − c/qsmust be a product of some of them, but this is impossible. 2 Proof of Theorem 3. Suppose that PG(s) ∈ Rpfor a prime p and let N be a minimal normal subgroup of G. As PG(s) = PG/N(s)PG,N(s), it is easy to see that both PG/N(s) and PG,N(s) belong to Rp. By induction on the order of G, it follows that G/N is a p-group, so it remains to prove that N is a p-group. Clearly we may assume that N is a non-Frattini subgroup, because |G| and |G/ Frat(G)| have the same prime divisors. This gives that PG,N(s) contains a non constant term c/pls, where pl = |G : H| = |N : H ∩ N | for a suitable supplement H of N in G. In particular, p divides |N |. Note that the proof is completed by showing that N is abelian, since in this case N is an elementary abelian p-group. Suppose, contrary to our claim, that N is nonabelian. Then N has a proper Sylow p-subgroup P 6= 1 and, by the same arguments used in the proof of Theorem 1 to discuss case a), we conclude that PG,N(s) /∈ Rp, a contradiction. 2
References
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[3] K. S. Brown, ‘The coset poset and probabilistic zeta function of a finite group’, J. Algebra 225 (2000) 989–1012.
[4] E. Detomi and A. Lucchini, ‘Crowns and factorization of the probabilistic zeta function of a finite group’, preprint.
[5] W. Gasch¨utz, ‘Die Eulersche Funktion endlicher aufl¨osbarer Gruppen’, Illinois J. Math. 3 (1959) 469–476.
[6] R. Guralnick, ‘Subgroups of prime power index in a simple group’, J. Algebra 81 (1983) 304–311.
[7] M. W. Liebeck, C. E. Prager and J. Saxl, ‘On the O’Nan-Scott theorem for finite primitive permutation groups’, J. Austral. Math. Soc. 44 (1988) 389–396.
[8] A. Mann, ‘Positively finitely generated groups’, Forum Math. 8 No. 4 (1996) 429–459.
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25133 Brescia, Italy [email protected] [email protected]
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