On the Probabilistic Zeta Function for Finite Groups

On the Probabilistic Zeta Function for Finite Groups

210, 703]707 Ž1998. JA987560 JOURNAL OF ALGEBRA ARTICLE NO. On the Probabilistic Zeta Function for Finite Groups John ShareshianU Department of Math...

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210, 703]707 Ž1998. JA987560

JOURNAL OF ALGEBRA ARTICLE NO.

On the Probabilistic Zeta Function for Finite Groups John ShareshianU Department of Mathematics, California Institute of Technology, Pasadena, California 91125 Communicated by Jan Saxl Received April 13, 1998

Let G be a finite group, and define the function P Ž G, s . [

Ý HFG

mŽ H , G.

w G: H x

s

,

function on the subgroup lattice of G. The function P Ž G, s . where m is the Mobius ¨ is the multiplicative inverse of a zeta function for G, as described by Mann and X Boston. Boston conjectured that P Ž G, 1. s 0 if G is a nonabelian simple. We will X prove a generalization of this conjecture, showing that P Ž G, 1. s 0 unless Ž . GrOp G is cyclic for some prime p. Q 1998 Academic Press

In wMax and wBox, respectively, Mann and Boston examined the function P Ž G, s . s

mŽ H .

Ý w G: H x s , H

where the sum runs over the set of subgroups of finite index in a finitely generated profinite group G, and m is the Mobius function on the poset of ¨ such subgroups. Here, and in the rest of this paper, m Ž H . means m Ž H, G .. In wBox, it was conjectured that P X Ž G, 1. s 0 when G is a nonabelian finite simple group. Exponentiating both sides of the equality allows one to write this conjecture as follows. CONJECTURE 1 wBox.

Let G be a nonabelian finite simple group. Then Ž . w x Ł w G: H x m H r G : H s 1.

HFG U

E-mail: [email protected]. 703 0021-8693r98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.

704

JOHN SHARESHIAN

In what follows, we will prove a generalization of this conjecture, which will imply that the conclusion of the conjecture holds for any finite group G such that there is no prime p with GrOp Ž G . cyclic. From now on, every group G will be assumed to be finite. DEFINITION 2. Let G be a group and let  p1 , . . . , pk 4 be the set of prime divisors of < G <. Define the rational numbers a 1 , . . . , a k by k

Ž . w x Ł w G: H x m H r G : H s Ł pia . i

HFG

is1

A formula for a i will be determined from which the next result will follow. If GrOp iŽ G . is not cyclic then a i s 0.

PROPOSITION 3.

The first step is to get a crude formula for a i directly from the definition. Fix from now on a prime p s pi and let a s a i . For a positive integer m, we denote by m p the largest power of p dividing m. DEFINITION 4. 1. For H F G set a p Ž H . s log p Ž< H < p .. 2. For 1 F a F a p Ž G . set GaŽ G . s  H F G: a p Ž H . s a4 . 3. For 1 F a F a p Ž G . define f aŽ G . s Ý H g GaŽG. < H < m Ž H .. The next proposition follows from the definition by an easy calculation. PROPOSITION 5.

as

ap Ž G.
Ý

< H < mŽ H . y

HFG

1
a pŽ G .

Ý

af a Ž G . .

as1

Now we will calculate the two sums in the difference above. The first term was examined by Philip Hall in wHax, where he defined the Mobius ¨ function m and proved the Mobius inversion formula in order to show, ¨ among other things, that if n is a positive integer then P Ž G, n. gives the probability that a randomly chosen ordered n-tuple Ž g 1 , . . . , g n . from G satisfies G s ² g 1 , . . . , g n :. The following proposition is a special case of that result. PROPOSITION 6.

Let f be Euler’s function. Then

Ý HFG

< H < mŽ H . s

½

f Ž < G <. , 0,

if G is cyclic, otherwise.

705

ZETA FUNCTION FOR GROUPS

The second sum in Proposition 5 will be determined by calculating each f aŽ G .. DEFINITION 7. Fix a with 1 F a F a p Ž G .. 4. 1. Set D aŽ G . s  R F G: < R < s p a4 and DUaŽ G . s  R g D aŽ G .: ReG } Ž . Ž .  Ž .4 2. For R g D a G set V R s H F G: R g Syl p H . LEMMA 8.

Let R be a nontri¨ ial p-subgroup of G. Then

Ý

NH Ž R . m Ž H . s

¡ ¢0,

<

Ý H ~ HgV Ž R.

HgV Ž R .

< mŽ H . ,

ReG, } otherwise.

Proof. If ReG then NH Ž R . s H for all H g V Ž R . and the claim of } the lemma holds, so assume NG Ž R . - G. Note that for H F G we have H g V Ž R . if and only if NH Ž R . g V Ž R .. Indeed, if P g Syl p Ž H . strictly contains R, then so does NP Ž R . F NH Ž R .. Let VU Ž R . s  L g V Ž R .: R 1 L4 . It follows from the preceding argument that NH Ž R . m Ž H . s

Ý HgV Ž R .

< L<

Ý

LgV U Ž R .

Ý

mŽ H . .

HlNGŽ R .sL

Fix L g VU Ž R .. Since NG Ž R . / G, it follows from Weisner’s theorem Žsee, e.g., Corollary 3.9.3 of wStx. that

Ý

m Ž H . s 0,

HlNGŽ R .sL

and the lemma follows. COROLLARY 9.

For 1 F a F a p Ž G ., faŽ G. s

Ý

Ý

RgDUaŽ G . HgV Ž R .

< H < mŽ H . .

Proof. By Sylow’s theorem, we have faŽ G. s

Ý

Ý

NH Ž R . m Ž H . .

RgD aŽ G . HgV Ž R .

The corollary now follows immediately from Lemma 8. Now we can prove Proposition 3. By Propositions 5 and 6, it suffices to show that if GrOp Ž G . is not cyclic then f aŽ G . s 0 for 1 F a F a p Ž G .. If

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JOHN SHARESHIAN

a ) a p Ž Op Ž G .. then DUaŽ G . s B and f aŽ G . s 0 by Corollary 9. If a s a p Ž Op Ž G .. then DUaŽ G . s  Op Ž G .4 and by Corollary 9 we have faŽ G. s

< H < mŽ H . .

Ý HgV Ž O pŽ G ..

Set G s GrOp Ž G . and let m be the Mobius function on the subgroup ¨ lattice of G. Also, let JŽ G . be the set of subgroups of G whose order is not divisible by p. Then < H < m Ž H . s Op Ž G .

Ý HgV Ž O pŽ G ..

< H < mŽ H . .

Ý HgJ Ž G .

Since G is not cyclic, Proposition 6 gives

Ý

< H < mŽ H . s y

HgJ Ž G .

< H < mŽ H . .

Ý HfJ Ž G .

Now Sylow’s theorem gives a pŽ G .

Ý

< H < mŽ H . s

Ý

Ý

Ý

NK Ž R . m Ž K . .

bs1 RgD bŽ G . KgV Ž R .

HfJ Ž G .

Since Op Ž G . s 1, NK Ž R . m Ž K . s 0

Ý KgV Ž R .

for each nontrivial p-subgroup R F G by Lemma 8. Thus f aŽ G . s 0 for a s a p Ž Op Ž G ... We now proceed by downward induction on a, so we assume that a - a p Ž Op Ž G .. and that if Q g DUb Ž G . with a - b then Ý H g V ŽQ. < H < m Ž H . s 0. By Corollary 9 we have faŽ G. s

Ý

Ý

RgDUaŽ G . HgV Ž R .

< H < mŽ H . .

˜ s GrR. Let m Fix R g DUaŽ G . and set G function on the ˜ be the Mobius ¨ ˜ Since G is not cyclic, neither is G. ˜ Using the same subgroup lattice of G. arguments we used when examining a s a p Ž Op Ž G .., we see that, to show ˜ is a nontrivial that Ý H g V Ž R. < H < m Ž H . s 0, it suffices to show that if Q ˜ then Ý H˜ g V ŽQ. ˜ ˜ < < Ž . normal p-subgroup of G H m H s 0. Let Q be the full ˜ ˜ ˜ in G. Then Q g DUb Ž G. for some b ) a, and by inductive preimage of Q

ZETA FUNCTION FOR GROUPS

707

hypothesis we have < R<

Ý ˜ Ž Q˜. HgV

< H˜ < m ˜ Ž H˜ . s

Ý

< H < m Ž H . s 0.

HgV Ž Q .

This completes the proof of Proposition 3. If GrOp Ž G . is cyclic for some prime p, then G is solvable and P Ž G, s . is quite well understood Žsee Section 3 of wBox..

ACKNOWLEDGMENT The idea of examining Boston’s conjecture one prime at a time was suggested by Richard Lyons, whom I thank for that suggestion and for his encouragement. I thank Nigel Boston and Avinoam Mann for their helpful comments, and I thank the anonymous referee for clarifying the history of the study of P Ž G, s ..

REFERENCES wBox N. Boston, A probabilistic generalization of the Riemann zeta function, in ‘‘Analytic Number Theory, Proceedings of a Conference in Honor of Heini Halberstam,’’ Vol. 1, Progress in Mathematics, Vol. 138, Birkhauser, Boston, 1996. wHax P. Hall, The Eulerian functions of a finite group, Quart. J. Math. 7 Ž1936., 134]151. wMax A. Mann, Positively generated finite groups, Forum Math. 8 Ž1996., 429]459. wStx R. P. Stanley, ‘‘Enumerative Combinatorics,’’ Vol. 1, Wadsworth and BrooksrCole, Monterey, 1986.