On the (non-)contractibility of the order complex of the coset poset of an alternating group

On the (non-)Contractibility of the Order Complex of the Coset Poset of an Alternating Group

MASSIMILIANOPATASSINI

ABSTRACT- Let Altkbe the alternating group of degreek. In this paper we prove that the order complex of the coset poset of Altkis non-contractiblefor a big family of k2N, including thenumbers of theformkˆp‡mwherem2 f3;. . .;35gand p>k=2. In order to prove this result, we show thatPG( 1) does not vanish, wherePG(s) is the Dirichlet polynomial associated to the groupG. Moreover, we extend the result to some monolithic primitive groups whose socle is a direct product of alternating groups.

MATHEMATICSSUBJECTCLASSIFICATION(2010). 20D30, 20P05, 11M41.

KEYWORDS. Probabilistic zeta functions; simplicial complexes; order complexes;

contractibility; coset posets; alternating groups; Brown conjecture.

1. Introduction

For a finitegroupG, le tC(G) be the set of (right) cosets of all proper subgroups of G, partially ordered by inclusion. Let DˆD(C(G)) bethe order complex ofC(G), so thek-dimensional faces ofDarechains of lengthk fromC(G). Thestudy ofDwas initiated in a paper of K. S. Brown (see [2]), who attributed therein to S. Bouc the observation that there exists a connection between the reduced Euler characteristic ~x(C(G)) and theDi- richlet polynomialPG(s) ofG, defined by

P_G(s)ˆX¹

nˆ1

an(G)

n^s ; where an(G)ˆ X

HG;jG:Hjˆn

m_G(H):

Here m_G is theMoÈbius function of thesubgroup latticeof G, which is

(*) Indirizzo dell'A.: Via Rovede, 13, 31020 Vidor (TV), Italy.

E-mail: frapmass@gmail.com

defined inductively bym_G(G)ˆ1, m_G(H)ˆ P

K>Hm_G(K). The counterpart of the Dirichlet polynomial is called the probabilistic zeta function of G (see [1] and [6]).

In particular, Brown ([2], § 3) showed that PG( 1)ˆ ~x(C(G)):

It is a well-known fact that ifD(C(G)) is contractible, then its reduced Euler characteristic ~x(C(G)) is zero. Hence, ifP_G( 1)6ˆ0, then the simplicial complex associated to the groupGis non-contractible.

Moreover, in [2], Brown conjectured the following.

CONJECTURE1. If G is a finite group, then P_G( 1)6ˆ0. Hence the order complex of the coset poset of G is non-contractible.

The conjecture was proved for some families of groups, as the following theorem shows.

THEOREM2. Let G be a finite group.

(1) If G is soluble, then PG( 1)6ˆ0(see[2]).

(2) If G is a classical group which does not contain non-trivial graph automorphisms, then P_G( 1)6ˆ0([9, Theorem 2]).

(3) If G is a Suzuki simple group or a Ree simple group, then P_G( 1)6ˆ0 ([7]).

Moreover, in our PhD thesis (see [8, Theorem 7.1]), we proved Con- jecture 1 for a class of monolithic primitive groups with socle isomorphic to a direct product of copies of a simple classical group with some conditions on the rank and the number of factors of the direct product.

In this paper we show the following.

THEOREM3. Let G be Sym_k orAlt_k. Let p be a prime number such thatk

25p5k 2.

(1) If k p35, then P_G( 1)6ˆ0.

(2) If p>2((k p)!)³, then P_G( 1)6ˆ0.

A proof of this result proceeds as follows (the general strategy is the same which we employed in [9]). Let r be a prime number and let f(s)ˆ P

n1

a_n

n^s be a Dirichlet polynomial. We denote byf^(r)(s) the Dirichlet

polynomial X

(n;r)ˆ1

a_n n^s: Letpbe a prime number such thatk

25p5k 2. We have that P_G(s)ˆP^(p)_G(s)‡X

pjk

a_k(G) k^s :

The first summandP^(p)_G (s) collects the contribution given by the subgroups Hof X such thatHcontains a Sylow p-subgroup, which are intransitive subgroups of G (see [10, Lemma 9]). In [10, Proposition 12] an explicit formula for the Dirichlet polynomialP^(p)_G (s) is given. So, a careful analysis of the valueP^(p)_G ( 1), shows thatjP^(p)_G ( 1)j_pˆpin many cases (see Section 3), wherejaj_pis the greatest power ofpthat divides the integer numbera (see Section 2 for a more precise definition).

Sincendividesa_n(G) forn2N f0g(see Lemma 5), we have that X

pjn

an(G)n pp²; hence

jPG( 1)j_pˆ jP^(p)_G ( 1)‡X

pjn

an(G)nj_pˆp wheneverjP^(p)_G( 1)j_pˆp. So we concludeP_G( 1)6ˆ0.

With a little more work we can obtain a more general result. Let us note that if N is a normal subgroup of a finite group G, then P_G(s)ˆP_G=N(s)P_G;N(s) (see [2]), where

P_G;N(s)ˆX

n1

a_n(G;N)

n^s ; with a_n(G;N)ˆ X

HG;jG:Hjˆn;

NHˆG;

m_G(H):

Thus, if 1ˆN05N15. . .5NlˆG is a chief series of G, applying the above formula repeatedly, we obtain

P_G(s)ˆY^{l 1}

iˆ0

P_{G=Ni;Ni‡1=Ni}(s):

If the chief factorN_i‡1=N_iis abelian, by a result of [4], we have that P_{G=Ni;Ni‡1=Ni}(s)ˆ1 c_i

jN_ij^s

where ci is the number of complements of Ni‡1=Ni in G=Ni. Hence P_{G=Ni;Ni‡1=Ni}( 1)6ˆ0.

If the chief factorN_i‡1=N_iis non-abelian, then there exists a monolithic primitive groupL_isuch that

P^(r)_Li;soc(Li)(s)ˆP^(r)_{G=Ni;Ni‡1=Ni}(s)

for each prime divisor r of the order of the socle soc(Li) of Li (see, for example, [3, Proposition 16]).

The above discussion suggests that in order to prove Conjecture 1, we can focus our attention on the monolithic primitive groups. In particular, here we prove the following.

THEOREM 4. Let G be a primitive monolithic group with socle N isomorphic to a direct product of n copies of the alternating groupAlt_k with k8. Let p be a prime number such that k

25p5k 2. If p>2n((k p)!)^2n‡1, then P_G( 1)6ˆ0.

2. Some useful results and definitions

Letabe an integer and letrbe a prime number. We denote byjaj_rthe r-part ofa, i.e.jaj_rˆrⁱwherei2Nsuch thatrⁱdividesabutr^i‡1does not dividea. Moreover, we setj0j_rˆ0.

Ifbˆc=dis a rational number for somec2Z;d2N f0g, then we let jbj_rˆjcj_r

jdj_r.

We record here an important result on the MoÈbius function of the subgroup lattice ofG.

LEMMA5 ([5], Theorem; 4.5). Let A be a finite group and B a subgroup of A. The indexjN_A(B):Bjdividesm_A(B)jA:BA⁰j.

In particular, if X is an almost simple group with socleGand His a subgroup ofG, then Lemma 5 yieldsjG:Hjdividesm_G(H)jG:N_G(H)j. So, in particular,ndividesan(X;G).

We can say some more words on the Dirichlet polynomial of a mono- lithic primitive groupLwith non-abelian socleN. Assume thatSis a simple component of L, define XˆN_L(S)=C_L(S) and nˆ jL:N_L(S)j. We have that NSⁿ. SinceSsoc(X), assume that SX. The following result shows a connection between the Dirichlet polynomialsPL;N(s) andPX;S(s).

THEOREM6 (See [11, Theorem 5]).

P^(r)_L;N(s)ˆP^(r)_X;S(ns n‡1) for each prime divisor r of the order of S.

LetXbe Sym_kor Alt_k withk8. A key role in our paper is played by the Dirichlet polynomial P^(p)_X;S(s), which can be expressed by an explicit formula.

PROPOSITION7 (See [10, Proposition 12]). Let k8. Let X be either Alt_k orSym_k, and let p be a prime number such thatk

25p5k 2. Then P^(p)_X;S(s)ˆ X

v2Ck:klp

m(v) n(v)i(v)^{s 1}:

HereC_kis the set of partitions of k, i.e. non-decreasing sequences of pos- itive integers whose sum is k, we have vˆ(k₁;. . .;k_l)2C_k for some lk;k1;. . .;kl2N f0gsuch that k1‡. . .‡klˆk, and we define

m(v)ˆ( 1)^{l 1}(l 1)!; i(v)ˆ k!

Q^l

iˆ1k_i!

; n(v)ˆY^k

iˆ1

v_i!;

wherev_iˆ jfj:k_j ˆigj.

3. The Dirichlet polynomialP^(p)(s)

Let XˆSym_k or Alt_k for k8. For this section, we let P(s)ˆP_X;soc(X)(s). The aim of this section is to find the valuejP^(p)(1 n)j_p forn2 a natural number.

Letpbe a prime number such thatk

25p5k 2. By Proposition 7, we have that

P^(p)(1 n)ˆ X

v2Ck:klp

m(v)

n(v)i(v) ⁿ ˆ X

vˆ(k1;...;kl)2Ck:klp

( 1)^{l 1}(l 1)!(k!)ⁿ Q^k

iˆ1v_i!(Q^l

iˆ1k_i!)ⁿ

ˆ

ˆX^{k p}

jˆ0

g(j;n) k!

(k j)!

_n

;

where

g(j;n)ˆ X

vˆ(k1;...;kl 1)2Cj

( 1)^{l 1}(l 1)!

Q

i1v_i! ^lQ¹

iˆ1k_i! n

forj0 (the setC0consists of the empty partitionvˆ( ), sog(0;n)ˆ1).

Consider the polynomial fn(x)ˆg(0;n)‡X^{k p}

jˆ1

g(j;n) Y^{j 1}

iˆ0

(x‡k p i)

!_n

inQ[x]. Clearlyf_n(p)ˆP^(p)(1 n). We want to show thatxdividesf_n(x) in Q[x], butx²does not dividef_n(x) inQ[x]. This implies thatjP^(p)(1 n)j_pˆp if the coefficienta1ofxinfn(x) is such thatja1j_pˆ1 (see the proof of The- orem 11).

First of all, we prove the following formula, which is very useful in order to find some properties ofg(j;n).

PROPOSITION8. If m1, then X^m

jˆ0

g(j;n) ((m j)!)ⁿˆ0

PROOF. Let us rewrite the sum in another way. Let

vˆ(k₁;. . .;k_l)2C_m. For each h2 f1;. . .;lg there exists v^hˆ

(k1;. . .;kh 1;kh‡1;. . .;kl)2Cm kh. Clearlyvappears in the termg(m;n) andv^happears in the termg(m kh;n). In particular, the contribution of v^h ing(m k_h;n) is

( 1)^{l 1}(l 1)!

n(v^h) Q

i6ˆhk_i!_n:

Moreover, since n(v)ˆ ji:kiˆkhjn(v^h) we have that the previous ex- pression becomes

( 1)^{l 1}(l 1)!ji:k_iˆk_hj n(v) Q

i6ˆhk_i!_n :

Furthermore, it is clear that if (k₁;. . .;k_l⁰)ˆt2C_m⁰ for somem⁰5m, then there exists a unique v2C_m such that tˆv^h for some

h2 f1;. . .;l⁰‡1g: indeed, there exists h2 f1;. . .;l⁰‡1g such that kh 1m m⁰kh(with the convention thatk0ˆ0 andkl⁰‡1ˆk) and so vˆ(k1;. . .;k_{h 1};m m⁰;k_h;. . .;k_l⁰).

Let H_vˆ fh2 f1;. . .;lg:hˆ1 or k_{h 1}5k_hg (so fk₁;. . .;k_lg ˆ fk_h:h2H_vg and if h₁;h₂2H_v, then k_h1ˆk_h2 if and only if h₁ˆh₂).

We get X^m

jˆ0

g(j;n)

((m j)!)ⁿˆ X

(k1;...;kl)ˆv2Cm

( 1)^ll!

n(v) Y^l

iˆ1

ki!

!n‡X

h2Hv

( 1)^{l 1}(l 1)!ji:kiˆkhj n(v) Y

i6ˆh

ki!

!_n 1 (ki!)ⁿ 0

BB BB B@

1 CC CC CA

ˆ X

(k1;...;kl)ˆv2Cm

( 1)^l(l 1)!(l X

h2Hv

ji:kiˆkhj)

n(v) Y^l

iˆ1

ki!

!n ˆ0

since P

h2Hv

ji:k_iˆk_hj ˆlby definition. p

Thanks to the recursive formula we obtained forg(j;n), we can prove the following inequalities.

PROPOSITION9. We have that

05( 1)^{m 1}g(m 1;n)=25( 1)^mg(m;n)( 1)^{m 1}g(m 1;n)1 for m1and n2.

PROOF. Let us prove the proposition by induction onm. Ifmˆ1, the claim holds by definition ofg(m;n). Assume thatm>1. By induction, it is enough to prove that

( 1)^mg(m;n)=25( 1)^m‡1g(m‡1;n)( 1)^mg(m;n) By Proposition 8, we have that

g(m‡1;n)ˆ X^m

jˆ0

g(j;n) ((m‡1 j)!)ⁿ; hence

( 1)^m‡1g(m‡1;n)ˆ( 1)^mg(m;n)‡X^{m 1}

jˆ0

( 1)^{m j} ( 1)^jg(j;n) ((m‡1 j)!)ⁿ: (y)

Letg⁰(j;n)ˆ( 1)^{m j} ( 1)^jg(j;n)

((m‡1 j)!)ⁿ. Now, by induction we have ( 1)^jg(j;n)5 ( 1)^{j 1}g(j 1;n)=250

for 1jm. Hence, form jodd, we get

g⁰(j;n)‡g⁰(j 1;n)5( 1)^{j 1}g(j 1;n)2 (m j‡2)ⁿ 2((m j‡2)!)ⁿ 0 forn1. This implies that

X

m 1

jˆ0

( 1)^{m j} ( 1)^jg(j;n)

((m‡1 j)!)ⁿˆag⁰(0;n)

‡ X

1jm 1;m jodd

g⁰(j;n)‡g⁰(j 1;n)50 where aˆ0 if m is even, aˆ1 otherwise (when m is odd, note that g⁰(0;n)50). So this proves that ( 1)^m‡1g(m‡1;n)( 1)^mg(m;n).

Now, by (y), it remains to prove that ( 1)^m‡1g(m‡1;n) ( 1)^mg(m;n)=2

ˆ( 1)^mg(m;n)=2‡X^{m 1}

jˆ0

( 1)^{m j} ( 1)^jg(j;n) ((m‡1 j)!)ⁿ >0:

Again, by induction we have that ( 1)^jg(j;n)=25( 1)^j‡1g(j‡1;n) for 0jm 1. Hence, form jeven, we have that

g⁰(j;n)‡g⁰(j 1;n)>( 1)^jg(j;n)(m‡2 j)ⁿ 2 ((m‡2 j)!)ⁿ 0 forn1. Moreover, we get

( 1)^mg(m;n)=2‡g⁰(m 1;n)>( 1)^mg(m;n)2^{n 2} 1 2^{n 1} 0 forn2. This implies that

( 1)^mg(m;n)=2‡X^{m 1}

jˆ0

( 1)^{m j} ( 1)^jg(j;n) ((m‡1 j)!)ⁿ

ˆbg⁰(0;n)‡( 1)^mg(m;n)=2‡g⁰(m 1;n)

‡ X

1jm 2;m jeven

g⁰(j;n)‡g⁰(j 1;n)>0

where bˆ1 if m is even, bˆ0 otherwise (when m is even, note that g⁰(0;n)>0). So this proves that ( 1)^m‡1g(m‡1;n)>( 1)^mg(m;n)=2. p

Now, we are ready to prove the main result of this section.

PROPOSITION10. Let n2. We have that x divides f_n(x)inQ[x], but x² does not divide f_n(x)inQ[x].

Letmˆk p. To prove thatxdividesfn(x) is enough to show that fn(0)ˆX^m

jˆ0

g(j;n) m!

(m j)!

_n

ˆ0;

which follows from Proposition 8. In order to prove thatx²does not divide fn(x) we may show that the coefficient ofxinfn(x) does not vanish. It is easy to realize that the coefficient ofxinf_n(x) is:

X^m

jˆ0

ng(j;n) m!

(m j)!

_nX^{j 1}

iˆ0

1 m i (y) By Proposition 8, we can subtract

X^m

jˆ0

ng(j;n) m!

(m j)!

_n^mX¹

iˆ0

1 m iˆ0

from (y), hence proving that (y) does not vanish is equivalent to prove that X

m 1

jˆ0

g(j;n) m!

(m j)!

_n^mX¹

iˆj

1 m i6ˆ0:

Let

h_a;b(m;n)ˆX^b

jˆa

g(j;n) m!

(m j)!

_nX^{m 1}

iˆj

1 m i: Let 0jm 2. Note that

m!

(m j 1)!

_n X^{m 1}

iˆj‡1

1

m i2 m!

(m j)!

_n X^{m 1}

iˆj

1 m i>0 since

(m j)^{n m}X¹

iˆj‡1

1

m i2X^{m 1}

iˆj

1 m i>0:

Hence, by Proposition 9 we get ( 1)^j‡1g(j‡1;n) m!

(m j 1)!

_n X^{m 1}

iˆj‡1

1 m i>

( 1)^jg(j;n) m!

(m j)!

_nX^{m 1}

iˆj

1 m i; i.e. ( 1)^j‡1h_j;j‡1(m;n)>0. This implies thath_{0;m 1}(m;n)50 ifmis even, and h_{3;m 1}(m;n)>0 if m is odd. Since g(0;n)ˆ g(1;n)ˆ1 and g(2;n)ˆ1 2 ⁿ, it is easy to see thath_0;2(m;n)>0 (forn2 andm3).

Thus we conclude that

h_{0;m 1}(m;n)ˆh_0;2(m;n)‡h_{3;m 1}(m;n)>0

ifmis odd. The proof is complete. p

We can finally prove the main theorem.

THEOREM 11. Let k8 and let p be a prime number such that k=25p5k 2. Assume that p>n((k p)!)^n‡1. Then

jP^(p)( n‡1)j_pˆp:

PROOF. Let mˆk p. Let fn(x)ˆa0‡a1x‡. . .‡atx^t for some a_iˆa_i=b_i2Qwith (a_i;b_i)ˆ1 (ifa_iˆ0, letb_iˆ1) andt2N.

By the proof of Proposition 10 we have that a₁ˆ nh_{0;m 1}(m;n).

Moreover, we have seen that ( 1)^{m 1}h_{0;m 1}(m;n)>0 and ( 1)^{m 1}h_{0;m 2}(m;n)50, hence jh_{m 1;m 1}(m;n)j>jh_{0;m 1}(m;n)j. So we have

ja1j njhm 1;m 1(m;n)j ˆnjg(m 1;n)j(m!)ⁿn(m!)ⁿ beingjg(m 1;n)j 1 (by Proposition 9).

Now, we want to show that jP^(p)( n‡1)j_pˆp. Note that since p>k pandb_idivides (k p)!ˆm!(by definition off_n(x)), we have that ja_ij_pˆ ja_ij_p. By Proposition 10 we have thata₀ˆ0, hencejP^(p)( n‡1)j_pˆ jf_n(p)j_pp. For a contradiction, assume that jP^(p)( n‡1)j_p>p. Then jfn(p)j_pp², henceja1j_pp, so

ja₁j ja₁j_pp>n(m!)^n‡1 ja₁jm! ja₁kb₁j ˆ ja₁j;

a contradiction. p

When the numberk pis small, we know explicitly the coefficient ofx infn(x). For example, we have thatjP^(p)( n‡1)j_pˆpifpdoes not appear in Table 1 (for k p7 and n2 f2;4;6g). Further calculations can be done, but the coefficient ofxin f_n(x) grows very quickly and it has very large prime factors (for example, ifk pˆ35, the coefficient ofxinf₂(x) has a prime divisor of 50 digits).

TABLE1. Exceptions forp.

k p nˆ2 nˆ4 nˆ6

3 23 31, 53 109, 617

4 677 1047469 19, 31, 5051

5 71, 103 643, 148579 3889, 595523689

6 7, 13 23, 269, 89714671 2357, 4584299, 35430211 7 863897 181, 2005732476817 70353197, 1633443829219

However, there are many prime numbers betweenk=2 andk 2 ifkis big enough. For example, we have the following.

PROPOSITION12. Let8k10⁶. If k6ˆ13, then there exists a prime number p such thatjP^(p)( 1)j_pˆp.

4. The main result

We can now prove the main result.

THEOREM13. Let G be a monolithic primitive group with socle iso- morphic to Altⁿ_k for k8. Let p be a prime number such that k=25p5k 2.

(1) If p>2n((k p)!)^2n‡1, then P_G;soc(G)( 1)6ˆ0.

(2) If nˆ1and k p35, then P_G;soc(G)( 1)6ˆ0.

(3) If nˆ1and8k10⁷, then P_G;soc(G)( 1)6ˆ0.

PROOF. By Theorem 6, we have that

jP^(p)_G;soc(G)( 1)j_pˆ jP^(p)_X;soc(X)(1 2n)j_p;

whereXˆN_G(S)=C_G(S) andSis a simple component ofG. By Theorem 11, ifp>2n((k p)!)^2n‡1, thenjP^(p)_X;soc(X)(1 2n)j_pˆp. By Lemma 5, we have

thatldividesal(G;soc(G)) forl1, hence

jPG;soc(G)( 1) P^(p)_G;soc(G)( 1)j_pp²; thus we get the claim.

Ifnˆ1 and k p35, the coefficient ofx in f_n(x) is known, so by direct computation, there exists a prime number p⁰ (not necessarily dif- ferent from p) such that k=25p⁰5k 2 such that jP^(p_X;soc(X)⁰⁾ ( 1)j_p0ˆp⁰ (except whenkˆ13). Arguing as above, we get the claim (using a direct computation for kˆ13). Finally, if nˆ1 and 8k10⁶, arguing as

above, by Proposition 12 we have the claim. p

REFERENCES

[1] N. BOSTON, A probabilistic generalization of the Riemann zeta function, Analytic Number Theory,1(1996), pp. 155-162.

[2] K. S. BROWN,The coset poset and the probabilistic zeta function of a finite group, J. Algebra,225(2000), pp. 989-1012.

[3] E. DETOMI- A. LUCCHINI,Crowns and factorization of the probabilistic zeta function of a finite group, J. Algebra,265(2003), pp. 651-668.

[4] W. GASCHUÈTZ, Zu einem von B. H. und H. Neumann gestellten Problem, Math. Nachr.,14(1955), pp. 249-252.

[5] T. HAWKES- M. ISAACS- M. OÈZAYDIN, On the MoÈbius function of a finite group, Rocky Mountain Journal,19(1989), pp. 1003-1034.

[6] A. MANN, Positively finitely generated groups, Forum Math., 8 (1996), pp. 429-459.

[7] M. PATASSINI, The Probabilistic Zeta function of PSL2(q), of the Suzuki groups²B2(q)and of the Ree groups²G2(q), Pacific J. Math.,240(2009), pp.

185-200.

[8] M. PATASSINI,On the Dirichlet polynomial of the simple group of Lie type, UniversitaÁ di Padova, 2011, http://paduaresearch.cab.unipd.it/3272/1/Phd_The- sis.pdf,

[9] M. PATASSINI,On the (non-)contractibility of the order complex of the coset poset of a classical group, J. Algebra,343(2011), pp. 37-77.

[10] M. PATASSINI,Recognizing the non-Frattini abelian chief factors of a finite group from its Probabilistic Zeta function, Accepted by Comm. Algeb., 2011.

[11] P. JIMEÂNEZ SERAL, Coefficient of the Probabilistic Zeta function of a monolithic group, Glasgow J. Math.,50(2008), pp. 75-81.

Manoscritto pervenuto in redazione il 9 Settembre 2011.