A relation between m and the Euler characteristic of the G,N nerve space of some class poset of G Heguo Liu1, Xingzhong Xu∗,1,2 , Jiping Zhang3 Abstract. LetGbeafinitegroupandN✂Gwith|G:N|=pforsomeprime p. Inthis note, tocompute mG,N directly, weconstruct aclass poset TC(G) of G for somecyclic subgroup C. And wefind a relationbetween mG,N and 7 theEulercharacteristicofthenervespace|N(TC(G))|(seetheTheorem1.3). 1 Asanapplication,wecomputemS5,A5 =0directly,andgetS5 isaB-group. 0 2 Key Words: B-group;nervespace;posetofgroup. n 2000 MathematicsSubject Classification: 18B99· 19A22· 20J15 a J 7 2 ] 1. Introduction R G In [4], Bouc proposed the following conjecture: . h Conjecture 1.1. [4,ConjectureA] LetG beafinitegroup. Then β(G) is nilpotent t a if and only if G is nilpotent. m [ Here, β(G) a largest quotient of a finite group which is a B-group and the 1 definition of B-group can be found in [4, 5] or in the Section 2. Bouc has proven v the Conjecture 1.1 under the additional assumption that finite groupG is solvable 8 in [4]. In [16], Xu and Zhang consider some special cases when the finite group G 1 is not solvable. But this result relies on the proposition of Baumann [3], and his 0 8 proposition relies on the Conlon theorem [7, (80.51)]. If we want to generalize the 0 resultof[16], we needusethe newmethod to compute mG,N directly. Here, N is a . normal subgroup of G. And the definition of m can be fined in [4, 5] or in the 1 G,N 0 Section 2. 7 Forthis aim, someattempts havebeendone inthe paper. We beginto compute 1 : mG,N when |G : N| = p for some prime number. But it is not easy to compute v m directly even if |G : N| = p. When |G : N| = p, we have the following i G,N X r a ∗ Date: 27/01/2017. 1. DepartmentofMathematics,HubeiUniversity,Wuhan, 430062, China 2. Departament de Matema`tiques, Universitat Aut`onoma de Barcelona, E-08193 Bellaterra, Spain 3. SchoolofMathematicalSciences, PekingUniversity,Beijing,100871, China HeguoLiu’sE-mail: [email protected] XingzhongXu’sE-mail: [email protected], [email protected] ∗Correspondingauthor JipingZhang’sE-mail: [email protected] SupportedbyNational973Project(2011CB808003) andNSFCgrant(11371124, 11501183). 1 2 observation: 1 m + |X|µ(X,G) G,N |G| X X≤N 1 1 = |X|µ(X,G)+ |X|µ(X,G) |G| X |G| X XN=G,X≤G X≤N 1 1 = |X|µ(X,G)+ |X|µ(X,G) |G| X |G| X XN=G,X≤G XN6=G,X≤G 1 = |X|µ(X,G) |G| X X≤G = m =0,if G is not cyclic. G,G So to compute m , we can compute 1 |X|µ(X,G) first. And we find G,N |G|PX≤N there is a formula (see the proof of the Proposition 4.1) about 1 1 |X|µ(X,G)+ |X|µ(X,N) |G| X |G| X X≤N X≤N 1 (= |X|µ(X,G)+m ). |G| X N,N X≤N Now, we set 1 1 m′ := |X|µ(X,G)= |X|µ(X,G); G,N |G| X |G| X XN6=G,X≤G X≤N and set M′ := |X|µ(X,G)=|G|m′ . G,N X G,N X≤N We get the following theorem about M′ . G,N Proposition 1.2. Let G be a finite group and N ✂G such that |G : N| = p for some prime number p. Then M′ =− ϕ(|C|)− M′ . G,N X X Y,Y∩N C≤N, C is cyclic Y(cid:12)G,Y(cid:2)N Here, ϕ is the Euler totient function. To compute M′ , we need compute M′ for every Y (cid:12) G. Since Y (cid:12) G, G,N Y,Y∩N thuswecangetM′ byfinitesteps. Here,wedefineanewclassposetofsubgroups G,N of G as following: Let C be a cyclic subgroup of N, define T (G):={X|C ≤X (cid:12)G,X (cid:2)N}. C We can see that T (G) is a poset ordered by inclusion. We can consider poset C T (G) as a category with one morphism Y → Z if Y is a subgroup of Z. We set C N(T (G)) is the nerve of the category T (G) and |N(T (G))| is the geometric C C C realization of N(T (G)). Then we get a following computation about m . C G,N Theorem 1.3. Let G be a finite group and G not cyclic. Let N ✂G such that |G:N|=p for some prime number p. Then 1 m = ( (1−χ(|N(T (G))|)·ϕ(|C|))). G,N |G| X C C≤N, C is cyclic Here,|N(T (G))|isasimplicialcomplexassociatedtotheposetT (G),andχ(|N(T (G))|) C C C is the Euler characteristic of the space |N(T (G))|. C 3 Proof. Since m +m′ = m = 0 when G is not cyclic, thus we prove this G,N G,N G,G theorem by using the Proposition5.4 (cid:3) Byusing the methodofthe Theorem1.2-3,we compute m directly,andwe S5,A5 get the following result. Proposition 1.4. S is a B-group. 5 In fact, [6] had proved that S is a B-group. And by using [3], we also get that n S is a B-group when n≥5. n After recalling the basic definitions and properties of B-groupsin the Section 2, we prove some lemmas about Mo¨bius function in the Section 3. And this lemmas will be used in the Section 4 to prove the Proposition 1.2. In the Section 5, we construct a class poset T (G) of G for some cyclic subgroup of C and prove the C Theorem 1.3. As an application, we compute m =0 and get S is a B-group S5,A5 5 in the Section 6. 2. Burnside rings and B-groups In this section we collect some known results that will be needed later. For the background theory of Burnside rings and B-groups, we refer to [4], [5]. Definition 2.1. [5, Notation 5.2.2] Let G be a finite group and N ✂G. Denote by m the rational number defined by: G,N 1 m = |X|µ(X,G), G,N |G| X XN=G whereµ is the Mo¨bius function of the poset of subgroups of G. Remark 2.2. If N =1, we have 1 1 m = |X|µ(X,G)= |G|µ(G,G)=16=0. G,1 |G| X |G| X1=G Definition2.3. [4,Definiton2.2]ThefinitegroupGiscalled aB-groupifm = G,N 0 for any non-trivial normal subgroup N of G. Proposition 2.4. [5, Proposition 5.4.10] Let G be a finite group. If N ,N ✂G 1 2 are maximal such that m 6=0, then G/N ∼=G/N . G,N 1 2 Definition 2.5. [4,Notation2.3] When G is a finite group, and N✂G is maximal such that m 6=0, set β(G)=G/N. G,N Theorem 2.6. [5, Theorem 5.4.11] Let G be a finite group. 1. β(G) is a B-group. 2. If a B-group H is isomorphic to a quotient of G, then H is isomorphic to a quotient of β(G). 3. Let M ✂G. The following conditions are equivalent: (a) m 6=0. G,N (b) The group β(G) is isomorphic to a quotient of G/M. (c) β(G)∼=β(G/N). We collect some properties of m that will be needed later. G,N 4 Proposition2.7. [4,Proposition2.5]LetGbeafinitegroup. ThenGisaB-group if and only if m =0 for any minimal (non-trivial) normal subgroup of G. G,N Proposition 2.8. [5, Proposition 5.6.1] Let G be a finite group. Then m = 0 G,G if and only if G is not cyclic. If P be cyclic of order p and p be a prime number, then m = p−1. P,P p Remark 2.9. If G is a finite simple group, then G is a B-group if and only if G is not abelian. Proposition 2.10. [5, Proposition5.3.1] Let G be a finite group. If M and N are normal subgroup of G with N ≤M, then m =m m . G,M G,N G/N,M/N We collect two results that will be needed later. When p is a prime number, recall that a finite group G is called cyclic modulo p (or p-hypo-elementary) if G/O (G) is cyclic. And M. Baumann has proven the p Conjecture under the additional assumption that finite group G is cyclic modulo p in [3]. Theorem 2.11. [3, Theorem 3] Let p be a prime number and G be a finite group. Then β(G) is cyclic modulo p if and only if G is cyclic modulo p. In [4], S. Bouc has proven the Conjecture under the additional assumption that finite group G is solvable. Theorem 2.12. [4, Theorem 3.1] Let G be a solvable finite group. Then β(G) is nilpotent if and only if G is nilpotent. 3. Some lemmas about the M¨obius function In this section, we prove some lemmas about the Mo¨bius function. The main lemma is the Lemma 3.4, and the Lemma 3.2-3 are prepared for it. In fact, the Lemma 3.4 will be used in computing m where G is a finite group and N ✂G. G,N LetGbe afinitegroupandletµdenotethe Mo¨biusfunctionofsubgrouplattice of G. We refer to [1],[17, p.94]: Let K,D ≤G, recall the Zeta function of G as following: 1, if K ≤D;  ζ(K,D)= 0, if K (cid:2)D.  Set n:=|{K|K ≤G}|, we have a n×n matrix A as following: A:=(ζ(K,D)) . K,D≤G It is easy to find that A is an invertible matrix, so there exists A−1 such that AA−1 =E, Here, E is an identity element. Recall the Mo¨bius function as following: (µ(K,D)) =A−1. K,D≤G Now, we set the subgroup lattice of G as following: {K|K ≤G}:={1=K ,K ,...,K =G} 1 2 n where n=|{K|K ≤G}|. 5 Definition3.1. LetK ≤G,thereexistsapropersubgroupseriesofK asfollowing: σ :1=K (cid:12)K (cid:12)K (cid:12)···(cid:12)K =K 1 2 3 t where K are subgroups of G and K is a proper subgroup of K for all i. Set X i i i+1 K is the set of elements like above σ. And we call t is the length of σ and set t:=l(σ) We define the height of K as following: ht(K)=max{l(σ)|σ ∈X }. K It is easy to see that ht(1)=1. Lemma 3.2. Let G bea finitegroup and K,Lbe subgroups of G. If ht(K)≥ht(L) and K 6=L, then ζ(K,L)=0. Proof. Suppose that ζ(K,L) 6= 0, by the definition of Zeta function, we have that K ≤L. Set ht(K)=t, there exists a proper subgroup series of K as following: 1=K (cid:12)K (cid:12)K (cid:12)···(cid:12)K =K 1 2 3 t Also K ≤L and K 6=L, thus we have the following series of L: 1=K (cid:12)K (cid:12)K (cid:12)···(cid:12)K =K (cid:12)L. 1 2 3 t Hence ht(L)≥t+1(cid:13)t=ht(K). That is a contradiction to ht(K)≥ht(L). (cid:3) Lemma 3.3. Let G be a finite group. Let {K |i = 1,2,...,n} be the set of i all subgroups of G. And Set K = 1,K = G. Then we can reorder the se- 1 n quence 1 = K ,K ,...,K = G as the sequence 1 = K ,K ,...,K = 1 2 n l(1) l(2) l(n) G such that (ζ(K ,K )) is an invertible upper triangular matrix. Here, l(j) l(k) n×n {l(1),l(2),...,l(n)}={1,2,...,n}. Proof. As the above definition of height of subgroups, we can set T ={K ≤G|ht(K)=1}={1=K }; 1 1 T ={K ≤G|ht(K)=2}:={K ,K ,...K }; 2 21 22 2t2 ········· T ={K ≤G|ht(K)=m−1}:={K ,K ,...K }; m−1 (m−1)1 (m−1)2 (m−1)tm−1 T ={K ≤G|ht(K)=m}={K =G}; m n T ={K ≤G|ht(K)=m+1}=∅. m+1 Now, we can reorder 1=K ,K ,...,K =G as 1 2 n 1=K ; 1 K ,K ,...K ; 21 22 2t2 ········· K ,K ,...K ; (m−1)1 (m−1)2 (m−1)tm−1 K =G. n We can set l(1) :=1,l(2) :=2 ,...,(m−1) :=l(n−1),l(n) :=n. 1 tm−1 Let s(cid:13)r, we want to prove that ζ(K ,K )=0. l(s) l(r) Here, we can set K = K and K = K for 1 ≤ a ≤ t , 1 ≤ b ≤ t . Since l(s) ka l(r) jb k j s(cid:13)r, thus k ≥j. Then by the Lemma 3.2, we have ζ(K ,K )=0 ka jb 6 for 1≤a≤t , 1≤b≤t . k j So, we reorder 1 = K ,K ,...,K = G as 1 = K ,K ,..., K = G and 1 2 n l(1) l(2) l(n) we have A:=(ζ(K ,K )) l(j) l(k) n×n is an invertible upper triangular matrix. (cid:3) Let us list the main lemma as following, and this lemma is used to prove the Proposition 4.1 in the Section 4. Lemma 3.4. Let G be a finite group. Let {K |i = 1,2,...,n} be the set of all i subgroups of G. And Set K1 = 1,Kn = G. Then we have µ(Ki,Ki′) = 0 if Ki (cid:2)Ki′. Proof. By the proof of the above lemma, we can suppose that {K |i=1,2,...,n} i be the set ofall subgroupsof G andset K =1,K =G suchthat (ζ(K ,K )) 1 n j k n×n is an invertible upper triangular matrix. Here, 1≤j,k ≤n. To prove the lemma, we need to adjust the positions of some K such that the j sequence 1=K ,K ,...,K =Gare reorderedas 1=K ,K ,...,K =G, 1 2 n 1(1) 2(1) n(1) and we can get that (ζ(K ,K ) is an invertible upper triangular matrix. j(1) k(1) n×n Here, j(1),k(1) ∈{1(1),2(1),...,n(1)}={1,2,...,n}. First, we can set ht(Ki) = k and ht(Ki′) = j. Moreover, we set Ki = Kka and Ki′ = Kjb for 1 ≤ a ≤ tk, 1 ≤ b ≤ tj. If k (cid:13) j, it is easy to see that µ(Kka,Kjb) = µ(Ki,Ki′) = 0. Now, we will consider the cases when k = j and k (cid:12)j as following. Case 1. k =j. We will prove this case when k (cid:13)k and k (cid:12)k as following. a b a b Case 1.1. If k (cid:13)k =j , we can set a b b T ={K ≤G|ht(K)=1}={1=K }; 1 1 T ={K ≤G|ht(K)=2}:={K ,K ,...K }; 2 21 22 2t2 ········· T ={K ≤G|ht(K)=k}:={K ,K ,...K }; k k1 k2 ktk ········· T ={K ≤G|ht(K)=m−1}:={K ,K ,...K }; m−1 (m−1)1 (m−1)2 (m−1)tm−1 T ={K ≤G|ht(K)=m}={K =G}; m n T ={K ≤G|ht(K)=m+1}=∅. m+1 It is easy to see that we can reorder of 1=K ,K ,...,K =G as 1 2 n K (=1), 1 K ,K ,...K , 21 22 2t2 ········· K ,K ,...,K ,K ...,K ,K ,...,K , k1 k2 kb kb+1 ka ka+1 ktk ········· K ,K ,...K , (m−1)1 (m−1)2 (m−1)tm−1 K (=G). n And we set this order of subgroup series of G as 1=K ,K ,...,K =G. r(1) r(2) r(n) 7 We have B :=(ζ(K ,K )) r(j) r(k) n×n is an invertible upper triangular matrix by the proof of the Lemma 3.2. Hence B−1 is also invertible upper triangular matrix, thus µ(K ,K ) = 0. That is jb ka µ(Ki,Ki′)=0. Case 1.2. If k (cid:12)k =j , we can set a b b T ={K ≤G|ht(K)=1}={1=K }; 1 1 T ={K ≤G|ht(K)=2}:={K ,K ,...K }; 2 21 22 2t2 ········· T ={K ≤G|ht(K)=k}:={K ,K ,...K }; k k1 k2 ktk ········· T ={K ≤G|ht(K)=m−1}:={K ,K ,...K }; m−1 (m−1)1 (m−1)2 (m−1)tm−1 T ={K ≤G|ht(K)=m}={K =G}; m n T ={K ≤G|ht(K)=m+1}=∅. m+1 It is easy to see that we can reorder 1=K ,K ,...,K =G as 1 2 n K (=1), 1 K ,K ,...K , 21 22 2t2 ········· K ,K ,...,K ,K ...,K ,K ,...,K , k1 k2 kb ka+1 ka kb+1 ktk ········· K ,K ,...K , (m−1)1 (m−1)2 (m−1)tm−1 K (=G). n That is, K and K switch places. And we set this order of subgroups of G as ka kb 1=K ,K ,...,K =G. r(1) r(2) r(n) We have B :=(ζ(K ,K )) r(j) r(k) n×n is an invertible upper triangular matrix by the Lemma 3.2. Hence B−1 is also invertible upper triangular matrix, thus µ(Kka,Kkb)=0. That is µ(Ki,Ki′)=0. Case 2. k (cid:12) j. We will prove this case when j −k = 1 and j −k ≥ 2 as following. Case 2.1. If j−k =1, we can set T ={K ≤G|ht(K)=1}={1=K }; 1 1 T ={K ≤G|ht(K)=2}:={K ,K ,...K }; 2 21 22 2t2 ········· T ={K ≤G|ht(K)=k}:={K ,K ,...K }; k k1 k2 ktk T ={K ≤G|ht(K)=k+1}:={K ,K ,...K }; k+1 (k+1)1 (k+1)2 (k+1)tk+1 ········· T ={K ≤G|ht(K)=m−1}:={K ,K ,...K }; m−1 (m−1)1 (m−1)2 (m−1)tm−1 T ={K ≤G|ht(K)=m}={K =G}; m n T ={K ≤G|ht(K)=m+1}=∅. m+1 Wecanseta=t ,thatisK =K . Andwecansetb=1,thatisK =K . k ka ktk jb (k+1)1 8 Now, we reorder the sequence 1=K ,K ,...,K =G as 1 2 n K (=1), 1 K ,K ,...K , 21 22 2t2 ········· K ,K ,...,K ,K ,K (=K ), k1 k2 ktk−2 ktk−1 (k+1)1 jb K (=K ),K ,K ,...K , ktk ka (k+1)2 (k+1)3 (k+1)tk+1 ········· K ,K ,...K , (m−1)1 (m−1)2 (m−1)tm−1 K (=G). n That is, K and K switch places. And we set the above sequence as ka jb 1=K ,K ,...,K =G. r(1) r(2) r(n) That is r(1) =1,r(2) =2 ,...,r(n−1) =(m−1) ,r(n) =n. 1 tm−1 SinceKka (cid:2)Kjb,wecansetKjb =Kr(l),Kka =Kr(l+1),wehaveB :=(ζ(Kr(j),Kr(j′)))n×n is an invertible upper triangular matrix by the Lemma 3.2. Hence B−1 is also in- vertible upper triangular matrix, thus µ(Kka,Kjb)=0. That is µ(Ki,Ki′)=0. Case 2.2. If j−k ≥2, set c:=j−k and we have j =k+c. Here, we can set T ={K ≤G|ht(K)=1}={1=K }; 1 1 T ={K ≤G|ht(K)=2}:={K ,K ,...K }; 2 21 22 2t2 ········· T ={K ≤G|ht(K)=k}:={K ,K ,...K }; k k1 k2 ktk T ={K ≤G|ht(K)=k+1}:={K ,K ,...K }; k+1 (k+1)1 (k+1)2 (k+1)tk+1 ········· T ={K ≤G|ht(K)=k+c−1}:={K ,K ,...K }; k+c−1 (k+c−1)1 (k+c−1)2 (k+c−1)tk+c−1 T ={K ≤G|ht(K)=k+c}:={K ,K ,...K }; k+c (k+c)1 (k+c)2 (k+c)tk+c ········· T ={K ≤G|ht(K)=m−1}:={K ,K ,...K }; m−1 (m−1)1 (m−1)2 (m−1)tm−1 T ={K ≤G|ht(K)=m}={K =G}; m n T ={K ≤G|ht(K)=m+1}=∅. m+1 Wecanseta=t ,thatisK =K . Andwecansetb=1,thatisK =K . For eachk+1≤kl≤k+c−k1a,we ckotknsiderT andwe cansuppose thajtbthere(ke+xics)t1s l 1≤s ≤t such that l l (cid:12)K , if 1≤d≤s ;  (k+c)1 l Kld (cid:2)K , if s +1≤d≤t .  (k+c)1 l l 9 First, we reorder 1=K ,K ,...,K =G as 1 2 n K (=1), 1 K ,K ,...K , 21 22 2t2 ········· K ,K ,...,K ,K , k1 k2 ktk−2 ktk−1 K ,K ,...,K , (k+1)1 (k+1)2 (k+1)sk+1 ········· K ,K ,...,K , (k+c−1)1 (k+c−1)2 (k+c−1)sk+c−1 K (=K ),K (=K ), (k+c)1 jb ktk ka K ,K ,...,K , (k+1)sk+1+1 (k+1)sk+1+2 (k+1)tk+1 ········· K ,K ,...,K , (k+c−1)sk+c−1+1 (k+1)sk+c−1+2 (k+1)tk+1 K ,K ,...K , (k+c)2 (k+c)3 (k+c)tk+c ········· K ,K ,...K , (m−1)1 (m−1)2 (m−1)tm−1 K (=G). n Now, we set the above sequence as 1=K ,K ,...,K =G. r(1) r(2) r(n) To prove B :=(ζ(Kr(j),Kr(j′)))n×n is an invertible upper triangular matrix, we will prove the following (1)-(3) first: (1) For each K ∈ {K ,K ,...,K , v (k+1)sk+1+1 (k+1)sk+1+2 (k+1)tk+1 ········· K ,K ,...,K } (k+c−1)sk+c−1+1 (k+1)sk+c−1+2 (k+1)tk+1 and K ∈ {K ,K ,...,K , u (k+1)1 (k+1)2 (k+1)sk+1 ········· K ,K ,...,K }, (k+c−1)1 (k+c−1)2 (k+c−1)sk+c−1 we can see that ζ(K ,K )=0. v u Suppose ζ(K ,K )6=0, we have K ≤K . But K (cid:12)K and K (cid:2)K , v u v u u (k+c)1 v (k+c)1 that is a contradiction. So ζ(K ,K )=0. v u (2) For each K ∈ {K ,K ,...,K , v (k+1)sk+1+1 (k+1)sk+1+2 (k+1)tk+1 ········· K ,K ,...,K }, (k+c−1)sk+c−1+1 (k+1)sk+c−1+2 (k+1)tk+1 we have K (cid:2)K , thus v (k+c)1 ζ(K ,K )=0. v (k+c)1 10 (3) For each K ∈ {K ,K ,...,K , u (k+1)1 (k+1)2 (k+1)sk+1 ········· K ,K ,...,K }, (k+c−1)1 (k+c−1)2 (k+c−1)sk+c−1 we can see that ζ(K ,K )=0. ktk u Suppose ζ(K ,K ) 6= 0, that is K ≤ K . But K (cid:12) K , thus K (cid:12) ktk u ktk u u (k+c)1 ktk K . And we know K = K ,K = K and K = K (cid:2) K = K . (k+c)1 (k+c)1 jb ktk ka ka i j jb That is a contradiction. So ζ(K ,K )=0. ktk u Since (1)−(3) hold, we have B :=(ζ(Kr(j),Kr(j′)))n×n is an invertible upper triangular matrix. It implies B−1 is also an invertible upper triangular matrix. We can set K =K ,K =K . jb r(l) ka r(l+1) Thatisµ(K ,K )=0because B−1 is aninvertibleupper triangularmatrix.. r(l+1) r(l) So µ(Kka,Kjb)=0, it implies µ(Ki,Ki′)=0. (cid:3) 4. Computing the m when |G:N|=p for some prime number p G,N Let G be a finite group and N ✂G with |G : N| = p. In this section we will compute m . First, we set G,N 1 1 m′ := |X|µ(X,G)= |X|µ(X,G); G,N |G| X |G| X XN6=G,X≤G X≤N and set M′ := |X|µ(X,G)=|G|m′ . G,N X G,N X≤N We can see that 1 1 m +m′ = |X|µ(X,G)+ |X|µ(X,G) G,N G,N |G| X |G| X XN=G,X≤G XN6=G,X≤G 1 = |X|µ(X,G) |G| X X≤G = m . G,G Sotocompute m ,we onlyneedto computeM′ . Thuswe havethe follow- G,N G,N ing propositions. Proposition 4.1. Let G be a finite group and N ✂G such that |G : N| = p for some prime number p. Then M′ =− |X|µ(X,Y) G,N X X Y(cid:12)GX≤N∩Y