A relation between $m_{G, N}$ and the Euler characteristic of the nerve space of some class poset of $G$

0.2 MB·

by  Heguo Liu

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