191, 371]381 Ž1997.
JA976869
JOURNAL OF ALGEBRA
ARTICLE NO.
A Simplicity Criterion for Finite GroupsU
Huiwen Deng
Department of Computer Science, Southwest-China Normal Uni¨ ersity,
Chongqing 630715, Peoples Republic of China
and
Wujie Shi
Institute of Mathematics, Southwest-China Normal Uni¨ ersity, Chongqing 630715,
Peoples Republic of China
Communicated by Gernot Stroth
Received February 14, 1996
Given a finite group G, let pe Ž G . denote the set of all orders of elements in G.
In this paper, we investigate the relation between the number of primes and that of
composite numbers in pe Ž G . and obtain a criterion for the simplicity of finite
groups. Q 1997 Academic Press
1. INTRODUCTION
Throughout this paper, all groups are finite and all simple groups are
nonabelian.
Given a group G, we denote the set of all prime divisors of order G by
p Ž G . and the set of all element orders of G by pe Ž G .. Clearly, p Ž G . is the
set consisting of primes that lies in pe Ž G .. Usually < X < denotes the number
of elements of a set X. For convenience, we define
c Ž G . s the number of composite numbers in pe Ž G . ,
which in fact is the number of elements of pe Ž G . y p Ž G . y 14 . All
further unexplained notation is standard and can be found, for example, in
w4x.
* Project supported by the National Natural Science Foundation of China.
371
0021-8693r97 $25.00
Copyright Q 1997 by Academic Press
All rights of reproduction in any form reserved.
372
DENG AND SHI
The groups G such that c Ž G . s 0 were classified in w5, 17x. Especially,
such a group G satisfies <p Ž G .< F 3 and if <p Ž G .< s 3, then G is a simple
group. In w18x, it was proved that if c Ž G . s 1, then <p Ž G .< F 4, and
further <p Ž G .< s 4 only if G is simple. w16x proposed the following
Problem. Let G be a finite group. Does there exist a general relation
among <p Ž G .<, c Ž G ., and the simplicity of G?
In this paper, we solve the problem and prove the following.
THEOREM A. Suppose that G is simple. Then we ha¨ e <p Ž G .< F c Ž G . q
3, and further <p Ž G .< s c Ž G . q 3 holds if and only if G is one of the
following simple groups.
A 5 , L2 Ž11., L2 Ž13., L2 Ž2 4 ., L3 Ž4., J1;
II. Sz Ž q ., where q s 2 2 nq1 satisfies that each of q y 1, q y 2 q q 1,
and q q 2 q q 1 is either a prime or a product of two distinct primes;
III. L2 Ž2 n ., where n Ž n G 5. is an odd prime and satisfies both
Ž2 n q 1.r3 is a prime and 2 n y 1 is either a prime or a product of two
distinct primes;
IV. L2 Ž3 n ., where n is an odd prime and satisfies both Ž3 n q 1.r4 is a
prime and Ž3 n y 1.r2 is either a prime or a product of two distinct primes;
V. L2 Ž5 n ., where n is an odd prime satisfying both Ž5 n y 1.r4 and
n
Ž5 q 1.r6 are primes;
VI. L2 Ž p ., where p is a prime greater than 13 and one of the following
holds
I.
'
'
Ž1. Ž p y 1.r4 and Ž p q 1.r6 are primes;
Ž2. Ž p y 1.r6 and Ž p q 1.r4 are primes.
THEOREM B. Suppose that G is a finite group. Then, we ha¨ e an
inequality <p Ž G .< F c Ž G . q 3, and if <p Ž G .< s c Ž G . q 3, then G is a
simple group.
We use the classification of all finite simple groups.
2. SEVERAL LEMMAS
Our arguments depend on the prime graph components of simple
groups Žsee w11, 21x.. The prime graph G Ž G . of a group G is a graph whose
vertex set is the set p Ž G . and two distinct primes p, q are linked by an
edge if and only if G contains an element of order pq. Denote the
connected components of the graph G Ž G . by p i , i s 1, 2, . . . , t Ž G ., where
A SIMPLICITY CRITERION
373
t Ž G . is the number of connected components, and if < G < is even, denote
the component containing 2 by p 1.
In fact, p i w i s 1, 2, . . . , t Ž G .x are the vertex sets of the connected
< <
components of p Ž G .. It is clear that <p Ž G .< s ÝtŽG.
is1 p i . For every i, since
p i is a connected component, p i has a spanning tree Ža connected graph
that contains no cycles is called a tree.. It is well known that any tree with
n vertices has exactly n y 1 edges. Put
c X Ž G. s c Ž G. y
t ŽG .
Ý Ž <p i < y 1 . .
is1
Then c X Ž G . is greater than or equal to the number of such composite
numbers that are in pe Ž G ., but not equal to pq Ž p and q are two distinct
primes.. The following lemma and its corollary are obvious.
LEMMA 1.
For any group G, <p Ž G .< s c Ž G . q t Ž G . y c X Ž G . holds.
COROLLARY 2. Ž1. If t Ž G . F 2, then <p Ž G .< - c Ž G . q 3. Ž2. If t Ž G . s 3,
then <p Ž G .< F c Ž G . q 3. Furthermore the equality holds if and only if
c X Ž G . s 0.
Next we will prove the following.
LEMMA 3. Suppose that G is a connected graph with n ¨ ertices. If we
remo¨ e m Ž m - n. ¨ ertices from the graph G, then we remo¨ e m edges at
least.
Proof. Since any connected graph has a spanning tree, we may assume
that the graph G is tree without loss of generality. The removal of m
Ž m - n. vertices results in a subgraph, which has obviously at most
n y m y 1 edges. Suppose the conclusion of the lemma were not true.
Then the number of edges in G is at most Ž m y 1. q Ž n y m y 1. s n y 2.
This contradicts with the fact that G has exactly n y 1 edges as G is a tree
by our assumption.
In our proof of Theorem B, we will use the following unpublished result
of Gruenberg and Kegel w8x.
LEMMA 4 Žsee w21x.. If G is a group such that t Ž G . G 2, then G has one
of the following structures: Ž1. Frobenius or 2-Frobenius, Ž2. simple, Ž3. an
extension of a p 1-group by a simple group, Ž4. a simple group extended by a
p 1-group, or Ž5. an extension of a p 1-group by a simple group that itself is
extended by a p 1-group.
A group G is called 2-Frobenius if there exists a normal series 1 F H F
K F G of G such that H is the Frobenius kernel of K and KrH is the
Frobenius kernel of GrH.
374
DENG AND SHI
The following two results, which can be found in w12x, will also be used
several times in our proof of Theorem B.
LEMMA 5 ŽThompson.. If G admits a fixed-point-free automorphism of
prime order, then G is nilpotent.
LEMMA 6. Assume that a p-group H acts fixed-point-freely on a nontri¨ ial
p X-group G. Then any Sylow subgroup of H is either cyclic or generalized
quaternion.
About the number t Ž G . of connected components of p Ž G ., we prove
the following result.
LEMMA 7. Let G be a group. Suppose that either G is sol¨ able or the
Sylow 2-subgroups of G are generalized quaternion. Then t Ž G . F 2.
Proof. Ž1. When G is solvable, it is almost clear that t Ž G . F 2 by w8x.
We may prove this as follows. Suppose t Ž G . G 3. We could get a set
consisting of primes pi , which is some prime in p i , i s 1, 2, . . . , t Ž G ., i.e.,
p s pi ¬ 1 F i F t Ž G .4 . We clearly have that <p < s t Ž G . G 3. Since G is
solvable, G has a Hall p-group H. By our choice of p , H is a group in
which every element has prime power order. It is impossible by w9x.
Ž2. Assume that a Sylow 2-subgroup of G is generalized quaternion.
By the Brauer]Suzuki theorem w3x, GrOŽ G . has an element of order 2
that lies in its center. Hence t Ž GrOŽ G .. s 1. Suppose t Ž G . G 3. Since
O Ž G . is solvable, t Ž O Ž G .. F 2. Therefore, t Ž O Ž G .. s 2 and an element of
order 2 acts fixed-point-freely on O Ž G .. By Lemma 5, O Ž G . is nilpotent.
So t Ž O Ž G .. s 1, a contradiction.
3. PROOFS OF THE THEOREMS
Proof of Theorem A. By Corollary 2Ž1., we may assume that t Ž G . G 3.
The prime graph components of simple groups are listed in w11, 21x. It is
clear that t Ž G . F 6 for any group. By our assumption, we will discuss the
four cases t Ž G . s 6, 5, 4, 3 separately.
Ža. t Ž G . s 6. In this case, G is isomorphic to J4 . From w4x, we have
<p Ž J4 .< s 10 and c Ž J4 . G 8. Hence, it is clear that <p Ž J4 .< - c Ž J4 . q 3
holds.
Žb. t Ž G . s 5. From w11, 21x, we have that G is isomorphic to E8 Ž q .
for some q. Suppose that W is the Weyl group of G. Then W has a simple
Ž . Žsee w1x, pp. 228]232.. This gives 4, 9, 12 g pe Ž G . by w4x.
section Oq
8 2
Furthermore c X Ž G . G 3. Hence, we have that <p Ž G .< F c Ž G . q 2 by
Lemma 1. So <p Ž G .< - c Ž G . q 3.
375
A SIMPLICITY CRITERION
Žc. t Ž G . s 4. From w11, 21x, it is easy to see that G is one of the
following simple groups.
Ž1.
Ž2.
Ž3.
E8 Ž q . for some q;
Sz Ž q ., q s 2 2 nq1 ;
2
E6 Ž2., L3 Ž4., M22 , J1 , OX N, Ly, FiX24 , M.
For group 1, we have that <p Ž G .< - c Ž G . q 3 as seen in case b.
Suppose G is SzŽ q .. Since 4 g pe Ž G ., <p Ž G .< F c Ž G . q 3 holds by
Lemma 1. By w4, 11x, < G < s q 2 Ž q y 1.Ž q y 2 q q 1.Ž q q 2 q q 1., p 1 s
24 , p 2 s p Ž q y 1., p 3 s p Ž q y 2 q q 1., and p4 s p Ž q q 2 q q 1..
From the structure of G, we know that G contains cyclic Hall p i-subgroup
for i G 2. The equality <p Ž G .< s c Ž G . q 3 forces c X Ž G . s 1. Therefore,
we have assertion II of Theorem A.
From w4x, we may find some information in Table 1 about the simple
groups listed in group 3. It is evident from the table that <p Ž G .< F c Ž G . q
3, and if the equality holds, then G is either L3 Ž4. or J1 , which are listed in
I.
Žd. t Ž G . s 3. By Corollary 2Ž2., the inequality <p Ž G .< F c Ž G . q 3
always holds, and <p Ž G .< s c Ž G . q 3 if and only if c X Ž G . s 0. In particular, Sylow 2-subgroups of G are elementary abelian. By w6, p. 485x, G is
one of the following:
'
'
'
'
Ž1. J1;
Ž2. a simple group of Ree type;
Ž3. L2 Ž2 n ., n G 2;
Ž4. L2 Ž q ., q ) 3, q ' 3, 5 Žmod 8..
Since t Ž J1 . s 4 and since any simple group of Ree type contains an
element of order 9 w20x, i.e., c X Ž G . G 1, we exclude cases 1 and 2 immediately.
TABLE 1
Group
2
E6 Ž2.
L3 Ž4.
M22
J1
X
ON
Ly
X
Fi 24
M
<p Ž G .<
c Ž G.
8
4
5
6
7
8
9
15
18
1
3
3
10
19
25
57
376
DENG AND SHI
Suppose that G is isomorphic to L2 Ž2 n ., n G 2. From w11x, we have
p 1 s 24 , p 2 s p Ž2 n q 1., and p 3 s p Ž2 n y 1.. Since G contains elements of order 2 n q 1 and order 2 n y 1, and since c X Ž G . s 0, 2 n q 1 and
2 n y 1 are either primes or the product of two distinct primes. First, we
assume that n is a composite number. When n F 6, the group satisfying
c X Ž G . s 0 must be L2 Ž2 4 ., which is listed in I of Theorem A. If n ) 6,
then 2 n q 1 or 2 n y 1 contains at least three prime factors. This does not
occur. Second, we assume that n is a prime. If n s 2, then G is isomorphic
to A 5 , which is in I. If n is an odd prime, then 3 divides 2 n q 1. Hence
Ž2 n q 1.r3 must be a prime. Of course, 2 n y 1 may be either a prime or a
product of two distinct primes. This is III.
Suppose G is isomorphic to L2 Ž q ., q ) 3, and q ' 5 Žmod 8.. Put
q s p n, n G 1. From w21x, we have p 1 s p Ž q y 1., p 2 s q4 , and p 3 s
p ŽŽ q q 1.r2.. That 4 Žbut 8 does not. divides q y 1 implies q s 4 k q 1, k
odd. Since Ž q q 1.r2, Ž q y 1.r2 g pe Ž G . and c X Ž G . s 0, we have that
either k s 1 or k is an odd prime. When k s 1, G is isomorphic to A 5 as
q s 5. When k is an odd prime, consider
4 k s q y 1 s Ž p y 1 . Ž p ny 1 q ??? qp q 1 . .
Ž ).
If n s 1, then q s p is an odd prime of type 4 k q 1, k an odd prime.
Assume that 3 does not divide q q 1. Then 3 divides q y 1. So k s 3, that
is, q s 13, which is in I. Assume that 3 divides q q 1. In this case, we have
that both Ž q q 1.r6 and Ž q y 1.r4 are primes. This is the case VIŽ1.. If
n G 2, from the identity Ž). we have p s 3 or p s 5. When p s 3, n must
be even, and further 8 divides 3 n y 1, a contradiction. Hence we get
p s 5. Since 3 divides 5 n q 1 and c X Ž G . s 0, we have that Ž5 n y 1.r4 and
Ž5 n q 1.r6 are primes. Clearly, n is an odd prime. This is the case V.
Finally, suppose G is isomorphic to L2 Ž q ., q ) 3, and q ' 3 Žmod 8..
From w21x, we arrive at p 1 s p Ž q q 1., p 2 s q4 , and p 3 s p ŽŽ q y 1.r2..
Since Ž q q 1.r2, Ž q y 1.r2 g pe Ž G . and p X Ž G . s 0, we have q s 4 k y 1,
k an odd prime. If k s 3, then G is L2 Ž11., which is listed in I. Consider
the case k ) 3. That n is even implies 2 divides Ž q y 1.r2, which is a
contradiction. When n is odd, we have the identity
4 k s q q 1 s Ž p q 1 . Ž p ny 1 y p ny 2 q ??? yp q 1 . ,
q s n n . Ž )) .
If n s 1, then q s p is a prime of type 4 k y 1, where k is an odd prime.
Clearly, 3 divides p y 1. Hence, Ž p y 1.r6 and Ž p q 1.r4 must be
primes. This is the case VIŽ2.. If n G 2, then p s 3 from the identity Ž))..
Since Ž q q 1.r2, Ž q y 1.r2 g pe Ž G ., we have that both Ž3 n q 1.r4 is a
prime and Ž3 n y 1.r2 is either a prime or a product of two distinct primes.
Furthermore, n is an odd prime. This is the case IV of Theorem A.
A SIMPLICITY CRITERION
377
Remark. In recent years some papers have dealt with the question of
characterizing groups G by the set pe Ž G .. For example, in w2, 14x it was
proved that if pe Ž G . s pe Ž L2 Ž q .., q / 9, then G ( L2 Ž q .; if pe Ž G . s
pe Ž J1 ., then G ( J1. In w15x it was proved that if pe Ž G . s pe ŽSzŽ q .., then
G ( SzŽ q ... An analogous result in the case pe Ž G . s pe Ž L3 Ž4.. was proved
in w13x. Therefore all these simple groups listed in Theorem A can be
characterized only by pe Ž G ..
Next we prove Theorem B.
Proof of Theorem B. Use induction on < G <. By Corollary 2Ž2., we only
need to assume that t Ž G . G 3. According to Lemma 4, we may divide the
proof into five cases.
Case 1. G is either Frobenius or 2-Frobenius.
Suppose G is Frobenius. By Lemmas 5 and 6, we know that either G is
solvable or a Sylow 2-subgroup of G is generalized quaternion. We arrive
at t Ž G . F 2 by Lemma 7. This contradicts our assumption.
The case when G is 2-Frobenius is similar.
Case 2. G is simple.
The assertion follows with Theorem A.
Case 3. G is an extension of a nontrivial p 1-group by a simple group.
In this case, there exists a normal p 1-subgroup N of G such that GrN
is simple. Since t Ž G . G 3, N admits a fixed-point-free automorphism of
prime order. By Lemma 5, N is nilpotent. Note that <p Ž G . y p Ž GrN .< is
the number of vertices that lie in p 1Ž G . but not in p 1Ž GrN .. Since
2 f p Ž G . y p Ž GrN ., by Lemma 3 we have that <p Ž G . y p Ž GrN .< is at
most the number of edges that are in the component p 1Ž G . but not in
p 1Ž GrN .. Evidently the latter is at most the number c Ž G . y c Ž GrN ..
Hence we have an inequality
<p Ž G . y p Ž GrN . < F c Ž G . y c Ž GrN . .
Ž 1.
By Theorem A we have another inequality
<p Ž GrN . < F c Ž GrN . q 3.
So
<p Ž G . < s <p Ž G . y p Ž GrN . < q <p Ž GrN . <
F Ž c Ž G . y c Ž GrN . . q Ž c Ž GrN . q 3 . s c Ž G . q 3.
Ž 2.
378
DENG AND SHI
It is clear that the equality <p Ž G .< s c Ž G . q 3 holds if and only if both
equalities in Ž1. and Ž2. hold.
If p Ž N . : p Ž GrN ., then p Ž G . s p Ž GrN .. Hence c Ž G . s c Ž GrN .,
and further pe Ž G . s pe Ž GrN .. That the inequality in Ž2. holds implies
that GrN is one of the simple groups listed in Theorem A. From w2,
13]15x, we have that G is isomorphic to GrN, i.e., N s 1, a contradiction
as N / 1 by assumption.
If p Ž N . ­ p Ž GrN ., we will get a contradiction too.
Ž1. If r g p Ž G . y p Ž GrN . then 2 r g pe Ž G .. Otherwise a Sylow
2-subgroup of G is cyclic or a generalized quaternion group. By Lemma 7
we now get t Ž G . F 2, a contradiction.
Ž2. p Ž G . y p Ž GrN . s r 4 , r / 2. Let r g p Ž G . y p Ž GrN .. Suppose <p Ž G . y p Ž GrN .< G 2. Put r / s g p Ž G . y p Ž GrN .. Consider
GrS, where S is a Sylow s-subgroup of N Žalso G .. On the one hand,
<p Ž G .< s <p Ž GrS .< q 1. By statement 1, 2 s g pe Ž G .. Since N is nilpotent,
rs g pe Ž G .. Furthermore, we have c Ž G . G c Ž GrS . q 2. On the other
hand, <p Ž GrS .< F c Ž GrS . q 3 holds by induction on GrS. Hence
<p Ž G . < s <p Ž GrS . < q 1 F c Ž GrS . q 4 F c Ž G . q 2,
which contradicts < c Ž G .< s c Ž G . q 3.
Ž3. pe Ž G . y pe Ž GrN . s r, 2 r 4 . From statement 2, we have c Ž G . y
c Ž GrN . s 1. By statement 1, 2 r g pe Ž G .. From statement 2 we have the
conclusion statement 3.
Ž4. CG Ž R . : R, where R is a Sylow r-subgroup of N. Clearly R is
also a Sylow r-subgroup of G. Considering GrR, from statements 2 and 3
we get that <p Ž GrR.< s c Ž GrR. q 3. By induction on GrR, we have that
GrR is simple. Since GrN is simple, we know that N s R. So we have
CG Ž R . : R.
Ž5. We finally get a contradiction. Assume there is some elementary
abelian 2-subgroup H in GrR on which some element x of prime order
acts irreducibly. Then by statement 3, x acts fixed-point-freely on R.
Hence it acts fixed-point-freely on the preimage of H, which is nilpotent
by Lemma 5. However, this contradicts CG Ž R . : R. The groups L2 Ž2 n .
and SzŽ2 n . contain groups of order 2 n Ž2 n y 1. by w10, Chap. 11x. The group
J1 contains a group of order 8 ? 7 by w4x and L3 Ž4. and L2 Ž q . contain A 4 ,
so we are done.
Case 4. G is an extension of a simple group by a nontrivial p 1-group.
In this case, there exists a simple normal subgroup N of G such that
GrN is a p 1-group. Since t Ž G . G 3, we have t Ž N . G 2. Hence CG Ž N . s 1.
A SIMPLICITY CRITERION
379
Therefore, N F G F AutŽ N .. By induction on N, we have an inequality
<p Ž N . < F c Ž N . q 3
Ž 3.
and the equality holds if and only if N is one of the simple groups listed.
According to a reason similar to the first lines of the proof in Case 3, we
have an inequality
<p Ž G . y p Ž N . < F c Ž G . y c Ž N .
Ž 4.
by Lemma 3. So
<p Ž G . < s <p Ž G . y p Ž N . < q <p Ž N . < F c Ž G . q 3,
and <p Ž G .< s c Ž G . q 3 holds if and only if the equalities in Ž3. and Ž4.
hold.
First, assume that p Ž G . s p Ž N .. We have then that <p Ž G .< s <p Ž N .< F
c Ž N . q 3 F c Ž G . q 3. If <p Ž G .< s c Ž G . q 3 holds, then c Ž N . s c Ž G ..
Hence pe Ž N . s pe Ž G .. By w2, 13]15x, we have that G is isomorphic to N, a
contradiction.
Second, assume that p Ž G . / p Ž N .. Note that the simple groups mentioned below are from Theorem A.
Ž1. N is isomorphic to either J1 or L3 Ž4.. From w4x, OutŽ J1 . s 1, we
have a contradiction. If N is isomorphic to L3 Ž4., then OutŽ L3 Ž4.. s Z 2 =
S3 by w4x. Clearly p Ž G . s p Ž N . Žsee w4, p. 23x., a contradiction as p Ž G . /
p Ž N ..
Ž2. N is isomorphic to L2 Ž q ., q s p n. It is well known that
Ž
Aut L2 Ž p n .. s PGL2 Ž p n . : Z n . First, assume p ) 2. When n s 1, AutŽ N .
s PGL2 Ž q .. So p Ž G . s p Ž N .. This contradicts p Ž G . / p Ž N .. When
n G 2, p s 3 or 5 by Theorem A. By w7, Theorem 4.23, p. 304x, the group
of diagonal automorphisms of order 2 is normal in OutŽ N .. Hence, the
diagonal automorphism commutes with the field automorphism. Since n is
odd, OutŽ N . s Z 2 = Z n . As p Ž G . / p Ž N . there is some field automorphism of prime order r in G. However, this automorphism centralizes
some element of order p in G. So r is connected to p in the prime graph,
i.e., p g p 1Ž G .. As p f p 1Ž N . this now implies t Ž G . F 2, a contradiction.
Second, assume p s 2. We have AutŽ N . s L2 Ž2 n . : Z n . Let g s Ž 11 10 . g
L2 Ž2 n .. Clearly the order of g is 3. On the one hand, as p Ž G . / p Ž N .
there is some field automorphism of prime order r in G. On the other
hand, his automorphism centralizes the element g. Hence 3r g pe Ž G .. We
have t Ž G . F 2 as above, a contradiction.
380
DENG AND SHI
Ž3. N is isomorphic to Sz Ž q ., q s 2 2 nq1. It is well known that
AutŽ Sz Ž q .. s Sz Ž q . : Z 2 nq1. Without loss of generaity, we may assume
G s Sz Ž q . : Z 2 nq1. Take
1
1
xs
0
1
0
1
1
1
0
0
1
1
0
0
.
0
1
0
From w6x, x g Sz Ž q .. Clearly < x < s 4. Let t : a ª a 2 be a field automorphism of GF Ž q .. Since x t s x and the order of t is odd, G has an element
of order 4Ž2 n q 1.. G also has an element of order 2Ž2 n q 1.. So c X Ž G . G
2. By Lemma 1, <p Ž G .< - c Ž G . q 3, a contradiction.
Case 5. G is an extension of a nontrivial p 1-group by a simple group by
a nontrivial p 1-group.
In this case, there exists normal subgroups N and G1 of G such that
1 - N - G1 - G, N is a p 1-group, G1rN is simple, and GrG1 is a
p 1-group. By the inductive hypothesis, <p Ž G1 .< F c Ž G1 . q 3, and further
as G1 is not simple we have <p Ž G1 .< - c Ž G1 . q 3. If p Ž G . s p Ž G1 ., then
<p Ž G .< - c Ž G1 . q 3 F c Ž G . q 3 holds. If p Ž G . / p Ž G1 ., then <p Ž G . y
p Ž G1 .< F c Ž G . y c Ž G1 . by Lemma 3. So
<p Ž G . < s <p Ž G . y p Ž G1 . < q <p Ž G1 . <
- Ž c Ž G . y c Ž G1 . . q Ž c Ž G1 . q 3 .
s c Ž G . q 3,
i.e., in Case 5 we always have inequality <p Ž G .< - c Ž G . q 3.
This completes the proof of Theorem B.
ACKNOWLEDGMENT
Both authors are grateful to the referee for kind help in modifying the original version of
the paper.
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