THE HYPERQUASECENTER OF
A FINITE GROUP

Thesis for the Degree of Ph. D.

MICHIGAN STATE UNIVERSITY
NARAYAN P. MUKHERJEE
1968

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ABSTRACT

THE HYPERQUASICENTER OF A FINITE GROUP

by Narayan P. Mukherjee

The concept of a quasicentral element of a group G is
a generalization of the idea of an element in the center.
An element 9 belonging to a group G is called a quasi—
central element (QC-element) of G if <g><x> = <x><g>
\fx 6 G. This leads to the definition of the quasicenter
Q(G) which is defined to be the subgroup generated by all
quasicentral elements. It can be proved very easily that
the quasicenter is a characteristic subgroup. Some ques-
tions naturally arise in this connection: (1) What kind
of structure has Q(G)? (2) Does the knowledge about the
quasicenter of a subgroup of G help to say anything about
Q(G)? (3) Are the well known subgroups like the Kern
group or the Hypercenter related to Q(G)?

In Chapter I, we try to answer these questions and
the principal results there-in may be summarized as follows:
(1) If x e G is a QC—element then Xr is a QC-element
for all integers r. (2) Q(G) is nilpotent. (3) Suppose
G = H x K, where H and K have relatively prime orders.
Then Q(G) = Q(H) x Q(K), where Q(G), Q(H), Q(K) are the
quasicenters of G, H and K respectively. (4) Q(G)
neither always contains nor is always included in the Hyper—
center H(G) though of course the Kern group is always

contained in Q(G).

Narayan P. Mukherjee
The idea of the quasicenter leads to the definition of

the Hyperquasicenter and it is defined like the Hypercenter.

 

0' 1
1+ be the quasicenter of g—u Then
Qi Qi

we have an ascending chain of characteristic normal sub-
groups as follows: 1 = 00 Cin C 02 C --- CiQn = Q*. The
terminal member Q* of this series (termed upperquasicen-
tral series) is called the Hyperquasicenter of G.

The Hypercenter H(G) and the hypercentral subgroups
of G have been studied in detail by R. Baer in a series
of papers in which he gave seven principal characterizations
of the Hypercenter. A question naturally arises in this
connection: Can we generalize these results for the Hyper-
quasicenter?

In Chapter II we show that three of these results of
Baer can be completely generalized, and we obtain partial
generalizations of two others. No equivalent could be ob-
tained for one of the remaining two results while in the
case of the other one, an obvious natural generalization is
not possible and we provide an example to confirm this.

Besides these, we also investigate the structure of Q*
and this investigation helps us to find some properties of
the Hyperquasicenter and to characterize supersolvable
groups in certain new ways. Some of the results obtained
in this connection are as follows:

(1) The Hyperquasicenter Q* of a group G is super-

solvable. (2) A group G is supersolvable if and only if
2

Narayan P. Mukherjee
G = Q*. (3) If N is a normal subgroup of G contained
in Q* then G is supersolvable if and only if g- is
supersolvable. (4) For any supersolvable subgroup S of
G, Q*S is supersolvable. (5) Every supersolvably im-
mersed subgroup of G is contained in Q*.

In Chapter III, some additional properties of the Hyper-
quasicenter are obtained and we also observe, that if we
restrict our attention to groups of a certain nature then
the partial generalizations of Baer's results can be com—
pleted and it is also possible to characterize the Hyper—
quasicenter in an entirely different manner. It is shown,
that this incidentally leads in a natural fashion to a new
characterization of the Hypercenter. Some of the additional

properties of the Hyperquasicenter included in Chapter III

may be stated as follows: (1) If the smallest prime di—

visor of [GI divides |Q(G)|, then G has a nontrivial cen-
ter. (2) If G has a normal Sylow p—subgroup correspond-
ing to a prime divisor p of |Q*| then Q* has a cyclic

normal subgroup of G.

For groups of special nature, some of the results ob-
tained are as follows: (1) Q* is the intersection of all
maximal supersolvable subgroups of G. (2) Q* is the

largest supersolvably immersed subgroup.

THE HYPERQUASICENTER OF A FINITE GROUP

BY

>

, 50'
Narayan Pinukherjee

A THESIS

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
Department of Mathematics

1968

PREFACE

I am deeply indebted to Professor W. E. Deskins for
suggesting this investigation and for his helpful guidance
during the preparation of the thesis. To say that, I ap-
preciate all the time he has taken for me, all the things
he has taught me and all the kindly advice he has given me
is certainly an understatement and I can only express my
deep gratitude for his extreme patience during the prepara-
tion of the final draft. Words can hardly describe, how
grateful I am for everything he has done for me.

I wish to express my gratitude also to Dr. J. Adney
for the interest he has displayed, for the discussions I
have had with him and for all the advice he has given me.

My thanks are also due to Professor L. M. Sonneborne,

who has helped me a lot by giving many useful suggestions.

ii

TABLE OF CONTENTS

Page
INTRODUCTION . . . . . . . . . . . . . . . . . . . . 1
CHAPTER

I. THE QUASICENTER OF A GROUP . . . . . . . . . 3

II. THE HYPERQUASICENTER OF A GROUP . . . . . . 19
III. SOME ADDITIONAL PROPERTIES OF THE HYPER-

QUASICENTER . . . . . . . . . . . . . . . 61

INDEX OF NOTATION . . . . . . . . . . . . . . . . . 83

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . 85

iii

INTRODUCTION

The concept of a quasicentral element of a group G
is a generalization of the idea of an element in the center.
This leads to the definition of the quasicenter which is
defined to be the subgroup generated by all quasicentral
elements. It can be proved that the quasicenter is char-
acteristic and nilpotent.

The idea of the quasicenter leads to the definition
of Hyperquasicenter and it is defined like the Hypercenter.

Qi+1

Let Q0 = 1 and -af—- be the quasicenter of ET” Then

1 i
we have an ascending chain of characteristic normal sub—
groups as follows: 1 = Q0 C Ql C Q2 C °-- C Qn = Q*. The
terminal member Q* of this series @ermed upperquasicentral
series) is called the Hyperquasicenter of G.

The Hypercenter H(G) and the hypercentral subgroups
of G have been studied in detail by R. Baer in a series
of papers in which he gave seven principal characterizations
of the Hypercenter. A question naturally arises in this
connection: Can we generalize these results for the Hyper—
quasicenter?

We show that three of these results of Baer can be
completely generalized and obtain partial generalizations
of two others. No equivalent could be obtained for one of
the remaining two results while in the case of the other

one, an obvious natural generalization is not possible and

we provide an example to confirm this.

1

2

Besides these, we also investigate the structure of
the Hyperquasicenter and prove it to be supersolvable. This
enables us to find some properties of the Hyperquasicenter
and to characterize supersolvable groups in certain new ways.

We observe finally, that if we restrict our attention
to groups of certain nature, then the partial generalizations
of Baer's results can be completed and it is also possible
to characterize the Hyperquasicenter in an entirely different
manner. It is shown, that this incidentally leads in a
natural fashion to a new characterization of the Hypercenter.

Chapter I contains some basic results on the quasi—
center while in Chapter II we investigate the structure of
the Hyperquasicenter and derive some of its properties to-
gether with the equivalents and partial generalizations
mentioned earlier. In Chapter III, some additional prop-
erties of the Hyperquasicenter are obtained and the partial
generalizations of Baer's results are completed for groups
of certain nature.

An index of notations is given in the Appendix.

CHAPTER I
THE QUASICENTER OF A GROUP

The concept of a quasicentral element of a group G
is a generalization of the idea of an element in the center
of G. This leads to the definition of the quasicenter
Q(G) of G which is defined to be the subgroup generated
by all quasicentral elements. Some questions naturally
arise in this connection: (1) What kind of structure has
Q(G)? (2) Does the knowledge of the quasicenter of a
subgroup of G help to say anything about Q(G)? (3) Are
the well known subgroups of G like the Kern group or the
Hypercenter related to Q(G)?

The aim here is to answer these questions and derive
some basic properties of Q(G). For the sake of complete—
ness and easy reference this chapter contains some basic
definitions and properties of quasinormal subgroups. Some
proofs are included for which no immediate references are

available. Throughout, the groups are finite.

Definition 1.1

Subgroups H and K of the group G permute iff

<H,K> = HK = KH.

Definition 1.2

The subgroup H of G is quasinormal iff H permutes

with every subgroup of G.

Definition 1.3

 

An element x e G is a quasicentral element (QC-ele-
ment) iff (x) is quasinormal in G.

Since a quasinormal subgroup H of G permutes with
every subgroup of G, it certainly permutes with the cyclic
subgroups of G and it is easily seen that this property

characterizes quasinormality.

Lemma 1.4

 

H is quasinormal in G iff H permutes with every
cyclic subgroup of G.

Proof:-

From the definition of quasinormality, H permutes
with every cyclic subgroup of G. Again, let K be any
subgroup of G. Consider HK and let hk e HK where
h e H and k e K. If H permutes with every cyclic sub-
group of G, then H<k> = <k>H and so hk = knh' where
h, h3 are in H and n is some integer. Hence HK.E
KH. By similar arguments KH E.HK and so HK = KH. There-

fore, H is quasinormal in G.

Lemma 1.5

 

x e G is a QC—element iff <x><y> = <y><x> for
all y in G. V
By definition, x is a QC—element iff (x) is quasi-

normal in G. And from Lemma 1.4, (x) is quasinormal iff

5

(x) permutes with every cyclic subgroup of G. Hence the

assertion in the lemma follows.

Lemma 1.6

 

If H is quasinormal in G and e is a homomorphism

of G then He is quasinormal in Ge.

Proof:-

From Lemma 1.4, H is quasinormal iff for each g in
G and h in H there is an h' in H and an integer n

such that gh = h'gn. This equation under 6 has the

n
image gehe = h'e(ge) , Thus He permutes with every cyclic

subgroup of G9 and so is quasinormal in Ge

Theorem 1.7

 

If x e G is a QC-element, then xr is a QC-element
for all integers r.

Let y be an arbitrary but fixed element of G. Then
we must prove (xr><y> = <y><xr> for all r. The proof
is by induction on [G] and so we need only consider the

a1 a2 an

case where G = <x)<y> = <y><x>. If in a p1 .p2 ... pn

where pi for i = 1,2,...n is a prime then <y> = <y1> X
<y2> x --- x (yn> where yi is a power of y and lyil
a .
= p.1- Then if <xr><yi> = <yi><xr> \f i: we have
1

<xr><y>

y is a p-element and G = <x><y> = <y><x>. Let ly] = p

<y><xr>. So we need only consider the case where
b

If Ix] = paq, (p,q) = 1 then (x) = (x1) x <x2> where

6

a

x. for i = 1,2 is a power of x and lxll = p and

i
|x2| = q.

Since a quasinormal subgroup of a group is subnormal
by a theorem of Ore [Proposition 1.3, [13]] it follows
that <x> is subnormal in G.

If (x) is normal in G then it is trivially sub—
normal in G and since every subgroup of a cyclic normal
subgroup is also normal it follows that (xr>‘fl G and
so (Xr><y> = <y><xr>.

If (x) is not normal in G, we have a subnormal
chain as follows:

<x> <1 6 ----- 4 HQG, where H is a subgroup of
G. Since x e H and any element in H can be written
as xrys where r, s are integers it follows that
(yslxrys e H) is a subgroup (yd> of (y). Hence Hf;
<x><yd> and so H = <x><yd> = (yd><x>, since H is a group.

Since H SLG, by induction it follows that xr is
a QC-element of H for every integer r. This implies
x2 = xpa is a QC—element of H. We claim that (x2>4 H.
Otherwise T = (x2>yd # <x2> is a cyclic subgroup of H

of order q. But x2 is a QC-element of H and so it

follows that (x2>T = T<x2> = R, R is a subgroup of H,

IT I |<X2> l

and IR] = TESfl <x;§l = q-£ where g is a diVisor of q

 

different from 1. Since (p,q) = 1, it follows that (p,£)
= 1. Now R being a subgroup of H, )R] must divide 1H].
This is impossible since [H] = pcq, where c is some in-

teger and 2 does not divide p. Hence <x2> <3 H.

7
Since 1x2] is prime to p, every Sylow subgroup of
(x2> is also a Sylow subgroup of H and is therefore nor-
mal in H. This implies every Sylow subgroup of (X2) is
characteristic in H and so (x2) is characteristic in
H. Therefore <x2>AG as H 4G.

Since <X> = <X1> X <x2>, it follows that <xr> =

t1 t2 ' . ' t1
(x1 > x <x2 > where t1, t2 are some integers and x1
t2 _.
and x2 are powers of x. Applying induction to G =

Z_%;;. we have, <X11><§> = <§><X1 > where X11 = <X2 >X11
x2 .

_ t2 . ' . t2 t1

and y = (x2 >y. This equation yields (x2 >°<x1 ><y> =
t2 t1 . t2 t1 ' t2 A t1. .

<X2 ><Y><X1 >. l-e- <X2 >°<X1 ><Y> = <Y><X2 >'<X1 . l-e-
<Xr><y> = <y><xr>-

Thus we need only consider the case when <x2> = 1,
i.e. when, G = <x><y> = <y><x> is a p-group.

Since G is a p-group it has a nontrivial center.
Z(G) contains an element z, [z] = p, with 2 being either
a power of x or of y or possibly both. Let z = xmyn

where either m or n may be zero. We wish to show both

xm and yn are in Z(G). Now x.mxmynxm = men and hence
m n n m . —m+1 m n m-1 m n n m
x y = y x . Again, x x y x = x y = y x . There-
n n n . .
fore xy = y x. Thus y is in the center of G and
so xm is also in the center of G. If m # 0 then
lxmkl = p for some integer k and xmk is in the center
of G. Since in a cyclic group there is exactly one sub-

group of each order, <ka> Ei<xr>.

8

 

Applying induction to El: §k> we have <§T><§> =
(x

<§><§x> where Er = <ka>xr. This equation yields
<xmk><xr><y> = (ka><y><xr>. Since <ka> <3 G and
<ka> E <xr> it follows that (xr><y> = <y><xr>.

If m = 0 then applying induction to 5': g>
(y

we have again as before <21><y> = <y><xr> where' XI

 

: <yn>xr and §'= <yn>y. This equation yields
<Yn><xr><y> = <yn><y><xr>, i-e- <xr><y> = <y><xr> since
<Yn> <3 G and <yn> C (y). Therefore the theorem is proved
completely.

The definition of a quasicentral element leads natur-
ally to the idea of the quasicenter of a group and we

have the following definition.

Definition 1.8

 

The quasicenter Q(G) of a group G is the subgroup
generated by all QC—elements of G.

The following example shows that every element of
Q(G) need not be a QC-element and also that the product

of any two QC-elements is not necessarily a QC-element.

 

Example 1.
Let G = <a,b><x> - <x><a,b> where a3 = 1, b3 = 1,
ab = ba, aX = a2, bX = b and x2 = 1. Q(G) = <a,b> and

a, b are QC-elements of G. But ab 6 Q(G) is not a
QC—element of G. For, otherwise we must have <ab><x> =

<x><ab>, which implies (ab) must be normalized by x.

9

But, x abx = x-lax-x-lbx = azb and azb é (ab). Hence,

ab is not a QC-element of G and our assertion follows.

Remark

Let G be a group and Q(G) be its quasicenter. Then
the quasicenter of Q(G) is Q(G) itself. For, the QC-
elements of G being in Q(G) are also QC-elements of
Q(G). Therefore the quasicenter of Q(G) is Q(G) itself.

In this connection, a question now naturally arises:
What sort of structure has Q(G)? This we investigate next
and prove that the quasicenter is a nilpotent subgroup.

We begin with the following lemma.

Lemma 1.9

 

Let G be a group and Q(G) be its quasicenter.
Then each Sylow subgroup of Q(G) is generated by QC-ele-
ments of G.

m:—

Let X = (X1. x2, ’°', xt) be the set of all QC—

elements of G and note that (X) = Q(G). Each xi can

be written as xi = yizi = ziyi, where yi is a p—element,
zi is a p'—element, i = 1,2, --- t and yi, zi are QC-
elements of G, being powers of Xi'

Define Xp = [y1. y2, ~-- , yt) and let 9 be any
automorphism of G. The image of the QC-element xi under

6
6 is Yizi and by Lemma 1.6 it is a QC-element of G.

e e 9 . .
Therefore yizi is in X and yi is in X . Consequently,
Xp is a normal set. Now; yi V i being a QC—element of

10
G, it follows that <Xp> is a p-group and therefore if P
is a Sylow p-subgroup of Q(G), then <Xp> C P.
But Q(G) = (x1><x2> ---- <Xt> = <y1>-<y2> ....
<yt>-<z.>-<z.> ---- <zt>- Hence, IF] = |<91><y2> --- <yt>|
and so P e <Xp>' The assertion in the lemma therefore

follows.

Theorem 1.10

 

The Quasicenter of a group G is nilpotent.

Let P1 and P2 be two Sylow p—subgroups of Q(G)
corresponding to the prime divisor p of [Q(G)]. Let
P1 = (g1, g2, ---, gn>, where gi is a QC-element of G
for i = 1,2, °°° n (by Lemma 1.9). If gi é P2 for some
i, then consider P2<gi> = (gi>P2. It is a p-group whose
order is larger than 1P2), a contradication. Hence
P1 gng and so P1 = P2, since [P1] = IP21. Thus there
is a unique Sylow subgroup of Q(G) corresponding to each

prime divisor of |Q(G)|. Hence Q(G) is nilpotent.

Remark

By the definition of the Fitting subgroup F of a
group G, it follows that Q(G) Q;F. Also, every element
of the Kern group K of G being a QC-element, K is
contained in Q(G). Hence K E Q(G) E F. However, in gen-
eral the quasicenter Q(G) neither contains nor is included
in the Hypercenter H(G). The following example confirms

this assertion.

11

Example 2

 

Let G = $3 = <a><b> = <b><a> where a3 3 1, b2 = 1

and ba = a2b. Q(G) Q(S3) = <a>. But Z(G) = Z(Sa) = 1

and so H(G) = H(Sa) 1. Thus in this case H(G) ${Q(G).
On the other hand, consider the following group G
of order 16:
G = <a,b,c><x> = <x><a,b,c>, where a2 = b2 = c2 = 1,
ab = ba, bc = cb, ac = ca, a = ab, bX = b, c = c, x2 = 1.
It may be verified that QC-elements of G are b, c,
ax, abx, acx, abcx. Therefore Q(G) may be written as
<ax,b,c>. Since <b,c> <3 G, <ax,b,c> = <ax><b,c>. But
)(ax>| = 4 and (b) C (ax). Hence, |Q(G)l =I8. However,

G being a p-group, H(G) = G. Therefore, in this case

Q(G) S H(G).

Corollary

 

Let Q(G) be the quasicenter of a group G. If p1
is the largest prime divisor of [Q(G)] then every pi-ele-
ment in Q(G) centralizes every pl-element of G. In
general, if pi is any prime divisor of |Q(G)| then
every pi-element in Q(G) involving primes less than pi
centralizes every pi-element of G.

E£22£="

Let x e Q(G) be a q-element, where q is a prime

and q < p1. Since every Sylow subgroup of Q(G) is gen-

a a
erated by QC-elements of G, it follows that x = 911922...
am bi
gm where gi is a QC-element of G and lgil = q ,

where bi is some integer and i = 1,2, ---, n.

12

If y e G is a pl-element, then by Theorem 1.7,

<g:i><y> = <y><g:i> = H. Now, if the product of two finite
cyclic groups is again a group then it is supersolvable.
[Theorem 13.3.1, Scott [12]]. H is therefore supersolvable
and so it follows that (y) 4 H. However (g:i>4 H by
Theorem 1.10. Therefore y is centralized by x and hence
the assertion in the corollary follows.

By definition 1.8 and Lemma 1.6 it follows that Q(G)
is a characteristic subgroup of G. Each Sylow subgroup
of Q(G) is therefore characteristic in G.

In the following we shall use Q(G) and Q1 inter—
changeably for the quasicenter of a group G. If we use
any other symbol S for a group then Q(S) will stand
for the quasicenter of S.

We observe that if H is a normal subgroup of G
then its quasicenter Q(H), in general, neither contains
nor is included in Q(G).

The quasicenter of G = A4 is 1 but the Sylow sub-
group H of order 4 is normal and it coincides with its
quasicenter. In this case therefore Q(G) $§Q(H).

On the other hand, the quasicenter of the Quaternion
group G is G itself while the quasicenter of a normal
subgroup H of order 4 is H itself. In this case there—
fore, Q(H) ; Q(G).

This leads us to investigate the relationship of the

quasicenter of a group G and the quasicenters of subgroups

13
of G with additional restrictions besides normality im-
posed on the subgroups. In the following we investigate

the quasicenter of direct products.

Theorem 1.11

 

Suppose a group G : H X K. Then Q(G) S Q(H) x Q(K).
If Q(H) and Q(K) are both cyclic or if each QC—element
of H and of K generates a normal subgroup of H and of K
respectively then Q(G) 3 Q(H) x Q(K).

23.92::—

Let aobo e Q(G) be a QC-element of G where a0 6 H
and b0 e K. By the definition of a QC-element we have,
(aob0><x> = <x><a0bo> x e H. This implies that for arbi-

trary integers r, s 3 integers u, v such that agngs =

xuaXbX. It follows therefore, that agxS = xuaX and
b3 = bx. From agxS = xuag, we conclude, that any element

in (ac/(x) is also an element in <x><ao>. By similar
arguments, for arbitrary integers r1, s1,3 integers ul,
v1 such that xglail = aglxvl, so that any element in
<x><ao> is also an element of (ao)<x>. Thus (ao><x> =
<x><ao) ‘d x e H. Therefore a0 is a QC-element of H.

Again, since aobo is a QC—element of G for an
arbitrary element y in K, (aobo><y> = <y><a0bo>. By
an argument identical with the above, it follows that ho
is a QC-element of K. Hence Q(G) E Q(H) >< Q(K).

If Q(H) = <c> and Q(K) = <d> then Q(H) ‘3 H im-

plies <c> <3 G and similarly (d) 43 G. Thus both c

14
and d are QC-elements of G and so Q(H) and Q(K) are
both contained in Q(G). Therefore Q(G) = Q(H) x Q(K).
If e is a QC-element of H and (e) <d H then <e>
<3 G and similarly if f is a QC-element of K and (f)
.4 K then (f) ‘3 G. Hence as before it follows that

Q(G) = Q(H) x Q(K)

Example 3

 

Let G = H1 x H2 where Hi for i = 1,2, is a copy
of 83. The quasicenters of both H1 and H2 consist of
the cyclic subgroups of order 3 and Q(G) = Q(H) x Q(K).

Also the QC—elements of H1 and H2 generate normal
subgroups of H1 and H2 (cyclic subgroups of order 3 are
normal) and again Q(G) = Q(Hl) x Q(Hz).

If G = H x K, it is not in general true that a QC-
element of H or K will be a QC-element of G. That is
why the equality of Q(G) and Q(H) x Q(K) seems improb-

able. The following example confirms the above assertion.

Example 4

 

Let G - <a><b> = <b><a>, where [a] = 3, [bl = 9
and a—lba = b4. The elements of order 3 in G are
a, a2, b3, b5, ab3, ab6, a2b3, azb6 and they all permute
with each other. G is of order 27 and so every sub—
group of order 9 is normal. Hence every element of order

3 is a QC-element of G. In particular, 'a' is a QC-ele-

ment of G.

15

Let S = (a1>-<b1> x (a2>-<b2> = G1 X G2, where G1
and G2 are copies of G, a1 and a2 correspond to a
and b1, b2 correspond to b.

If a1 were a QC-element of S then we must have
<a1><b1b2> = (b1b2><a1>. This implies blbzal = angbZ
where u and v are non-negative integers, u < 3, v < 9.
From this equality we have, blal = agb: and b2 = bZ.
Now blal = albi. This implies v = 4 and so b: = 1, a
contradiction since b2 is of order 9.

With some restriction on the orders of H and K how-

ever, the equality of Q(G) and Q(H) x Q(K) follows.

Theorem 1.12

 

Suppose a group G = H x K. If H and K have rela-
tively prime orders then Q(G) = Q(H) x Q(K).

emf:-

By Theorem 1.11, Q(G) EEQ(H) xQ(K). Let a0 6 H be
a QC-element of H. We would like to show that a0 is a
QC-element of G also. Let xy be an element of G
where x e H and y e K. From the definition of a QC-
element we have (a0><x> = <x><ao>. Therefore, it follows
that <ao><X><Y> = <X><ao><Y> = <X><Y><ao>l since aoY =
yao. As xy = yx, <x><y> = <y><x> is a group of order
mn where [X] = m and |y| = n. But <x><y> contains
the element xy and its order is also mn, since (m,n) = 1.
This implies <x><y> = <y><x> = <xy>. Hence <ao><xy> =

<xy><ao> and any QC—element of H is therefore a QC-element

16
of G too. Similarly, any QC—element of K is a QC-ele-
ment of G. Therefore both Q(H) and Q(K) are contained

in Q(G). So Q(G) = Q(H) x Q(K).

Remarks

1. If G = H x K and Q(G) = Q(H) x Q(K), it is not
true in general that H and K are of relatively prime
orders. This is evident from Example 3. The condition on
the orders of H and K as laid down in Theorem 1.12 is
therefore a condition of sufficiency for the equality of
Q(G) and Q(H) x Q(K) and is not a necessary one.

2. If G = Q(G) x H and ([Q(G)[, [H[) = 1 then H
is centerless, nonabelian and has no normal cyclic subgroup.

By Theorem 1.12, Q(G) = Q(Q(G)) x Q(H) = Q(G) X Q(H).
Hence Q(H) = 1 and so the conclusions for H as stated
above are true. In particular G is not supersolvable as
otherwise H will be supersolvable and so shall contain a
cyclic normal subgroup.

3. Turning our attention to semidirect products we

AB, where A 4G, AflB = 1 then the

observe that if G
statement Q(G) = Q(A) ° Q(B) is false. Even when
(lAl, [B]) = 1, Q(G) is not equal to Q(A) ' Q(B) in gen-
eral, as is evident from S3.

S3 = <a>-<b> = <b><a>, a3 = 1, b2 = 1, ba = a2b.
Q(S3) = (a). Also Q(<a>) = (a) and Q(<b>) = (b). Thus
Q(S3) # Q(<a>) ° Q(<b>). However we observe that in 33

Core (a) = (a) and Core (b) = 1 and Q(Sa) = (a) x Core <b>.

17

Generalizing this observation we have the following theorem.

Theorem 1.13

 

Suppose a group G = A-B, where A and B are cyclic
subgroups of G of orders pa and qb respectively, p
and q are different primes and p > q. Then Q(G) =
A x Core B.

By Theorem 13.3.1 of Scott [12], G is supersolvable.
Since in a supersolvable group the Sylow subgroup correspond-
ing to the largest prime divisor of the order of the group
is normal, it follows that, A QG. Since A is cyclic,
A is in Q(G). Now, Q(G) being nilpotent, is the di-
rect product of Sylow subgroups and so Q(G) = A x C, where
C is a subgroup of G of order qt, t is some integer.

Since G = A'B and A C Q(G), it follows that C is
a subgroup of B. Now Core B 22C and Core B being a cyclic
normal subgroup of G, will be contained in Q(G). Hence
Q(G) = A x Core B.

We conclude this chapter with the generalization of a
theorem of Baer which states that the socle of a group G
is part of the centralizer of the Hypercenter. However,
the Hypercenter of a group may be the identity alone with—
out the quasicenter being so. This is evident from S3.

The method of proof is very much the same as in Baer's re-

sult.

18

Theorem 1.14

The socle of a group G is part of the centralizer
of the quasicenter of G.

Proof:-

Let M be a minimal normal subgroup of G. If
Q(G) n M # 1, then M E Q(G) by the minimality of M.
Hence M must be contained in the center of some Sylow
subgroup of the quasicenter. This is so because Q(G) is
the direct product of Sylow subgroups each of which is nor—
mal in G and hence M must be contained in some Sylow
subgroup. But the center of this Sylow subgroup is nor-
mal in G and as every normal subgroup in a p-group inter-
sects the center nontrivially it follows that M must be
contained in the center of the Sylow subgroup that contains
it. Thus M is contained in the centralizer of the quasi-
center. If M H Q(G) = 1 then every element of M cen—
tralizes every element of Q(G). Thus the assertion in

the theorem is proved completely.

CHAPTER II
THE HYPERQUASICENTER OF A GROUP

The Hypercenter H(G) and the hypercentral subgroups

of a finite group G have been studied in detail by R. Baer
in a series of papers. Some of the principal results about
the Hypercenter are as follows: (1) H(G) is the terminal

member of the uppercentral series of G. (2) H(G) is the
intersection of all normal subgroups N such that Z(fi) - I-
where 'I is the identity of Sn (3) H(G) is the product
of all normal hypercentral subgroups of G. (4) If x e H(G)

and y e G and if the orders of x and y are relatively
prime then xy = yx. (5) If N is a normal subgroup of G
such that N-U is nilpotent for any nilpotent subgroup U
of G then N g H(G). (6) H(G) is the intersection of all
maximal nilpotent subgroups of G. (7) H(G) is the inter-
section of the normalizers of all Sylow subgroups.

In this chapter, we generalize the idea of the Hyper-
center of a group G to that of the Hypwerquasicenter Q*
of G and try to obtain corresponding generalizations of

the above results of Baer. We succeed in generalizing (1),

(2), (3) completely and obtain partial generalizations of
(5) and (6). No equivalent of (7) could be obtained and in
the case of (4), an obvious natural generalization is not
possible. We provide an example to confirm this.

Besides these, in this chapter, we also investigate
the structure of Q* and prove that the Hyperquasicenter

19

20
is supersolvable. This enables us, as will be seen, to
characterize supersolvable groups in certain new ways. We
conclude the chapter with a discussion of supersolvably im-
mersed subgroups and show how they are related to the Hyper-

quasicenter.

Qi+1

Q.

l

 

Let G be a group and suppose QO 2 1 and is

the quasicenter of §—-. Since Qi is characteristic in

Q 01
i+1 U I I n G u
is characteristic in -—— , it follows that

Q- Q.

l l

G and

 

Qi+1 is characteristic in G. We thus get an ascending
chain of characteristic normal subgroups as follows:

1 = Q0 C Q1 C 02 C ------ C Qn = Q*. This series and its
terminal member 0* will be called the upperquasicentral
series and the Hyperquasicenter of G respectively. If we
use Q(S) and Q*(S) in relation to a symbol S that stands
for a group, then they will denote the quasicenter and the

Hyperquasicenter of S respectively. If H <3 G then the

identity of %- shall be denoted by I2

Remarks:

1. From the above definition it is evident that the

Hypercenter is contained in the Hyperquasicenter. For, let
1 = 20 c 21 c 22 c ---- c zm = H(G) be the upper central
Zi+1 G
series of G, where '7Z__ = Z(E—fi, i = O,1,°--, m—1. Since
i i

every element in the center of a group G is a QC-element,
it follows that, 21 CiQ1 C 0*. We shall assume Zi CIQ*

*
and prove that Zi+1 C Q .

21

Let z e Zi+1 and suppose z p 21' If 2 does not

belong to Q*, then we obtain a contradiction as follows:

We have, i2 = 2;, where 2 = Ziz and i = Zix is any ele—

ment of ‘37 - This implies Zixz = Zizx, so that Q* Zixz
i

= Q* Z zx and Q*zx = Q*xz. Therefore 22': i2 where

now 2 = Q*z and i = Q*x is any element of g;-. Hence

2 e Z(%;) and so Q(gr) % I} a contradiction. Hence

zi+1 c 0*. This implies H(G) E Q*.

2. The Hypercenter H(G) is always nilpotent but the
Hyperquasicenter is not necessarily so. The Hyperquasi-
center of G = $3 is G itself but it is not nilpotent.

As remarked earlier, the Hypercenter of a group is
known to be the intersection of all normal subgroups N

G

such that Z(fi) =.I. We begin this chapter by establishing

a generalization of this result.

Theorem 2.1

 

Let G be a group and Q* be its Hyperquasicenter.
Then, 0* = fl [NIN4G. Q(g‘) =T).

Proof:-

Let 1 = 00 C,Ql C 02 C -°-- C Q Q* be the upper

n

quasicentral series of G. Suppose T n (NINW<JG. Q(S): I].

Now, Q(——) = II implies Q* 6 [N|N<a G, Q(%) = I). Hence
T.E 0*. We now prove that Qj EQN V j = 1,2,---,n when-
ever N <3G and 0(3) = I} We observe that if Q(%) =II
then Ql.g N. For if Ql.$ N then at least one QC-element

g of G does not belong to N. By the definition of a

22
QC-element, <g><x> = <x><g> ‘V x e G. Hence N<g><x> =
N<X><g>. i-e- N<g>°N<X>

<§><§> = <§><§> where 5 = Ng and i = Nx is any element

N<X>°N<g> V x e G. Therefore,

of g» Hence a is a QC-element of g- and so Q(S) # I:
a contradiction. Therefore 01 SEN.
Now assume Qi EQN. We shall prove Qi+1 EEN. Let
G
(Qial, Qiaz, o-o, Qiam] be a set of QC elements of Qi
. Qi+1
which generates -75—— . If each a2 for g = 1,2,-v°,m
i
is in N then Qi+1 SQN and we are done. So for some k

between 1 and m, let ak f N.

Since Qiak is a QC—element of , it follows that

(10

i
<5k><§> = <x><5k> Where 5k = Qiak and i = Qix is any

element of §- Therefore, for arbitrary pairs of integers

Q..

i
r, s and r1, s1 3 pairs of integers u, v and ul, v1

r _s _ _u _V -r —S
x - x a and x 1 a 1 =
k k k
s u V
x = Q. x a and

such that the following hold: 5

_ul -vl
a x . From these we have Qi a

W
W‘H
[.1
w

x = Q a H e N-Q ar x8 = N-Q xu v a d
Qi ak . x . enc i k i ak n

V . . r S u V
S u Similarly, N x 1 ak1 - N a 1 1.

Therefore, (ak><x> = <x><ak> where now ak = Nak and

i = N x is any element of 3» Hence 0(3) # 11 since ak
is a QC-element of g- and this is a contradiction. So
' ' *-
ak e N and Qi C N implies Qi+1 C N. Therefore Q C N
G __
and T = n{N|N<1G, Q(fi) — 1}

Q* n [H(NIN'éJG, N # Q*, Q(gj = I}] : Q*.

23

Corollary

If N is a minimal normal subgroup and 3- has
trivial quasicenter then 0* is elementary abelian.

Proof:-

Since Q(S) = I} it follows that Q* g;N and N
being minimal, Q* = N. The elements of order p in the
center of some Sylow p-subgroup P of Q1 generates a
normal subgroup of G and N being minimal must be equal
to this subgroup. Hence the assertion in the corollary
follows.

In general, knowledge about the Hyperquasicenter of a

group G does not seem to help to say anything about the

Hyperquasicenter of a factor group 33 N AfiG.

In G = A4, Q* = 1, but Q*(%) = N" where N is the
Sylow subgroup of order 4. Again, consider G = A4 x 83.
Q*(G) = s3 but mg?) = 1.

G

233'. We

observe that (a) is normal in G = Q* and this leads us

However in 83 = G, 0* = G and Q*(Zgji

to examine the Hyperquasicenter of a factor group '3

where T <flG and T C Q*. In Theorem 2.3, we prove that

*-
Q*(TJ : %—-. To prove this we need the following lemma.

Lemma 2.2

 

Let 0 be an isomorphism between two groups 81 and

82. Then Q(Sl) and Q(S2) are isomorphic under 0 also.

24
Proof:-
If t e 81 is a QC—element then t o is also a QC-
element of 82 and if w e 82 is a QC—element, then w 0—1
is also a QC-element of 81 by Lemma 1.6. Thus, there is

a 1-1 correspondence between the QC-elements of SI and 52.

Let Q(Sl) = <t1><t2> --- <tm> where ti for i =
1,2,---, m is a QC-element of 81. Any element of Q(Sl)
. a1 a2 am . .
can be written as t1 t2 --- tm where ai is some integer.
a.

Since til is a QC-element of 81 it follows that the

. a1 a2 am . .
image of t1 t2 --- tm is an element of Q(SZ). Sim-

ilarly thegxe-image of any element of Q(Sz) is an element
of Q(Sl). o is therefore a 1—1, onto map from Q(Sl)

to Q(Sz). As o is an isomorphism between $1 and $2

it follows therefore that Q(Sl) and Q(Sz) are isomorphic

under 0.

Theorem 2.3

 

Let G be a group and T be a normal subgroup of G

contained in the Hyperquasicenter Q*. Then the Hyperquasi—
*
center of g- is ‘%—.
Proof:-
Let the upperquasicentral series for g- be as follows:
G G
T T T T 39. Wm HE
T T
it follows by Lemma 2.2 that, Q(%—) = I: Therefore by

m

Theorem 2.1, Q* ggwm. Now suppose the quasicenter 'T_ of

25

E

bk f Q* for some k between 1 and 2 then we have a

%- is generated by QC-elements Tbl, sz. -°°. Tb . If

contradiction as follows:

Since Tbk is a QC-element of g» <5k><§> = <§><Bk>,

where bk = Tbk and i = Tx is any element of %u This

implies, for arbitrary pairs of integers r, s and r1, 51 3
s

r _
pairs of integers u, v and ul, v1 such that Bk x -

.11 _V _r1 S]. _ 1.11 _V1 r S _
x bk and x 5k — 6k x . From these we have Tbk x —
u v r s = * u v . r s _
Tx bk , so that Q* T bk x Q T X bk’ i.e. Q* bk x —

r s u v
Q* xu b: and similarly Q* x 1bk1 = Q* bk1 x 1. Thus

<Bk><x> = <§><5k> where now Bk = Q* bk and i = Q* x

is any element of %?" Bk is therefore a QC-element of
gh contradicting the fact that, ngk) = I: Hence W1 g;Q*.

Assume now Wj is contained in Q*. We wish to prove,

*-

wj+1 C Q °
G wg+1
T. T “I W'+1

Observe that Q W_' = —fi_' = -1%—- ‘while

.1. _l J
T T

9 w

I_.;'§_. E. :..111

W. W. . Hence Q(W.) W. by Lemma 2.2. Suppose

.1. 3 J J

T

chl, ch2,-o--, chr are QC-elements of %—- which generate

j
W

-%g3n Let CE for some integer E between 1 and r
j

be not in Q*. Now, <E£><§> = <§><E£>, where 52 = chg

26

and i = ij is any element of %—-. As above we can show
j

that Q*c is a QC—element of g;- and obtain a contradic-

2

tion. Hence, Wj+1 g Q*. This implies Wm‘g Q* and so

Wm a Q*. Thus the Hyperquasicenter of is

F-JIQ
arc;

Corollary

 

If 81 and 82 are two groups isomorphic to each
other then Q*(Sl) and Q*(SZ) are isomorphic under the
same map.

Follows from Theorem 2.3 and Lemma 2.2 by induction.

If H is a normal subgroup of G then its Hyperquasi-
center Q*(H) in general neither contains nor is included
in Q*(G).

The Hyperquasicenter of G = A4 is 1 but the Sylow
subgroup H of order 4 is normal and it coincides with
its Hyperquasicenter. In this case therefore Q*(G) $;Q*(H).

On the other hand, the Hyperquasicenter of the Quatern-
ion group G is G itself while the Hyperquasicenter of
a normal subgroup H of order 4 is H itself. In this
case therefore Q*(H) ; Q*(G).

This leads us to investigate the relationship of the
Hyperquasicenter of a group G and the Hyperquasicenter
of subgroups of G with additional restrictions besides
normality imposed on the subgroups and we investigate the

Hyperquasicenters of direct products.

27
It follows by induction on |G| from Theorem 1.12
and Theorem 2.3 that when G = H X K and (]H|,|K|) = 1
then Q*(G) = Q*(H) X Q*(K). However, Theorem 2.3 alone
also enables us to establishtfle:equality when instead of
(|H|,|K|) = 1, we have, at least one of the factors H or

K supersolvable.

Theorem 2.4

 

Suppose a group G = H X K where either H or K is
supersolvable. Then Q*(G) = Q*(H) X Q*(K).

2222::-

Let H be supersolvable. We proceed by induction.
For [G] = 1, the assertion is trivially true. Assume it
for all direct products with order less than [GI and
one of the direct factors supersolvable. Since H is
supersolvable 3 a cyclic subgroup <b>AH of prime order.

As H and K centralize each other, it follows that

<b> <‘G- Consider 3%? = 3%“; X (Ii; '33; being super‘
solvable induction applies to ZESF Note that b is a

QC-element of G and <b> C Q*(G). For identical reasons

<b> c 0* (H).

G - ’ H K - Q*G =
Hence Q*(zg) - Q*\ 357) x Q*(Zfi ) , l.e. b
Q*bH X Q*bK by Theorem 2.3. Therefore Q*(G) =

Q*(H) X Q*(K).
We next investigate the structure of the Hyperquasi-

center and in Theorem 2.5 we prove it to be supersolvable.

28

Theorem 2.3 together with this result enables us to char-
acterize all supersolvable groups in two new ways. Theorem
2.5, also helps to establish that the product of the Hyper—
quasicenter and a supersolvable subgroup is supersolvable.
This shows that every maximal supersolvable subgroup of a
group contains the Hyperquasicenter. This is a partial
generalization of a result of Baer on Hypercenter which
states that the Hypercenter is the intersection of all maxi—

mal nilpotent subgroups.

Theorem 2.5

 

Let G be a group and Q* be its Hyperquasicenter.
Then Q* is supersolvable.

Let Q1 be the quasicenter of G. If Q1 = 1 then
Q* = 1 and the theorem is trivally true. Also if 01 3 Q*
then too 0* is supersolvable since Q1 is nilpotent. To
prove Q* is supersolvable in general, we proceed by induc-

tion on [G].

*-
Consider %- and note its Hyperquasicenter is g—- by
1 1
*-
Theorem 2.3. By induction hypothesis ‘g— is supersolvable.
1
Let M be a maximal subgroup of Q*. If M contains
*
01 then g—- is a maximal subgroup of ‘g— as otherwise
1 1
g- :>%—- implies S :>M and maximality of M is violated.
1 1

9:.fl : . 2:.M - -
Now [51 . Q;] [Q* . M] . But [51 . Qi is prime

since the index of every maximal subgroup in a supersolvable

29

group is a prime. Thus M is of prime index in 0*.

Suppose now M does not contain 01. Since 01 is a
direct product of its Sylow subgroups, at least one Sylow
subgroup P of 01 corresponding to some prime divisor p
of [01] is not contained in M. As P <3G, we have
0* = PM = MP.

But P is generated by QC-elements of G and so there
is at least one such QC-element g which is not in M.
Hence 0* = <g>M = M<g>. There are therefore two possibil-
ities: either (9) n M = 1 or <g> n M # 1. Consider the
first case: If g is of order p, then M is of index p
in 0*. If g is of order pn then gpn-1 is of order p

and by Theorem 1.7 it is a QC-element of G. Hence 0* =

n-i n
M<gp > = <gp >M. So the index of M in 0* is again

p. a prime.
Now consider the second case, <g> n M # 1. Then
<g> U M is a subgroup Of <9> Of order pa. The unique

subgroup of order pa of <g> is contained in the unique

n-a—1
1
subgroup of order pa+ generated by gp . As before
n-a-1 n—a—i
9p . being a 0C-element of G, 0* = M<gp > =
n-a-1
<gp >M and

n—a-l
lM<qp >l : IMI 3+1 =
TNT» [Ml . pa p-

Thus every maximal subgroup of 0* is of prime index.

 

Hence 0* is supersolvable by Huppert's theorem [Theorem

9.3.8, Scott [12]].

30

Theorem 2.6

A group G is supersolvable if and only if G = 0*,
where 0* is the Hyperquasicenter of G.

am:—

If 0* = G then G is supersolvable by Theorem 2.5.
Let now G be supersolvable. For [G] = 1, the Hyperquasi-
center coincides with G, so we assume the result true for
all groups of order less than |G|.

Since G is supersolvable, G has a cyclic subgroup
(b) of prime order which is normal. Consider ZES'. It

is supersolvable and b is a 0C-element of G. Therefore

(b) C 0*. By induction hypothesis 0*(Z%§) - ZESI' But

*-
by Theorem 2.3, 0*(ZES) = Z%§-. Hence 0* 3 G and the

theorem is proved completely.
If N is a normal subgroup of a group G, then super-

solvability of g- does nOt imply in general that G is

supersolvable. This is evident from G = A4. N" where N

is the Sylow subgroup of G of order 4, is supersolvable
but G is not. However, if N is contained in the Hyper-
quasicenter then the supersolvability of E- does imply

that G is supersolvable.

Theorem 2.7

 

Let N be a normal subgroup of a group G contained
in the Hyperquasicenter 0*. Then G is supersolvable if

and only if g- is supersolvable.

31

Proof:-

If G is supersolvable then certainly 3- is super-
solvable. On the otherhand if g- is supersolvable then the
Hyperquasicenter of g- coincides with g- by Theorem 2.6.
Also by Theorem 2.3, the Hyperquasicenter of 3- is £13
Hence %:-= fin i.e. 0* = G. Since 0* is supersolvable

it follows that G is supersolvable.

Corollary
Let G be a group and 0* be its Hyperquasicenter.

If 0* # G then for N.4]G, %- supersolvable implies

Ngot

Proof:-

If N'g;0* then by Theorem 2.8, supersolvability of
3- implies G is supersolvable and therefore 0* = G, a
contradiction. Hence the corollary follows.

We remarked earlier that Baer has proved that the Hyper-
center of a group G is the intersection of all maximal
nilpotent subgroups of G. Since the Hyperquasicenter 0*
of G is supersolvable, the question that naturally arises
is whether analogously 0* is the intersection of all maxi—
mal supersolvable subgroups. We are able to answer the ques-
tion partially and prove that 0* is contained in the inter-

section of all maximal supersolvable subgroups of G. We

begin with the following theorem.

32

Theorem 2.8

 

Let G be a nonsupersolvable group. Then the Hyper-
quasicenter 0* is not a maximal subgroup and a maximal

subgroup that does not contain 0* is not supersolvable.

Proof:-
Suppose 0* is a maximal subgroup of G. Then -g;
is of prime order and so 0(%;) = g;- which contradicts the

fact that 0(g;) ='I and 0* # G. Thus 0* is not a
maximal subgroup of G.

If M is a maximal subgroup and 0* ggM then G = Q*M.

 

 

 

 

. G _*M~ M
Suppose M is supersolvable. Now 5;— — 26:. = Q*OM and
as 0(§-) 2 it follows by Lemma 2.2 that 0( M ) ='I.
0* ’ ' 0*nM '
But, since Q*NM is supersolvable, its quasicenter
Q(anM) has to be different from identity by Theorem 2.6.

So we have a contradiction and M is therefore not super-
solvable.

We next prove that the product of the Hyperquasicenter
and a supersolvable subgroup is supersolvable. As an aid

to this we need the following lemma.

Lemma 2.9

 

If a subgroup H of a group G contains the Hyper—
quasicenter 0* of G then the Hyperquasicenter 0} of
H contains 0*.

Proof:-

We establish the lemma by induction on [G]. For [G] = 1

it is trivially true, and so we assume the result true for

33
all groups of order less than |G|. Note, that if 0* = 1
then we are done and so we assume 0* + 1.

Consider and its subgroup §—- where 01 is the

EL.
Q1 01

quasicenter of G. Observe that 01 is contained in Q(H)
_._*

and so by Theorem 2.3 the Hyperquasicenter of QIL is 8—.
1 1
By the same theorem the Hyperquasicenter of g—- is '——.

Therefore, by induction, it follows that
*

Q Z>Q*-

Theorem 2.10

 

Let 0* be the Hyperquasicenter of a group G. Then
for any supersolvable subgroup S of G, 0*S is super-
solvable.

trees:-

We use induction on |G|. For [G] = 1, the theorem is
trivially true. If [0*S] < IGI, induction applies to 0*S
and 0*S is supersolvable Where 0* is the Hyperquasi-

—*
center of 0*S. But, by Lemma 2.9, 0 :>0*. Hence it

follows that 0*S is supersolvable.

 

 

G _. Q*S 9,1 S . G "-
If 0 S G then 0* 0* 0*flS . Since, Q(0* 1,
. S _-— S
it follows by Lemma 2.2, that Q(0*nS) — 1. But Q*HS

being supersolvable its quasicenter cannot be identity un-
less 8 = Q*US. Therefore S C10* and so G = 0*S = 0*.
G is therefore supersolvable and so 0*S is supersolvable.

The theorem is thus completely proved.

34

Corollary 1

 

0* is contained in the intersection of all maximal
supersolvable subgroups of G.

m:-

Let M be any maximal supersolvable subgroup of G.
If 0*.$ M then Q*M being supersolvable by Theorem 2.10
violates the maximality of M unless Q*M = G. This im-
plies G is supersolvable and G = 0*. The corollary
therefore follows since a maximal supersolvable subgroup

of G is G itself.

Corollary 2

 

The Hyperquasicenter 0* of a group G is contained
in the intersection of all maximal normal supersolvable
subgroups of G.

Proof:-

If G is supersolvable then 0* = G and also a maxi-
mal normal supersolvable subgroup of G is G itself.
Hence, the truth of the assertion follows. Suppose G is
not supersolvable and let N be any maximal normal super—
solvable subgroup of G. If 0*.; N then consider Q*N.
By Theorem 2.10, it is supersolvable and being normal it
violates the maximality of N. Hence 0* is contained in

N and the corollary follows.

Corollagy 3

 

Let T ’ fl{M|M is a maximal supersolvable subgroup of G}

R = H(NIN is a maximal normal supersolvable sub-
group of G}

35
m = Frattini subgroup of G.
0* = Hyperquasicenter of G.

Then 0* ngTang.

Proof:-

Observe that T is a characteristic and therefore
normal subgroup of G. Also if S is any supersolvable
subgroup of G then TS is supersolvable. For, S ggm,
where M is a maximal subgroup of G and T ggm. This
however implies T ggR. For, suppose N is any maximal
normal supersolvable subgroup of G. If T is not contain—
ed in N then TN violates the maximality of N unless
G = TN. But then G is supersolvable and T = R.

All that remains for us to prove is that m ggR. For
this purpose we prove that if H is a normal nilpotent
subgroup and N is a normal supersolvable subgroup of G
then HN is supersolvable.

If H n N = 1 then HN 3 H x N and a direct product
of supersolvable groups is supersolvable. Hence HN is
supersolvable. So let V = H 0 N # 1. Let P be a
Sylow subgroup of V corresponding to a prime divisor p
of [V]. P being characteristic in V is normal in G.
Now P C'P' where 'P is the Sylow subgroup of H correSpond-
ing to the prime divisor p of [H] and '5 being char-
acteristic in H it follows that Pd G. Thus Z(P) 4G
since Z(P) is characteristic in P1 Therefore K =
P 0 Z(P) is a nontrivial normal subgroup of G since a

normal subgroup of a p—group intersects the center of the

36
group nontrivially. Thus we have the following normal
series for N : N :>V :>P :>K :>1. Also, N being super-
solvable, we have a normal series for N as follows:
N :>U1 :>U2 D --- D‘Un = 1 where Ui <3 N" V i = 1,2,--- n and
U

i+1

U.
1

[Theorem 2.10.1, Scott [12]], it follows that, we can ob-

 

is of prime order. By Schreirer's refinement theorem

tain (b) Ck, <b>4N and lb] is aprime. But b e 2(5)
and so b e Z(H). Hence (b) is normal in HN.

To prove HN is supersolvable we use induction on |G|.
If |HN| < [G] then HN is supersolvable by induction.

If HN = G then we have to take care of two cases: (1)
HnN=1, (2) HON#1.

If H 0 N = 1 then G = HN = H X N and G is super-
solvable, as argued before. If H 0 N # 1 then, as in
above, HN contains a cyclic normal subgroup (b) where
b e Z(H). Note that ZgS- is normal, nilpotent and Zgg-is
normal, supersolvable. By induction Z%§-= Zgg- is super—
solvable. Therefore HN is supersolvable.

Suppose N is any maximal normal supersolvable sub-
group of G. If m ggN then a N violates the maximality
of N unless G = 0N. But this implies G = N and so G
is supersolvable. Hence R = G and o EQRJ Thus in any

case m is contained in R and so the corollary follows.

Corollary 4
A normal subgroup N of G is contained in T if

for each supersolvable subgroup S of G, N8 is supersolvable

37
and conversely this is true if N.g T.

Proof:-

If NS is supersolvable for each supersolvable sub—
group S of G then in particular, NM is supersolvable
where M is a maximal supersolvable subgroup of G. If
N is not contained in M then NM violates the maxi-
mality of M unless G = NM. But, then G is supersolvable
and G = 0* - T and this implies N ggT.

Conversely if N g;T then since TS is supersolvable,

it follows that NS is supersolvable.

Remarks

1. A4 provides an example where 0* is strictly con-
tained in R. The Hyperquasicenter of A4 is 1 but it
has a maximal supersolvable subgroup namely the Klein group
of order 4, and this is the only maximal normal supersolv-
able subgroup.

2. The Fitting subgroup, in general, neither contains
nor is included in the Hyperquasicenter. In A4, the Hyper-
quasicenter, being 1, is contained in the Klein four group
which is also the Fitting subgroup. On the other hand, in
S3, the Hyperquasicenter, being the whole group (83 is super—
solvable) properly contains the normal subgroup of order 3
which is the Fitting subgroup.

The idea of a hypercentral subgroup was introduced by
Baer. A subgroup N of a group G is called Hypercentral

if for each M S N and M 4G, %0 2(3) # I. In a series

38
of papers, he studied in detail, various properties of hyper—
central subgroups. Generalizing the idea of a hypercentral
subgroup we now introduce the notion of the hyperquasicen-
tral subgroup of a group and study its various properties.
We require every hyperquasicentral subgroup to be normal
but we may very well do away with this restriction and most
of the theorems we prove, will still be true.

If the hyperquasicentral subgroups are not normal then
the characterization of the Hyperquasicenter reads as fol-
lows:

The Hyperquasicenter of any group G is the product
of all normal hyperquasicentral subgroups.

To establish this we need Lemma 2.15 and Lemma 2.16
and they require that the hyperquasicentral subgroups should

be normal.

Definition 2.11

 

A subgroup N of a group G is called hyperquasicen-
tral if N4G and if for MAG and MgN, -3-00(%)#I
From the definition it follows by taking M = 1 that
N n Q(G) # 1. If N is a minimal normal subgroup then this
is the only condition that has to be satisfied by a hyper—

quasicentral subgroup.

Theorem 2.12

 

If N is a hyperquasicentral subgroup then every fac—

tor group % , where T A G is also a hyperquasicentral

subgroup of

em

39

Proof:-
N . . . G .
If 5- is a minimal normal subgroup of 5- then in
order to prove that E- is hyperquasicentral we need to

T
show that g-n 0(gd # 1' and this follows from the fact

that N is a hyperquasicentral subgroup. Otherwise, let

S G S N

T 4 T and T g T and we need to show that,
9 1‘1
T T -'

Q a ”a 7‘ 1'
T T

m
m

mlz

Since N is

Observe that g— and

elmlalo
elmlalz

a hyperquasicentral subgroup of G, we have, g-n 0(3) # 1.

But this implies by Lemma 2.2 that # 1. And

elmlelz
I)
o
almlu-llo

so the assertion in the theorem follows.

Theorem 2.13

If a normal subgroup N of a group G is part of
the quasicenter Q(G) then N is a hyperquasicentral sub—
group of G.

Proof:-

We need to show that N J Q(G) # 1 and this is trivi-
ally satisfied since N E Q(G) . Also if T E N and T AG

then we are required to prove g-fl 0(%) # I:

If b is a 0C—element of G then <b><x> = <x><b>

\7’ x e G. This implies, T<b><x> = T<x><b>, i.e.

40

<B><§> = <§><b> where b = Tb and i

of %u Hence Tb is a QC-element of

TX is any element

elm

. Thus if 0(%) =

%- then 0':)01. But T C N C 01 and T # 01. Hence
N91_N§ E. 9. - -
T n T _ T C T . Therefore T n Q(T) # 1 and the assertion
follows.

Remarks

1. A minimal normal subgroup N is hyperquasicentral
if and only if N E Q(G), the quasicenter of G.

2.1-22::-

Let N be a minimal normal subgroup of G. If N is
hyperquasicentral then since N 0 Q(G) # 1, it follows that
N S Q(G). Again if N is minimal and is contained in Q(G)

then it is hyperquasicentral by Theorem 2.13.

2. From Theorem 2.13 it follows that the quasicenter

Q(G) of a group G is a hyperquasicentral subgroup of G.

3. If N is a hyperquasicentral subgroup of G then
Theorem 2.13 implies that a normal subgroup S of G, con-
tained in N n Q(G), is also hyperquasicentral. However,

S <JG and S C N does not seem to imply S is a hyper—
quasicentral subgroup of G as N 0 Q(G) # 1 apparently
does not guarantee S 0 Q(G) # 1. We do not have an example
to corroborate this yet and we shall avoid this difficulty
later with an altered definition of hyperquasicentral sub-

group.

41

Theorem 2.14
A normal subgroup N of G is a hyperquasicentral

subgroup if and only if V M 4 G, M S N, 3- includes a non-
trivial hyperquasicentral subgroup of 33
Proof:-

Suppose N is a hyperquasicentral subgroup of G.

Then it follows that % = fin Q(fi) ,5 '1‘ and by Theorem 2.13,

%- is a hyperquasicentral subgroup of

3H3

Conversely, suppose, for M 4G, MgN, % includes

a nontrivial hyperquasicentral subgroup R: of $7 This

implies, 3-0 0(%) #'I and therefore it follows that

EnQ(§) #

M M IN Hence the theorem follows.

We now proceed to prove that the Hyperquasicenter of a
group G is the product of all hyperquasicentral subgroups

with the help of the following three lemmas.

Lemma 2.15

 

Let 1 = 00 C 01 C 02 C --. C 0n = 0* be the upper
quasicentral series of a group G. Then Qi’ V i = 1,2,...n
is a hyperquasicentral subgroup of G and if P is the
product of all hyperquasicentral subgroups of G then
0* g P.

We know Qi is normal in G Vi = 1,2,---n. Since

01 C 0i ‘d i, obviously 01 n 01 # 1. Let 1 75 M$0i and

42

Q- _
M4G. We need to show that 0(-%) Oil—75 1. If 01 ZM

Q- _.
then it follows that 09%) n 34—1 75 1. For 01 g M implies

that at least one 0C-element b of G does not belong to
M. Hence <b><x> = <x><b> V X e G and so M<b><x> =
M<x><b>, i.e. <B><§> = <§><b> where 5 = Mb and i = MX

is any element of Eu Therefore 5 is a 0C—element of
Q.

G G i -—
M and Q(M) n M # 1.

Now assume that 0j (for 1 j.j < i) is contained in
M but . M. Let . a , . a , °°' . a be a set

QJ+1$ Q] 1 Q: 2 Q. O] k
of 0C-elements of 8- which generate -7%il . Not every
j j

ai for 1 :.i :.k is in M as otherwise Qj+1 will be

contained in M. Suppose at 5 M where t lies between

1 and k. Then <5 ><§> =<§><a> where a =0.a e9
t t t J t Qj
and i = ij e g-. Hence for arbitrary pairs of integers
j
m,n and m1,n1 3 pairs of integers u,v and u1,v1 such
_m _n -u _v _m _n _u _v _
that a x = x a and x 1 a 1 = a 1 1' These equa
t t t t
, , m n u v m1 n1 ul V1
tions give Mat x = MX at and Mx at = Mat x

since 0j C M. Therefore <5t><§> = <§><5t> where now

at - Mat amd x - MX is any element of fin So at is

Q- __
a 0C—e1ement of and hence 0(%) fl fii'F 1 .

E.
M
Thus Qi is a hyperquasicentral subgroup. Consequently

0* is a hyperquasicentral subgroup and is contained in the

product P of all hyperquasicentral subgroups.

43
Lemma 2.16
The product of any two hyperquasicentral subgroups of
a group G is a hyperquasicentral subgroup.
322::-
Let N1 and N2 be two hyperquasicentral subgroups

of a group G and note that Nl'Nz <3G. We wish to prove that

NI‘NZ

)n——-—;£‘1‘, where MdG and MgNl-Nz.

Q( M

3H3

For M = <1), it follows that Q(G) fl Nl'Nz # 1.
This is so because N1 being a hyperquasicentral subgroup,

N n Q(G) # 1 and N1 is contained in Nl-Nz. Now suppose

M#<1>. If M$N1 or MgNz then Q(fi) {WEI-$74.1.

For example, if M $LN1 then since N1 is a hyperquasi-

N ._
central subgroup, Q(%) n fil-# 1 and so our above assertion

Ni‘Nz

N
follows since -—l is part of M . Otherwise, let

M
N1 0 M 3 W. For the present we assume W # N1. We shall
eventually take care of this case.
Since W 4G and W i N1, it follows that,

0(9) fl El f'f. Let a a 5 °-- 5 where a. = Wa.
W W 1’ 2’ 3’ s’ i 1'

be a set of QC—elements of g- which generate Q(go. Note

that, Q(S—q) = <51><ez> (as) and <5j><ek> = <5k><5j>

a a a
for all j,k 3 1,2, °-- 5. Suppose W a11a22--- asS is a

N
nontrivial element belonging to Q(%) n‘—l and observe

W
01 02 Q a1 a2 as . .
a1 ° a2 °'-° assfi M. For a1 a2 -~- aS e M Will imply
a a a . d1 a a
all a22 ---- as8 e W Since a1 a22 --- ass 6 N1 and

44

a a a . . .
that will make W all a22 ---- asS the identity of the inter-

N
section of Q(%J and -—l, which is a contrdiction.

W
Now, since 51 for i = 1,2, --° s is a QC-element
of %” it follows that <51><§> = <§><ai>, where i = Wx
is any element of g» Hence for arbitrary pairs of integers

m,n and m1,n1 3 pairs of integers u,v and u1,v1 such

_m _n _ _u ...V _m1 _n1 _ _u1 _V1 .
that a. x - x a. and x a- e a. x . These imply,
i i i i
m n _ u v m1 n1 u1 v1
we have W a. x - W x a. and W x a. = W a. x so
i i i i
n 11 V m1 n1 111 V1
that we get MW a? x = MW x ai and MW x ai = MW ai x
. m n _ u V “11 I11 111 V1
i.e. M a. x - M x a. and M x a. = M a. x . Hence
1 i i i
<51><§> = <§><ai>, where now 5i = M ai and i = M x is
G _ - . .
any element of Ru Thus M ai — ai is a quaSicentral ele-
t f 9- f h '-12 -011 ’02 ’08
men 0 M or eac i - , , ---, s° Hence a1 a2 ~-- 5
. G . C11 C12 as .
is an element of Q(H), i.e. M a1 a2 °°° aS is an ele-

ment of Q(go. Note that, this is a nonidentity element

M
. 01 012 as O£1 G2 OS .
Since a1 a2 --- aS ¢ M. Also M a1 a2 °-- a8 is an
Ni'Nz . 01 C12 CIS
element of -—7E—— , Since a1 a2 --- as 6 N1. Hence

G N1°N2 -— _ _
Q(fi) nTyéi. If NICM then NlflM-W-Nl. But

M being a proper subgroup of N1°N2 it follows that M 0 N2
= R # N2. Therefore we can apply the above argument with

R taking the role of W, N2 that of N1 and prove in the
N1°N2
M

Hence the lemma is proved completely.

same manner that Q(%) n is not the identity alone.

45

Lemma 2.17

If N is a hyperquasicentral subgroup of G then N
is contained in the Hyperquasicenter Q* of G.

2122::-

Let 1 = Q0 CiQ1 C202 C.--- CZQn = Q* be the upper-
quasicentral series of G. Consider N~Qn and by Lemma
2.17 and Lemma 2.18, N‘Qn is a hyperquasicentral subgroup.

So, if N g Qn then Qn ; N-Qn and from the definition

N-Qn G -
0 Q(—) 7‘ 1,
On On

 

of a hyperquasicentral subgroup we have,

a contradiction Since Q(g—O ='1. Thus N gan = 0*. Note,
n

that if Qn = G, then the truth of the lemma is obvious and

we do not have anything to prove.

We can now state the theorem which we were aiming at,

as follows.

Theorem 2.18

 

The product of all hyperquasicentral subgroups of a
group is the Hyperquasicenter.
Proof:-

By Lemmas 2.15, 2.16, 2.17, the theorem is true.

Corollary

 

If 0* = 01 then any normal subgroup contained in a
hyperquasicentral subgroup is hyperquasicentral.
Proof:-

This follows immediately.

46

The idea of the hyperquasicentral subgroup of a group
G was introduced for the study of the normal subgroup of
G contained in the Hyperquasicenter Q* and for their
characterization. From the definition however, it does not
seem to follow that the property of being a hyperquasicen-
tral subgroup is hereditary in the sense that if N is a
hyperquasicentral subgroup and S is a normal subgroup of
G contained in N then it does not seem likely that S
is hyperquasicentral. For, N n Q(G) # 1 does not guaran-
tee S n Q(G) # 1 which is necessary if S is to be a
hyperquasicentral subgroup. We, however, obviate this dif-

ficulty by introducing a new definition.

Definition 2.19

 

A subgroup N of a group G is called extra hyper—

quasicentral if NdG and for M 4G, MSN,
5: G N _
_. _. 1.
Q (M) n M #
The definition implies that every hyperquasicentral
subgroup is also extra hyperquasicentral. We conjecture

that an extra hyperquasicentral subgroup is not necessarily

hyperquasicentral, for Q*(fi) 0 fi-# 1 does not seem to
imply Q(S) fl %-# T. No example is known yet but we shall

prove a theorem towards the end of this chapter which sug-
gests the line along which the example might be worked out.
From the definition of an extrahyperquasicentral sub-

group it follows that if N is an extrahyperquasicentral

47

subgroup then N 0 Q*(G) # 1. Since every hyperquasicentral
subgroup is extrahyperquasicentral, the product P of all
the extrahyperquasicentral subgroups contains the Hyper-
quasicenter Q*. We shall, however, Show that P is in
fact equal to Q*.

In the following, we first, derive some properties of
the extrahyperquasicentral subgroups and try to character—

ize them. For that, we need the following lemma.

Lemma 2.20

 

Let G be a group and T, S be normal subgroups of

G with T C S. If Q*(%0 =‘¥: and Q*(gfi = gf then
V* C R* .

Let 1 = Qo C Ql C Qz C ---- C Qn = Q* be the upper-
quasicentral series of G. If b is a QC-element of G

then <b><x> = <x><b> V x e G. This implies S<b><x> =
S<x><b>, i.e. <E><§> = <§><B> where b 3 Sb and i = Sx

is any element of Ex Hence 01 C R* and let

S
Qi a1, Qi a2, °°° Qiam be a set of QC—elements of g: which
i
Q .
generates -%i1 . If for some k between 1 and m, ak
i

does not belong to R* then we have a contradiction as
follows:

<5k><§> = <§2<ak> where 5k 3 Qi ak and i = Qix is

any element of g—u From this it follows that <§k><i> =
i
<§><§k> where i = R* x is any element of g?' and

48

k = R* ak. This, however, implies that 5k is a QC-ele—

DJII

ment of and this contradicts the fact that Q(§¥ =‘1.

E.
R'X'
* as 9(-
Thus Qi+1 C'R and so Q C R .
To establish the lemma, we use induction on [G]. If

T = 1 then it is true by what we showed above. Suppose

T # 1 so that induction hypotehsis applies to %u Now

 

 

G G
_I_.2'§. - * Q. = 31 - * T = k-3
§_ S and Since Q (S) S , it follows that Q .§

T T

RX-
-{§—, by corollary to Theorem 2.3. Hence by induction,

5' e.

* *-
¥ CI%— , i.e. V* C R*.

Theorem 2.21

 

A normal subgroup N of a group G is extra hyper-

quasicentral if and only if V M 4G, M g N, 31% contains

a nontrivial extra hyperquasicentral subgroup of 3-.
21952:
Suppose N is an extra hyperquasicentral subgroup and
let S = N n Q*(G) # 1. S is normal in G and if T $§S

§. *9. :§. .Q_*:§ — ‘
and TAG then TflQ (T) TOT T;£1. Hence if

M = 1, then also N contains a nontrivial extra hyper-

quasicentral subgroup.

Suppose now M # 1 and let 3" E-fl Q*(gfi # 1. If
M M M
Q. .3
X.C.5 .X E. ,, *.J!_ :E— '—
M‘I'M and MOM then by Lemma2.20, Q 2 my #1
M M

49

 

 

G E
G..U M H
*—__ ‘X- :—
For, suppose, Q (M) M and Q (.2 ‘2’. Then
M M
_ Q B.
E. H. H. * M ..fl. = " E, -
M CIM ClM and hence Q 2, n 2, ¢ 1. Thus M is
M M

w
3m zl<|zl

an extra hyperquasicentral subgroup of .
Conversely let N <JG satisfy the hypothesis of the

theorem. Since N contains an extra hyperquasicentral

subgroup V of G, we have V n Q*(G) # 1 and so N n Q*(G)

contains an

'_I‘. Q -
,MnQ*(M);£1.

#1. If MCN and MAG then since

Zflngflz

extra hyperquasicentral subgroup %- of

which implies fi-fl Q*(EJ # 1. This completes the proof of
the theorem.

In order to Show that the Hyperquasicenter is the pro-
duct of all extra hyperquasicentral subgroups we need the

following lemma.

Lemma 2.22

 

The product of two extra hyperquasicentral subgroups
of a group G is extra hyperquasicentral.

Proof:-

Let N1 and N2 be two extra hyperquasicentral sub-
groups. Consider N1°N2 and note that it is normal in G

Since N1 and N2 are both extra hyperquasicentral

subgroups, Nl-Nz fl Q*(G) # 1. Let S.<1G and S $;N1~N2.
N1°N2 G _.
We need to Show ——§——-fl Q*(gd # 1.

Since S is contained in Nl-Nz properly, S cannot

contain both N1 and N2. Suppose N1 is not contained in

 

 

T'“

N S N N
S and consider -—1- 35-i- =-—1. Let Q*(QO = 21- and
s NlflS T T T
*-
0* (981) =3. By Lemma 2.20, v* c R*.

Since N1 is extra hyperquasicentral it follows that

G N1 — G N1
Q*(EO n T—-# 1. Let Ta 6 Q*(Ev 0 E—-# T and note a ¢ T implies

. .. R* .
a 4 S. Hence Sa is a nontriVial element of -— , Since

S
G le
a e V* and V* C'R*. Therefore Q*(gfl 0 -§—- contains Sa
G N1.S ._
which is not the identity. Thus, Q*(EJ 0 —§— # 1 and so
G N1 °N2 - . .
Q*(gfi H S # 1. Hence Nl-Nz is extra hyperquaSicentral.

 

Theorem 2.23

 

The product of all extra hyperquasicentral subgroups
of a group G is the Hyperquasicenter Q* of G.

2122::-

Since there can be only a finite number of extra hyper-
quasicentral subgroups, their product W is again an extra
hyperquasicentral subgroup. Also every hyperquasicentral
subgroup being extra hyperquasicentral, it follows from
Theorem 2.18, that Q* g W.

If Q* Sgw then by definition of extra hyperquasicen-

tral subgroups, Q* %;0 n g¥ # 1' which is impossible since
Q*(g') =‘I He W = *
0* . nce Q

Corollary

All extra hyperquasicentral subgroups are supersolvable.

51
Proof:-

This follows immediately.

Theorem 2.24

 

Let N be a normal subgroup of G with N n Q*(G) # 1.
Then N is an extra hyperquasicentral subgroup if and only
if VMdG and MCN, M and 31% are both extra hyper-
quasicentral subgroups of G and %- respectively.

REESE:-

Let N be an extra hyperquasicentral subgroup of G
and let 175MCN and MAG. Since NCQ*(G), we have
MCQ*(G) and so MflQ*(G) =M#1. If 84G and S$M

M M-

* _
= %.n —-— g-# 1. Hence M is an extra

Then Q*(EJ n S

ml

*-

3K3

hyperquasicentral subgroup of G. Also, firm Q*(EJ = 3:0

 

_ E, -' - X. E. X. E. ' E. * Q. :
M 75 1 and if M g M' MAM then Since V 0 Q (V)
E E
E. 'I ' .JE. * M " E. '
V f , it follows that y. 0 Q E. # 1. Thus M is
M M

an extra hyperquasicentral subgroup of fin

Suppose now that N 43G and N n Q*(G) # 1 are given.
If there is no normal subgroup M # <1) in N then
N H Q*(G) # 1 guarantees N is an extra hyperquasicentral

contains a nontrivial extra

3B2

subgroup. Otherwise, Since

of , it follows that,

3ND

hyperquasicentral subgroup say :3

§. * §_ : é. —' E. * g. _' ' - -
M 0 Q (M) M # 1. Therefore M n Q (M) # 1. This implies
N

is an extra hyperquasicentral subgroup of G.

52

Corollary

 

A normal subgroup N of a group G is an extra hyper-
quasicentral subgroup if and only if N C Q*(G) and if S
is a normal subgroup of G contained in the extra hyper-
quasicentral subgroup N then S and E- are extra hyper-

S

quasicentral subgroups of G and g- respectively.
Proof:-

This follows immediately by Theorems 2.23 and 2.24.

As remarked earlier, we conjecture, that an extra
hyperquasicentral subgroup is not necessarily a hyperquasi-
central subgroup. The following theorem indicates the line

along which the example might be worked out.

Theorem 2.25

 

Let G be a group and Q1, 0* be its quasicenter and
Hyperquasicenter respectively. Then if Q* is the direct
product of subgroups N and Q1 where N 43G, then N
and Q1 cannot be of relatively prime order.

m:-

Suppose N and Q1 are of relatively prime order.

Since Q(gh) = gi, it follows that Q2 = Q1 x S where

S 3 02 n N. Note that S is normal in G since Q2 and

N are both normal in G. Since %2_ is nilpotent, it fol-
1

lows that S is nilpotent. If N is a Sylow q-subgroup

of S then N is characteristic in S and this implies N

Q1 Xfi Qz

is normal in G. Now is a Sylow q-subgroup of a:

and it is generated by QC-elements of g-. Let ng be a
1

53
QC-element of 8- where g e N. Observe that N C N where
1
N is a Sylow q-subgroup of G and N, Q1 are of relatively

<§><§> where i =

le and g = ng. This implies that for arbitrary pair

prime order. If x e N then <§><§>

of integers r, s 3 pair of integers u,v such that

-I' -S -u -V

_ . r s _ u v r s
g x - x g , i.e. ng x

- Q1 x g . Hence 9 x =

S.
g X = Yo-

yoxugV where yo 6 Q1. From this we have grx

But the left hand side of this equation is an element of

N and its order is prime to that of yo. Hence grxs =

xugv. By identical arguments, for arbitrary pair of inte-

r S
gers r1,51 3 pair of integers u1,v1 such that x 1g 1 =

ul v1
g x . Hence <g><x> = <x><g> i.e. g is a QC-element
of N. If L is another Sylow q—subgroup of G then since
N 4G, it follows that N C L and we can repeat the same
argument as in above to Show that g is a QC-element of
L too. Thus <g> permutes with the cyclic subgroup gen-
erated by each q—element of G.
Let x e G be a t—element where t is a prime and
t > q. Consider N<x> = <x>N. It is supersolvable by
Theorem 2.10 since N<x> C Q*(x). Hence <X>4N<x> and
so N<x> = <x>N = (x) x N. Therefore again <g><x> = <x><g>.
Suppose now, that x e G is a t-element where t is

a prime and t < q. Note that if x e Q1 then gx = xg

and so <g><x> = <x><g>. Assume therefore x f Q1. Since

g = ng is a QC-element of g-, <§><§> = <§><§> where
1 .
i = le. Since <§><§> = <§><§> is supersolvable [Theorem

,54

13.3.1, Scott [12]] , it follows that ((3)4 <§><§>. Hence

-1__ _w _
x gx = g where w is some integer. This implies x 1gx =

ygw where y 6 Q1 and hence x-lgxg—w = y. But N and

Q1 are of relatively prime order and x_1gxg"w e N since

N ‘3G. Hence x-J‘gxg-w =.y = 1, i.e. x 1gx = gw. Thus
<g> is normalized by x and so <g><x> = <x><g>.

Let now c e G be any element and suppose (c) =

j.
(c1) X (c2>x '°°°x <Ck> where |ci| = a.1

l , ai is some

prime and ji is some positive integer for i = 1,2, °-- k.

Since <g><ci> (ci><g> it follows that <g><c> = <c><g>.

Therefore g is a QC-element of G and this is a contra-
diction Since g e N and by assumption ([Nl, [Q1|) = 1.
Therefore, the assumption is false and the theorem is proved
completely.

As remarked earlier, Baer has shown that, if y is

any element of a group G, x e H(G), the Hypercenter of

G and (|x|, |y|) = 1 then xy yx. In View of this,
we are naturally led to consider the question: IS it true,
that if x e Q*(G), y 6 G and LIX], |y|) = 1 then
<x><y> = <y><x>? The answer is false and the counter
example is provided by a supersolvable group G which we
shall describe later. The Hyperquasicenter of G being
G itself, it is shown that in G there are two elements
le with (|x|, lyl) = 1 but <X><Y> # <y><X>-

It may be observed, however, that the permutability

of elements of relatively prime order in a group is a

55
sufficient criterion for its supersolvability, as is evident

from the following theorem.

Theorem 2.26

If elements of relatively prime order permute in a
group G then G is supersolvable.

Proof:—

Let M be a maximal subgroup of G and p1, p2,... pn
be the prime divisors of ]M]. Suppose Pi is a Sylow pi—

subgroup of M for i = 1,2, --- n. Then M = Pl-Pz- ----Pn

where Pij = PkPj and j,k = 1,2, --- n. This is true,

Since if gx e Pij. g e Pj and x 6 Pk’ then gx =
xrgS e PkPj for some integers r and 5, since <g><x> =
<x><g>. Similarly if hy e P P. then hy = yuhV e P.P

k j j k

for identical reasons, where h e P y E Pj and u,v are

k’
some integers.

If p is a prime divisor of ]G] and p does not di-
vide ]M| then consider an element x e G of order p. We
have G = <x>M = M<x>, since <X>Pi = Pi<X>’ i = 1,2,"° n
for the reasons given above. Hence [G : M] = p, a prime.
Now suppose all the prime divisors of ]G] divide [M]. We
observe that at most one Pi is not a Sylow pi-subgroup
of G. For, otherwise, suppose P1 and P2 are two Sylow
subgroups of M which are not Sylow subgroups of G. Then
P1 S 131 and P2 g P2 where Pl, P2 are two Sylow sub-
groups of G. Now, consider S = §1P2 °-- Pn' Note that

P1 permutes with each Pi for i = 2,3, ’-- n and also

56
Pij = PkPj for j, k = 2,3,°--,n. Thus S is a subgroup
and S # G. Therefore S contains M properly and maxi-
mality of M is violated. Thus exactly one of the Pi is
not a Sylow subgroup of G and we denote this subgroup by
P1 (relabelling if necessary).

Now P1 is contained in some Sylow subgroup P} of G.

_ m m—1
Let ]P1] = p1. If |91|< p1 then P1 is contained in
= _ = m-i
a subgroup P1 of P1 and ]P1] = p1 . But then R =
Ele --- Pn will be a group by same arguments as before

and R will properly contain M violating the maximality
m-1

of M. Thus |P1] = p1 and hence [G:M] = p1, a prime.

Thus by Huppert's Theorem it follows that G is supersolv—

able. [Theorem 9.3.8, Scott [12]]

Corollary

 

If all subgroups of relatively prime order permute in
a group G then it is supersolvable.

However, relatively prime ordered elements do not neces-
sarily permute in a supersolvable group. This is confirmed

by the following example.

Example:
Let G = (a,b><x> = <x><a,b> where a3 = 1, b3 = 1,
ab = ba, aX = a2, bX = b and x2 = 1. G is supersolvable

because it has the normal series: G :><a,b> :><b> :31.
But, observe that (x) and (ab) do not permute though x

and ab have relatively prime orders.

57

It may be noted that the product of two normal, nilpotent
subgroups of a group G is normal and nilpotent and this
led to the definition of the Fitting subgroup. (However,
the answer to the question, that whether the product of two
normal supersolvable subgroups is normal and supersolvable,
is false. The counter example is provided in 9.2.19, Scott
[12]). This leads to the consideration of supersolvably
immersed subgroups which have been studied in detail by R.
Baer. The definition of a supersolvably immersed subgroup
is more restrictive than that of a supersolvable subgroup
and this enables us to establish the existence of a maximum
normal supersolvably immersed subgroup of a group analogous
to the Fitting subgroup.

In the following we derive some basic properties of the
supersolvably immersed subgroups of a group G and Show how
they are related to the Hyperquasicenter Q* of G. That

will be the conclusion of the present chapter.

Definition 2.27

A subgroup M of a group G is called supersolvably
immersed in G if for each homomorphism of G, the image of
M contains a cyclic subgroup which is normal in the homo—

morphic image of G.

Remarks:-
1. Note that the supersolvability of M is included

in the definition. This may be established by induction on

]M

 

58
9

I

Since Me contains a cyclic normal subgroup of G
under any homomorphism e of G, so in particular, if 9
is the identity homomorphism then it follows that M con-
tains a cyclic subgroup (a) which is normal in G. Con—

sider ZES- and let ¢ be the homomorphism: G ——> 2%;-

¢ M C G _ ¢ .
Z;§-__ZES- G . Under any homomorphism

¢°9

Then we have M

¢ ¢'9

9 of G , M ELG But ¢-6 is a homomorphism of G

and by definition of supersolvably immersed subgroup, M¢.e Eflfl

contains a cyclic normal subgroup of G¢'e. Hence Z2; is

a supersolvably immersed subgroup of Z23. and so by induc-

 

tion {$5. is supersolvable. Hence M is supersolvable. k”’

2. Every supersolvably immersed subgroup M of G
is normal. This is established by induction on |G|. For
]G] = 1, this is trivially true and assume the assertion
is true for all groups of order less than |G]. As in Re-

mark 1, we have, -Zg3- is a supersolvably immersed subgroup

of Z2: and so Zia-43.4% . Hence M is anormal subgroup
of G.

In order to establish the existence of a unique maximum
supersolvably immersed subgroup of a group we need the fol—

lowing lemma.

Lemma 2.28

 

If N1 and N2 are two supersolvably immersed sub-

groups of a group G then so is NI'NZ.

59

Proof:-

Note that Nl'Nz is a normal subgroup of G since
N1 and N2 are both normal in G. Also Nl-Nz contains a
cyclic normal subgroup of G Since for example N1 being
a supersolvably immersed subgroup of G contains a cyclic
normal subgroup of G. Now, suppose a is any homomorphism
of G. Nl'Nz under 9 has the image Ni°NS. But, Ni

contains a cyclic normal subgroup of G6 and so it follows

that (NI-N2)e contains a cyclic normal subgroup of Ge.

Hence Nl-NZ is a supersolvably immersed subgroup of G.

Corollary
The product of all supersolvably immersed subgroups of
G is a supersolvably immersed subgroup.

This follows immediately from Lemma 2.28.

We now conclude the chapter by Showing how, the super-
solvably immersed subgroups of a group G are related to

the Hyperquasicenter Q* of G.

Theorem 2.29

 

Every supersolvably immersed subgroup of a group G
is contained in the Hyperquasicenter of G.

me:—

Let the Hyperquasicenter of G be Q* and M be any
supersolvably immersed subgroup of G. We proceed by induc-

tion on ]G] to prove that M is contained in Q*. For

A

 

60
]G] = 1, the truth of the theorem is obvious and assume it
for all groups of order less than ]G].

AS remarked earlier, M being a supersolvably immersed
subgroup, it contains a cyclic subgroup (a) which is nor-
mal in G and ZES- is a supersolvably immersed subgroup
of ($53 Since (a) is a normal subgroup of G, it follows
that, <a><x> = <x><a> V x e G. Therefore 'a', is a QC-
element of G and (a) C Q*. Hence by induction,

M

G M x-
Z?)— E Q*(@) and so by Theorem 2.3 we have, Z5)— g 2%}— .

This implies M g;Q*.

Corollary

 

The Hyperquasicenter Q* contains the largest super—
solvably immersed subgroup of G.
Proof:—

This follows immediately from Theorem 2.29.

It seems unlikely that Q* is always the largest
supersolvably immersed subgroup of a group G but under
some restriction on G, we Shall Show later, that Q* is,

in fact, so.

 

CHAPTER III
SOME ADDITIONAL PROPERTIES OF THE HYPERQUASICENTER

In Chapter II we obtained the following results:

(1) The Hyperquasicenter Q* of a group G is contained
in the intersection T of all maximal supersolvable sub—
groups of G. (2) A normal subgroup N of G is con;
tained in T if for each supersolvable subgroup S of 6.
NS is supersolvable and conversely this is true if N ggT.
(3) Each supersolvably immersed subgroup of a group G is
contained in Q*.

The first two results are partial generalizations of
the following results of Baer: (1) The Hypercenter H(G)
of a group G is the intersection of all maximal nilpotent
subgroups of G. (2) If N is a normal subgroup of G
then N g H(G) iff for each nilpotent subgroup U of G,
NU is nilpotent.

In this chapter, it is proved that if G is of a cer-
tain nature then the Hyperquasicenter Q* of G contains
a cyclic normal subgroup of G. This enables us to estab-
lish complete generalizations of the above results of Baer
and also the fact that for all such groups G, Q* is the
largest supersolvably immersed subgroup.

Our quest for a cyclic normal subgroup in Q* led us

to examine the groups referred to above. AS a corollary

61

-.‘ LI”-.5.._I

 

1'”

'1‘»

62
to this investigation, a new characterization of the Hyper-
center of a group is given and the Hyperkern and the Hyper-
center are Shown identical.

The principal difficulty in showing that a cyclic nor-
mal subgroup of G is contained in Q*, arises when some
of the prime divisors of ]G] do not divide ]Q*] or if
there is no normal Sylow p-Subgroup of G corresponding

to a prime divisor p of ]Q* . Under suitable conditions,

 

however, Q* does contain a cyclic normal subgroup of G

and that is precisely what we Show in the following theorems.

Theorem 3.1

 

If the smallest prime divisor p1 of the order of a
group G divides the order of the quasicenter then G has
a nontrivial center.

Let Q1 be the quasicenter of G and P1 be the
Sylow subgroup of Q1 corresponding to the prime divisor
p1 of ]QI]. P1 is contained in some Sylow pl-subgroup
P1 of G. Suppose, corresponding to each prime divisor
pi of ]G], Pi is a Sylow pi-Subgroup where i = 1,2,-°-n
and n is the number of distinct prime divisors of ]G].
Then G = <P1, P2, ---, Pm). We know P1 is generated by
QC-elements of G and if h is a QC-element of G con-
tained in P1 then hi, for some integer 'a' is a QC-

element of G of order p1 and is of course in P1. The

set of all QC—elements of G of order p1 are in P1 and

63
generates a characteristic subgroup N of G. Note that,
N = (h1><h2> -~- <hv> where hi for i = 1,2, ---, v is
a QC-element of G or order p1 and <hi><hj> = <hj><hi>
for i,j = 1,2,'°- v. Since a normal subgroup of a p-group
intersects the center of the group nontrivially, it follows
that N n Z(Pl) # 1. Let 1 ¢ gl-gz----gv e N n Z(Pl)
where gi is a power of hi and is therefore also a QC-
element of G. Suppose x is any element in Pk where
k = 2,3,--- n.

We have, Ti = <gi)<x> = <x><gi)‘v i and Ti is super—
solvable by Theorem 13.3.1 of Scott [12]. Therefore <x><a
Ti since (x) is a Sylow subgroup of Ti corresponding
to the largest prime divisor of ]Ti]’ Also (gi> is a
Sylow subgroup of the quasicenter of Ti and therefore

<gi>‘<1Ti' And, Since <9i> n (x) = 1, it follows that

gix = xgi V i. Thus gl-g2 --- gV is centralized by every
element in Pi for i = 1,2, --- n. Hence gl-gz --- gv
lies in the center of <P1, P2, °'° Pm) = G.

Theorem 3.2

 

If Pm is a normal Sylow pm—subgroup of a group G
and pm divides the order of Q(G) then the quasicenter
has a cyclic normal subgroup of G of prime order.
Corresponding to each prime divisor pi of ]G] let
P; be a Sylow pi—subgroup of G where i = 1,2, --- n and

n is the number of distinct prime divisors of ]G]. Without

64

any loss of generality assume Pj for j = m + 1, m + 2, ---n
is a Sylow pj-subgroup with pj > pm and let Pm be the
Sylow pm-subgroup of the quasicenter. The set of all QC-
elements of order pm are in Pm and it generates a char-
acteristic subgroup N of G (as argued in Theorem 3.1)
which can be written as equal to (h1><h2> --- <hv> where
hi for i = 1,2, --- v is a QC—element of order pm. Note
that, <hi><hj> = <hj><hi> = H is a group of order p; and
therefore it follows that N is an abelian characteristic
subgroup of G. Suppose 1 # gl'gz-g3---- gv e N n Z(Pm),
where gi is a power of hi and is therefore a QC-element
too. By identical arguments as in Theorem 3.1, we can Show
that if x e Pj where j = m + 1, m + 2, -~- n, then x
commutes with gi V i. This implies that gl-g2---- g is
centralized by every element x e Pj for j = m + 1, m + 2.
- n.

Now, consider an element x e Pf where f = 1,2, --- m—l.
Note, that Ti = <gi><X/ = <x><gi> is supersolvable and
since pf < pm, it follows that (gi><fl Ti' Thgs x
X 19192 ‘°' 9 X = X—191X'X_192X "‘ X g X = 9f1°gfz "' 92 I

V V
X

- ' x x
where x 1gix = gil and 2i is some integer. If all 1i,

i = 1,2,°:- v, are equal for each x then it follows that

91-92 -~- gV generates a cyclic subgroup of order pm which

is normalized by (P1, P2, --- Pn) = G. Otherwise, without

any loss of generality, suppose for some x, g = E =

= EX = Z and

v E: for k = 1,2, --- t—1 are different from g.

65

x X x
21 22 BV _
Since g1 ~g2 --- gv is also in N 0 ZéPm) it follows
x X x
21 17,2 2: -E -£ -E S}; 5:; St_1
(91 —92 °°° gv )-(g1-92 °°° 9V ) = 91 '92 '°- gt_1 e
X" x x
_ $1 82 S
N n Z(P ) where SX = EX -2 Takin - t_1
m I i i ' g 91 'gZ gt-l I

we repeat the argument so that either it generates a cyclic

a1 32 ak

subgroup of order pm normal in G or else g1 g2 -- gk
6 N n Z(Pm) where k < t~1 and a1, a2, °°' ak are in—
tegers. Proceeding thus and repeating the argument we ob-
b b b
1 2 z

tain finally that either g1 g2 -.- gz e N O Z(Pm) gener—

ates a cyclic normal subgroup of G or else g? is in

N n Z(Pm) where b1, b2, --- b2 and d are integers. But
this implies g? generates a normal cyclic subgroup of
order pm. For g? is centralized by x e fij for j =
m+1, m+2, -~- n and <gg> is normalized by x e Pf for
f = 1,2, °-- m—l. And, since g? e Z(Pm) our assertion
follows.

Corollary

 

If P is a normal Sylow p-subgroup of G and p di—
vides the order of the Hyperquasicenter Q* then Q* has a cyclic
subgroup of prime order which is normal in G.

25.00;:-

Let 1 = Q0 CLQ1 CZQZ C --- C Qn = Q* be the upper—
quasicentral series of G. If p divides the order of 01
then the corollary follows from Theorem 3.2. We therefore

assume, p divides ]Qj+1] but doesn't divide ]Qj] where

66
j lies between 1 and n-1.

Let P be a Sylow p-subgroup of Qj+1 corresponding

Q.P
to the prime divisor p of ]Qj+1]° Then 61—. is a
j

.+1 0.5
Sylow p-subgroup of -%——- and 61—. is the normal Sylow

3 3'

Q.
p—subgroup of E—u Since Q (g-) = —Ji1-, it follows by
Qj Qj Qj

Theorem 3.2, that Qja, where a e P generates a normal

subgroup of Q—u Hence if Q.x is any element of ‘Q
Q- 3 Q.

J r 3
then we have i'15§ = 5 , where i = ij and a = Qja.
This implies ij_1ax = Qjar and therefore x-lax =
yar, i.e. x-lax-a_r = y, where y e Qj' But x-laX°a“r

is in P since P'<JG and a e P C P. AS p does not

"1 -r _ 1

divide ]Qj]’ it follows that, x ax-x = y - and

so x-lax = ar. Also, since Qj O P = 1, Qj and P
centralize each other and in particular every element of
Qj will centralize 'a'. Thus (a) is a normal subgroup
of G and (a) C Q* is of order p.

By imposing some restrictions on the prime divisors of
]G], we can still prove that Q* contains a cyclic normal

subgroup of G. We begin with the following well-known

theorem.

Theorem 3.3
Let G be a supersolvable group of order paqb where

p,q are different primes and p > q. Then, if q does

 

67
not divide p-1, every p—element centralizes every q—element.
P_ro_o£=-
Let P be a Sylow p-Subgroup of G of order pa. It

is normal and contains a cyclic subgroup <d>¢d G of order

p Since G is supersolvable. Note, that if a 1 then
the number of Sylow q-subgroups, 1 + mg, of G can divide
qu only if n = O. The theorem therefore follows in that
case. In general, we can establish the theorem by induc-

. . G
tion on ]G] and conSider Z35"

If x,y are two elements of G where ]x] = pa and

<d>y.

Let H be a Sylow q—subgroup of G containing y and note

]Y] = qC then §§ = 9; where i = <d>x and 9

that <d>H = H<d> is a subgroup of G of order qu. By
induction, it follows that <d>H = H<d> = (d) x H.
From the equation Q; = yx, we have <d>xy = <d>yx, so
-1

that xy = dryx where r is some integer. Hence xyx =

dry. But xyx-1 has same order as y and so it is a q-

element while dry has order pqc. Hence we must have dr= 1
and so xyx.1 = y, i.e. xy = yz. The theorem therefore
follows.

Remark

If we remove the restriction, q does not divide p-l,

than a p—element does not even permute with a q—element in

general. The following example confirms this.

<a,b><x> = <x><a,b> where a3 = 1, b3 = 1,

LEt G
ab = ba, aX = a2, b = ab, x2 = 1. G is supersolvable but

<b> and (x) do not permute with each other.

68

Corollary

 

Let G be a group, Q1 be its quasicenter and p
be the smallest prime divisor of ]Ql]. If for each prime
divisor q less than p of ]G], q does not divide p-l
then Q1 contains a cyclic normal subgroup of G: in fact
G has a nontrivial center.

Let P be the Sylow p-subgroup of Q1 and P C P
where P is a Sylow p-Subgroup of G. P is generated by
QC-elements of G and all QC-elements of order p form a
normal abelian subgroup N of. G. N intersects Z(P)
nontrivially and let 1 # 91-92 °°° gV e N n Z(P), where
gi for i = 1,2, --- v is a QC-element of G of order p.
Suppose x e R where R is a Sylow q-subgroup of G and
q > p. Since (gi><x> = <x><gi> = H is supersolvable
[Theorem 13.3.1, Scott [12]], it follows that <x>efliH.
Also <gi> is a Sylow subgroup of Q(H) and hence <gf><fl
H. Thus H = (gi><x> = <x><gi> = <gi> x <x> and no.
gix = xgi.

Again, if x is any element of a Sylow q-subgroup S
of G and q < p then since H is supersolvable and q
does not divide p—l, it follows that gix = xgi. Thus
91.92.... gv is centralized by G and so <g1-g2-o-o gv>

is a cyclic normal subgroup of G.

Remark
1. Baer showed that the Hypercenter H(G) of a group

G is the intersection of all maximal nilpotent subgroups

69
of G. In Chapter II, we obtained a partial generalization
of this result when we showed that the Hyperquasicenter Q*
is contained in the intersection T of all maximal super—
solvable subgroups of G. However, if the prime divisors of
]G] are such that for any two prime divisors p,q with
p > q, q does not divide p—1 then, in fact, 0* = T =
H(G). Before proving this, we give the following example
which confirms the existence of nonsupersolvable groups with

restrictions on the prime divisors of the order, as described

above.
Let, Zp = Group of integers modulo p, where p is
a prime
Zq = Group of integers modulo q, where q is

a prime.
Choose p > q such that q does not divide p-l. Form
the wreath product G = prrzq. rNote that ]G] = pq-q and
G is not supersolvable by a result of J. R. Durbin [Pro-
ceedings American Mathematical Society, 17(1966), pp. 215-
218] which implies that G is supersolvable if and only
if p E 1 mod q.

For the proof of Q* = T = H(G), we Show that a super-
solvable subgroup of a group G with restrictions on the
prime divisors of ]G] as mentioned before, is also a nil-
potent subgroup. Let M be a supersolvable subgroup of G
and so M = P1-P2 '°° Pn where Pi is a Sylow pi-subgroup
of M and P.P. = PjPi i = 1,2, "‘ n. Nilpotency of

1 J
M is established by induction on n. For n = 2, M is

70
nilpotent by Theorem 3.3 and by induction, the subgroup
S = P1'P2 "° Pn_1 Of M is nilpotent. Hence P1°P2 "'

Pn_1 = P1 x P2 x P3 x --° x Pn-1' Also, since PiPn = PnPi
V i = 1,2, --:- n-l, it follows by Theorem 3.3, that

PiPn = Pi x Pn and this implies M is nilpotent. The in-
tersection of all maximal supersolvable subgroups is there-
fore, the same as the intersection of all maximal nilpotent
subgroups and hence our assertion follows.

2. If G is a group with prime divisors of ]G] re—
stricted as described above, then a normal subgroup N of
G is contained in Q* if for each supersolvable subgroup
S of G, NS is supersolvable and conversely this is true
if N g Q*.

Proof:-

If NS is supersolvable for each supersolvable sub-
group S of G, then by Corollary 4 to Theorem 2.10, N g;T
and T = Q* by the remark above. Conversely, if N g;Q*
then supersolvability of Q*S implies N5 is supersolvable.

We now turn our attention to another class of groups
and Show that the partial generalizations of Baer's results
which we obtained in Chapter II can be completed for these
groups.

It is not in general true that the product of a normal
supersolvable subgroup N and any supersolvable subgroup S
of a group G is supersolvable. Such an example is pro—

vided by A4. The partial generalizations of Baer's results

which we obtained, however, can be completed if a p-element

71
of S permutes with a q—element of N, whenever NS is
supersolvable and p < q. -Note that $3 is such a group
and an example of a nonsupersolvable such group would be

G =53 ><A4.»

Lemma 3.4

Let G be a group with the property that the super-
solvability of a product NS where N,S are supersolvable
subgroups of G and NQ4JG implies for p < q, a p-element
of S permutes with a q-element of N where p,q are dif-
ferent primes. Then the Hyperquasicenter Q* of G con—
tains a cyclic subgroup which is normal in G.

Let p1 be the smallest prime divisor of ‘Ql| and
P1 be the Sylow pl-subgroup of Ql-Pl is contained in 51
where 51 is a Sylow pl-subgroup of G. Let Ei for
i = 1,2, °°° m be a Sylow pi-subgroup of G and m be

the number of distinct prime divisors of iGl. Then G

<51: P2, ... §m>'

The set of all QC-elements of G, each of order p1
generates a characteristic abelian subgroup N of G con-
tained in P1. N H Z(El) is nontrivial and let gl-g2°-°gn
be a nonidentity element in N n Z(El) where gi for
j = 1,2, --- n is a QC-element of G of order p1.

Suppose x is any element in 5k where pk > p1.

Since Ti = (gi><x> = <x><gi> is supersolvable, it follows

that, <x>«<]Ti. Also, (gi> is normal in Ti’ since it is

72
a Sylow subgroup of Q(Ti). Hence gix = xgi and it fol-
lows therefore, that gl-g2 --- 9V is centralized by each
x e Pk where pk > p1.
If pk < p1 then since Nfik is supersolvable (Nfik C

Q*fi and Q*Pk is supersolvable by Theorem 2.10), it fol—

k
lows by the condition on G that x-1<g1-g2 --- gn)x =

<gl-gz --- gn) V x e fik' Since gl-gz --- gn e Z(fil) it
follows therefore, that <g1-g2 °'° gn><fl G.

Theorem 3.5

 

Let G be a group as in Lemma 3.4. Then the Hyper—
quasicenter Q* is the intersection of all maximal super-
solvable subgroups of G.

2222::-

Let T = fl{M|M is a maximal supersolvable subgroup of
G]. We know, that Q*‘g T and to establish the theorem
we first prove that Q* = 1 implies T = 1. Suppose T # 1.

T being supersolvable, the Sylow subgroup P1 of T
corresponding to the largest prime divisor p1 of IT] is
normal in T and hence P1<Q G. P1 is contained in some
Sylow pl-subgroup 51 of G and P1 n Z(il) # 1. Suppose
z is an element different from identity contained in
P 0 2(51) and let fii for i = 1,2, ... m be a Sylow
pi—subgroup of G where pi is a prime divisor of [GI

and m is the number of such distinct prime divisors. Then

G = <51, fiz, °°° §m>' Let pk > p1 and consider Tfik which
being supersolvable, it follows that §k<fl Tfik and Tfik =

T X 5k. If pk < p1 then by the condition on G, we have

73
H = <x><z> = <z><x> V x e 5k, and H being supersolvable,
it follows that x-1<z>x = (2). Hence <z>«d G and so 2
is a QC-element of G which is a contradiction since 0* = 1.
Now, suppose Q* # 1. We proceed to establish the
theorem by induction on |G|. By Lemma 3.4, Q* has a cyclic

subgroup <b> 43G and consider Zg§u Let ZESn-ng' be
supersolvable subgroups of 2%3- and 2%;- be normal in

(ES. ‘We wish to establish that the supersolvability of

 

<§> - <i§ implies for p < q, a p-element § = <b>y of
ng- permutes with a q—element i = <b>x of ngl°
m
Let ]b] = t, a prime and |§| = pm. yp e (b) and
m t m t
so [Y] = p °t- Hence <y> = (y > x <yp > = <y > x (b)-

5
Similarly if |§| = qS then <x> = <xt> X (Xq > :

<xt> x <b>. Now Z3;- 43% implies R46 and £7 2%;

_RS . RS. . '
- Z55” Since Z3; is supersolvable, 1t follows that R5

is supersolvable and also supersolvability of ng-and 2%;-
implies R,S are supersolvable. By the condition on G,

t t t t s m
(x ><y > = (y ><x > and since (xq >, (yp > are normal in
G being both equal to (b), it follows that <x><y> = <y><x>.
If t equals p or q then also, the same arguments ap-

ply and we still have <x><y> = <y><x>. Therefore under

the homomorphism : G —-> Z%§-, we get <§><§> = <§><§>.

74

Induction therefore applies to 2%3- and we have,

Q*(ngq - n {515 is a maximal supersolvable subgroup of ngfl

{Z%§-| M is a maximal supersolvable subgroup of G] 3
(23-. By Theorem 2.3 this implies ZQ§-= ZIS- i e. 0* = T
b I b b I ‘

as desired.

Corollary

 

Let G be a group as in the theorem above. Then a
normal subgroup N of G is contained in 0* if for each
supersolvable subgroup S of G, NS is supersolvable and
conversely this is true if N g 0*.

Proof:-

This follows by exactly the same arguments as in re-
mark 2 following the Corollary to Theorem 3.3.

We are now able to prove a characterization of the
Hyperquasicenter Q* of a group G as described in Lemma
3.4, which relates the supersolvably immersed subgroups of

G to 0*.

Theorem 3.6

 

Let G be a group as described in Lemma 3.4. Then
the Hyperquasicenter Q* of G is a supersolvably immersed
subgroup and contains every supersolvably immersed subgroup
of G.

329.2:-

We know from Chapter II that every supersolvably im-

mersed subgroup of G is contained in Q*. We now proceed

75
to prove that 0* is a supersolvably immersed subgroup of
G by induction on |G|.

For [G] = 1, this is obvious so assume the Hyperquasi-
center is a supersolvably immersed subgroup for all groups
of order less than |G|.

Note, that if 0* # 1 then it contains a cyclic sub-
group <b> G and induction applies to 2&3“ Let 9 be

any homomorphism of G and K be its kernel. If b é K

then <b6>‘d G9 and <b9> C Q*e. Otherwise, suppose

*
<b> C K. By induction, Q*(IZES) = ng- is a supersolv—
*9 2, Q*K

 

 

 

 

. Gr .
ably immersed subgroup of ZBS-and Since Q K , all
*-
we need to show is that QKK contains a cyclic normal sub-
group of E
K'
Under the homomorphism : G >ZgSy 0* has the image
0* . Let ¢ be the homomorphism : G > §_ and note
(b3 2b; K
. . K 0* . Q*K
. d
that its kernel is 233- 233- has the image K an

*

. Q . . G
Since 333- is a supersolvably immersed subgroup of Z333

*
it follows that 9E5 has a cyclic normal subgroup of g»

Hence 0* is a supersolvably immersed subgroup of G and
the assertion of the theorem follows.

The fact that the Hypercenter H(G) of a group G
contains a cyclic normal subgroup of G leads naturally to
the consideration of whether we can obtain a characteriza-
tion of the Hypercenter, analogous to that obtained for the

Hyperquasicenter 0* in Theorem 3.6. For that matter, let

 

76

us recall the definition of the Kern group of G.

The Kern of a group G is the totality of elements
each of which normalizes every subgroup of G, i.e. HX = H,
where H is a subgroup of G. Then the Kern of G is a
subgroup. If x is a Kern element then (x) permutes
with every subgroup of G and hence the Kern of a group
G is contained in the quasicenter of G and is therefore
nilpotent. Observe also that the Kern of G is a charac-
teristic subgroup.
51.31

K be the Kern of -— . Then, in
i i

5V6)

Let K0 = 1 and

the ascending series of characteristic normal subgroups

1 = kO C‘kl C‘kz C ... Ckn = K*, the terminal member K*

is the Hyperkern of G. We will call this series the upper-
kernal series of G and shall use K(S) to denote the Kern
of a group S.

In the following, we show that the Hyperkern and the
Hypercenter of a group coincide and obtain a new characteri-
zation of the Hypercenter. In Chapter II we remarked about
the existence of the largest supersolvably immersed subgroup
L which is of course normal and supersolvable. A question
arises naturally: How much of L is normal and nilpotent
or can we say anything about its intersection with the
Fitting subgroup F? The new characterization of the Hyper—
center which we establish in the following implies that the
Hypercenter H(G) is contained in F n L and that will

be the conclusion of the present chapter. However, in order

to prove that, we need the following theorem and lemma.

77

Theorem 3.7

 

Let G be a group and K* be its Hyperkern. Then,
_ G _ ‘
K* - n {NIN.4G, ME) - 1}.

Proof:-
Let 1 = Kb C'Ki C K2 C °-- C Eh = K* be the upper—
kernal series of G and suppose T = fl[N|K(§O = 1, N 41G}

Since K(%}) = 1, it follows that T ggxfi. Suppose x 6 K1

and x 4 N where K(§J 1. Then, x-lyx = yr, y e G

where r is some integer. Hence Nx_1yx = Nyr, i.e.
_r _ _
x yx = y where x = Nx and y = Ny is any element of

. Therefore, K(§O # 1, a contradiction. Thus KliggN.

We now assume Ki C N and prove that Ki+1 is also con-

tained in N where i lies between 1 and n-1.

K.
Let x = Kix e jfiii— and x 4 N. As before, we have
i
_-1__ _S ._ ' G
x yx = y where y = Kiy is any element of E. and s
i
is some integer. From this it follows, as before, that
-1 s . —-1__ —S _ -
Nx yx = Ny , i.e. x yx = y where now x = Nx and y =
. G G -
Ny is any element of fin Therefore K<NJ + 1, a contra-
diction. Thus Ki+1 g N and so K* g N. Hence K* g T
and we have K* = T.

Lemma 3.8

 

Let G be a group and K* be its Hyperkern. If N is

a normal subgroup of G contained in K* then the Hyperkern

78

G - K*
of N is N

Proof:-

We shall only sketch the lines of proof since it very

W1

much follows the same pattern as in Theorem 2.3. Let N_. C
W m G

2 O O o ' . o
N_.CZ fir- be the upperkernal series of N. As in

Lemma 2.2, it can be easily verified that if SI and 52
are two groups isomorphic to each other then K(Sl) and

K(Sz) are also isomorphic under the same map.

E. .9
Since K EF- - 1 and E—- E'§—- it follows that
Wm Wm Wm
EF' BF"
K(§—J = 1. Hence by Theorem 3.7, K* C‘W . It can be
W - m

W
proved, as in Theorem 2.3, that if Nx e §1- and x 4 K*

then K*X is a kern-element of £0 contradicting the

fact that K(%¥) = 1. Inductively, by assuming Wi C K*,

it can be proved that Wi+1 C K*, following the same lines

of argument as in Theorem 2.3. Thus Wm C K* and so
_ G . K*
Wm - K*. Therefore, the Hyperkern of E' is fi"
Remark

By exactly the same arguments as above, we can show

that if N is a normal subgroup of G contained in the

Hypercenter H(G) of G then the Hypercenter of g- is
(
HNG This, however, helps to establish easily that

H(G) g K*, by induction on |G|.

79
If H(G) = 1, then certainly it is contained in K*.

Otherwise, if z e Z(G) then by induction on ]G], it

follows from G , that H G C K* , i.e. H G) C K*.
223 z 225 —

Corollary

 

The Hyperkern K* and the Hypercenter H(G) of a group
G coincide.
Proof:-

Baer has shown [composite Mathematica, 1(1935)] that

G >

if K(G) # 1 then Z(G) 7! 1. Consider Mm and note

that if K(E%ETJ # 1 then Z<fi%§7) # 1, a contradiction.

Hence K(fi%570 3 1 and so by Theorem 3.8, we have K*‘g
H(G). It therefore follows by the remark above that K* =

H(G).
We now give the following definition which we require
for the establishment of the new characterization of the

Hypercenter.

Definition 3.9

A subgroup M of a group G is called k-supersolvably
immersed if for every homomorphism of G, the homomorphic
image of M contains a cyclic normal subgroup generated

by a kern—element of the image of G.

Remark:
From the definition it follows that a k-supersolvably

immersed subgroup is also a supersolvably immersed subgroup

80
and is therefore normal. The center of a p-group is an
example of a k-supersolvably immersed subgroup. In the
same manner, as in the case of supersolvably immersed sub-
groups, it can be shown that the product of two k-super-
solvably immersed subgroups is also a k—supersolvably im-
mersed subgroup. The product of all k—supersolvably.im-
mersed subgroups is therefore a k-supersolvably immersed
subgroup and the following theorem says that the Hyper- Inf?

center (Hyperkern) is precisely equal to this product.

Theorem 3.10

 

 

The Hyperkern K* of a group G is the largest

I'W

k-supersolvably immersed subgroup.

2.122;:-

Let M be a k—supersolvably immersed subgroup of G.
Observe that M contains a cyclic subgroup <b>‘<1G where
b is a kern-element. We proceed by induction to show that
every k-supersolvably immersed subgroup of G is contained
in the Hyperkern K*. Let ¢ be the homomorphism :

G ¢ _ M . . . M
? ZBSu Then M — Z5;- and by definition 25;- has

a normal cyclic subgroup of 2%3- generated by a kern-ele-

 

G

ment of 3%; and if 9 is any homomorphism of ZgS- then
«we . . ‘ awe

M contains a cyclic normal subgroup of G generated

by a kern-element of G¢.e. By induction therefore,

G K*

|
|
M _ |
Z—Sb CK*(Z—>—b ) ..Zjb . Hence M_C_K*. E

81

We now prove that K* is a k—supersolvably immersed
subgroup. Note that if K* = 1, then the theorem evidently
follows and so we assume K* # 1. To establish the above
assertion we use induction on |G|. For |G| = 1 it is
obvious and so assume it to be true for all groups of order
less than ]G]. Since K* = H(G), the Hyperkern contains
a cyclic normal subgroup of G generated by an element
2 e Z(G). Let 0 be any homomorphism of G and N be

its kernel. Then §_g GO. If N n K* = 1 then <zo>.<]GG

and (20> C K*O. Otherwise let N n K* = T and consider

*
. By Lemma 3.9, K*(%§ = %- and

 

G K*N f; *O’
T' Note that, N K

*-

by induction 5 3 Ta 6 %- generates a cyclic normal sub—

group of g- and Ta is a kern-element of %u Now,

—-1——

a ya

and so it

BIG)

_r _
y where y = Ty is any element of

follows that, Ta_1ya = Tyr, so that NTa-lya NTyr and

therefore Na_1ya - Nyr.

G K*N
N and Na 6 N .

 

Thus Na is a kern-element of

Note that 'a' 4 N as otherwise 'a? will belong to T.
v

also i 1ax = a where i = Tx is any element of g- and

a = Ta. From this equation as before, we obtain, Nx-lax =

NaV and so Na generates a normal subgroup of En Thus

K*N
N

 

contains a cyclic normal subgroup of 3' generated by

a kern-element of fin

82
Hence K* is a k-supersolvably immersed subgroup of

G and therefore the theorem follows.

Corollary

 

If L and F are the largest supersolvably immersed
subgroup and the Fitting subgroup of a group of a group G
respectively then the Hypercenter H(G) is contained in
F n L.

Since H(G) is normal and nilpotent, H(G) ggF. By
the theorem we proved above, it also follows that H(G) ggL.

Hence H(G) g F n L.

Remark
In general, however, H(G) # F n L. Take the case of

’F.

S3 = G and observe that H(G) = 1 while F n L

INDEX OF NOTATIONS

I. Relations:

 

Q Is a subgroup of

C Is a proper subgroup of (C for emphasis)
‘d Is a normal subgroup of

4&3 Is a subnormal subgroup of

e Is an element of

4 is not an element of

IQ

Isomorphic to

II. Operations:

 

GX x_1Gx
x -1
a x ax
%- Factor group
]G] The number of elements in G or order of G.
]x] The order of the element x
(m,n) = 1 m and n are two numbers relatively

prime to each other.

< > Subgroup generated by
X Direct product of groups
0 Intersection
[ } Set whose elements are
[G:M] Index of a subgroup M in G.

83

III.
1
i

p-element

p'-element

84

Groups and Elements:

Identity of a group G

Identity of a factor group %

An element of
of some prime

An element of

G whose order is power
p.

G whose order is not

divisible by prime p.

Largest normal subgroup of G

con—

tained in subgroup A.

The center of a group G.

The Frattini subgroup of G.

The quasicenter of G.

The Hyperquasicenter of G.

The Hypercenter of G.

The kern group of G.

The Hyperkern of G.

The derived subgroup of G.

The symmetric group of degree 3.

The alternating group of degree 4.

For each.

There exists.

BIBLIOGRAPHY

1. R. Baer, Der Kern, eine Charakteristische Untergruppe,
Composito Mathematica, 1(1935), pp. 254-283.

2. , Zentrum Und Kern Von Gruppen mit elementen
unendlicher Ordnung, Composito Mathematica,
2(1935), pp. 247-249.

3. , The Hypercenter of a group I, Acta Mathematica
89(1953), pp. 165-208.

4. , Das HyperZentrum einer Gruppe II, Archiv Der
Mathematik, 4(1953), pp. 86-96.

5. , Das HyperZentrum einer Gruppe III, Mathematische
Zeitschrift, 59(1953), pp. 299-338.

6. , Das HyperZentrum einer Gruppe IV, Archiv Der
Mathematik, 5(1954), pp. 56-59.

7. '-—---, Group elements of prime power index, Transac-
tions of the American Mathematical Society, 75(1953)
pp. 20-47.

8. , Supersolvable Immersion, Canadian Journal of
Mathematics, 11(1959), pp. 353-369.

9. w. E. Deskins, On Qausinormal subgroups of Finite
Groups, Mathematische Zeitschrift, 82(1963), pp.
125-132.

10. J. R. Durbin, Finite supersolvable wreath products,
Proceedings of the American Mathematical Society,
17(1966), pp. 215-218.

11. N. Ito und J. Szep, Uber die Quasinormalteiler von
endlichen Gruppen, Acta Scientiarum, 23(1962),
pp. 168-170.

12. W. R. Scott, Group Theory, (Englewood Cliffs, New Jersey:
Prentic Hall, 1964).

13. M. Suzuki, Structure of a group and the structure of
its lattice of subgroups, (Springer-Verlag, 1956).

14. H. Wielandt, fiber das Produkt paarweise Vertauschbaren

nilpotenter Gruppen, Mathematische Zeitschrift,
55(1951), pp. 1-7.

85