R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
A VINOAM M ANN
Extreme elements of finite p-groups
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Extreme Elements of Finite p- Groups.
AVINOAM MANN
(*)
In this paper we consider
only
finite p-groups. Such a group is said to haveequal centralizers,
if all non central elements have cen-tralizers of the same size. It is said to have
independent centralizers,
if of any two
centralizers,
which are distinct and different from the whole group, none contains the other. These two classes(of
whichthe second
obviously
contains thefirst)
were first consideredby
N. Ito
[I].
In[M1]
it wasshown, sharpening
a result of[I],
that if Ghas
independent centralizers,
then either G has an Abelian maximalsubgroup,
orGIZ(G)
is ofexponent
p. Another subclass of the groups withindependent centralizers, namely
groups with Abelian centralizers(each
centralizer different from G isAbelian),
wasinvestigated by
D. M. Rocke
[R]. Recently,
the aforementioned result of[Ml]
wasreproved by
L. Verardi for groups withequal
centralizers[V] (this
author seems to be unaware of
[M1]), using
a differentmethod,
whichemployes
a certaininequality
for centralizers. InProposition
1below,
we
generalize
thisinequality
to all p-groups, and then consider thecase of
equality.
This turns out to be a verystrong assumption:
ingroups with
equal
centralizers itimplies
that thenilpotency
class istwo,
and ingeneral
such a group ismetabelian, usually
of class atmost p, the central factor group has Abelian
centralizers,
etc.(Pro- position
2 and Theorem3).
(*)
Indirizzo dell’A.:Department
of Mathematics, HebrewUniversity
Jerusalem
(Israele).
This paper was written while the author was
visiting
theDepartment
ofMathematics of the
University
of Trento. The author wishes to thank thisDepartment,
and inparticular
C.Scoppola,
for their warmhospitality.
The paper
[V]
contains also a result of H.Heineken, showing
thatelements
centralizing
G’ in groups withequal
centralizers mustbelong
to the second centre.
Proceeding by contradiction,
thisproof shows,
on the way, that if one of the relevant elements is not in
Z2(G),
thenits centralizer in the central factor group is
just
theimage
of theoriginal
centralizer. It turns out that theargument showing
thisremarkable
property
can also begeneralized
to all p-groups, for anappropriate type
of an element the extreme elements of the title which are defined as elements whose centralizers have a certain maxi-mality property.
We open the paperby defining
extreme elementsand
developing
theirproperties.
We can. e.g.,generalize
Heineken’sresult, by showing,
that in groups withindependent
centralizersC(G’)
is often contained in the
p-th
centre. Extreme elements are also ap-plied
in theproofs
of the results mentioned in theprevious paragraph.
An
interesting
feature of the situation described in Theorem 3is,
that one is led to consider two naturalgeneralizations
of earlierconcepts, namely
groups in which the centralizers have class at most2,
and groups in which the centralizers have at most two distinct sizes.
We do not pursue this line here.
The notation is
mostly
standard. Weusually
writeC(x)
forCa(x).
When we say that an element has a
centralizer,
we meanthat the element is not
central,
and the centralizer is maximal relative to othercentralizers,
notnecessarily
among allsubgroups.
The breadthb(x)
of an element x is definedby = pb(x),
and we denoteZ(x)
=1. - DEFINITION. An element x of a
p-group G
is called extreme ifa) Z(G).
b) If y
=[x, tl, ... , tk]
is not central(for
some elementstl, ... , tk),
then =
Non central elements of
Z2(G)
aretrivially
extreme. As anotherexample,
let x be a non central element with a maximal centralizer among the elements ofC(G’).
ThenC(x) a G,
so alsoZ(x) v G.
There-fore all
commutators y
as inb)
are inZ(x),
soC(y) D C(x),
and themaximality
ofCCx)
shows thatC(y)
=C(x),
and x is extreme. Inthe same way we see that if x is an element whose centralizer is both maximal and
normal,
then x isextreme,
but we shall see below thatthe extra
generality
of thisexample
isonly apparent,
and theprevious
47
example
is rathertypical.
If G hasindependent centralizers,
all ele-ments of
C(G’)
are of thistype.
If x is
extreme,
and x E H çG,
then x is either central or extreme in .H.Also,
anelement y
as inb )
of the definition is extreme.An extreme
element,
which does notbelong
toZ2(G),
will be calleda proper extreme element.
If G = it is easy to see that a
(non central)
element x == (0153l, x~)
is extreme if anonly
if it eitherbelongs
toZ2(G),
or one ofthe
components
isextreme,
and the other one is central. Inparti- cular,
y if say, contains proper extremeelements, and G2
is notAbelian,
then the extreme elements of G(together
withZ(G))
do notform a
subgroup.
Pl.
If
x isextreme,
theC(x)
is a normalsubgroup.
PROOF. Let t E
G,
and consider xt =x[x, t].
If[x, t]
EZG),
thencertainly C(xt)
=C(x).
If[x, t] 0 Z(G),
then conditionb)
of the defi- nitionimplies
thatC(xt),
and henceC(x)
=C(xt).
In eithercase,
C(X)l
=C(xl)
=C(X).
P2.
I f x is extreme,
there exists an element z EZ2(G),
such that xand z have the same centralizer. In
particular,
x EC(G’).
PROOF. Let x E
Zi(G),
with i minimal. If i >2,
there exists anelement t,
suchthat y
=[x, t] g~ Z(G).
Then0(y)
=C(x), and y
EE
Zi-,,(G),
so induction on iyields
our claim.COROLLARY.
I f G
hasindependent centralizers,
the extreme elementsare
exactly
the non central element.of C(G’).
Property
P2justifies
ourprevious claim,
that the secondexample given
after the definition of extreme elements istypical.
In thatexample
we considered elements ofC(G’)
with a maximalcentralizer,
and it is easy to see that the
maximality
isequivalent
to all elements ofZ(x)-Z(G) having
the same centralizerC(x),
so P2 shows that the definition of extreme elementsjust generalizes slightly
thisexample.
P3.
If
x is properextreme,
thenPROOF. Choose an element t such
that y
=[x, t]
is notcentral,
and assume that
P4.
If
x is properextreme,
thenxZ(G)
is extreme inG jZ(G).
PROOF. Part
a)
of the definition isassumed,
andb)
follows fromP3, applied
for both x and y.P5.
I f x is extreme,
andx 0 Zi(G),
then x centralizesZi(G).
PROOF.
Apply
P3 and P4 severaltimes,
until we reachP6.
If
x is a proper extremeelement,
thenZ(G) ~ pbcx>.
PROOF. As in the
proof
ofP2,
we may assume that x EZ3(G).
By P3,
x hasexactly conjugates (mod Z(G)~ .
Theseconjugates
constitute a
subgroup
ofZ2(G)¡Z(G),
and thissubgroup
is containedin the normal
subgroup Z(x),
and does not contain xitself,
hence ourassertion.
Now if
even x 0 Z3(G),
we can assume x EZ4(G),
and use a similarargument, working (mod Z2(G)),
toget
nZ3(G) : Z(x)
(1etc. Thus we obtain
P7. Let x be a proper extreme
element,
such thatx 0 Zi+I(G).
ThenOn the other hand we have
P8. Let x be as in
P7,
and let y not commute with x. Thenb(y)
>PROOF.
First,
let x EZ3(G),
so i = 1. Then we consider the com-mutators of x. All of these have the same centralizer as x, so
they
do not commute with y. Thus the
subgroup
that these elements form(mod Z(G)~
actsregularly
on theconjugacy
classof y,
andgives
distinct
conjugates
of y. All of theseconjugates
arecongruent to y
(mod Z(G)), but yx
isnot, by
P3.Thus y
has more thanconjugates.
For the
general
case, weget, by
induction inthat y
has morethan
p(1-1)b(m) conjugates,
which are distinct(mod Z(G)),
while each of these hasconjugates congruent
to it(mod Z(G)), by
the pre- viousargument.
COROLLARY. In a group with
equal centralizers,
all extreme elementsbelong
toZ2(G),
and thusC(G’)
=Z2(G).
(This
is the result of Heineken referred to in theintroduction).
49
P9. A proper extreme element commutes with all elements whose breadth is not more than its own. In
particular,
a proper extreme element commutes with all extreme elements.Indeed,
of any two proper extremeelements,
one has a breadthat most as
large
as theother’s,
sothey commute,
andthey
commutealso with elements of
Z2(G), by
P5.COROLLARY. Let G have
independent
centralizers.If C(G’) =1= Z2(G),
then
C(G’)
is Abelian.Indeed,
in that caseC(G’)
isgenerated by
its elements outsideZ2(G),
which are all proper extreme(Corollary
toP2),
and so com-mute with each other.
P10. Let x be an extreme element
satisfying x 0 Zk(G), for
somek,
and let
PROOF. If k =
1,
this is obvious. Let k > 1. Theninduction, using
P3 andP4,
shows that y =[x, t1,
... ,t,,-,] 0 Z(G),
soC(y)
=C(x)
and
[y, tk]
=1= 1.P11. In a group G
of exponent
p, all extreme elements lie in PROOF. Let x beextreme,
and let y not commute with x.Suppose
that x E
Zc(G) - ZC-I(G).
Let H =~x, y). Letting
allti
beequal
to y inPI 0 ,
we see that cl H = c, and that if r c, then the commutatorz =
[x,
y, ... ,y],
with r - 1y’s,
lies inHr
but not inHr+t.
But forr = p, we know that z lies in
Hl’+l [H, 111.9.7].
Thus c p.COROLLARY.
If
G hasindependent centralizers,
and does not containan Abelian maximal
subgroup,
thenZ~(G).
This follows from P11
by combining
the facts that in such a group all elements ofC(G’)
areextreme,
and thatG/Z
hasexponent
p.2. - We now pass to the
inequality
mentioned in the introduction.PROPOSITION 1. Let x be a non central element
of
the p-groupG,
such that
C(x)
ismaximal,
and lety 0 C(x).
ThenIZ(x): Z(G) ~
Equality
holdsif
andonly if
G =Z(x) C(y).
PROOF. The
maximality
ofC(x) implies that,
if t EZ(x) - Z(G),
then
C(t)
=C(x),
and soZ(x)
mC(y)
=Z(G).
ThusjZ(x):Z(G)/
=with
equality
if andonly
if G =We consider the case of
equality
in thisProposition, starting
withgroups with
equal centralizers,
for which the result isparticularly simple.
PROPOSITION 2. Let G be a group with
equal centralizers,
and sup- pose thatfor
some non central elements x and y.Then G has class 2.
PROOF. Of course, the
equality
holds for all elements y, becauseb(y)
is constant. Inparticular,
let y be any element outsideC(x).
Then G =
Z(x) C(y),
so allconjugates
of y areconjugates
underZ(x).
It follows
that,
like y, none of theseconjugates
lies inC(x).
Thusthe
complement
ofC(x)
is a union ofconjugacy
classes. But then sois
C(x),
i.e.C(x) a G. By
Heineken’s result(Corollary
toP8),
x EZ2(G), and,
since all elements ofZ(x)
have the samecentralizer, Z(x) 9 Z2(G).
We then have
[y, G] = [y, [y, Z2(G)] ç Z(G),
i.e. y EZ2(G).
Since the elements of
type
y, i.e. elements outsideC(x), generate G,
we have G =
Z2(G).
THEOREM 3. Let x be an element with a maximal centralizer
of
thep-group
G,
let y have maximal breadth among all el,ements notcommuting
with x, and suppose that
(Z(x) : Z(G) (
= Forsimplicity,
assume alsothat cl G > 2 . T hen
a)
G ismetabelian;
indeedZ(x)
is au normal Abeliansubgroup,
with an Abelian
factor
group.b) If
thencIGp,
andG/Z(G)
hasexponent
p.c) If t 0 Z2(G),
then either . Ineither case, el
C(t)
2.d) G¡Z(G)
is a group with Abeliancentralizers,
in which the noncentral elements have at most two distinct breadths.
PROOF. Denote b =
b(y). By assumption,
x commutes with allelements of breadth
greater
thanb,
andby Proposition 1,
it com-mutes with all those of smaller breadth. Thus each element t outside
51
C(x)
has breadthb,
and for each such element G =Z(x) C(t).
Theargument
ofProposition
2 shows thatC(x)
isnormal,
hence so isZ(x),
and we have
[t, G] _ [t, Z(z)] C Z(x).
Since the elements t outsideC(x) generate G,
we see that is Abelian. This provesa).
Note that as
C(x)
ismaximal,
thenormality
ofC(x)
shows that x is extreme. IfZ(x) ç Z,(G),
we have above[t, Z(z)] C Z(G),
and clG c 2.
Since we have excluded this case,
Z(x)
is not contained in~3(6~
andwe may as well assume
x 0 Z2(G),
so x is a proper extreme element.Let .K =
C(x)
r1C(y),
then and K centralizesZ(x),
so.g a G.
Next,
and thus K C
Z,(G).
The inclusion is proper, becauseZ,(G)
also inter- sectsZ(x)
outsideZ(G).
AsC(x), by P5,
we haveC(x} _
Z2(G) === Z(x)),
andLet
again t 0 C(x).
Then we have seen that G -Z(x)C(t)
andZ(x)
r1C(t)
=Z(G).
ButC(t)
nZ(x),
so clC(t)
2. Now lett E
C(x) - Z2(G).
Write t = uz, with u EZ(x), z
E ..g.Then u 0 Z2(G),
and so u is a proper extreme element.
Suppose
that[s, t]
_-
[s, u] [s, z]
EZ(G),
for some s, then[s, u]
EZ(G),
so P3 shows thats E
C(x).
Inparticular, C(t) C C(x).
The last
argument
showsalso,
that in H =GIZ(G),
thesubgroup
C =
C(x)JZ(G) (which
isAbelian)
is the centralizer of each of its noncentral elements. Let ac =
tZ(G)
C. Then thisimplies
thatCC(a)
=Z(H).
Since G =C(t) C(x),
we see thatCH(a)
=Z(H). CG(t)IZ(G)
is Abelian. This proves d. Now
[R,
3.13 and3.16]
shows that ifI G: C(x)
> p, then cl+
1 and expGjZ2(G)
= p, but these claimscan be
improved.
We startby employing
a variation of theargument
of
[Ml]
to show that expGIZ(G)
= p(similar
variations appear in[M2]
and [M3 ]) .
To this
end,
considerZ(y).
If u EZ(y)
r1K,
then u centralizes bothC(y)
andZ(x),
and thus u EZ(G),
soZ(y)
r1C(x) C Z(y)
r1 K ==
Z(G). Moreover,
we have alsoZ(t)
r1C(x)
=Z(G)
forany t 0 C(x),
as we have remarked that such a t behaves
exactly
like y. This in turnshows,
that if s EZ(t) - Z(G),
thenC(s)
=C(t)
andZ(s)
=Z(t).
Therefore for any two
elements t, s
outsideC(x),
eitherZ(t)
=Z(s)
or
Z(t)
r1Z(s)
=Z(G).
Thus thesubgroup C(x), together
with thevarious
subgroups Z(t), partition
g =GIZ(G).
Now suppose
exp H
> p.By [B, 5.6],
one of thecomponents
ofthe
partition
contains all elements of H of ordergreater
than p, aswell as
Z(g).
Thiscomponent
is thenC(x)/Z(G).
But in the meta- belian groupH,
all elements of ordergreater than p generate
asubgroup
of index at mostp [G],
so if we haveexp
G/Z(G)
= p.Let cl G = c. We have seen that G’ _
[Z(x), G],
and thereforeZ(x)
is not contained in and we may as well assume thatx 0
Then P10 shows that[x,
y, ...,y] =1= 1,
where y is re-peated
c - 1times,
so cl(s, y)
= c. But in the metabelian groupH,
of
exponent
p, theimages
of xand y generate
asubgroup
of class atmost p - 1,
and thus clx,y>p.
This ends theproof.
COROLLARY 1. Under the
assumptions
and notationsof
Theorem3,
the
subgroup
K is the coreof C(y),
andZ(K)
=Z(G).
PROOF. Since
C(x)
=Z(s)K,
we haveZ(K) C Z(C(x)),
soZ(K) =
= Z(G). Next,
if L is a normalsubgroup
contained inC(y),
then P3shows that L C
C(x),
so L c K.COROLLARY 2. A group such as in Theorem 3 has
independent
cen-tralizers
if
andonly if C(x)
isAbelian,
which isequivalent
to K =Z(G).
In this case all centralizers are
of
class2,
and haveexactly
two distinctindices.
Conversely, i f
all centralizers areof
class2,
then the centralizersare
independent.
PROOF. If the centralizers are
independent,
thenC(x)
is Abelianby
thecorollary
to P9.Conversely,
ifC(x)
=Z(x)
isAbelian,
it isthe centralizer of each of its non central
elements,
andthis, together
with Theorem 3
c),
shows that the centralizers areindependent,
ofclass 2,
and of indicespb (z)
or which aredistinct, by
P8. Corol-lary
1 shows thatC(x)
is Abelian if andonly
if .~ =Z(G).
Now suppose that the centralizers are not
independent,
and let ube a non central element of g. Then C =
C(u)
contains bothZ(x)
and y, and therefore C’ D
[Z(x), y]
=[G, y].
If C is of class2,
thenC’C
C(y),
so[G, y] C C(y)
t1Z(x) and y
EZ2(G),
which does not hold.We note
also,
that thedefining property
of G in Theorem 3 is sharedby G/Z
andG/g,
the last grouphaving
also Abeliancentralizers,
as K n G’ C K t1
Z(x)
=Z(G),
and so centralizers(mod .g)
and(mod Z(G))
coincide.Finally,
wegive
the conditions that the abstract groupsZ(x)
andC(y)
have tosatisfy,
in order for a group G of the relevanttype
to53
exist. We do not
attempt
the more difficultproblem
ofdescribing
«
concretely»
all suchpairs,
y and thus «lassifying » completely
ourgroups. The
following
may rather bethought
of as the firststep
insuch a classification.
For this
end,
let us denote A =Z(x),
B =C(y), Z
= N =Z(G),
where we use Z if we consider
Z(G)
as asubgroup
ofA,
and N if weconsider it as a
subgroup
ofB,
and let K be as above. Then the fol-lowing properties
hold:I. A is an Abelian group with a
subgroup
Z.II. B is a gloup with
subgroups N ç K,
andZ(B).
III. There is an
isomorphism
between Z and N.IV. B acts on
A,
with kernelK,
in such a way, that for each b we haveCA(b)
=CA(B)
= Z.V. =
N,
but N is a propersubgroup
ofZ(B).
(Condition V)
holds inG,
because y is inZ(B)).
The group G is obtained from A and Bby forming
first thesplit
extension of Aby B,
and then
amalgamating Z
and Naccording
to g~.Given any two groups A and B
satisfying I)
toIV),
y we can forman «
amalgamated split
extension» G as above. Thegiven
conditionsimply
thatC(x)
=AK,
whileZ(x)
= A follows fromZ(K)
= N. Thesecond half of TT
gives
us anelement y
such thatC(y)
= B. It remainsto control the size of the centralizers. Let then t E
G,
withThese commutators are, of course, calculated inside the group G.
Note that VI
implies
that B has class 2. Condition VIshows,
thatgiven
any element b ofB,
we can find an element a inA,
such that t commutes with ab.By IV, b
determines auniquely (mod Z),
and so~ C(t) ~
=I C(y) 1.
Thus allassumptions
of Theorem 3 areguaranteed.
A more
elegant
condition than VI would be the same condition statedonly
for elements ofB,
because this can be formulatedusing only
the groupsA, B, Z, N,
and theisomorphism
99. It is not clear to us if this weaker version of VIimplies
the full one.To
conclude,
let us draw attention to twointeresting special
cases:when K =
N,
which is the case ofindependent
centralizers(Co-
rollary 2),
and when N is a direct factor of B : the groupsG/g
andGlZi(G) (for i
>1)
are of thistype.
REFERENCES
[B]
R. BAER, Partitionen endlicherGruppen,
Math. Z., 75(1901),
pp. 333-372.[G]
J. A. GALLIAN, TheHughes problem
andgeneralizations,
Proc. Conf.Finite
Groups (Park City, 1975),
pp. 521-528, New York, 1976.[H]
B. HUPPERT, EndlicheGruppen -
I, Berlin, 1967.[I]
N. ITO, Onfinite
groups withgiven conjugate
types - I,Nagoya
Math. J., 6 (1953), pp. 17-28.[M1]
A. MANN,Conjugacy
classes infinite
groups, Israel J. Math., 31(1978),
pp. 78-84.
[M2]
A. MANN,Groups
with small Abeliansubgroups,
Arch. Math., 50 (1988), pp. 210-213.[M3]
A. MANN, Somefinite
groups withlarge conjugacy
classes, inpreparation.
[R]
D. M. ROCKE, p-groups with Abelian centralizers, Proc. London Math.Soc., (3) 30 (1975), pp. 55-75.
[V]
L.VERARDI,
On groups whose non central elements have the samefinite
number
of conjugates,
Boll. U.M.I.,(7)
2-A (1988), pp. 391-400.Manoscritto pervenuto in redazione il 26