R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
G IORGIO B USETTO
F EDERICO M ENEGAZZO
Groups with soluble factor groups and projectivities
Rendiconti del Seminario Matematico della Università di Padova, tome 73 (1985), p. 249-260
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Groups Groups Projectivities.
GIORGIO BUSETTO - FEDERICO MENEGAZZO
1. Introduction.
In
[1] (Problem 40)
thefollowing question
was asked: if_G is a sol- uble group and n is aprojectivity
from 6 to some groupG (i.e. n
isa lattice
isomorphism
from the lattice.L(G)
ofsubgroups
of G to thelattice
L(G)
ofsubgroups
of(G),
is G soluble as well? In the finite casethe answer was obtained
by
Suzuki([7],
Theorem12)
andZappa ([11]).
The
general
answer wasgiven by
Yakovlev([9]),
y who also gave abound for the derived
length
of G in terms of the one of G(namely
4n3
+
14n2 - 8n if n is the derivedlength
ofG) .
In thepresent
paperwe deal with the
following
moregeneral question:
ifN«G,
issoluble and ~: G - G is a
projectivity,
does some term of the derived series of G lie inside_N(= N~) ~
We provethat,
if n is the derivedlength
of then(the
6n-th term of the derived series ofG)
lies inside N
(see
Theorem4.1).
Our main tool is a detailedstudy
of the
structure,
andparticularly
of the derivedlength,
of finite p-groups of the form G = wherehaving
aprojective image
Gsuch that the
image
of .g is core-free in G(we
shall refer to this situa- tion as to the « reduced case»).
As acorollary
of Theorem 4.1 we areable to
improve
Yakovlev’s bound to 6n-4(see Corollary 4.2).
(*)
Indirizzodegli
AA.: Istituto diAlgebra
e Geometria, Università diPadova, Via Belzoni 7, 35100 Padova.
The authors wish to thank C.N.R. and M.P.I. for the
partial
financial support.2.
Preliminary
results.Our notation will be standard. For lattice theoretical definitions
see
[8].
LEMMA 2.1. Let G be a
finite
modular p-groupof exponent
pr, where p is aprime. If
G is not Hamiltonian and is notcyclic,
thenG contains ac characteristic abelian
subgroup
A such thatG/A
iscyclic
and every
automorphism of
G induces theidentity
onG/A.
PROOF. Assume that G is not abelian.
By
Iwasawa’s theoremon the structure of finite modular p-groups
([8],
ch.~,
Th.18),
y G ===
N~t~
where N is abelian and t induces on N some power 1+ pz,
say, with
pÂ
> 2.By hypothesis
N hasexponent
pr. Set A =Ca(N) ;
A is
abelian, G/A
iscyclic
andCa(A)
= A. We nowdistingush
twocases.
(i)
is notcyclic.
Let a = xti E
A,
where x E N and let cc E Aut G. We shall show that a()(, E A. Since aa E A if andonly
if xa E A. Asbeing isomorphic
to is notcyclic,
there existsof order such that
~u~
r1 = 1.u~
and(ce")
are bothnormal
subgroups
ofG,
thereforethey
commute: it follows that the powerautomorphism
inducedby
xa on N is theidentity.
Thus xa E Aand A is characteristic.
Moreover, ta
induces on Na the power1-~- px
and it induces a power
automorphism
on N as well. These powerautomorphisms
coincide on N r1Na,
which hasexponent (otherwise N/Qr_i(N)
would be aquotient
of thecyclic
groupN/N
r1Na),
y andtherefore ta induces on N the power
1 + p-z.
Thus EA,
asrequired.
(ii) N/SZr_1(N)
iscyclic.
In this case t must have order
Moreover,
since N hasexponent pur
and~t>
n A =C~t~(N),
it follows thatt>
r1 A =(tpr-A)
and A ===
Then,
let xtip" E A of orderpr,
where x E N and IÀ > r - ~,.Since -
XpA
we haveTherefore t
operates
on A as the power 1-f- pz. Hence, recalling
that the group of power
automorphisms
of an abelian group lies in the centre of the wholeautomorphism
group, in order tocomplete
case
(ii)
it is sufficiente to prove that A is characteristic in G. In order to show this we shall prove that A coincides with thesubgroup
Bof G
generated by
thecyclic
normalsubgroups
of G of order pr.Clearly
A
B. Conversely,
letb~
be acyclic
normalsubgroup
of G of order pr.We can write b =
tpvy,
where and v > 0. Assume v = 0. Then G =N~b~
and so,using
the well-known fact that« raising
to thepower is an
endomorphism
ofG,
since is notcyclic
whereas is
cyclic,
it follows that there exists h E N of order pr such thatb~
nh~ _ ~ .
Thus[b, h]
= 1 and this forces b to centralizeN, contradicting
ourassumption
that G is not abelian.Hence and Since we have
and,
moreover,for some
integer
k. It follows that Et>
ny~ =
1. Thereforev ~ r - Â and so =
A,
asrequired.
0In the
following
lemma wecollect,
for the convenience of thereader,
some basic factsoccurring
inprojectivities
of someparticular
finite p-groups
(what
we call the « reducedcase ») .
Similar resultscan be found in
[3].
All these facts are easy consequences of resultsby
I’. Gross([4],
Lemma3.1). Also,
it seems useful to introducea
special
notation. If n: G - G is aprojectivity,
and .bt is asubgroup
of
G,
we will set HHn; and,
if H KG,
we often set HKand
HE
=LEr>rA 2 . 2 . Let G = and
G
befinite
p-groups, where 1 ~ and let n: G ~G
be aprojectivity
such thatH
iscore- f ree
inG.
Thenthe
following
hold:(i) (a)
r1 H = 1 andS2i(G) _ ,S2i(H) for
every(ii)
and areelementary
abelian and there exist bases_
(iii)
1l induces aprojectivity from GjQi(G)
to andcore-free
in(iv)
The(pi-’)th
power determines anendomorphism of Qi(G)
andof D,(0), for
every(V) C Z(G) .
(vi) If
p = 2 and H hasexponent 2r,
thenPROOF. The
arguments
are the same as those in[3],
Lemma 1. C7THEOREM 2.3. Let G and
G
befinite
p-groups, n: G -G
aproject- ivity,
H a normal abeliansubgroup of exponent
pr such that G =H(a)
and
H
iscore-free
inG.
Then(a) Qr(G)
is a modular p-group ;(b) G
is metabelian.PROOF. Let
f eo,
... ,em~, f f o,
... ,fm~
be bases of and ofrespectively,
chosen as in Lemma2.2,
and set~a~~ _ ~a;~.
In ordert o prove
(a)
we show first that(1 )
induces a group of powerautomorphisms
on Hwhere =
12,,«a».
This is obvious if .H iscyclic. Thus,
suppose that 1~ is notcyclic
and write s = min i E N andH/Qi(H)
iscycliel.
U,-,(H)
is a non-trivial normalsubgroup
of G andtherefore, by
Lemma 2.2
(ii),
it contains eland, by (iv),
it coincides with the set Therefore there exists h E H such thathpr-l
- e1. Thenwhere 1
c y, ~ c p -1. Hence, using
Lemma 2.2(ii),
it follows that(h)al1T.-l ==
for some h’ EH. SetQ
= H n iscyclic
of order at mostpr
and r’1Q
=1, by (2).
ThusMoreover,
since is a modular p-group(as
theimage
of the abel-ian p-group .H via the
projectivity na,n-1)
we havefor all In other words
aPh’,
and thereforea-6,
induces a powerautomorphism
onQ.
Then(aP)an
induces a powerautomorphism
on
Qa
and moreover, aO induces the same power onQ
and onQa"
forevery n, i.e. a,6 induces a power
automorphism
onQ,7.
SinceQ
andH/Q
iscyclic,
we have H =Q X h>
and alsothat Q
is core-freein G.
Thus,
sinceQ
hasexponent
ps, wehave QG
= =and
(1)
isproved
if s = r. If s r, we may assume that aa induces the power /o onand, using
Lemma 2.2(iii)
and induction onthe
power À
on Hence where xhtpr-s). = hpr-8!-t and mod. ps. is normalized
by
a,O
(because
and therefore aft acts as the power À onIf x E
(~)y
i.e. x =hvpr-1
for someinteger
v, set Â.’ = À+
as
for À, À’ m p
mod. p-,. Forall y
= whereZ E Q,
we haveyas
=(hiz)ao
= =y-%, namely
aO acts as the power ~,’on .H. On the other
hand, if oe w h~,
then r1 = 1.Also,
aO acts as the
power À
both on and on Inparticolar
= hAX
gives
= = and so x Ex~a,
a contra-diction. This
completes
theproof
of(1). Thus, by
Lemma 2.2(i),
is a modular p-group if
p ~
2. On the otherhand,
if p =2,
is abelian
([2],
Lemma1),
and therefore[H,
~ ,5~~_~(G)
n .H~ _Qr-2«a»)
rl H =~Gr-2(lZ ).
This shows that a~ induces on ,H~ a powercongruent
to 1 mod.4, namely
thatis modular. As far as
(b)
isconcerned,
observe thatQr(G)
is a modularnon-Hamiltonian p-group
(see
e.g. Lemma 2.2(iv)),
andis not
cyclic (since (h, Then,
as a result of Lemma2.1,
contains an abelian
subgroup
A_ normal in G such thatTherefore,
since iscyclic, {b)
follows. C73. The « reduced case ».
For the convenience of the reader we state the
following theorem,
which was
proved by
the second author for p odd([5]),
y andby
thefirst one for p = 2
([3]) .
THEOREM 3.1. Let G and G be
finite
p-groups, n: G - G aprojec- tivity,
.ga G such that G =~H~a~
andH
iscore-free
inG.
T hen(a) if p ~ 2,
then H is abelian(hence
thehypothesis
that H is abe-tian in the statement
of
T heorem 2.3 is unnecessaryi f p =1= 2).
is abelian and is a
metacyclic
modular non-.,g’amiltonian group(*).
Making
a combined use of Theorems 2.3 and 3.1 we obtain THEOREM 3.2..Let G andG
be G -+G
a pro-jectivity,
suck that G =.H~a~
andH
iscore-free
inG.
ThenG
has derived
length
at most 6.PROOF. If p is
odd,
thenG
is metabelianby
Theorems 2.3 and3.1(a). Thus,
suppose that p = 2. LetSuppose
first that r = 0. Then H is ametacyclic
modular non Hamil-tonian group,
by
Theorem3.1(b).
We showthat,
in this case(4)
G has derivedlength
at most 4.Set X =
(H’)$
and suppose that = 28.By [9],
Lemma6,
for all Hence X is the
join
ofcyclic
normal sub- groups of G of order 2S. ThereforeG’,
andconsequently H’,
centraliseX. Moreover X has
exponent IH’l
andthen,
since X ismetacyclic, X/H’
iscyclic.
It follows that X is abelian. Now considern induces a
projectivity
fromO/Qs(G)
to iscore-free in
(Lemma
2.2(iii))
and is abelian.Then, by
Theorem 2.3(b),
(5)
is metabelianand so
(4)
isproved
if s = 0.Suppose s
> 0 and consider the groupX_a~
=Y,
induces aprojectivity
from Y toY, X
is core-free inY
and X is abelian.Therefore, by
Theorem 2.3(a_), QiY)
is a modularp-group.
Also, by
Lemma2.2 (iii), applied
toY,
is non-trivial and core-free in and therefore
is not
cyclic. Moreover,
since =(Lemma
2.2(i))
-_X~ = Xoand,
since iscyclic,
it followsalso that
Q~(G) /Q~( Y)
iscyclic. Then, applying
Lemma 2.1 to(*)
Theproof
of(b)
makes use of anunpublished
result due to the second author. Forcompleteness
reasons we shallgive
aproof
here, in theAppendix.
it follows that is metabelian.
Hence, recalling (5), (4)
follows.Suppose
now that r > induces aprojectivity
fromto
fÏDr(G)/Dr(G)
is core-free in(Lemma
2.2(iii))
and =
I c 4. Therefore, by applying
what we have
just proved
to we see thatG/Qr(G)
has derivedlength
at most 4.Also, by
Theorem 3.1(b),
is abelian.Hence, by applying
Theorem 2.3 toDr(H)(a),
it follows that£Jr(G)
(=
is a modular p-group. Inparticular
it is metabelian.Consequently G
has derivedlength
at most6,
asrequired.
Il4. On the derived series of a
projective image
of a grouphaving
anormal
subgroup
with soluble factor group.THEOREM G be
projectivity,
and suppose tha,t
G/N
is solubleof
derivedlengthn.
T henG(6n) N.
PRooF.
Clearly
we may assume that G isfinitely generated. Using
induction on the
derived length
ofG~N,
we may assume thatcN. G/G’
is afinitely generated
abelian group, therefore- cl~G’/G’ >C ~c2~G’/G’ ...
>C for some suitable elements ci E G such that r’1 G’ is either infinitecyclic
or a p-group. Set.gi = G’,
cx, ... , c;i, ... ,et),
for1 ~ i c i.
If r1_G’ is infinitecyclic, then,
as a result of[10],
Corollario1, HiaG.
If( ~ci~/_~ci_~
r1 is aprime
power, thenIG:Ïlil
is finite([10], Teorema A_)
and is a chain.
Therefore, assuming
thatHi
is notnormal
inG, according
to[6],
Lemmas 2 and3,
thefollowing
twopossibilities
canoccur:
(a)
is ac non-abeliangroup of
order pq where p and q areprime
numbers. Inparticular
is metabelian.(b) G/(Hi)G is a finite
non-abelian p-group.(Hi)a
is normal inG ([2 ])
and n induces aprojectivity f rom
to..B_‘y [7],
Theorem
3,
is also a p-group. Theorem 3.2implies
thathas derived
length
6._ _ _ i _
Now,
in ail cases forall 1 i t. Therefore G(6) n Hi
It follows that as
required.
0 -COROLLARY 4.2. Let G and
G
be groups, n: G ~G
aprojectivity,
and suppose that G is soluble
of
derivedlength
n. ThenG
is solubleof
derivedlength
6n - 4.PROOF.
By
Theorem4.1, G~g~n-~’~ c G~n-1~. Moreover,
since G(n-1) isabelian,
is metabelian([8],
ch.1,
Theorems 17 and18).
Itfollows that
G~g~n-~~+2~ = 1,
asrequired.
05.
Appendix.
As it was mentioned in the statement of Theorem
3.1,
wegive
here a
proof
of thefollowing result,
which was needed in theproof
of Theorem 2.4
(b).
PROposzTroN 5.1. In the
hypotheses of
Theorem 3.1 and with p =2, i f
2 n is theexponent of H,
thenPROOF. We first show
(a). Suppose false,
and choose a count-erexample
G with minimal. Let ... ,em}, {f0,
... ,fm)
be basesof
£2i(G)
andrespectively,
as in Lemma 2.2(ii)
and set ==
~a~.
Then_
Set = and it does not contain el. Thus r1.H~ = 1 and this
implies
that andconsequently
.K arecyclic. Moreover,
since el eo EÏialn-l
r1Z( G) (Lemma 2.2), K =1=
1 and g is normal in G([2] applied
to theprojectivity
na, G -G).
Then the
minimality
ofyields
= 1. Inparticular
Now is a non-trivial normal
subgroup
of G contained inThus
Also, by
Lemma 2.2(iv), Un-l(H)=
- It follows that there exists h E .g of order 2n such that
h2"-’
= ei. Then = 1 and so, since nis
cyclic,
weget
h~
is normalisedby (7),
therefore is normalisedby
as well.
Moreover,
since it isquasi-normal
inG,
is nor-malised
by Hence,
from(8)
Using
thedécomposition ,S2n(G) = (see
Lemma 2.2(i)),
we get
Let
~b~
be anycyclic subgroup
of G of order 2ncontaining e,,>.
Consi-dering that,
ifb~
is normalisedby Q1(G),
y then it is centralisedby Ql(G),
it follows from(9)
and from thedecomposition Qn(G)
=(see
Lemma 2.2(i))
that(11) b~
is not normalisedby
Set
(hi) = ~h~.
is normalisedby Q1(G). Suppose
that>
fo, Il,
... ,Im-1).
Thenla
=t~
mod. and so eitherfm
centralises both and or it induces on and the same
power
1-~--
2n-1. In both casesby
Lemma 2.2(iv).
It follows that is normalisedby Ql(G)
and contains eo ,
contradicting (11).
~ C~ 1 (G) ( ~hl~ ) ,
and so we can find such thatIf we let
~~~
=9,,«a» ::= (say)
in(11),
it follows from the actionof a on that
The 21 elements form a basis of Therefore
Moreover, by
Lemma 2.2(iv)
and( vi ) ,
_(eo)
for all z E H.Hence
a2~)
=za2~)
=(eo)
X Inparticular
and so,
by (12),
yBut then el E
a2))
r1 .,g =~x?~x,a$~, contradicting (14).
Thiscontradiction
completes
theproof
of(a).
_In arder to show
(b)
observethat, by (a),
everyelement y
eQi(G)
induces a power
automorphism
onThen,
from the fact that contains twocyclic subgroups
of order 2nintersecting trivially, using
Lemma 2.2(iv),
it follows that the powerautomorphism
inducedby y
onDn(G)
isuniversal,
and it is either theidentity
or the power 1+
2n-1. Hence2
and so, since£J1(G)
r~r1 we
get
and
~eo , ... , em~
are the usual bases of and!21(G), accord,ing
to Lemma 2.2(ii). Assume, by
way of contradic-tion, that lm
induces the power 1+
2n-l on Set =Then
[lm, ==/0 implies
2£. The 2elements em (0 c 2l -1 )
form a basis of Therefore
Moreover, by
Lemma 2.2(iv)
and(vi), Qi((a2z))
=eo~
for allIn
particular ~a2 z~
r1 .g = 1. Thereforeand
Then
for all zi E H. As we have seen in
proving (a),
thereexists h1
E H suchthat -
Il.
Thusby (16).
Thereforecontradicting (15).
This
completes
theproof
ofProposition
5.1. CIWe
finally point
outthat, using
Lemma2.1,
y we canimprove
aresult in
[3] (Theorem)
as follows:THEOREM 5.2. Let G be a group
projectivity. If
then
HIHU
andHIHU
are soluble groupsof
derivedlength
atmost 3.
(In [3]
the upper bound for the derivedlength
ofH/Ha
was4).
PROOF.
Looking
at theproof
of[3], Theorem,
it isimmediately
seen that the critical case is when G =
.H~~ac~, H
is core-free inG and G,
G are finite2-groups.
In this case,applying
Theorem 3.1(b),
itfollows that there exists r such that is
metacyclic
andis a non-Hamiltonian modular
2-group
such that isnon-cyclic. Then,
as a result of Lemma2.1,
H~3~ =1,
asrequired.
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subgroups
andprojectivities,
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(1971),
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