Some automorphism groups of finite p-groups

0.09 MB·English

by  Yassine Guerboussa

Some automorphism groups of finite p-groups
YassineGuerboussaa,MiloudReguiatb
aUniversityKasdiMerbahOuargla,Ouargla,Algeria
Email: yassine [email protected]
bUniversityKasdiMerbahOuargla,Ouargla,Algeria
Email: Miloud [email protected]
3
1
0
2
Abstract
n
a AconjectureofBerkovichassertsthateverynon-simplefinite p-grouphasanon-innerautomor-
J
phismoforder p. Thisconjectureisfarfrombeingproveddespitethegreateffortdevotedtoit.
1
Inthispeperweproveitfor p-groupsofcoclass2,providedthat pisodd. Somerelatedresults

arealsoproved,andmaybeconsideredasinterestingindependently.
]
R
Keywords: automorphisms,finite p-groups,radicalrings,adjointgroups
G
.
h
t
a 1. Introduction
m
[ Automorphismsof finite p-groupsare quite mysterious, and manyoldproblemsin this area

are still open. Maybe the mostattractiveare the divisibility conjecturestating that|G| divides
1
v |Aut(G)|, if |G| is greaterthan p2. And the questionorthe conjectureofBerkovichstating that
5 everynon-simplefinite p-grouphasanon-innerautomorphismoforder p.
8 Thisconjectureisconfirmedforvariousclassesof p-groups,regular p-groupsbySchmid[12],
0
see also Jamali [8]for p = 2, powerful p-groupsby Abdollahi[2], and specially for p-groups
0
ofclassnotexceeding3byLiebeck[10],Abdollahi[1]andAbdollahietal[3]. Notably,Dea-
.
1
conescuandSilberberg[4]havereducedittothecasewhenthe p-groupGisstronglyfrattinan,
0
thatisC (Z(Φ(G)))=Φ(G).
3 G
IfGisap-groupoforderpnandclassc,thenthecoclassofGisequaltor =n−c. Classifyingp-
1
: groupsbyconsideringthecoclassastheprimaryinvariant,hasprovedagreatsuccess. Werefer
v
thereadertoC.Leedham-Green[9],forthedetails. Atthistime,It’snotclearhowtostudying
i
X the conjectureof Berkovichin the frameworkof coclasstheory. And this conjectureis almost
r studiedforgroupsofsmallclassorsomeclosefamilies,asforexampleconsideringG/Z(G)to
a
bepowerfulasdonebyA.Abdollahiin[2].
It’s natural to consider this question for p-groups in the opposite side, namely for those with
small coclass. it’s easy to see thatis true in the maximalclass case. So the nextcandidateare
p-groupsofalmostmaximalclass,orthoseofcoclass2. Insection5,ofthispaperweprovethat
theconjectureistrueforthese p-groupsprovidedthat pisodd.
Thereadermaynotconsiderthepreviousasthemainresultinthepaper.Insection3,andundera
suitablecondition,weprovearesultaboutthestructureofsomeautomorphismgroups.Namely,
aboutthegroupofautomorphismscentralizingG/H,whereH denotesthefullinverseimagein
G of Ω (ζ (G)/ζ (G)). These automorphismsmay be consideredas the farthestremovedfrom
1 2 1
centralautomorphisms(see[6]).
PreprintsubmittedtoarXiv December11,2013
Before,andinsection2,wehavetolistsomepreliminaryresultswhichweshallneedinthefol-
lowingsections. ThemainideaistoconsidertheadditivegroupDer(G,A)ofderivationsfrom
GintoAasring,whenAisanormalabeliansubgroupofG,viewedasaG-moduleviaconjuga-
tion. Andalsoconsideringtheadjointgroupofthisringasasuitablegroupofautomorphisms.
This allows us to capture some of the propertiesof this automorphismgroup by means of the
propertiesoftheassociatedring.
Anaturalsequenceofringhomomorphisms
0→Der(G,B)→Der(G,A)→Der(G/B,A/B)
arisewhenBisanormalsubgroupofG,suchthatB≤ AandBδ ⊂ Bforeveryδ∈ Der(G,A).
Insection4,weshallcaracterizethe p-groups,podd,withanexactsequenceoftheobovetype,
undertheassumptionthatAiselementaryabelianofrank2.
Throughoutthispaper,theunexplanednatationisstandardintheleterature. Wedenotebyζ(G)
i
andγ(G),thetermsoftheupperandthelowercentralseriesofthegroupG,respectively. And
i
d(G) will denote the minimal number of generators of G. Following A. Mann, we say that a
p-groupG is p-centralifeveryelementoforder p(oforder4,if p = 2)inG iscentral. If N is
anormalsubgroupofG,wedenotebyAut (G)thegroupofautomorphismsσofG havingthe
N
propertyx−1σ(x)∈N,forallx∈G.
If R is a ring, thenwe donoteby R+ it’s addativegroup, andby R◦ it’s adjointgroup. The nth
powerofanelementxinthegroupR◦ isdenotedbyx(n).
2. Preliminaries
LetRbea(associative)ring,notnecessarilywithanidentityelement. Thesetofallelements
ofRformsamonoidwithidentityelement0∈Runderthecirclecompositionx◦y= x+y+xy,
this monoidiscalled the adjointmonoidorsemi-groupof theringR. TheadjointgroupR◦ of
the ringR is thegroupofinvertibleelementsinthe previousmonoid. FollowingJacobson, we
saythatRisradicalifhisadjointsemi-groupisagroup,orequivalentlyR◦ =R.
The nth power Rn of R is the additive subgroup generated by the set of all products of n
elementsofR. WesaythatRisnilpotentifRn+1 = 0forsomenon-negativeintegern,theleast
nsatisfyingRn+1 =0iscalledthedegreeofnilpotencyoftheringR. Noticethateverynilpotent
ringRisradicalsinceforeveryx∈Rwehave
x◦ (−1)ixi =( (−1)ixi)◦x=0.
Xi Xi
TheJacobsonradicaloftheringRisthelargestidealofRcontainedintheadjointgroupR◦. This
implythatRisradicalifandonlyifitcoincideswithitsJacobsonradical. Byaclassicalresult
the Jacobson radical of an artinian ring is nilpotent, hence every artinian (in particular finite)
radicalringisnilpotent.
Lemma 2.1. The adjointgroup of a nilpotentring R is nilpotentof class at most equalto the
degreeofnilpotencyofR.
Following[6],wesaythataringRisright(left,resp) p-nilifeveryelement x ∈ Rsatisfying
px = 0isaright(left,resp)annihilatorofR. TheringRissaidtobe p-nilifitisrightandleft
p-nil.
2
Forexample,foreveryringRthesubringS = pRis p-nil.Alsowecheckeasilythattheleftand
therightannihilatorsofΩ (R+)arerespectivelyrightandleft p-nil.
1
Thefollowingthreeresultsaretakenfrom[6].
Lemma2.2. LetRbearingwithanadditivegroupoffiniteexponentpm.IfRisleftorrightp-nil
thenRisnilpotentofdegreeatmostequaltom. InparticulartheadjointgroupR◦ isnilpotent
ofclassatmostequaltom.
Lemma2.3. LetRbea p-ring,podd.IfRisleftorrightp-nil,thenΩ (R◦)=Ω (R),forevery
{n} n
n≥1. InparticularwehaveΩ (R◦)=Ω (R◦).
n {n}
Now, letG be a groupand A be a normalabelian subgroupofG. Then A can be viewedas
aG-moduleviaconjugation. LetdenotebyDer(G,A)theadditivegroupofderivationsfromG
into A, thatarethemappingδ : G → Ahavingthepropertyδ(xy) = δ(x)yδ(y),forall x,y ∈ G.
Thisadditivegroupendowedwiththemultiplicationδ δ (x)=δ (δ (x))hasastructureofaring.
1 2 2 1
LetAut (G)bethegroupoftheautomorphismsσofGhavingthepropertyx−1σ(x)∈ A,forall
A
x∈G,thenwehavethefollowingrelation.
Proposition2.4. LetGbeagroupandAbeanabeliannormalsubgroupofG.Thenthemapping
σ 7→ δ ,withδ (x) = x−1σ(x)determinesanisomorphismbetweenthegroupAut (G)andthe
σ σ A
adjointgroupoftheringDer(G,A).
LetBbeanormalsubgroupofG,suchthatB≤ AandBδ ⊂ Bforeveryδ∈ Der(G,A). Then
wehaveacanonicalembeddingoftheringDer(G,B)inDer(G,A).
AlsoA/BisaG/B-moduleundertheusualactionandthemapδ7→δ˜definedfromDer(G,A)to
Der(G/B,A/B)byδ˜(xB)=δ(x)Bisringhomomorphism.
Lemma2.5. Undertheabovenotation,wehavethefollowingexactsequenceofringhomomor-
phisms
0→Der(G,B)→Der(G,A)→Der(G/B,A/B)
Proof. straightforward.
Wealsoneedthefollowingeasyobservations.
Lemma2.6. IfGisafinite p-group,theneverymaximalsubgroupMofGoccursasthekernel
ofahomomorphismr:G →Z .
p
Proof. Letbeg ∈ G−M. ThenG =< g > M andsoeveryelementofG canbewritteninthe
formgim,withm∈ M. Themapr(gim)=imod p,definesahomomorphismfromGtoZ with
p
akernelequaltoM.
Lemma2.7. IfGisapurelynon-abelianfinite p-group,thenΩ (Z(G))≤Φ(G).
1
Proof. Letz∈Ω (Z(G)). IfthereexistsamaximalsubgroupMsuchthatz< M,thenG =<z>
1
×M. ThusGisnotpurelynon-abelian.
3
3. Almostcentralautomorphisms
Let H be the fullinverse imageinG of Ω (ζ (G)/ζ (G)). The subgroup H arise naturalyin
1 2 1
thestudyofcentralautomorphismsofG,sincetheinnercentralautomorphismsoforder p,are
inducedexactelybyH.
It’s wellknownthatd(G)d(ζ(G)) , d(H/ζ (G))implythatG hasa non-innercentralautomor-
1
phismoforder p. Infactthisconditionisalsonecessaryprovidedp>2(see[6]).The p-groups
ofmaximalclassarestraightforwardexamplessatisfyingthiscondition. Notethat p-central p-
groupsalsosatisfythisinequality.ThisfollowseasilyfromatheoremofThompson(see[7],III,
12.2,Hilfssatz),andfromatheoremofA.Mann[11]fo p=2.
Here, we provestructuralresults aboutthe group Aut (G), undera suitable condition, namely
H
forfinite p-groupsG satisfyingC (Φ(G)) ≤ Φ(G). Itseemsthatthisconditionplaysthesame
G
roleinstudyingAut (G),asζ (G)≤Φ(G)forcentralautomorphisms(see[6]).
H 1
Theorem3.1. LetG beafinite pgroupsuchthatC (Φ(G)) ≤ Φ(G). Let H bethefullinverse
G
imageinGofΩ (inn(G))andletdenoteH =Ω(H). Then
1 i i
1. thegroupAut (G)isnilpotentofclassatmostmin{r,s}+1. Where pr =exp(G/γ (G))and
H 2
ps =exp(ζ(G)).
2. Aut (G)isnilpotentofclassatmost2.
H1
3. If p>2,thenAut (G)hasaregularpowerstructureinthesens
H
Ω(Aut (G))=Ω (Aut (G))= Aut (G).
i H {i} H Hi
Beforeprovingthistheorem,weneedthefollowinglemmas.
Lemma 3.2. The subgroup H lies in C (Φ(G)). Moreover, under the conditionC (Φ(G)) ≤
G G
Φ(G),thesubgroupHisabelian.
Proof. Everyelement h ∈ H defines a homomorphism x 7→ [x,h] fromG to its center. Since
[x,h]p = [x,hp] = 1, the image of the previous homomorphism is elementary abelian. Thus
G/C (h)iselementaryabelian,andsoΦ(G)≤C (h). Thereforeh∈C (Φ(G)). IfC (Φ(G))≤
G G G G
Φ(G),thenH ≤Z(Φ(G)).HenceHisabelian.
In the remaining part of this section D and D denote respectively the rings Der(G,H) and
1
Der(G,H ).
1
Lemma3.3. UndertheconditionC (Φ(G))≤Φ(G),wehave
G
1. D2 ⊂Hom(G,ζ(G)).
2. D 3 =0.
1
3. The factor ring D/D is right p-nil, and has exponent at most pmin{r,s}. Where pr =
1
exp(G/γ (G))and ps =exp(ζ(G)).
2
Proof. 1. Let take A = H and B = ζ(G), in lemma 2.5. Since H/ζ(G) lies in the center of
G/ζ(G), we have Der(G/ζ(G),H/ζ(G)) = Hom(G/ζ(G),H/ζ(G)). And since H/ζ(G) is
elementary abelian, H/ζ(G) ≤ Φ(G)/ζ(G) is contained in the kernel of every homomor-
phismin Hom(G/ζ(G),H/ζ(G)). ThisimplythatHom(G/ζ(G),H/ζ(G))isanullring. By
lemma2.5,theringD/Hom(G,ζ(G))canbeembeddedinHom(G/ζ(G),H/ζ(G)). Andso
D2 ⊂Hom(G,ζ(G)).
4
2. Let take A = H and B = Ω (ζ(G)) = Z . We have H /Z is central in G/Z and has
1 1 1 1 1 1
exponent p. A similarargumentto thatin(1)implythat D 2 ⊂ Hom(G,Z ). Sinceevery
1 1
homomorphisminHom(G,Z )containsΦ(G)initskernel,Hom(G,Z )isarightannihilator
1 1
ofD . ThusD 3 =0.
1 1
3. Letδ+D ∈ Ω (D/D ). Then p2δ = 0. Andsince pδ(x) = δ(x)p ∈ ζ(G),forallx ∈G,we
1 1 1
havepδ∈Hom(G,Z ).Thuspδ′δ=0forallδ′ ∈ D,sinceHom(G,Z )isarightannihilator
1 1
of the ring D. This imply that δ′δ ∈ D , and so the ring D/D is right p-nil. On the
1 1
otherhand,thehomomorphismofgroupD → Hom(G,ζ(G))whichmapsδto pδ,induces
a monomorphism D/D → Hom(G,ζ(G)). Thus exp(D/D ) ≤ exp(Hom(G,ζ(G))) =
1 1
pmin{r,s}. Theresultfollows.
Proofoftheorem3.1. 1. Since D/D is a right p-nil ring, by lemma 2.2, Dmin{r,s}+1 ⊂ D .
1 1
And by lemma 3.3, 1., we have Dmin{r,s}+1 ⊂ Hom(G,ζ(G)). It follows that Dmin{r,s}+1 ⊂
Hom(G,Z ). SinceHom(G,Z )isarightannihilatoroftheringD,wehaveDmin{r,s}+2 = 0.
1 1
HenceDhasnilpotencydegreeatmostmin{r,s}+1,andso Aut (G) (cid:27) D◦ hasnilpotency
H
classatmostmin{r,s}+1.
2. SinceAut (G)(cid:27) D ◦,itfollowsimmediatelyfromlemma3.3,2. thatAut (G)hasnilpo-
H1 1 H1
tencyclassatmost2.
3. We begin by showing that D/Ω (D+) is a right p-nil ring, for n ≥ 1. If δ + Ω (D+) ∈
n n
Ω (D/Ω (D+)), then pδ ∈ Ω (D+). Hence pnδ ∈ D ∩ Hom(G,ζ(G)), and so pnδ ∈
1 n n 1
Hom(G,Z ). Itfollowsthat pnδ′δ=0,forallδ′ ∈ D. Thusδ′δ∈Ω (D+).
1 n
Now,weclaimthatΩ (D◦)=Ω (D+),foreveryn≥1. Forn=1wehaveδ∈Ω (D+),or
{n} n 1
equivalentlyδ∈ D ,implythatδ3 =0. Thus
1
p p
δ(p) = δi = pδ+ δ2 =0
i! 2!
Xi≥1
hence δ ∈ Ω (D◦). Conversely, if δ(p) = 0, then in particular δ(p) = 0modD . Since
{1} 1
D/D isright p-nil,bylemma2.3wehave pδ= 0modD . Henceδ2 ∈ D ,andbylemma
1 1 1
3.3, 1. δ2 ∈ Hom(G,ζ(G)). It follows that δ2 ∈ Hom(G,Z ), which imply δ3 = 0. Thus
1
0=δ(p) = pδ+ p δ2 = pδ,andsoδ∈Ω (D+).
2 1
Now we procee(cid:16)d(cid:17)by induction on n. We have δ ∈ Ω (D+) imply δ + Ω (D+) ∈
n n−1
Ω (D/Ω (D+)). Since D/Ω (D+) is a rightp-nilring, (δ+Ω (D+))(p) = Ω (D+).
1 n−1 n−1 n−1 n−1
Henceδ(p) ∈Ω (D+),andbyinductionδ(p) ∈Ω (D◦). Thusδ(pn) =0. Thisshowsthat
n−1 {n−1}
Ω (D+)⊂Ω (D◦). Theinverseinclusionfollowssimilarly.
n {n}
Finally,sinceΩ (D+)isanidealintheringD,itfollowsthatΩ (D◦)isasubgroupofD◦.
n {n}
ThusΩ (D◦)=Ω (D◦). AndsinceΩ (D+)=Der(G,H )◦ = Aut (G),theresultfollows.
n {n} n n Hn
4. p-groupswithanexactsequence
Throughoutthissection,unlessotherwisestated,Gdenotesafinitep-group,wherepisanodd
prime. And AdenotesanormalsubgroupofG, whichiselementaryabelianofrank2,andnot
5
central.
wewishtocaracterizetheclassofthegroupsGsuchthatthefollowingsequenceisexact.
0→Der(G,Z )→Der(G,Z )→Der(G/Z ,A/Z )→0
1 1 1 1
WhereZ denotestheintersectionofAwiththecenter.
1
Noticethattheabovesequencecanbewrittenas
0→Hom(G,Z )→Der(G,Z )→Hom(G/Z ,A/Z )→0
1 1 1 1
andbylemma2.5wehaveonlytofocusonitsexactenessinHom(G/Z ,A/Z )
1 1
Toworkinageneralframework,it’sconvenienttoconsiderthefollowingnotion.
Definition 4.1. LetC be a maximal subgroup ofG. We say thatG is full with respect to C if
everymaximalsubgroupM ,CcontainsasubgroupK suchthat:
1. Kisnotnormalandhasindex p2inG
2. K∩CisanormalsubgroupofGwhichcontainsthesubgroupGp.
Letusnotesomeeasyfactsrelatedtotheabovedefinition.
1. Clearly,ifGisfullwithrespecttooneofitsmaximalsubgroups,thenGcannotbepowerful.
2. InorderforGtobefullwithrespecttoamaximalsubgroupC it’snecessaryandsufficient
thatG/γ (G)GpisfullwithrespecttoC/γ (G)Gp.
3 3
3. IfGisfullwithrespecttoC,thenGisfullwithrespecttoCσforeveryσ∈ Aut(G).
Theclassofp-groupsdefinedaboveisverylargeasshowsthefollowingproposition.
Proposition 4.2. If G is 2-generated and not powerful, then G is full with respect to all its
maximalsubgroups.
Proof. LetCbeamaximalsubgroupofG.
Let N = γ (G)Gp, and let denote by G the sectionG/N. It follows immediately thatG has
3 1 1
exponentpandclassatmost2,andsoΦ(G )≤Z(G ),thatisΦ(G)/N ≤Z(G/N).
1 1
If G is abelian, then γ (G) ≤ γ (G)Gp. This imply that γ (G) ≤ Gp, which means thatG is
1 2 3 2
powerful,acontradiction.ThusGhasclassexactely2.
Nowlet M , C beamaximalsubgroupofG. If M = M/N iscyclic,then|M | = p, sinceG
1 1 1
hasexponent p. Thisimlpythat|G | = p2,thusG isabelian,acontradiction. Therefore M is
1 1 1
notcyclic,andsoitcontainsatleast p+1maximalsubgroup.LetK/N bemaximalsubgroupin
M/N whichisdistinctfromΦ(G)/N.
i) ItfollowsimmediatelythatK ≤ M hasindex p2. Andif K isnormalinG, thenγ (G) ≤ K.
2
AndsinceGp ≤ K weobtainΦ(G)= K,acontradiction.ThusK isnotnormalinG.
ii) We have K∩C ≤ M∩C = Φ(G),hence(K∩C)/N ≤ Φ(G)/N ≤ Z(G/N). Thus K ∩C is
normalinG.
Theresultfollows.
Nowwestatethemaintheoremofthissection.
6
Theorem 4.3. Assume thatG is purely non-abelian,C is the centralizer of A in G, and Z =
1
A∩Z(G). Thenthesequence
0→Hom(G,Z )→Der(G,Z )→Hom(G/Z ,A/Z )→0
1 1 1 1
isexactifandonlyifGisfullwithrespecttoC.
We needthefollowingthreelemmasbeforeembarkingontheproof. Throughoutthese,G is
assumedtobefullwithrespecttoC =C (A). AndKwilldenoteasubgroupofafixedmaximal
G
subgroupM ,C,whichsatisfytheconditions1. and2. ofdefinition4.1.
Lemma4.4. Lety∈C−Mandx ∈ M−K. Thenthemapk 7→α(k)definedfromK intoZ by
p
therelation
[k,y]= xα(k) modK
isagrouphomomorphism,whosekernelisequaltoK∩C.
Proof. Clearly,K , K∩C andK/K∩C (cid:27) KC/C hasorder p. ItfollowsthatK∩C hasindex
p3inG,andsoγ (G)≤K∩C.
3
Letk,k′ ∈ K. Then
[kk′,y]=[k,y][k,y,k′][k′,y]
hence
[kk′,y]=[k,y][k′,y]modK
thatis
xα(kk′) = xα(k)xα(k′)modK.
Thisshowsthatαisahomomorphism.
If α is null, then < y >≤ N (K). And since K is normalizedby M, this imply that K ⊳G, a
G
contradiction. We claimthat K ∩C ≤ ker(α), whichimplythat K ∩C = ker(α). Thisfollows
immediatelyfromthefactK∩CisnormalinG.
Lemma4.5. Foreveryu∈A−Z ,thereexistsz∈Z suchthat[k,u]=zα(k) forallk∈ K. Where
1 1
α:K →Z isthehomomorphismdefinedinlemma3.4,andZ = A∩Z(G).
p 1
Proof. Since |A| = p2, we have A ≤ ζ (G). Therefore [k,u] ∈ A∩Z(G) = Z forallk ∈ K.
2 1
Itfollowsthattherelationk 7→ [k,u]isa homomorphismfrom K intoZ , andwhosekernelis
1
K∩C.
Ifwe fix anon-trivialelementz ofZ , we obtaina homomorphismβ : K → Z with[k,u] =
0 1 p
z β(k).
0
NowwehavetwohomomorphismsαandβdefinedfromKontoZ andwhosekernelsareequal
p
toK∩C. WecanidentifythemtotheirinducedhomomorphismsinHom(K/(K∩C),Z ),which
p
isisomorphictoZ . Hencethereexistsapositiveintegerisuchthatβ = iα,andsowehavefor
p
z=zi,
0
[k,u]=zβ(k) =zα(k) forallk∈K
0
Theresultfollows.
Thefollowinglemmamaybeconsideredasthekeytoprovetheorem4.3.
7
Lemma 4.6. Let be y ∈ C − M and x ∈ Φ(G)− K. Then for every u ∈ C −Z , there exists
1
z ∈ Z suchthatδ :G → Adefinedbyδ(kxjyi) = zjui isaderivation. Wherek ∈ K andi, jare
1
non-negativeintegers.
Proof. Letzbeasdefinedinlemma4.5.
First we show that δ is a well defined map. Clearly, everyelementg ∈ G can be written as
g=kxjyi,wherek ∈ K andi, j∈N. Ifg=kxjyi =k′xj′yi′,thenyi =yi′ modM. Thus pdivides
i−i′,thatisi−i′ = plforsomeintegerl. Wehavekxjypl =k′xj′. Sinceypl ∈Gp ≤ K,wehave
xj = xj′ modK,thus pdivides j− j′. Since Aiselementaryabelian,wehaveui−i′ = zj−j′ = 1,
andsozjui =zj′ui′.
Nowweprovethatδisaderivation.Letg=kxjyiandg′ =k′xj′yi′. Wehave
gg′ = kxj[y−i,x−j′k′−1]k′xj′yi+i′
= kxj[y−i,k′−1][y−i,x−j′]k′−1k′xj′yi+i′
sinceK∩CismaximalinK,K∩Chasindexp2inC. AndsinceK∩CisnormalinG,wehave
γ (C)≤ K∩C.
2
Thusk =[y−i,x−j′]k′−1 ∈K∩C.
1
Also,wehave
0
[y−i,k′−1] = [y−1,k′−1]y−r
rY=i−1
0
= [y−1,k′−1][y−1,k′−1,y−r]
rY=i−1
= [y−1,k′−1]ik
2
forsomek ∈ K
2
Andwehave
[y−1,k′−1]i = ([y,k′−1]y−1)−i =([k′−1,y]i)y−1
= [k′−1,y]i[[k′−1,y]i,y−1]
= x−iα(k′)k
3
forsomek ∈ K. Itfollowsthat
3
gg′ = kxjx−α(k′)k k k k′xj′yi+i′
3 2 1
= k xj+j′−iα(k′)yi+i′
4
forsomek ∈ K. Andthusδ(gg′)=zj+j′−iα(k′)ui+i′
4
Ontheotherhand,
δ(g)g′δ(g′)=(zjui)k′xj′yi′zj′ui′
usingthefactthatxj′yi′ ∈C (A)=C,weobtain
G
δ(g)g′δ(g′) = (zjui)k′zj′ui′
= zj(uk′)izj′ui′
= zjui[u,k′]izj′ui′
8
bylemma3.5,[u,k′]i =z−iα(k′),hence
δ(g)g′δ(g′)=zj+j′−iα(k′)ui+i′.
Theresultfollows.
Proofoftheorem4.3. FirstwehavetoshowthatifGisfullwithrespecttoC =C (A),thenthe
G
followingsequenceisexact.
Der(G,A)→Hom(G/Z ,A/Z )→0
1 1
Let f ∈ Hom(G/Z ,A/Z ). If f = 0, thenitcanbeliftedtothe trivialderivationin Der(G,A).
1 1
Henceweshallassumethat f , 0. Since|A/Z | = p,thekernelof f isamaximalsubgroupin
1
G/Z ,andsohavetheformM/Z ,whereMisamaximalsubgroupinG. wehavetoexaminethe
1 1
twocases:
1. M ,C
Let y ∈ C − M, and let u ∈ A−Z such that uZ = f(yZ ). Let K be a subgroup of M
1 1 1
satisfyingtheconditions(1)and(2)ofdefinition4.1,andletx ∈φ(G)−K. Bylemma4.6,
thereexistsz∈Z suchthatδ:G →Adefinedbyδ(kxjyi)=zjuiisaderivation,andclearly
1
f isinducedbythisderivation.
2. M =C. Lett∈G−C,andletu∈ A−Z suchthatuZ = f(tZ ).Itfollowsfromtheidentity
1 1 1
(xu)p = xpup[x,u](2p)forallx∈G
that
u1+t+...+tp−1 =1
It’sstraightforwardnow,tocheckthat
δ(tim)=u1+t+...+ti−1
isderivationfromGintoA,whichinduces f.
Conversely,assumethatthepreceedingsequenceisexact. Let M , C bea maximalsubgroup
in G. By lemma 2.7, M/Z is maximal in G/Z . And by lemma 2.6, M/Z is the kernel of a
1 1 1
homomorphismr:G/Z →Z . Ifu∈ A−Z ,wecandefineahomomorphism f :G/Z →A/Z
1 p 1 1 1
with
f(xZ1)=(uZ1)r(xZ1)
This homomorphism has kernel M/Z , and can be lifted, by our assumption, to a derivation
1
δ:G →A. ConsiderK =kerδ={x∈G,δ(x)=1}.
WehaveK ≤ M,otherwiseG = MKandthusG/Z ≤ker(f),acontradiction.ThusKcoincides
1
withthekerneloftherestrictionδ/M : M → A. Clearly,δmapsMtoZ ,andsoδ/M : M →Z
1 1
isahomomorphismofgroup.ThisimplythatK hasindexatmost p2inG.
IfK isnormalinGthenγ (G)≤ K. Lety∈C−Msuchthatδ(y)=u,withu∈A−Z . Thenfor
2 1
everym∈ Mwehave
δ(my)=δ(m)yδ(y)=δ(m)u.
9
Ontheotherhandwehave
δ(my)=δ(ym[m,y])=δ(ym)[m,y]δ([m,y])=δ(ym)=umδ(m)
Thisimplythatum = u,forallm ∈ M, hence M ≤ C (u) = C, acontradiction. Thus K isnot
G
normalinG,andhasindexexactely p2inG.
It’sstraightforwardtochekthatδ(kg)=1,forallk∈ K∩Candg∈G. ThisimplythatK∩C⊳G.
AndfinallywehaveGp ≤ K,sinceforeveryx∈Gwehave
p−1 p−1
δ(xp) = δ(x)xi = δ(x)[δ(x),xi]
Yr=o Yr=o
= δ(x)p[δ(x),x](2p) =1.
Thetheoremfollows.
Corollary4.7. LetGbeap-groupwhichisfullwithrespecttoallitsmaximalsubgroupsandhas
cycliccenter. ThenforeveryelementaryabeliannormalsubgroupAofG suchthatA ≤ ζ (G),
2
thesequence
0→Hom(G,Z )→Der(G,A)→Hom(G/Z ,A/Z )→0
1 1 1
isexact. WhereZ = A∩ζ(G).
1
Proof. Letbe f ∈ Hom(G/Z ,A/Z ), and (v¯ ,...,v¯ ) a basis forthe Z -vectorspace A/Z . For
1 1 1 s p 1
eachi∈1,s;letA bethesubgroupofAdefinedbyA/Z (cid:27)<v¯ >. ItfollowsimediatelythatA
i i 1 i i
isanelementaryabeliannormalsubgroupofGofrank2,and
Hom(G/Z ,A/Z )(cid:27) Hom(G/Z ,A/Z )
1 1 1 i 1
Mi
Thus f canbewrittenas f = f,where f ∈Hom(G/Z ,A/Z ). Theabovetheoremimplythat
i i i 1 i 1
each fi can be lifted to a deriPvationδi : G → Ai ⊂ A. It followsthat δ = iδi : G → A is a
derivation,whichinduces f. P
5. Non-innerautomorphismsof p-groupsofcoclass2
TheresultsdeveloppedintheprevioussectionsallowustoprovetheconjectureofBerkovich
for p-groupsofalmostmaximalclass.
Theorem5.1. IfGisafinitep-groupofcoclass2, podd,thenGhasanon-innerautomorphism
oforder p.
Proof. LetG beafinite p-groupofcoclass2. Clearaly,wehavetoassumethatd(G)d(ζ(G)) =
d(ζ (G)/ζ (G)).SinceGhascoclass2,ζ (G)/ζ (G)hasatmostorderp2.Sod(ζ (G)/ζ (G))≤2.
2 1 2 1 2 1
Thiscombinedwiththeaboveequalityimplythat|ζ (G)/ζ (G)|= p2,|ζ (G)|= pandd(G)=2.
2 1 1
SoGis2-generatedandhascycliccenter,andbythemainresultin[4],wehavetoassumethat
C (Z(Φ(G)))=Φ(G),andsoC (Φ(G))≤Φ(G).
G G
10