The free ommutative automorphi
2-generated loop of nilpoteny lass 3
Dylene Agda Souza de Barros,
Alexander Grishkov, Petr
Vojtehovsk
y
Abstrat.Aloop isautomorphiifallitsinnermappings areautomorphisms.
Weonstrutthefreeommutativeautomorphi2-generatedloopofnilpoteny
lass3. Ithasdimension8overtheintegers.
Keywords:free ommutative automorphi loop, automorphi loop, assoiator
alulus
Classiation: 20N05
1. Introdution
AloopisanonemptysetQwithabinaryoperationsuhthatforeverya2Q
the left and right translations L
a , R
a
: Q ! Q, bL
a
= ab, bR
a
= ba are
bijetionsofQ,andthereisanidentityelement12Qsatisfying1a=a1=a
for all a 2 Q. We will also write the multipliation asjuxtaposition, and we
willusetoindiate thepriorityofmultipliations. Forinstane, a(bd)means
a((b)d).
ThemultipliationgroupofQisthepermutationgroupMlt(Q)=hL
a
;R
a
; a2
Qi generated by all left and right translations. The stabilizer of 1 in Mlt(Q)
is the inner mapping group Inn(Q). It is well known, f. [3℄, that Inn (Q) =
hL
a;b
;R
a;b
; T
a
; a;b 2 Qi, where L
a;b
= L
a L
b L
1
ba , R
a;b
= R
a R
b R
1
ab , T
a
=
R
a L
1
a .
AloopQ isautomorphi ifInn(Q)Aut (Q),that is,ifeveryinner mapping
ofQisanautomorphismofQ. Notethatgroupsareautomorphiloops.
Automorphiloopswererststudiedin[4℄,whereitwasproved,amongother
results, that automorphi loops form a variety and are power-assoiative, that
is, every element generates a group. It was shown in [9℄ that automorphi
loops have the antiautomorphi inverse property (ab) 1
= b 1
a 1
. In parti-
ular,ommutativeautomorphiloopshavetheautomorphiinverse property, or
Dylene Agda Souza de Barros supported by FAPESP - proess number 2010/16112-4.
Alexander Grishkov supported by FAPESP and CNPq (Brazil). The researh stay of Dy-
leneAgdaSouza deBarrosandAlexanderGrishkovattheUniversityofDenverwaspartially
AIP,(ab) =a b . Foranintrodutionto thestruturaltheoryandthehis-
toryofautomorphiloops,see [11℄. Foranintrodutiontothestruturaltheory
ofommutativeautomorphiloops,see[6℄,[7℄.
Thispaperisonernedwithfreeobjetsinthevarietyof ommutativeauto-
morphiloops,inpartiularwiththefreeommutativeautomorphi2-generated
loopofnilpotenylass3.
Theenter ofaloopQistheassoiativesubloopZ(Q)=fa2Q; a'=afor
every'2Inn (Q)g. ThusZ(Q)onsistsofallelementsa2Qthat ommuteand
assoiatewithallotherelementsofQ. DeneZ
0
(Q)=1,Z
1
(Q)=Z(Q),andfor
i1letZ
i+1
(Q)bethepreimageofZ(Q=Z
i
(Q))undertheanonialprojetion
Q!Q=Z
i
(Q). ThenaloopQisnilpotent oflass nifZ
n 1
(Q)6=Q=Z
n (Q).
Itwasshownindependently in[5℄and[8℄that foranoddprimepeveryom-
mutativeautomorphi loopoforderp k
isnilpotent. By[8℄,aommutativeauto-
morphiloopof orderp 2
is aommutativegroup,butthere exist nonassoiative
ommutativeautomorphi loopsof order p 3
|these were onstruted in [7℄ and
lassieduptoisomorphismin[2℄.
Oneofthemaintoolsusedinthelassiation[2℄wasthedesriptionofthefree
ommutativeautomorphi2-generatedloopofnilpotenylass2. Thispaperan
thereforebeseenasanaturalontinuationoftheprogrambegunin[2℄. Arelated
projetis [10℄, where heavyassoiatoraluluswasused todetermine thebases
andordersoffreeommutativeMoufangloopswithupto sevengenerators.
Forn2,letF
n
(x;y)bethefreeommutativeautomorphiloopofnilpoteny
lassnonfreegeneratorsx,y.
Forelementsa,b,ofaloopQdenoteby(a;b;)theassoiator ofa,b,,that
is,theuniqueelementsatisfyingtheequationab=(ab)(a;b;).
WeobtainedthefollowingdesriptionofF
2
(x;y)in [2℄:
Theorem1.1([2,Theorem2.3℄). LetF
2
(x;y)bethefreeommutativeautomor-
philoopofnilpotenylass2withfreegeneratorsx,y,andletu
1
=(x;x;y),u
2
=
(x;y;y). Theneveryelementof F
2
(x;y)anbewrittenuniquelyasx a
1
y a
2
u a3
1 u
a4
2
forsomea
1
;a
2
;a
3
;a
4
2Z,andthemultipliationinF
2
(x;y)isgivenby
(x a1
y a2
u a3
1 u
a4
2 )(x
b1
y b2
u b3
1 u
b4
2 )=x
a1+b1
y a2+b2
u
a3+b3 a1b1(a2+b2)
1
u
a4+b4+a2b2(a1+b1)
2
:
Aswearegoingtosee,to desribeF
3
(x;y)isonsiderablymorediÆult.
Letusallanassoiatorompounded ifitisoftheform (a;b;)whereatleast
oneofa,b,isagainanassoiator(u;v;w). Itiseasytosee, f.Proposition2.1,
that aommutativeloop isof nilpoteny lassat most3ifand only ifall om-
pounded assoiatorsareentral.
Ultimately we provein Theorem 5.4 that everyelement of F
3
(x;y) is of the
anonialform
(x a
1
y a
2
u a3
u a4
)v a5
v a6
v a7
v a8
;
where
u
1
=(x;x;y); u
2
=(x;y;y); v
1
=(x;x;u
1 );
v
2
=(x;x;u
2
); v
3
=(y;y;u
1
); v
4
=(y;y;u
2 );
and where themultipliation formula is asin Lemma 5.3. (The anonial form
anbeparsed unequivoally beausethe ompounded assoiatorsv
1 , v
2 , v
3 , v
4
areentral.) Thisisaomplishedin aseriesofsteps:
InSetion2westudysymmetriesand linearpropertiesoftheassoiatormap
(; ; )inommutativeautomorphiloopsofnilpotenylass3. Weonludethat
inF
3
(x;y)itsuÆestolookatompoundedassoiatorsoftheform(a;b;(;d;e))
where eah a, b, , d, e is either x or y. InSetion 3 we study powers within
assoiators and derive a formula for (a i
;b j
; k
). In Setion 4 we disover sev-
eral nontrivialrelationsamong ompounded assoiatorsof F
3
(x;y),reduingall
ompoundedassoiatorstojust v
1 ,v
2 ,v
3 ,v
4 .
ThemultipliationformulaforF
3
(x;y)isderivedinLemma5.3. Aritialstep
inprovingthemainresult,Theorem5.4, onsistsofshowingthatthemultiplia-
tionformulaofLemma 5.3atuallyyieldsanautomorphiloop. Thisfollowsby
straightforwardalulation(onemerelyneedstoverifythatthegeneratorsL
a;b of
theinnermappinggroupareautomorphisms),butthealulationisextremelyte-
diousanderror-proneandwehavethereforedeidedtodelegateittoaomputer.
TheMathematia[12℄odethataomplishesthealulationanbedownloaded
fromthewebsiteofthethird-namedauthor,www.math.du.edu/~petr. Onewe
knowthattheformulaofLemma 5.3yieldsanautomorphiloopQ,itiseasyto
showthatF
3
(x;y)isfreeandisomorphitoQ.
Reallthattheassoiator subloop A(Q)ofQistheleastnormalsubloopofQ
ontainingallassoiators(soQ=A(Q)isagroup). TheleftnuleusN
(Q),middle
nuleus N
(Q) andright nuleus N
(Q) onsist ofall elementsa2Qsuh that
(a;b;) =1, (b;a;) =1, (b;;a)=1 foreveryb, 2Q, respetively. Thenthe
nuleus N(Q) isdened byN(Q)=N
(Q)\N
(Q)\N
(Q). Weonludethe
paperbyalulatingtheassoiatorsubloop,nuleiandtheenterofQ=F
3 (x;y).
Remark 1.2. Inthebeginningofthispapertheproofsweoergiveallthedetails,
butlateronwegraduallyrelymoreandmoreonthereadertoprovideintermediate
stepsinalulations. Allsuhstepsanbeobtainedinastraightforwardfashion,
albeitsometimeswithonsiderabletimeommitment. Moredetailswillbefound
inthedissertation[1℄ oftherst-namedauthor.
2. Symmetryand linearity in assoiators
Reall that the assoiator in any loop Q is well-dened modulo Z(Q), that
is, (a;b;) = (az
1
;bz
2
;z
3
) for any a, b, 2 Q and z
1
;z
2
;z
3
2 Z(Q). In any
ommutativelooptheidentity
holdsbeauseaba=aab=aba. Itiswellknownthatinanyommutativeloop
ofnilpotenylass2wehave(a;b;)=(;b;a) 1
,(a;b;)(b;;a)(;a;b)=1. We
willuse allthese observationsand thefollowingwell-knownpropositionwithout
referene.
Proposition 2.1. LetQbeaommutativeloop.
(i) Qhasnilpotenylassatmost2ifandonlyifallassoiatorsareentral.
(ii) Qhasnilpotenylassatmost3ifandonlyifallompoundedassoiators
((a;b;);d;e),(a;(b;;d);e),(a;b;(;d;e))areentral.
Proof: Suppose that Q has nilpoteny lass at most 2. Then Q=Z(Q) is an
abelian group. Sine A(Q) is the least normal subloop S suh that Q=S is an
abeliangroup,itfollowsthatA(Q)Z(Q). Theonverseisprovedbyreversing
theargument.
Let us write a for aZ(Q) 2 Q=Z(Q). Suppose that Q has nilpoteny lass
at most 3. Then Q=Z(Q) has nilpoteny lass at most 2 and thus (a;b;) 2
Z(Q=Z(Q)) for every a;b; 2 Q by (i). This is equivalent to ((a ;b;);d ;e) =
(d ;(a;b;);e)=(d;e;(a;b;))=1
Q=Z(Q)
andthusto((a;b;);d;e),(d;(a;b;);e),
(d;e;(a;b;))2Z(Q). Theonverseis againproved by reversingtheargument.
We proeed to show that in a ommutative automorphi loop of nilpoteny
lass2theassoiatorislinearin alloordinates.
Lemma 2.2. Let Qbealoopand leta;b;2 Qbesuh that (a;b;) 2Z(Q).
ThenL
b;a
=(a;b;) 1
,aR
b;
=a(a;b;), andbL
a R
L
1
a R
1
=b(a;b;).
Proof: Sine(a;b;)isentral,wehaveab(a;b;) 1
=(ab)(a;b;) 1
=ab,
orL
b L
a
=((a;b;) 1
)L
ab ,orL
b;a
=(a;b;) 1
. Also,ab=a(a;b;)b,or
aR
b R
=(a(a;b;))R
b ,oraR
b;
=a(a;b;). Finally,ab=a(b(a;b;))yields
thelastequality.
Proposition 2.3. Let Q be an automorphi loop of nilpoteny lass 2. Then
(ab;;d) =(a;;d)(b;;d),(a;b;d) =(a;b;d)(a;;d), (a;b;d)=(a;b;)(a;b;d)
foreverya;b;;d2Q.
Proof: Sine Q is automorphi, the inner mapping R
;d
is an automorphism.
By Lemma 2.2, ab(ab;;d) = (ab)R
;d
= aR
;d bR
;d
= a(a;;d)b(b;;d) =
ab(a;;d)(b;;d)and(ab;;d)=(a;;d)(b;;d)follows. Consequently,(a;b;d)=
(d;b;a) 1
=((;b;a)(d;b;a)) 1
=(;b;a) 1
(d;b;a) 1
=(a;b;)(a;b;d). Finally,
1L
a R
d L
1
a R
1
d
=1showsthatL
a R
d L
1
a R
1
d
isalsoaninnermapping,soLemma
2.2 implies b(a;b;d) =(b)L
a R
d L
1
a R
1
d
= bL
a R
d L
1
a R
1
d L
a R
d L
1
a R
1
d
=
b(a;b;d)(a;;d)=b(a;b;d)(a;;d),andwearedoneuponanelingb.
HereisaloalversionofProposition2.3:
Lemma 2.4. Let Q be an automorphi loop of nilpoteny lass 3, and let
(i) If(a;;d),(b;;d)2Z(Q)then(ab;;d)=(a;;d)(b;;d).
(ii) If(a;b;d),(a;;d)2Z(Q)then(a;b;d)=(a;b;d)(a;;d).
(iii) If(a;b;), (a;b;d)2Z(Q) then(a;b;d)=(a;b;)(a;b;d).
Proof: Let us prove (ii). Sine Q=Z(Q) is automorphi of nilpoteny lass 2,
Proposition 2.3 implies (a;b;d) = (a;b;d)(a;;d)z for some z 2 Z(Q). This
meansthat(a;b;d) isentral,too. Thenthealulationattheendoftheproof
ofProposition2.3isstillvalid(sineallassoiatorsinvolvedinthealulationare
entral). Theproofsfor(i)and(iii)aresimilar.
Lemma 2.5. Let Q be aommutativeautomorphi loop of nilpoteny lass 3.
Then
((a;b;);d;e) 1
=(e;d;(a;b;));
(2.2)
(a;(b;;d);e)=(a;e;(b;;d))((b;;d);a;e):
(2.3)
foreverya;b;;d;e2Q.
Proof: Note that z = ((a;b;);d;e) 2 Z(Q) beause Q has nilpoteny lass
3. The identity (a;b;)de = ((a;b;)de)z hene yields ((a;b;)de)z 1
=
(a;b;)de, and we get ((a;b;)de)z 1
= (a;b;)de = ed(a;b;) = (e
d(a;b;))(e;d;(a;b;))=((a;b;)de)(e;d;(a;b;)),whihimplies(2.2). Weal-
ulate
(ae(b;;d))(a;e;(b;;d)) 1
=ae(b;;d)=a(b;;d)e
=(a(b;;d)e)(a;(b;;d);e) 1
=((b;;d)ae)((b;;d);a;e)(a;(b;;d);e) 1
=(ae(b;;d))((b;;d);a;e)(a;(b;;d);e) 1
;
whih implies(a;e;(b;;d)) 1
=((b;;d);a;e)(a;(b;;d);e) 1
,or(2.3).
Lemma 2.6. Let Q be aommutativeautomorphi loop of nilpoteny lass 3.
Then
(2.4) (a;(b;;d);(e;f;g))=((a;b;);d;(e;f;g))=((a;b;);(d;e;f);g)=1
foreverya;b;;d;e;f;g2Q.
Proof: SineQ=Z(Q)isofnilpotenylass2,wehave(efg)(efg) 1
=(e;f;g)z
forsomez2Z(Q). Then
(a;(b;;d);(e;f;g))=(a;(b;;d);(efg)(efg) 1
):
The automorphi inverse property and Lemma 2.4 yield (a;(b;;d);(e;f;g)) =
1. The identity ((a;b;);(d;e;f);g) = 1 follows by (2.2). The argument for
((a;b;);d;(e;f;g))=1issimilar.
Lemma 2.7. Let Q be aommutativeautomorphi loop of nilpoteny lass 3.
Then
(a;b;)=(;b;a) 1
; (2.5)
(a;b;)=(a;;b)(b;a;) (2.6)
foreverya;b;2Q.
Proof: Wehaveab=(ab)(a;b;)=(ba)(a;b;)=(ba)(;b;a)(a;b;)=
(ab)(;b;a)(a;b;),andthelasttermanberewrittenas(ab)(;b;a)(a;b;)
by(2.4),so(2.5)follows. Similarly,ab=(ab)(a;b;)=(ba)(b;;a)(a;b;)=
((ab)(;a;b)(b;;a))(a;b;),thelasttermequals(ab)((;a;b)(b;;a)(a;b;))
by(2.4)andLemma 2.4,sowehave(;a;b)(b;;a)(a;b;)=1. ByLemma2.6,
theAIP and (2.5), weget(a;b;)= ((;a;b)(b;;a)) 1
=(;a;b) 1
(b;;a) 1
=
(b;a;)(a;;b),whihis(2.6).
Notethatin anyommutativeloopofnilpotenylass3wehave
(2.7) aL
b;
=a(a;b;)(b;a;(a;b;));
beause
aL
b;
=(ba)L 1
b
=(ab)L 1
b
=((ab)(a;b;))L 1
b
=((ba)(a;b;))L 1
b
=(ba(a;b;)(b;a;(a;b;)))L 1
b
=a(a;b;)(b;a;(a;b;)):
Proposition 2.8. Let Q be a ommutative automorphi loop of nilpoteny
lass3. Then
(ab;;d)=(a;;d)(b;;d)((a;;d);a;b)((b;;d);b;a)
((a;;d);b;)((b;;d);a;)((a;;d);b;d)((b;;d);a;d);
(a;b;d)=(a;b;)(a;b;d)((a;b;);;d)((a;b;d);d;)
((a;b;);d;b)((a;b;d);;b)((a;b;);d;a)((a;b;d);;a);
(a;b;d)=(a;b;d)(a;;d)((a;b;d);b;)((a;;d);;b)
((a;b;d);;a)((a;;d);b;a)((a;b;d);;d)((a;;d);b;d):
Proof: By(2.7),theidentity(ab)L
;d
=aL
;d bL
;d
anberewrittenas
ab(ab;;d)(d;ab;(ab;;d))=a(a;;d)(d;a;(a;;d))b(b;;d)(d;b;(b;;d))
=(a(a;;d)b(b;;d))(d;a;(a;;d))(d;b;(b;;d)):
ByLemma 2.4and(2.4),(2.5),wehave
a(a;;d)b(b;;d)=(a(a;;d)b)(b;;d)(a(a;;d);b;(b;;d)) 1
=(a(a;;d)b)(b;;d)(a;b;(b;;d)) 1
((a;;d);b;(b;;d)) 1
=(ba(a;;d))(b;;d)(b;a;(a;;d)) 1
((b;;d);b;a)
=(ba(a;;d))(b;;d)((a;;d);a;b)((b;;d);b;a):
Sine(ab;;d)=(a;;d)(b;;d)zforsomez2Z(Q),Lemma2.4yields
(d;ab;(ab;;d))=(d;ab;(a;;d)(b;;d))=(d;ab;(a;;d))(d;ab;(b;;d))
=(d;a;(a;;d))(d;b;(a;;d))(d;a;(b;;d))(d;b;(b;;d)):
Uponsubstituting andanelingab and likeassoiators,theidentity (ab)L
;d
=
aL
;d bL
;d
thereforebeomes
(ab;;d)(d;b;(a;;d))(d;a;(b;;d))=(a;;d)(b;;d)((a;;d);a;b)((b;;d);b;a):
Theformulafor(ab;;d) nowfollowsbyLemma2.4and(2.5).
Note that Lemma 2.4 and (2.5) imply ((a;b;);d;e) 1
= ((a;b;) 1
;d;e) =
((;b;a);d;e). This observation and (2.5) applied to the formula for (ab;;d)
yieldtheformulafor(a;b;d).
Using(2.6),wealulate
(a;b;d)=(a;d;b)(b;a;d)=(a;d;b)(a;d;)(b;a;d)(;a;d)
((a;d;b);b;)((a;d;);;b)((a;d;b);;d)((a;d;);b;d)((a;d;b);;a)((a;d;);b;a)
((b;a;d);b;)((;a;d);;b)((b;a;d);;a)((;a;d);b;a)((b;a;d);;d)((;a;d);b;d):
Therstfourassoiatorsassoiateby(2.4),so(a;d;b)(a;d;)(b;a;d)(;a;d) =
(a;d;b)(b;a;d)(a;d;)(;a;d)=(a;b;d)(a;;d). Weansimilarlypairtheom-
poundedassoiators,usingLemma2.4. Forinstane,((a;d;b);b;)((b;a;d);b;)=
((a;d;b)(b;a;d);b;)=((a;b;d);b;). Theformulafor(a;b;d) follows.
We an now deal with produts in all arguments of a ompounded assoia-
tor. Forprodutsoftheform (ab;;(d;e;f))weanuseLemma2.4(orProposi-
tion2.8),andforproduts(a;b;(d;e;f))wenotethat(d;e;f)=(;e;f)(d;e;f)z
for someentral elementz (the expliitform ofz follows from Proposition 2.8)
and alulate(a;b;(;e;f)(d;e;f))=(a;b;(;e;f))(a;b;(d;e;f))byLemma 2.4.
Fromnowon,wewillusetheseandsimilaridentities,oftenwithoutexpliitrefe-
rene.
3. Powerswithin assoiators
UsingProposition2.8,weproeedtoderiveformulaeforpowerswithinassoi-
ators. Dene:Z!Z,:Z!Zby
(3.1) (n)=(n
3
n)=3; (n)=n 2
n:
Note that (0) =(0) =0, (n+1) =(n)+n 2
+n, (n+1) =(n)+2n,
2
Lemma 3.1. Let Q be aommutativeautomorphi loop of nilpoteny lass 3.
Then
(a n
;b;)=(a;b;) n
((a;b;);a;a) (n)
((a;b;);a;b) (n)
((a;b;);a;) (n)
; (3.2)
(a;b n
;)=(a;b;) n
((a;b;);b;b) (n)
((a;b;);b;a) (n)
((a;b;);b;) (n)
; (3.3)
(a;b;
n
)=(a;b;) n
((a;b;);;) (n)
((a;b;);;a) (n)
((a;b;);;b) (n)
(3.4)
foreverya;b;2Qandeveryn2Z.
Proof: We prove (3.2); the equations (3.3), (3.4) are proven analogously. If
n = 0, (3.2) holds. Suppose that (3.2) holds for some n 0. Note that
((a i
;b;);a j
;d) = ((a;b;) i
;a j
;d) =((a;b;);a;d) ij
for everyi, j 0, by Lem-
ma2.4,usingourusualtrik(a i
;b;)=(a;b;) i
z forsomez2Z(Q). ByPropo-
sition2.8wethenhave
(a n+1
;b;)=(aa n
;b;)
=(a;b;)(a n
;b;)((a;b;);a;a n
)((a n
;b;);a n
;a)
((a;b;);a n
;b)((a n
;b;);a;b)((a;b;);a n
;)((a n
;b;);a;)
=(a;b;)(a n
;b;)((a;b;);a;a) n
2
+n
((a;b;);a;b) 2n
((a;b;);a;) 2n
=(a;b;) n+1
((a;b;);a;a) (n)+n
2
+n
((a;b;);a;b) (n)+2n
((a;b;);a;) (n)+2n
=(a;b;) n+1
((a;b;);a;a) (n+1)
((a;b;);a;b) (n+1)
((a;b;);a;) (n+1)
:
Asfornegativepowers,rstnotethat Proposition2.8gives
1=(1;b;)=(aa 1
;b;)=(a;b;)(a 1
;b;)((a;b;);a;a 1
)((a 1
;b;);a 1
;a)
((a;b;);a 1
;b)((a 1
;b;);a;b)((a;b;);a 1
;)((a 1
;b;);a;)
=(a;b;)(a 1
;b;)((a;b;);a;b) 2
((a;b;);a;) 2
:
Sineassoiatorsassoiatewithoneanotherby(2.4),wededue
(a 1
;b;)=(a;b;) 1
((a;b;);a;b) 2
((a;b;);a;) 2
:
Thenforeveryn>0wehave
(a n
;b;)=((a n
) 1
;b;)=(a n
;b;) 1
((a n
;b;);a n
;b) 2
((a n
;b;);a n
;) 2
=(a;b;) n
((a;b;);a;a) (n)
((a;b;);a;b) 2n
2
(n)
((a;b;);a;) 2n
2
(n)
=(a;b;) n
((a;b;);a;a) ( n)
((a;b;);a;b) ( n)
((a;b;);a;) ( n)
;
nishingtheproofof (3.2).
Lemma 3.2. Let Q be aommutativeautomorphi loop of nilpoteny lass 3.
Then
i j k ijk (i)jk (i)j
2
k (i)jk
2
((a;b;);b;a) i(j)k
((a;b;);b;b) i(j)k
((a;b;);b;) i(j)k
2
((a;b;);;a) ij(k )
((a;b;);;b) ij(k )
((a;b;);;) ij(k )
foreverya;b;2Qandi;j;k2Z.
Proof: ByLemmas2.4, 3.1andProposition2.8, (a i
;b j
; k
)isequalto
(a;b j
; k
) i
((a;b j
; k
);a;a) (i)
((a;b j
; k
);a;b j
) (i)
((a;b j
; k
)a;
k
) (i)
=(a;b j
; k
) i
((a;b;);a;a) (i)jk
((a;b;);a;b) (i)j
2
k
((a;b;);a;) (i)jk
2
;
theterm(a;b j
; k
) i
isequalto
[(a;b;
k
) j
((a;b;
k
);b;b) (j)
((a;b;
k
);b;a) (j)
((a;b;
k
);b;
k
) (j)
℄ i
=(a;b;
k
) ij
((a;b;);b;b) i(j)k
((a;b;);b;a) i(j)k
((a;b;);b;) i(j)k
2
;
andtheterm(a;b;
k
) ij
isequalto
[(a;b;) k
((a;b;);;) (k )
((a;b;);;a) (k )
((a;b;);;b) (k )
℄ ij
=(a;b;) ijk
((a;b;);;) ij(k )
((a;b;);;a) ij(k )
((a;b;);;b) ij(k )
:
4. Redution
SineweultimatelywanttodesribethefreeloopF
3
(x;y),wewillfromnowon
startfousingonformulaethatinvolveonlytwovariablesx,y. Forxedelements
x,y (not neessarilythegeneratorsofF
3
(x;y)),let
u
1
=(x;x;y); u
2
=(x;y;y);
z
1
=(x;x;u
1
); z
2
=(x;x;u
2
); z
3
=(x;y;u
1
); z
4
=(x;y;u
2 );
z
5
=(y;x;u
1
); z
6
=(y;x;u
2
); z
7
=(y;y;u
1
); z
8
=(y;y;u
2 ):
Noadditional assoiatorswill be needed sine(x;y;x) =(y;x;y)= 1by (2.1),
ompoundedassoiatorsoftheform((; ; ); ; )and(;(; ; ); )anberewritten
asprodutsofompoundedassoiatorsoftheform( ; ;(; ; ))by(2.2)and(2.3),
ompoundedassoiatorswithassoiatorsintwoomponentsvanishby(2.4),and
produtswithinassoiatorsanbehandled byProposition2.8.
Thereadermighthavenotiedthatinourprodutformulas(suhasinPropo-
sition 2.8) we aumulate assoiators on the left, but we hose the anonial
ompounded assoiatorswithaumulatedassoiatorsontheright. It iseasyto
onvertbetweenthese twoformats,sine
1 1 1
by(2.5)andLemma 2.4. Alsonote that
(a;b;(;d;e))=(a;b;(e;d;) 1
)=(a;b;(e;d;)) 1
thanksto (2.5)and Lemma2.4.
Lemma 4.1. Let Q be aommutativeautomorphi loop of nilpoteny lass 3.
Thenforeveryx;y2Qwehavez
2
=z
3
=z
5 andz
4
=z
6
=z
7 .
Proof: Fousingrstontheprodutinthethirdoordinate,wealulate,start-
ingwith(2.1):
1=(xy;x;xy)=(xy;x;x)(xy;x;y)((xy;x;x);x;y)((xy;x;y);y;x)
((xy;x;x);y;x)((xy;x;y);x;x)((xy;x;x);y;xy)((xy;x;y);x;x y)
=(xy;x;x)(xy;x;y)z
5 z
1
3 z
3 z
1
1 z
3 z
7 z
1
1 z
1
5
=(xy;x;x)(xy;x;y)z 2
1 z
3 z
7
=(y;x;x)((y;x;x);y;x)((y;x;x);x;x)((y;x;x);x ;x)
(x;x;y)((x;x;y);x;y)((x;x;y);y;x)((x;x;y);y;y)z 2
1 z
3 z
7
=(y;x;x)(x;x;y)z
3 z
1 z
1 z
1
5 z
1
3 z
1
7 z
2
1 z
3 z
7
=1z
3 z
1
5 :
Henez
3
=z
5
,or(x;y;(x;x;y))=(y;x;(x;x;y)). Interhangingx andy in this
identityyields(y;x;(y;y;x))=(x;y;(y;y;x)),whihisequivalenttoz
6
=z
4 .
In the following alulation we will also use (2.7) and Lemma 3.1. As Q is
automorphi,yL
x;x (xy)L
x;x
=(yxy)L
x;x
. Onthelefthandsideofthisidentity
wehave
yL
x;x
=y(y;x;x)(x 2
;y;(y;x;x))=y(y;x;x)z 2
3
and
(xy)L
x;x
=(xy)(xy;x;x)(x 2
;xy;(xy;x;x))=(xy)(xy;x;x)(x 2
;xy;(y;x;x))
=(xy)(y;x;x)((y;x;x);y;x)((y;x;x);x;x) 2
z 2
1 z
2
3
=(xy)(y;x;x)z
3 z
2
1 z
2
1 z
2
3
=(xy)(y;x;x)z 1
3
;
whileontherighthand sidewehave
(yxy)L
x;x
=(yxy)(yxy;x;x)(x 2
;yxy;(yxy;x;x))
=(yxy)(y 2
x(y;y;x) 1
;x;x)(x 2
;y 2
x;(y 2
;x;x))
=(yxy)(y 2
x;x;x)((x;y;y);x;x)z 8
3 z
4
1
=(xy 2
(x;y;y))(xy 2
;x;x)z 1
2 z
8
3 z
4
1
=(xy 2
(x;y;y))(y 2
;x;x)((y 2
;x;x);y 2
;x)((y 2
;x;x);x;x) 2
z 1
2 z
8
3 z
4
1
=(xy 2
(x;y;y))(y;x;x) 2
((y;x;x);y;y) 2
((y;x;x);y;x) 4
z 4
3 z
4
1 z
1
2 z
8
3 z
4
1
=xy 2
(x;y;y)(y;x;x) 2
z 2
z 4
z 4
z 1
=xy 2
(x;y;y)(y;x;x) 2
z 2
z 1
:
Returningtothelefthandside,werewriteitas
y(y;x;x)(xy)(y;x;x)z 3
3
=(y(y;x;x)xy)(y;x;x)(y(y;x;x);xy;(y;x;x)) 1
z 3
3
=(xyy(y;x;x))(y;x;x)(y;xy;(y;x;x)) 1
z 3
3
=(xyy)(y;x;x)(xy;y;(y;x;x)) 1
(y;x;x)z
5 z
7 z
3
3
=(xyy)(y;x;x) 2
z
3 z
7 z
5 z
7 z
3
3
=(xy 2
)(x;y;y)(y;x;x) 2
z 2
3 z
5 z
2
7
=xy 2
(x;y;y)(y;x;x) 2
z 2
3 z
5 z
2
7 :
Comparingthe twosidesnowyieldsz 1
2
=z 2
3 z
5 . Butz
2
3 z
5
=z 1
3
bytherst
partofthislemma,andhenez
2
=z
3
. Swithingxandy intheidentityz
2
=z
3
givesz
7
=z
6
.
WiththeredutionofLemma4.1inmind, wesetforanyxedx,y
u
1
=(x;x;y); u
2
=(x;y;y); v
1
=(x;x;u
1 );
v
2
=(x;x;u
2
); v
3
=(y;y;u
1
); v
4
=(y;y;u
2 ):
WearenowreadytodesribeanonialelementsofthefreeloopF
3 (x;y).
Lemma4.2. EveryelementofF
3
(x;y)an bewrittenin theanonialform
(x a1
y a2
u a
3
1 u
a
4
2 )v
a
5
1 v
a
6
2 v
a
7
3 v
a
8
4
;
wherea
i 2Z.
Proof: Let X = fx;x 1
;y;y 1
g. We rst note that any assoiator an be
written as u b
1
1 u
b
2
2 Q
v
i
i
. Indeed, sine F
3
(x;y) has nilpoteny lass three, no
ompounded assoiatorsappear within assoiators. Using Lemma 2.4, Proposi-
tion2.8and theironsequenes,everyassoiatoranbewritten asaprodutof
ompoundedassoiatorsandordinaryassoiatorswithallvariablesinX. Infat,
equations(2.4),(2.1),(2.2)and(2.3)imply thateveryassoiatorisaprodutof
u
1 , u
2 , the z
i
sand theirinverses. Sine assoiatorsassoiateamong themselves
by(2.4),thisprodutisoftheformu b
1
1 u
b
2
2 Q
z d
i
i
forsuitableexponentsinZ,and
heneoftheform u b1
1 u
b2
2 Q
v i
i
byLemma4.1.
To establish the lemma, it suÆes to show that a produt of two anonial
words is also anonial. First, [x a1
y a2
u a3
1 u
a4
2 Q
v i
i
℄[x b1
y b2
u b3
1 u
b4
2 Q
v di
i
℄ =
(x a1
y a2
u a
3
1 u
a
4
2 )(x
b1
y b2
u b
3
1 u
b
4
2 )
Q
v
i +d
i
i
,soitsuÆestoshowthattheprodut
(x a
1
y a
2
u a3
1 u
a4
2 )(x
b
1
y b
2
u b3
1 u
b4
2
)hasthedesiredform. Weanrewritethisword
as ((x a1
y a2
x b1
y b2
)u a
3
1 u
a
4
2 )u
b
3
1 u
b
4
2
w with some produt w of ompounded
assoiators,andfurtherto(x a1
y a2
x b1
y b2
)u a3+b3
1 u
a4+b4
2
w,usingLemma2.4and
(2.4). Now,x a
1
y a
2
x b
1
y b
2
anbewrittenas(((x a
1 +b
1
y a
2 +b
2
)t
1 )t
2 )t
k ,where
eaht
i
isanassoiator. UsingLemma2.4and(2.4)again,wefurtherrewritethis
as(x a1+b1
y a2+b2
)(t t t ). Therest iseasy.
5. The mainresult
The alulation desribed in the proof of Lemma 4.1 is straightforward but
rathertedious. Beforeweattemptit,wenote:
Lemma5.1. IntheloopF
3
(x;y)allassoiatorsarein themiddle nuleus.
Proof: Thanksto(2.1), (2.5)andLemmas2.4, 4.2and Proposition 2.8,itsuf-
es to show that (x;u
1
;y) = (x;u
2
;y) = 1. By (2.3) and (2.5), (x;u
1
;y) =
(x;y;u
1 )(u
1
;x;y) = (x;y;u
1
)(y;x;u
1 )
1
= z
3 z
1
5
= 1, and also (x;u
2
;y) =
(x;y;u
2 )(u
2
;x;y)=(x;y;u
2
)(y;x;u
2 )
1
=z
4 z
1
6
=1,whereweusedLemma4.1.
Reallthemappings, of (3.1).
Lemma5.2. InF
3
(x;y)wehaveforeverya
1
;a
2
;b
1
;b
2 2Z
x a
1
y a
2
x b
1
y b
2
=x a
1 +b
1
y a
2 +b
2
u
a1b1(a2+b2)
1
u
a2b2(a1+b1)
2
v
(a2+b2)(b1(a1)+a1(b1))+a2(a1(b1)+b 2
1
(a1))+b2(b1(a1)+a 2
1 (b1))
1
v 2a
1 a
2 b
1 b
2 (a
1 +b
1 )+(a
2 +b
2 )(a
1 (b
1 )+b
1 (a
1 ))+((a
2 )+(b
2 ))(a
1 b
2
1 +b
1 a
2
1 ) a
2 b
2 (a
1 +b
1 )
2
v
2a1a2b1b2(a2+b2) (a1+b1)(a2(b2)+b2(a2)) ((a1)+(b1))(a2b 2
2 +b2a
2
2
)+a1b1(a2+b2)
3
v (a
1 +b
1 )(a
2 (b
2 )+b
2 (a
2 )) a
1 (a
2 (b
2 )+b
2
2 (a
2 )) b
1 (b
2 (a
2 )+a
2
2 (b
2 ))
4
:
Proof: Using(2.5),wealulate
x a1
y a2
x b1
y b2
=(x a1
y a2
x b1
)y b2
(y b2
;x b1
;x a1
y a2
)
=((x b
1 +a
1
y a
2
)(y a
2
;x a
1
;x b
1
)y b
2
)(y b
2
;x b
1
;x a
1
y a
2
)
=(x a1+b1
y a2
(y a2
;x a1
;x b1
)y b2
)(x a1+b1
y a2
;(y a2
;x a1
;x b1
);y b2
)(y b2
;x b1
;x a1
y a2
):
ByLemma 5.1,weanignoretheompoundedassoiatorandontinue
[(x a
1 +b
1
y a
2
y b
2
)(y a
2
;x a
1
;x b
1
)℄((y a
2
;x a
1
;x b
1
);y b
2
;x a
1 +b
1
y a
2
)(y b
2
;x b
1
;x a
1
y a
2
)
=[(x a1+b1
y a2+b2
(x a1+b1
;y a2
;y b2
))(y a2
;x a1
;x b1
)℄
((y a2
;x a1
;x b1
);y b2
;x a1+b1
y a2
)(y b2
;x b1
;x a1
y a2
):
Beauseassoiatorsassoiatewithoneanother,weanrewritetheformulaas
x a
1
y a
2
x b
1
y b
2
=x a
1 +b
1
y a
2 +b
2
(x a
1 +b
1
;y a
2
;y b
2
)(y a
2
;x a
1
;x b
1
)(y b
2
;x b
1
;x a
1
y a
2
)
((y a2
;x a1
;x b1
);y b2
;x a1+b1
y a2
):
Now,usingLemmas2.4and 4.1freely,
((y a
2
;x a
1
;x b
1
);y b
2
;x a
1 +b
1
y a
2
)=v
a1b1a2b2(a1+b1)
v a
1 b
1 a
2
2 b
2
:
ByLemma 3.2,
(x a1+b1
;y a2
;y b2
)=u
(a1+b1)a2b2
2
v
(a1+b1)a2b2
2
v
(a1+b1)a 2
2 b2
3
v
(a1+b1)a2b 2
2
3
v
(a1+b1)(a2)b2
3
v
(a1+b1)(a2)b2
4
v (a
1 +b
1 )(a
2 )b
2
2
4
v (a
1 +b
1 )a
2 (b
2 )
3
v (a
1 +b
1 )a
2 (b
2 )
4
v (a
1 +b
1 )a
2 (b
2 )
4
;
and,similarly,
(y a2
;x a1
;x b1
)=u a1b1a2
1 v
(a2)a1b1
3
v (a2)a
2
1 b1
2
v (a2)a1b
2
1
2
v a
2 (a
1 )b
1
2
v a
2 (a
1 )b
1
1
v a
2 (a
1 )b
2
1
1
v a
2 a
1 (b
1 )
2
v a
2 a
1 (b
1 )
1
v a
2 a
1 (b
1 )
1
:
Finally, byProposition2.8and(2.3),weseethat
(y b
2
;x b
1
;x a
1
y a
2
)=(y b
2
;x b
1
;x a
1
)v a
2
1 b
1 a
2 b
2
2 v
a
1 b
2
1 a
2 b
2
2 v
a
1 b
1 a
2 b
2
2
3 :
Theassoiator(y b
2
;x b
1
;x a
1
)anbeobtainedfrom thealreadyalulatedassoi-
ator(y a
2
;x a
1
;x b
1
). Puttingalltheseassoiatorstogether, wearriveat
x a1
y a2
x b1
y b2
=x a1+b1
y a2+b2
u a
1 b
1 (a
2 +b
2 )
1
u a
2 b
2 (a
1 +b
1 )
2
v
1
1 v
2
2 v
3
3 v
4
4
;
where,after summinguptheexponentsoftherespetivev
i
sandsimplifying,
1
=(a
2 +b
2 )(b
1 (a
1 )+a
1 (b
1 ))+a
2 (a
1 (b
1 )+b
2
1 (a
1 ))
+b
2 (b
1 (a
1 )+a
2
1 (b
1 ));
2
=2a
1 a
2 b
1 b
2 (a
1 +b
1 )+(a
2 +b
2 )(a
1 (b
1 )+b
1 (a
1 ))
+((a
2 )+(b
2 ))(a
1 b
2
1 +b
1 a
2
1 ) a
2 b
2 (a
1 +b
1 );
3
= (a
1 +b
1 )(a
2
2 b
2 +a
2 b
2
2 ) (a
1 +b
1 )((a
2 )b
2 +a
2 (b
2 ))
+a
1 b
1 ((a
2 )+(b
2 ))+a
1 b
1 a
2 b
2 (a
2 +b
2 );
4
= (a
1 +b
1 )(a
2 )b
2 (a
1 +b
1 )(a
2 )b
2
2
(a
1 +b
1 )a
2 (b
2 ) (a
1 +b
1 )a
2 (b
2 ):
Theexponents
1 ,
2
alreadyhavethedesiredform. Tomath theexponents
3 ,
4
withtheformulaof thelemma, note that (a+b)=(a)+(b)+ab(a+b)
whilerewriting
3
,andsubstitute(a)=a 2
ainto
4
.
Lemma5.3. InF
3
(x;y)wehave
((x a
1
y a
2
u a3
1 u
a4
2 )v
a5
1 v
a6
2 v
a7
3 v
a8
4 )((x
b
1
y b
2
u b3
1 u
b4
2 )v
b5
1 v
b6
2 v
b7
3 v
b8
4 )
=(x a1+b1
y a2+b2
u a
3 +b
3 a
1 b
1 (a
2 +b
2 )
1
u a
4 +b
4 +a
2 b
2 (a
1 +b
1 )
2
)
v
a5+b5+(a2+b2)(b1(a1)+a1(b1))+a2(a1(b1)+b 2
1
(a1))+b2(b1(a1)+a 2
1
(b1)) a1b1(a3+b3)
1
v a
6 +b
6 +2a
1 a
2 b
1 b
2 (a
1 +b
1 )+(a
2 +b
2 )(a
1 (b
1 )+b
1 (a
1 ))+((a
2 )+(b
2 ))(a
1 b
2
1 +b
1 a
2
1 )
2 2 1 1 1 1 4 4 3 3 1 2 2 1
v
a7+b7 2a1a2b1b2(a2+b2) (a1+b1)(a2(b2)+b2(a2)) ((a1)+(b1))(a2b 2
2 +b2a
2
2 )
3
+a
1 b
1 (a
2 +b
2 ) a
2 b
2 (a
3 +b
3 ) (a
4 +b
4 )(a
1 b
2 +a
2 b
1 )
v
a8+b8 (a1+b1)(a2(b2)+b2(a2)) a1(a2(b2)+b 2
2
(a2)) b1(b2(a2)+a 2
2
(b2)) a2b2(a4+b4)
4
foreverya
i
;b
i 2Z.
Proof: UsingLemma5.1in therststepand(2.4)intheseond,wehave
(x a
1
y a
2
u a
3
1 u
a
4
2 )(x
b
1
y b
2
u b
3
1 u
b
4
2 )
=x a1
y a2
(u a3
1 u
a4
2 (x
b1
y b2
u b3
1 u
b4
2 ))
=x a1
y a2
(x b1
y b2
u a
3 +b
3
1 u
a
4 +b
4
2 )
=(x a1
y a2
x b1
y b2
)u a
3 +b
3
1 u
a
4 +b
4
2 (x
a1
y a2
;x b1
y b2
;u a
3 +b
3
1 u
a
4 +b
4
2 )
1
:
Nownotethat Lemma2.4yields
(x a
y b
;x
y d
;u e
1 u
f
2 )=v
ae
1 v
af+ade+be
2
v
adf+bf+bde
3
v bdf
3 :
WearethereforedonebyLemma5.2.
IntheproofofthemaintheoremwewilluseaMathematia[12℄odetoverify
ertainproperties ofthemultipliation formulaofLemma 5.3. Theodeanbe
downloadedfromthewebsiteof thethird-namedauthor.
Theorem5.4. LetF
3
(x;y)bethefreeommutativeautomorphiloopofnilpo-
teny lass 3 on free generators x, y. Let u
1
= (x;x;y), u
2
= (x;y;y), v
1
=
(x;x;u
1 ), v
2
= (x;x;u
2 ), v
3
=(y;y;u
1 ), v
4
= (y;y;u
2
). Then eah element of
F
3
(x;y)anbewrittenuniquelyas(x a1
y a2
u a3
1 u
a4
2 )v
a5
1 v
a6
2 v
a7
3 v
a8
4
,andF
3 (x;y)is
isomorphito(Z 8
;),wherethemultipliationofexponentsisasinLemma5.3.
Proof: Let F be dened on Z 8
with multipliation aording to Lemma 5.3.
Denote by e
i
the element of Z 8
whose only non-zero oordinate is equal to 1
and is loated in position i. Straightforwardalulation in Mathematia shows
thatF is aloopwith identityelement(0;0;0;0;0;0;0;0)suhthat (e
1
;e
1
;e
2 )=
e
3 , (e
1
;e
2
;e
2 ) = e
4 , (e
1
;e
1
;e
3 ) = e
5 , (e
1
;e
1
;e
4 ) = e
6 , (e
2
;e
2
;e
3 ) = e
7 and
(e
2
;e
2
;e
4 ) = e
8
. Moreover, F is a ommutative automorphi loop. (Toverify
that F isautomorphi, the ode merelyneeds to hekby symboli alulation
thattheinnermappings L
a;b
areautomorphismsofF.)
Welaim that F
3
(x;y) is isomorphi to F. Letf : F
3
(x;y)! F be theho-
momorphism determined by f(x) = e
1
, f(y) = e
2
. Beause homomorphisms
behave well on assoiators, namely f((a;b;)) = (f(a);f(b);f()), the alula-
tion in the previousparagraphshowsthat f(u
1 ) =e
3 , f(u
2 ) =e
4 , f(v
1 ) =e
5 ,
f(v
2 )=e
6 ,f(v
3 )=e
7
andf(v
4 )=e
8
. ByLemma4.2,anyelementwofF
3 (x;y)
an be written as w = (x a1
y a2
u a3
u a4
)v a5
v a6
v a7
v a8
, and it now follows that
f(w)=(a
1
;a
2
;a
3
;a
4
;a
5
;a
6
;a
7
;a
8
). This meansthat f is ontoF, andalso that
theexponentsa
i
inthedeompositionofwareuniquelydeterminedbyw. Hene
f :F
3
(x;y)!F isanisomorphism.
WeonludethepaperwithsomestruturalinformationaboutF
3 (x;y).
Proposition5.5. LetQ=F
3
(x;y)beidentiedwith(Z 8
;)asinTheorem5.4.
Then A(Q) = N
(Q) = 00Z 6
and N
(Q) = N
(Q) = N(Q) = Z(Q) =
0000Z 4
.
Proof: We already know from Lemma 5.1 that A(Q) N
(Q). ByProposi-
tion2.8andLemmas3.2,4.1,
(x;x a1
y a2
;y)=(x;x a1
;y)(x;y a2
;y)
((x;x a
1
;y);x a
1
;y a
2
)((x;y a
2
;y);y a
2
;x a
1
)((x;x a
1
;y);y a
2
;x)
((x;y a2
;y);x a1
;x)((x;x a1
;y);y a2
;y)((x;y a2
;y);x a1
;y)
=(x;x a
1
;y)(x;y a
2
;y)v a
2
1 a
2 2a
1 a
2
2
v a
1 a
2
2 2a
1 a
2
3
=u a1
1 (u
1
;x;x) (a
1 )+(a
1 )
(u
1
;x;y) (a
1 )
u a2
2 (u
2
;y;y) (a
2 )+(a
2 )
(u
2
;y;x) (a
2 )
v a
2
1
a2 2a1a2
2
v a1a
2
2 2a1a2
3
=u a1
1 u
a2
2 v
(a1) (a1)
1
v (a
1 ) a
2
1 a
2 2a
1 a
2
2
v (a
2 ) a
1 a
2
2 2a
1 a
2
3
v
(a2) (a2)
4
:
Thus, ifeither a
1
6= 0or a
2
6=0then r=(x;x a
1
y a
2
;y)6=1. In otherwords,if
r2N
(Q)thenr200Z 6
. Weonlude,A(Q)N
(Q)00Z
6
A(Q),
soN
(Q)=A(Q)=00Z 6
.
SineQhasnilpotenylass3,wehave0000Z 4
Z(Q)N
(Q)=
N
(Q). Now,
(x;x;x a1
y a2
u a
3
1 u
a
4
2
)=(x;x;x a1
y a2
)(x;x;u a
3
1 u
a
4
2
)=(x;x;x a1
y a2
)v a
3
1 v
a
4
2
=[(x;x;y a2
)((x;x;y a2
);y a2
;x a1
)((x;x;y a2
);x a1
;x)((x;x;y a2
);x a1
;x)℄v a
3
1 v
a
4
2
=(x;x;y a
2
)v
a3 2a1a2
1
v a
4 a
1 a
2
2
2
=u a2
2 (u
2
;y;y) (a2)
(u
2
;y;x) 2(a2)
v
a3 2a1a2
1
v a4 a1a
2
2
2
=u a
2
2 v
a
3 2a
1 a
2
1
v a4 a1a
2
2
2 v
2(a2)
3 v
(a2)
4 :
If (x;x;x a1
y a2
u a
3
1 u
a
4
2
) =1,a
2
must be zero. Then v a
3
1 v
a
4
2
=1and thus a
3
=
a
4
=0. Therefore,if(x;x;x a
1
y a
2
w a
3
t a
4
)=1thena
2
=a
3
=a
4
=0. Finally,
(y;x;x a1
)=u a
1
1 (u
1
1
;x;x)
(a1)+(a1)
(u 1
1
;x;y) (a1)
=u a
1
1 v
(a
1 )+(a
1 )
1
v (a
1 )
2 :
So,(y;x;x a1
)=1implies a
1
=0. Summarizing,x a1
y a2
u a3
1 u
a4
2 2N
(Q) ifand
onlyifa
1
=a
2
=a
3
=a
4
=0. HeneN
(Q)=Z(Q)=0000Z 4
.
Referenes
[1℄ BarrosD.,Commutativeautomorphiloops,PhDdissertation,UniversityofSaoPaulo,in
preparation.
[2℄ BarrosD.,GrishkovA.,Vojtehovsk yP.,Commutativeautomorphi loopsoforderp 3
,J.
AlgebraAppl.,toappear.
[3℄ BrukR.H.,ASurveyofBinarySystems,Springer,1971.
[4℄ BrukR.H.,PaigeL.J.,Loops whoseinnermappingsareautomorphisms,Ann.ofMath.
(2)63(1956),308{323.
[5℄ Csorg}oP.,Themultipliationgroupofaniteommutativeautomorphiloopof orderof
powerofanoddprimepisap-group,J.Algebra350(2012),no.1,77{83.
[6℄ JedlikaP.,KinyonM.,Vojtehovsk yP.,Thestrutureofommutativeautomorphiloops,
Trans.Amer.Math.So.363(2011),no.1,365{384.
[7℄ JedlikaP.,KinyonM.,Vojtehovsk yP.,Construtionsofommutativeautomorphiloops,
Comm.Algebra38(2010),no.9,3243{3267.
[8℄ JedlikaP.,KinyonM.,Vojtehovsk yP.,Nilpotenyinautomorphi loopsofprimepower
order,J.Algebra350(2012),no.1,64{76.
[9℄ Johnson K.W., Kinyon M.K., Nagy G.P., Vojtehovsk y P., Searhing for small simple
automorphiloops,LMSJ.Comput.Math.14(2011),200{213.
[10℄ GrishkovA.N.,ShestakovI.P., Commutative Moufang loops and alternative algebras,J.
Algebra333(2011),1{13.
[11℄ Kinyon M.K., Kunen K., Phillips J.D., Vojtehovsk y P., The struture of automorphi
loops,inpreparation.
[12℄ WolframResearh,In., Mathematia,version8.0, WolframResearh,In., Champaign,
Illinois,2010.
D.Barros,A.Grishkov:
Instituteof Mathematis and Statistis, Universityof Sao Paulo, Ruado
Mat~
ao, 1010,Cidade
Universit
aria, S~
aoPaulo,SP,Brazil,CEP 05508-090
E-mail: dyleneime.usp.br
shuragrigmail.om
P.Vojtehovsk y:
DepartmentofMathematis,UniversityofDenver,2360SGaylordSt,Den-
ver,Colorado80208, USA
E-mail: petrmath.du.edu
(Reeived Marh31,2012 , revised April26,2012 )
