Identities and the group of isostrophisms
Ale
s Dr
apal, Vitor Shherbaov
Abstrat.Inthispaperwereexaminetheoneptofisostrophy.Weonnetitto
the notionoftermequivalene,anddesribetheationofdihedralgroupsthat
areassoiatedwithloops bymeansof isostrophy. Wealsouseittoproveand
presentinanewwaysomewellknownfatsonm-inverseloopsandmiddleBol
loops.
Keywords:isostrophe,isostrophism,paratope,paratopism,middleBol
Classiation: Primary20N05;Seondary15A30
Let A, B and C bepenils of a3-net. If , and bijet aset Q upon A,
Band C,respetively, thenthere existsa(unique)quasigroup onQ()suh that
xy = z if and only if(x), (y) and (z) meet in aommon point. It is well
known thatif Qisoneof thelinesof the3-net, then, and anbedened
naturallyin suh a waythat adistinguished element of Q (say 1) beomes the
unitofQ. Thisonstrutionwillserveasthedepartingpointofthepaper.
Supposethus thatQ2Aandthat 12Q. Dene and in suh awaythat
both(a)2Band(a)2C areinidenttoa, foreverya2Q. IfQ()istobea
loopwithunit1,thenthere mustbea1=a,andhene(a)2Ahastobethe
linethatisinidenttotheintersetionof(1)and(a). Withthisdenitionof
weget(1)=Qsine(1)and(1)meetin1. Now,(1)=Qimplies1a=a
foreverya2Q,bythedenitionof and . WehaveobtainedaloopQ(;1).
ConsidernowaloopQ=Q(Æ;1) that isobtainedbythis method whenr^oles
ofB andC areexhanged. ThenxÆy =z ifandonly(x),(y)and(z)meet
in apoint. In partiular, (a), (1) and (a) have a ommon point, and that
denes . Theexisteneof theommon pointmeansthat 1
( (a))a=1for
everya2Q. Thus 1
((a)) =1=a, andtherefore(a)=(1=a). Wesee that
xÆy =z ,(1=x), (y)and(z)meetin aommonpoint, (1=x)z =y ,
z=(1=x)ny.
Wehavedesribedthegeometrialmeaningofoperation(1=x)ny. Theopera-
tionis induedbythetransposition (B C) oftheset fA;B;Cg. Infat,everyof
thesixpermutationsanbeusedtoinduealoop. Artzy [3℄ seemstohavebeen
therstwhosystematiallyinvestigatedthesetransformationsofloops. Healled
them isostrophisms. Theonept isreexamined in this paper. Our approah is
purelyalgebrai.
ForaloopQdenotebyl(Q)theloopwithoperation(1=x)nyandbyo(Q)the
loopwithoperationyx (the opposite loop|itorrespondsto thetransposition
of A and B). It is easy to verify that l(l(Q)) = Q = o(o(Q)). Nevertheless,
alternatingappliationsoflandoprodueasetofloopsI(Q)thatanbeinnite
(however,itontainsatmostsixisomorphismlasses).Operatorslandoatupon
I(Q)asinvolutionsandgenerateapermutationgroupI(Q). Thisgroupiseither
dihedral,orylioforders1or2. (TheKleinfour-groupisregardedasadihedral
group.)
Weshallobservethat I(Q)atsnearlyalwaysregularly. There areonlythree
exeptionalsituations,twoofwhihanbeonsideredasrelatedtoBolloops(and
thatiswhyweshalldisussthemiddleBolidentityaswell). Intheseexeptional
asesjI(Q)j2f3;6g.
Ourmainaimistopresenttheoneptofisostrophyinaoherentandompat
way. Therearesomenewresultsandtherearemanynewproofsofoldresults.
However,itshouldbestressedthatnoideasinthispaperareprinipallynew.
Furthermore, many statements that are new might have been present in some
formin mindsofthose whooinedandstudied theoneptsofthis paperinthe
sixties. Wehopethat this paperwill sueed in illustratingthat theseonepts
arerelevantto ontemporarylooptheoryandan motivatefurtherresearh.
Veryimportantamongtheobjetsofourstudyarethem-inverseloops dened
byKarklinsand Karklin[14℄. Theyarisein anaturalwayasageneralizationof
ross inverse [1℄, [2℄ and weak inverse properties [21℄. It wasobserved already
byArtzy in [3℄that CIand WIpropertiesanbeobtainedviaidentiationsof
ertain isostrophes. Weshall see that suh an approah anbe extended to all
m-inverseloops. In fat, ourdesription hasa parallel in the work of Karklins
andKarklin[14℄ andanberegardedasaninterpretationoftheirSetion2.
Theisostrophes ofQ(i.e.theelementsofI(Q))havebeenalledinverseloops
(ofQ)by Belousov[7℄. He alsomentions themin his book [6, p.19℄. Usingthe
terminologyofBelousovasinspiration,wesuggesttoalll(Q)theleftinverse of
Q(weshalldene theright inverse r(Q) asamirrorimage).
AthoroughgeometrialtreatmentofisostrophyanbefoundinChapterIIof
Pugfelder'sbook[22℄. (InthePrefaeto[22℄Pugfelderwrites\ToRafaelArtzy
Iamgratefulforhisenouragementandadvieandforwritingtheoriginaltextof
ChapterII.")Artzy himself oeredin[4, Setion2℄amorestrutured approah
to the material of [3℄. In Setion 3of the samepaper he dened net motions.
Thealgebraiexpressionofnetmotionsisparatopy,whih is,togetherwithloop
terms,amaintoolofthispaper.
Notethatm-inverseloopshavebeenreentlystudiedwithrespettoapossible
appliation in ryptography [17℄ and that Buhsteiner loopsweredisovered[8℄
tobe1-inverse(synonymously,doublyweak inverse).
ProblemsthatinvolvethestrutureofI(Q) mightbeofinterestin thefuture
sine this is an areawhere the algebraistruture (loops) gets mixed with the
ombinatorial struture (normalized latin squares). Hene a future appliation
toryptographyannotbeexluded, whileitsguiding priniplemaybedierent
thanthatexpressedbyKeedwellin[16℄(whihmotivated[17℄andthesubsequent
papers[18℄and[19℄).
Setion 1 desribes endomorphismsof amonogenerated free loop. Setion 2
shows that isostrophies anbe viewed as paratopiesthat yield term equivalent
loops. In Setion 3 we dene the group of isostrophisms I(Q) and disuss its
strutural properties. The impat upon nulei is presented in Setion 4. The
numberofisostrophiisomorphismlassesisstudiedinSetion5. Inthatsetion
wealsodeneloopsof odd type asloopsthatareeither ommutativeorhavean
automorphism I r
for an odd r. We showhow suh loopsan be desribed via
I(Q). Setion 6 presents the onept of isostrophi varieties and employs it to
interpretseveralstandardresultsonLIP,RIP,AAIPandBolloops.
Inthispaperthemappings areomposedfrom righttoleft.
1. Freeloops in one generator
Thissetionis ofanauxiliaryharater. Itprovesinanelementarywaythat
all automorphismsof afree loopgenerated by asingle element x (denote it by
F(x)) are those substitutions that map x to one of its iterated inverses. This
resultwaspublished already in 1953byEvans[10, Theorem 1℄. Weshall useit
inCorollary2.9.
TheproofofEvansisshortand elegant. It depends uponthetheoryofloops
that arerelativelyfreewith respet to aset of(dening) relationsthat are in a
losed form. This theory wasdeveloped by Evans in [9℄. A speial ase is the
aseof the voidset ofrelations, that is thease of afree loop. The assoiated
setofrewritingrules(f.Table1)beamepartofafolkloreknowledge. Infatit
isone offew resultsof loopand quasigroup theorythat is well known bymany
non-speialists. However,the generaltheory ofrelations in alosed form isnot
nearlyas well-known. That is why we oer aproof that uses nothing else but
the well understood struture of a free loop. As a bonus we prove that every
nontrivialendomorphismofF(x) isinjetive| afat that seemsto beevident,
butforwhihwedonotknowareferene.
Forasetof variablesX onsiderthetotallyfreealgebraoftermsW(X)over
thebinaryoperations,=,nandthenullaryoperation1. Anelementw2W(X)is
saidtoberedued ifnoneofitssubtermsanbesubjetedtooneoftherewriting
rulesthatappearinTable1.
t
1 (t
1 nt
2 )!t
2 t
1 n(t
1 t
2 )!t
2 t
1
=(t
2 nt
1 )!t
2 t
1 1!t
1 t
1
=1!t
1
(t
2
=t
1 )t
1
!t
2 (t
2 t
1 )=t
1
!t
2 (t
1
=t
2 )nt
1
!t
2 1t
1
!t
1 1nt
1
!t
1
Table 1. Therewritingrulesforloopterms
It is lear that eah term w 2 W(X) an be transformed by a sequene of
beause theabove systemof rewriting rulesis knownto beonuent [9℄,a ter-
minalelementofsuhasequenewillalwaysbethesamereduedterm(inother
wordstheterminaltermisindependentofthehosenpath). Weshalldenotethe
(terminal) redued termby (w). Theset of allredued termswill be denoted
F(X)alluringthustothefatthatthereduedtermsyieldamodelofafreeloop
forwhih X isthefreebase(f.[9℄, [10℄fordetails).
If u and v are redued, then their term produt uv need not be redued.
Hene the produt in F(X) is dened as (uv). Left and right division are
treatedsimilarly.
As a synonym for tn1 write I(t). Similarly interpret I 1
(t) as 1=t. Note
that (II 1
(t)) = (t) =(I 1
I(t))sine 1=(tn1) ! t and (1=t)n1 ! t. Thus
(I r
I s
(t))=(I r+s
(t))foranyr;s2Z.
Weshallwrite F(x) andW(x)inplae ofF(X)and W(X)whenX =fxg.
For t 2 F(x) dene a mapping
t
: F(x) ! F(x) so that it expresses the
substitutionx7!t. Thusfors=s(x)2F(x)weset
t
(s)=(s(t)). Forexample
x 2
(xn(1=x))=x 2
n(1=x 2
)and
I(x)
(xn(1=x))=(xn1)nx.
Itiseasytoseethatforeveryt2F(x)thereexistuniquek2Zandt
0
2W(x)
suhthat
(1.1) t=I
k
(t
0
) and t
0 6=I
1
(s) forall s2W(x):
Forexample,ift=1=(1=x 2
),thenk= 2and t
0
=x 2
.
Callt
0
theI-ore of t and k theI-depth oft. Forthe nextthree statements
letusassumethatt6=1isreduedandthatt
0
andkaretheI-oreandI-depth
oft,respetively.
Lemma1.1. I j
(t
0
)2F(x)foreveryj2Z.
Proof: Anysubterm ofareduedtermhasto beredued,and thus t
0
2F(x).
Weanproeedbyindutiononjsinethemirrorsymmetryallowsustoassume
j 1. Note that t
0
n1 is reduedunless t
0
=1ort
0
=1=sfor somes2 W(x).
The latter situation is exluded by the denition of the I-ore, while t
0
= 1
would imply t=1. Thestatementthus holds forj =1. Assume j 2and set
s=I j 2
(t
0
). ThenI j 1
(t
0
)=sn12F(x),andheneI j
(t
0
)=(sn1)n12F(x)as
well.
Corollary1.2.
t (I
j
(x))=I j+k
(t
0
)foreveryj2Z.
Proof: Wehave
t (I
j
(x)) =(I j
(t))=(I j
(I k
(t
0
))) =(I j+k
(t
0
)). However,
I j+k
(t
0
)isredued,byLemma1.1.
Fors2W(x)denetheweightjsjas2i+j,whereiisthenumberofourrenes
ofxandj isthenumberofourrenesof1. Forexample,j1=(1=x 2
)j=6.
Leta;b2W(x). Thenabanmeananyof ab,a=bandanb. If morethan
oneoperationisinvolved,weshallalsouseaÆb.
Lemma1.3. Lets
0
betheI-oreof s2F(x)andletj beitsI-depth. Then
t (s)=
8
<
:
1 if s=1;
I j+k
(t
0
) if s
0
=x;
I j
(
t (a)
t
(b)) if s
0
=ab:
Furthermore,themapping
t
:F(x)!F(x)isinjetive.
Proof: Itisobviousthat
t
(1)=1. Corollary1.2givestheformulafors=I j
(x).
Fortherestweshall proeedbyindution onjsj. Theindution steponsistsof
showingthat
(a)
t (s)=I
j
(
t (a)
t
(b))whereabistheI-oreofsandjistheI-depth
ofs;andthat
(b)
t (s)=
t (s
0
)impliess=s 0
ifs;s 0
2F(x)andjsjjs 0
j.
Ifjsj2,thens=1ors=x. Part(a)isvoidlytruesine(a)assumess=ab.
Part(b)isobvious.
Toprove(a)forjsj3weneedtoshowthatI j
(
t (a)
t
(b))isredued. Note
that
t
(a)=1impliesa=1bypart(b)andtheindutionassumption. However,
ifa=1,theneitherabisnotredued,orjisnottheI-depthofs. Henea6=1
and
t
(a)6=1. Similarlyb6=1and
t
(b)6=1. ThereforeI j
(
t (a)
t
(b))2F(x)
if
t (a)
t
(b) 2 F(x). That follows by indution if j 6= 0. Assume j = 0
and suppose that there is a rule in Table 1 that applies to
t (a)
t
(b). We
have observed that it anbe none of the four rules that involve 1. In view of
theleft-right(mirror) symmetrywean assumethat
t
(b) =uÆv andthat the
rewritingrule mathes
t
(a)(uÆv). (Henethe rewritingrule mustbeoneof
t
1 n(t
1 t
2 )!t
2 ,t
1
=(t
2 nt
1 )!t
2 andt
1 (t
1 nt
2 )!t
2 .) Letb
0
betheI-oreof b.
Weknowthatb
0
6=1. Assumeb
0
6=x. Bytheindutionassumptionthestruture
of
t
(b)opies thestrutureofb. Hene b=Ædwhereu=
t
()andv=
t (d).
From part (b) weknowthat if
t
(a) =
t
() then a =, and if
t
(a) =
t (d)
thena=d. Therewritingrulethatmathes
t (a)(
t ()Æ
t
(d))thusappliesto
s=a(Æd) aswell. That isaontraditionsinesisassumedto beredued.
Tonishtheproofof(a)itremainstotreattheaseofb
0
=x. Thenb=I r
(x)
forsomer2ZanduÆv =
t (b)=I
r+k
(t
0
),byCorollary1.2. Frompart(a)of
theindution assumptionandfrom Corollary1.2weseethatj
t
(a)jjt
0 j. Both
uandv aresubtermsoft
0
ifr+k=0. Insuhaasejuj<j
t
(a)j,jvj<j
t (a)j,
and none of the above mentionedthree rewriting rules mathes
t
(a)(uÆv).
Thus r+k6= 0and theoperation Æis equalto nor=. Noneof thethree rules
allowsthealternativeof=,and soÆequalsn. Thatmeans r+k>0and v=1.
Sine
t
(a) 6= 1, the onlypossibility for simpliation is that of u(un1) ! 1.
Fromun1=
t
(b)weseethat theweightoftheI-oreofuisequaltojt
0
j. Ifthe
I-oreofa isdierent from x,then theI-oreofu=
t
(a) is ofweightat least
2jt
0
j, by part(a) of the indution argument. Hene a= I q
(x) for someq 2 Z.
Then
t (a)=I
q+k
(t
0
)=uwhih yieldsr=q+1ands=I q
(x)I q+1
(x). This
Toprove(b)rstnotethatj
t
(s)j>1ifs 6=1. Henes 6=1anbeassumed.
Byonsidering againtheweightsof I-ores, thistime with respetto
t (s)and
t (s
0
),weeasily distinguishthe asewhen theI-oreof s isequal tox andthe
I-oreofs 0
isnotequaltox(orvieversa). Now,Corollary1.2anbeemployed
ifbothI-oresare equalto x. Suppose thatnone oftheI-oresequalsx. Then
theI-depth of
t
(s)agreeswiththeI-depthofs,andhene(b)followsfrom(a)
byadiretindution argument.
Theorem1.4. Amapping':F(x)!F(x)isanendomorphismofthefreeloop
F(x)ifandonlyifthereexistst2F(x)suhthat'=
t
. Theendomorphism
t
isinjetiveif andonly if t6=1. It isan automorphismif andonly if t=I k
(x)
forsomek2Z.
Proof: Beause fxg is thefree base of F(x) there exists for everyt 2 F(x) a
unique endomorphism ' with '(x) = t. This endomorphism fulls '(s(x)) =
(s(t))foranys2F(x)andheneitagreeswith
t
. Ift6=1,then
t
isinjetive
byLemma 1.3. Of ourse,
1
maps everyelement of F(x) to 1. Letus assume
t 6=1 and lett
0
bethe I-oreof t. FromLemma 1.3 we see that j
t
(s)j jt
0 j
for every s 6= 1. Note that the endomorphism
t
is an automorphism if and
only if x 2 Im(
t
). Sine this annot happen if jt
0
j > 2 there must be t
0
= x
and t =I k
(x), where k is theI-depth of t. Insuh aase x =
t (I
k
(x)), by
Lemma1.3.
Corollary 1.5 (Evans). Aut (F(x)) isan inniteyli groupthat is generated
bythesubstitution x7!1=x.
2. Paratopisms,isostrophismsand terms
Quasigroupsanbe seenassets oftriples (a
1
;a
2
;a
3
)suh that twoelements
ofthetripleanbehosenfreelyfrom thegivenset Qwhile thethird elementis
determineduniquelybythishoie. Itisusualtoseta
3
=a
1 a
2 ,a
2
=a
1 na
3 and
a
1
=a
3
=a
2
. Put alsoa
3
=a
2 Æa
1 ,a
2
=a
3 nna
1 anda
1
=a
2
==a
3
. Inthis waywe
get sixquasigroup operationsthat arealled parastrophes. Theyare relatedby
permutations2S
3
. SaythatQ()isa parastrophe ofQ=Q()ifa
1 a
2
=a
3
is equivalent to a
(1) a
(2)
= a
(3)
. In other words, if we start from triples
(a
1
;a
2
;a
3
) where a
3
= a
1 a
2
, then thenew triples are obtainedby sending a
i
fromtheposition itotheposition (i). Itfollowsthatthe parastropheofa
parastropheisthe parastrophe.
IfQ
1 and Q
2
arequasigroups,then =(
1
;
2
;
3
)isanisotopism Q
1
!Q
2
ifall
i
arebijetionsQ
1
!Q
2 and
1 (x)
2 (y)=
3
(xy)forallx;y2Q
1 .
By ombining the notions of parastrophy and isotopy we get the notion of
paratopy. This term wasoined by Sade [24℄. It provides an algebrai frame-
workfortheombinatorialnotionofmain lasses. (The alternativeisostrophy =
isotopy +parastrophy hasadierentmeaning in thispaper. Admittedly, there
mayexist authorswhouseitasasynonym forparatopy.)
LetQ
1 andQ
2
bequasigroups. Thepair(;)=(;(
1
;
2
;
3
))issaidtobe
aparatopism fromQ to Q if :Q !Q is abijetion foralli2f1;2;3g,if
2S
3 andif
1
(1) (a
1
(1) )
1
(2) (a
1
(2) )=
1
(3) (a
1
(3) )
whenever a
1 a
2
=a
3
holds in Q
1
. It is not diÆult to dedue that is an iso-
topismfromQ
1
tothe 1
parastropheofQ
2
,andthatbyomposingparatopisms
(;) :Q
1
!Q
2
and (;):Q
2
!Q
3
weobtainaparatopism Q
1
!Q
3 . The
ompositionfollowstherule
(;)(;)=(;
); where (
1
;
2
;
3 )
=(
(1)
;
(2)
;
(3) ):
Hene(;) 1
=( 1
;( 1
)
1
). Therefore
b
1 b
2
=b
3 inQ
2
,
1
1 (b
(1) )
1
2 (b
(2) )=
1
3 (b
(3) )inQ
1 :
For aquasigroup Q,a set S, apermutation 2 S
3
and bijetions
i
:Q! S,
i 2 f1;2;3g, there exists a unique quasigroup struture on S suh that (;)
is a paratopism Q ! S. It is alled the quasigroup paratopially indued by
(;). Themultipliationandtheleftandrightdivisionsofsuhquasigroupsare
expliitlyshowninTable2foreah 2S
3 .
2S
3
multipliation leftdivision rightdivision
id
3 (
1
1
(x) 1
2
(y))
2 (
1
1 (x)n
1
3
(y))
1 (
1
3 (x)=
1
2 (y))
(123)
2 (
1
1 (y)n
1
3
(x))
1 (
1
3 (x)=
1
2
(y))
3 (
1
1 (y)
1
2 (x))
(132)
1 (
1
3 (y)=
1
2
(x))
3 (
1
1
(y) 1
2
(x))
2 (
1
1 (x)n
1
3 (y))
(12)
3 (
1
1 (y)
1
2
(x))
1 (
1
3 (y)=
1
2
(x))
2 (
1
1 (y)n
1
3 (x))
(23)
2 (
1
1 (x)n
1
3
(y))
3 (
1
1
(x) 1
2
(y))
1 (
1
3 (y)=
1
2 (x))
(13)
1 (
1
3 (x)=
1
2
(y))
2 (
1
1 (y)n
1
3
(x))
3 (
1
1
(x) 1
2 (y))
Table 2. Paratopiquasigroupoperationsinduedby(;)
LetQbealoop. PutI(x)=xn1andJ(x)=1=xforeveryx2Q. Thenboth
I
Q
=I andJ
Q
=J permuteQ,andJ =I 1
. FurtherpermutationsofQarethe
left translations L
a
: x 7!ax and the right translations R
a
: x 7!xa, for every
a2Q.
An isotopismof loops(
1
;
2
;
3
):Q!
Qis alledprinipal if
3
=id
Q . In
suh aasethere exist e;f 2Qsuh that
1
=R
f and
2
=L
e
. Furthermore,
Q=Q(Æ) where xÆy =(x=f)(eny) for allx;y 2 Q. LoopsQ(Æ) are known as
theprinipal isotopes ofQ.
Everyisotopism of loops : Q !
Q anbe written as(R
f
;L
e
;) where
e;f 2 Q. Thus it an be expressed as a omposition of an isomorphism :
Q(Æ)!
Qwithaprinipal isotopism(R
f
;L
e
;id
Q
):Q!Q(Æ).
A paratopims (;(
1
;
2
;
3 )) : Q
1
! Q
2
of loopsQ
1 and Q
2
will be alled
unital if (1)= (1)= (1)=1.
Lemma2.1. Let(;):Q
1
!Q
2
beaparatopismof loops. Thenthere exists
aunitalparatopism(;):Q
1
!Q
3
andaprinipalisotopism:Q
3
!Q
2 suh
that(;)=(id ;)(;).
Proof: Theinverseofaprinipal isotopismisaprinipal isotopism. Therefore
itsuÆestondaprinipalisotopism:Q
2
!Q
3
suhthat(id;)(;)=(;)
isaunital paratopismofloops.
Theisotopism will be ofthe form (R
f
;L
e
;id
Q2
) forsome e;f 2Q
2 . Then
1
(1)
= R
f
1
(1) ,
1
(2)
= L
e
1
(2) and
1
(3)
=
1
(3)
. Put e =
1
(1)
(1)and f =
1
(2)
(1). Ineveryloop11=1. Thusef =
1
(3) (1) =
1
(3) (1)=
1
(2) (1)=
1
(1)
(1). Theelementef servesastheunitofQ
3
.
Lemma2.2. LetQbealoop,S aquasigroup,and(;) aparatopism Q!S
suh that
i
(1)=1foreveryi2f1;2;3g. Then S is aloopifandonly ifthere
existsabijetion:Q!S,(1)=1,suhthat
(a) =(;;)if =id or=(12);
(b) =(I;;)if=(123)or=(2 3);and
() =(;J;)if=(132)and =(1 3).
Proof: Assume,forexample, that =(1 23). ByTable2 theoperationin S
anbeexpressed as
2 (
1
1 (y)n
1
3
(x)). Setting y =1yields
2
=
3
. Denote
this mapping by . Setting x = 1 yields y = I 1
1
(y) for all y 2 Q. Thus
1
=I. Otherasesaresimilar.
Foreahunitalparatopism(;)ofloopstherethusexistsa(unique)bijetion
suhthatthereareatleasttwodistinti;j2f1;2;3gwith
i
=
j
=. Aunital
paratopism is fully desribed by the pair(;). We shall say that it is arried
by (;). In Table3 we reord expliitlythe multipliation in S when Q! S
isa unitalparatopism arried by(;). The table anbe obtainedbyapplying
Lemma2.2toTable2. ForeveryloopQthesearetheloopsparatopially indued
by(;).
id (
1
(x) 1
(y))
(123) (J 1
(y)n 1
(x))
(132) ( 1
(y)=I 1
(x))
(12) ( 1
(y) 1
(x))
(23) (J 1
(x)n 1
(y))
(13) ( 1
(x)=I 1
(y))
Table3. Paratopiallyinduedloopoperations
LetQbealoop. Theloopparatopiallyinduedby((12);id
Q
)istheopposite
loopQ op
, while ((23);id
Q
) and(1 3);id
Q
) indue theleft inverse loop and the
right inverseloop ofQ,respetively.
Leftandrightinverse loopsandtheoppositelooparespeialasesof isostro-
phes ofQ. A loopis saidto beanisostrophe ofQ ifitis paratopiallyindued
by(;I m
),for somem2Zand 2S
3
. A (unital)paratopismQ!S is alled
anisostrophismifitisarriedby(;I m
)forsomem2Zand2S .
Lemma 2.3. LetQ
1
! Q
2
be aunital paratopismthat is arried by(;#). If
':Q
0
!Q
1
and :Q
2
!Q
3
areisomorphismsofloops, then(; #') arries
aunitalparatopism Q
0
!Q
3 .
Proof: Thisfollowsdiretlyfrom theruleforompositionofparatopisms.
Corollary 2.4. Every unital paratopism anbe expressed asa omposition of
anisomorphismandof anisostrophismthat is arriedby(;id
Q
),where Qis a
loopand2S
3 .
Proof: CombineLemmas 2.2and2.3.
The set of all isostrophes of Q will be denoted by I(Q). We an thus say
that I(Q) onsistsofall possibletargets forisostrophismsstartingfrom Q. Iso-
strophisms from Qto Q ouldbe alled autostrophisms. However,weshallnot
usethis termin thispaper. AutostrophismsofQ orrespondto theelementsin
thepointstabilizerofQinthegroupI(Q)(thegroupisdenedin Setion3).
Forapermutation 2S
3
denethe sign sgn()=" sothat"=1if isan
evenpermutation and"= 1if isanoddpermutation(atransposition).
Lemma 2.5. Consider aunital paratopism of loops Q! S that is arried by
(;). PutI =I
Q
. ThenI
S
=I sgn()
1
.
Proof: Suppose that x;y 2 S aresuh that xy =1. We shall useTable3. If
= (1 2 3), then J 1
(y)= 1
(x) and so y = I 1
(x), asrequired. Other
asesaresimilar.
Lemma2.6. Aompositionoftwoisostrophismsisagainanisostrophism. The
inverseofanisostrophismisalsoanisostrophism.
Proof: Let (;) : Q
1
! Q
2
be a unital paratopism of loops. Put I = I
Q1 .
From Lemma 2.2 we see immediately that this paratopism is an isostrophism
if and only if there exist k
i
2 Zsuh that
i
= I ki
for all i 2 f1;2;3g. Let
(;) : Q
2
! Q
3
be another paratopism of loops. If (;) is an isostrophism,
thenI
Q
2
=I 1
byLemma 2.5. Ifboth(;) and(;) areisostrophisms,then
there exist `
i
suh that
i
=I
`
i
. Formulas for the omposition and inverse of
paratopismsyieldtherest.
Corollary 2.7. Let Q
1
and Q
2
be loops. Then Q
1
2 I(Q
2
) if and only if
Q
2 2I(Q
1 ).
Proof: For loops A and B on a set S write (A;B) 2 if and only if B 2
I(A). Therelation issymmetriand transitivebyLemma 2.6. Hene itisan
equivalene.
Weshallnowdesribeanotherapproahtoisostrophy. Itisinspiredbynotions
ofuniversalalgebra. AloopQ
2
issaidtobeatermparatope ofaloopQ
1
ifthere
exist terms t
i
2 F(x), 1 i 3, and 2 S
3
suh that (;) : Q
1
! Q
2 is a
paratopism,where
i (u)=t
i
(u)foreahu2Q
1
. (LoopsQ
2 andQ
1
areassumed
Term paratopyis aspeialaseof amoregeneralonept: LetQ
1
bealoop
with binary operationsxy, xny and x=y, theunit ofwhih is equalto 1. Let
Q
2
bealoopuponthesameunderlying set,andwith thesameunit 1. Suppose
that thethree binaryoperationsof Q
2
anbeexpressed ast
1 (x;y), t
2
(x;y)and
t
3
(x;y), wherethe termst
1
;t
2
;t
3
2F(x;y)are evaluated in Q
1
. If weanpass
fromQ
2 toQ
1
in asimilarway,weallQ
1 and Q
2
term equivalent.
FromCorollary2.7and fromTables2and3wesee thateveryisostropheofa
loopQisatermparatopeofQ. Henewehave:
Corollary 2.8. Let Q
1 and Q
2
beloops suh that Q
2 2I(Q
1
). Then Q
1 and
Q
2
aretermequivalent. Furthermore,Q
1
isaterm paratopeof Q
2 andQ
2 is a
termparatopeof Q
1 .
Corollary 2.9. AloopisatermparatopeofthefreeloopF(x) ifand onlyifit
isanisostropheof F(x).
Proof: This follows from Theorem 1.4 sineu7!t
i
(u) doesnotpermuteQ =
F(x)ift
i
isnotoftheformI m
(x).
Isostrophesarehenetheonlytermparatopesthatanbeonstrutedwithout
assumingsomeadditionalequationalpropertiesoftheloopQ.
Term equivalene is astandardnotion of universal algebra. Term equivalent
algebrasshare subalgebras and ongruenes. This is easy to verify, and in the
aseofloopstheproofiseveneasier. Weanhenestate:
Proposition2.10.LetQ
1 andQ
2
betermequivalentloops. ThenSisa(normal)
subloopof Q
1
ifandonlyifitisa(normal)subloopof Q
2
. Inpartiular,thisis
trueif Q
2
isanisostropheof Q
1 .
Afurtherdisussionofonnetionsbetweenlooptermsandisostrophyanbe
foundinSetions6and7.
3. Isostrophismsand their groups
Letusinvestigatewhat exatlyhappenswhenweomposetwoisostrophisms.
Asanexampleonsider'Æ ,whereQ,RandSareloopswiththesameunderlying
set,and =((1 3);(I;id;I)):Q!R and'=((123);(I;id ;id)):R !S are
paratopisms. Notethat I in meansI
Q
,whileI in'meansI
R
. ByLemma2.5,
tobase bothparatopisms inQwehavetoreplaeI in 'byJ =I 1
. Usingthe
formulaforomposingparatopismsweanexpress'Æ as
((23);(I;id;id))=((123);(I;id;id))Æ((13);(I;id ;I)):
Inthis equality I means I
Q
in the outer triples,and it meansI
R
in themiddle
triple. LoopsQ,RandS sharethesameunderlyingset, andheneid
Q
=id
R
=
id
S
. ThehoieofRisdeterminedby ,whilethehoieofSisdeterminedby'.
Theequalityanbethusseenastruerelativeto Q. Ofourse, itistrueforany
hoie of aloopQ. Thereforewean viewtheequality asarulethat expresses
at upon the lass of all loops. For set theoretial reasons we annot dene a
mappinguponthelassofallloops. However,weandene'and asmappings
uponanysetofloopsthatislosedunder isostrophes.
Ournextaimistodetermineageneralomposition rule,i.e.to desribebya
formulatheisostrophismthatisobtainedwhenthereisomposedanisostrophism
thatisarriedby(;I n
)withanisostrophismthatisarriedby(;I m
). Inevery
givenasetheresultanbeomputedsimilarlyasabove. Table4givestheresults
forsituations when m=n =0. The table usesanabbreviatedform in whih 0
representsid ,andspaes,ommasandouterparenthesesaresuppressed.
0(000) (123)(I00) (132)(0J0) (12)(000) (23)(I00) (13)(0J0)
(123)(I00) (132)(I0I) 0(000) (13)(0J0) (12)(000) (23)(0JJ)
(132)(0J0) 0(000) (123)(0JJ) (23)(I00) (13)(I0I) (12)(000)
(12)(000) (23)(I00) (13)(0J0) 0(000) (123)(I00) (132)(0J0)
(23)(I00) (13)(I0I) (12)(000) (132)(0J0) 0(000) (123)(0JJ)
(13)(0J0) (12)(000) (23)(0JJ) (123)(I00) (132)(I0I) 0(000)
Table 4. Compositionsofisostrophismsthat arearriedbytheidentity
Proposition 3.1. Let :Q!R and':R!S beisostrophismssuhthat
isarried by(;I m
Q
)and 'by(;I n
R
). Then ' isanisostrophismQ!S that
isarriedby(;I k
Q
)wherek=m+sgn()n+d(;)andwhered:S
3 S
3
!
f0; 1;1gisdeterminedbythefollowingtable:
id (123) (132) (12) (23) (13)
id 0 0 0 0 0 0
(123) 0 1 0 0 0 1
(132) 0 0 1 0 1 0
(12) 0 0 0 0 0 0
(23) 0 1 0 0 0 1
(13) 0 0 1 0 1 0
Proof: Inthis proof we shalldenote by
0
(T)the isostrophismthat is arried
by(;id
T
), for every 2 S
3
and everyloop T. We see (f. Corollary2.4) that
=
0 (
Q)I m
Q
, where
Q is dened so that I m
Q
: Q !
Q is an isomorphism.
Similarly ' =
0 (
R )I n
R
. Put n 0
= sgn()n. We anexpress ' as
0 (
R )I n
0
Q , by
Lemma 2.5. Considernow the isostrophism I n
0
Q
0 (
Q). It is arried by (;I n
0
Q )
and so it equals
0 (T)I
n 0
Q
, where T is the loop suh that I n
0
Q :
Q ! T is an
isomorphism. Note that I
T
= I
Q
= I
Q
, by Lemma 2.5. We an express '
as
0 (
R )
0 (T)I
m+n 0
Q
. Thefat that
0 (
R )
0
(T)isarriedby(;I d(;)
T
)follows
TheompositionruleforisostrophismsthusinduesagrouponS
3
Zinwhih
(;n)(;m)=(;sgn ()n+m+d(;)):
ThegroupwillbedenotedbyI. Itatsinanaturalwayuponanysetofloops
that islosedunder isostrophes. Writing(;m)(Q)=R meansthatthere exists
a(unique) isostrophismQ!Rthat isarriedby(;I m
Q
). If(;n)(R )=S,then
((;n)(;m))(Q)=S.
ItanbeeasilyveriedthatIatsfaithfullyuponI(F)whenF isafreeloop.
Beforeinvestigating possiblekernelsofthe ationuponI(Q) forother loopsQ,
we shallrst study theabstrat nature of I. It is quiteeasy to seethat I isan
innitedihedralgroup.
Indeed,puts=((123);0). FromthedenitionofIweseethats 2
=((132);1)
and s 3
=(id ;1). Hene s 3k
=(id ;k)for every k 2 Zand sohsi =f(;i) 2I;
sgn() =1g. Put also o=((1 2);0), l=((2 3);0) and r =((1 3);0). Further
omputationsinI yieldthefollowingresults:
Proposition 3.2 (Artzy). The group I satises dening relations ho;s; o 2
=
1;oso=s 1
i. Furthermore,l=os,r=os 1
,r=s 2
landI=ho;li=ho;ri.
Inthe rest of this paper we shall treatI as an(innite dihedral) group that
isdetermined bythe dening relationsof Proposition 3.2, andshall notusethe
identiationofelementsofI aspairs(;m).
Reallthat identities J(x)(xy)=y, (xy)I(y)=x,J(xy)x=y, J(x)(yx) =y
and J(xy)=J(y)J(x)dene what is known asLIP, RIP, WIP, CIP and AAIP
loops. Therespetive\InverseProperty"isthusLeftorRightorWeakorCross
orAnti-Automorphi. Loopsthat are both LIP and RIPare alled IP (inverse
property)loops. Notethat Qisommutativeifand onlyifo(Q)=Q.
Lemma3.3 (Artzy). LetQbeloop. ThenQisanLIPorRIPorAAIPloopif
and onlyif l(Q)=Q, r(Q) =Q, oros 3
(Q) =Q, respetively. In eah of these
asesI =J.
Proof: The ase of LIP is immediate sine LIP an be learly expressed in a
weakerformthat foreveryx2Qthereexistsx 0
2Qsuhthatx 0
(xy)=y forall
x2Q. UsingTables3and4weseethatos 3
(Q)=QifandonlyifI(J(y)J(x))=
xy for allx;y 2 Q. TheequalityI =J is well known and easy(in aseof LIP
useJ(x)=J(x)(xI(x))=I(x),forAAIPemploy1=J(xI(x))=xJ(x)).
We have observed that the isostrophism s 3
equals (id ;(I;I;I)), and so it is
in fat anisomorphism. This fat isreordedin the nextlemmafor thesakeof
referene. TheinversesofloopsQandS oinide,say,byLemma2.5.
Lemma 3.4. Let Q be a loop. Put S = s 3
(Q). Then I = I
Q
= I
S is an
isomorphismQ
= S.
Supposethatthesetofallfk2Z;s k
(Q)=Qgisnontrivial. Thenthereexists
k
divisibleby3,thenthere existsauniquem2Zsuhthat j3m+1j=t. Insuh
aases 3m+1
(Q)=Q.
The isostrophism s 3m+1
is arried by ((1 2 3);I m
). Table 3 implies that if
s 3m+1
(Q)=Q,thenxy =I m
(JI m
(y)nI m
(x)) forallx;y 2Q. This isequiv-
alent to J m+1
(y)J m
(xy) =J m
(x). Everyloop satisfying suh an lawis alled
m-inverse [14℄.
Them-inverse lawan beequivalentlyexpressed asI m
(yx)I m+1
(y)=I m
(x)
[14℄, [8℄. A proofalong the lines of thispresentationan be obtainedif we put
m 0
= m 1 and note that s 3m
0
+2
is arried by ((1 3 2);I m
0
+1
). We have
m 0
+1= mandj3m 0
+2j=t,andsoTable3yieldsxy=J m
(I m
(y)=I m+1
(x)).
ThatisthesameasI m
(xy)I m+1
(x)=I m
(y).
Note that 0-inverse loops are the CIP loops (t = 1), and that ( 1)-inverse
loopsaretheWIPloops(t=2).
Proposition 3.5. LetQ bea loop and lett >0be suh that s t
(Q) =Q and
thatt istheleastpossible.
(i) If t = 3k, then I k
2 Aut (Q) and Q is not n-inverse for any n 2 Z.
Furthermore,I
`
2Aut (Q)ifandonlyif kdivides`.
(ii) If t=3k1,putm=k. Thent=j3m+1j. TheloopQisann-inverse
loopif andonly if 3m+1divides 3n+1. Furthermore,I
`
2Aut (Q)if
andonlyif 3m+1divides`.
Proof: If Q is an n-inverse loop, then we an reverse the proess desribed
above to show that s 3n+1
(Q) = Q. This is possible if and only if 3n+1 is
divisiblebyt. FortherestusethefatthatI k
2AutQifandonlyofs 3k
(Q)=Q
(Lemma3.4).
Thevalueofminthedenitionofanm-inverseloopanbethusseenasaway
ofoding thepositiveintegert =3k1. Up to nowthere isno evidene ofan
interestingalgebraitheory thatwouldinvolvem-inverseloopsfor highervalues
of jmj. Known onnetionsto other lasses of loopsare restrited to situations
when t is a small power of two. If t = 2 k
, then m = (( 2) k
1)=3. In suh
asesanm-inverse loopisalled [8℄aW k
IPloop(ithasthek-fold weak inverse
property). NotethatthenI 2
k
2Aut (Q),byProposition3.5.
Note alsothat a CIP loopis m-inverse foranym 2Z. Inpartiular, the CI
propertyimpliestheWIproperty.
The group I ats upon I(Q). The image of this ation will be denoted by
I(Q). Hene I(Q)isapermutationgroupthatis eithertrivial,orylioforder
two,ortheKleinfour-grouporanonommutativedihedral group. Thus I(Q)is
ommutativeifandonlyifjI(Q)j isadivisorof4.
Proposition 3.6. LetQbealoopsuhthatjI(Q)j divides4. Thenexatlyone
ofthefollowingasestakesplae:
(1) QisanonommutativeWIPloopthatis notIP;jI(Q)j=4.
(2) QisanonommutativeIPloop;jI(Q)j=2.
(4) QisanonommutativeCIPloop;jI(Q)j=2.
(5) QisaommutativeIPloop;jI(Q)j=1.
Proof: Our assumption an be also expressed by saying that s 2
ats trivially
upon I(Q). Hene Q is a WIP loop. From r = s 2
l we see that l(Q) = Q is
equivalenttor(Q)=Q. Thathappensexatlywhen Qis anIPloop. Eahofs,
oandlatsuponI(Q) either triviallyorasaninvolution. If noneof themats
trivially,thenwegetase(1). Cases(2){(4)desribesituationswhenexatlyone
ofthematstrivially(notethat anIPCIPloopisommutative).
A permutation group G on is said to be regular if it is transitive and if
g =id
wheneverg 2Gis suh thatg(!)=! for some! 2. If Qis aloop,
then I(Q)is transitive, but notneessarily regular. Weshall see that there are
onlyfewnonregularases. Atransitiveommutativepermutationgroupisalways
regular. Thereforeanite nonregularI(Q)hasto beisomorphito thedihedral
groupD
2n
forsomen3. Weshallseethatn2f3;6g. NotethatD
6
=S
3 .
A loop Q is said to haveautomorphi inverse property (AIP) if I 2 Aut (Q)
(i.e. I(xy) =I(x)I(y) for all x;y 2 Q. Equivalently J(xy)=J(x)J(y).) For a
looptohavetheAIpropertyitisnotneessarythatI =J. However,AIPloops
ourringinProposition3.7haveI =J (thenI(x)=J(x)iswritten asx 1
).
NotethatifQhastheAIP,theneveryelementofI(Q)hastheAIP.
NotealsothatI 2Aut (Q)ifandonlyifs 3
(Q)=Q,byProposition3.5.
Proposition 3.7. Let Q be aloop suh that I(Q) is notregular. Then oneof
thefollowingasestakesplae:
(1) I(Q)
= S
3
, jI(Q)j=3andthere exists auniqueommutativeAIPloop
Q
1
2I(Q) suhthat I(Q)=fQ
1
;s(Q
1 );s
1
(Q
1
)g. Thens(Q
1
)hasthe
LIPand the AIP, and s 1
(Q
1
) hasthe RIPand theAIP. On theother
hand,I(Q)
= S
3
andjI(Q)j=3wheneverQisanAIPloopthat isnot
IP,andisommutativeorRIPorLIP.
(2) I(Q)
= D
12
,jI(Q)j=6andthereexistinI(Q)twodierentommutative
loops Q
1 and Q
2
suh that I: Q
1
= Q
2
and I(Q) = fQ
i
;l(Q
i );r(Q
i );
i2 f1;2gg. On theother hand if Qis aommutativeloop withoutthe
AIP, thenjI(Q)j=6andI(Q)
=D
12 .
(3) I(Q)
= D
12
, jI(Q)j = 6and I(Q) onsists of two LIP loops, two RIP
loopsand twoAAIP loops,noneof whih isommutativeoranIPloop
oranAIPloop. Ontheotherhand,ifQisneitheranIPloopnoranAIP
loop,butitisanLIPlooporanRIPlooporanAAIPloop,thenitisnot
ommutative,I(Q)
= D
12
andjI(Q)j=6.
Proof: SupposethatI(Q)isnotregular. ThenitisisomorphitoD
2n
forsome
n 3. If I = J, then I 2
= id
Q
, and hene s 6
(Q) = Q. Thus n 2 f3;6g if
I = J. If nis odd, then I(Q) ontainsonly oneonjugay lass of involutions.
If n is even, then in I(Q) there are two lasses of nonentral involutions. One
ofthelassesontainsthenonentral involutionsthat arexedpointfree,while
thenland oyieldinvolutionsthat arenotonjugate. Weanthus assumethat
Qfulls l(Q)=Qoro(Q)=Q. BothasesimplyI =J (f.Lemma3.3),andso
n2f3;6g.
Suppose rst that n = 3. Then all elements of I(Q) are AIP loops sine
I =J2Aut (Q),byLemma3.4. Thereisonlyonelassofinvolutions,andsowe
anassumethatQisommutative. Thenls(Q)=os 2
(Q)=os 2
o(Q)=s 2
(Q)=
s(Q)sine l=os andsines 3
(Q)=Q. Thusl xess(Q)and, similarly, rxes
s 1
(Q). Thisprovesase(1),byLemma3.3.
Suppose nowthat n=6. If o(Q)=Q, then os 3
(Q) =os 3
o(Q)=s 3
(Q) =
s 3
(Q). We an thus put Q
1
= Q and Q
2
= s 3
(Q). Then I: Q
1
= Q
2 , by
Lemma 3.4, andtherest of ase(2)followsfrom l(Q
i
)=lo(Q
i )=s
1
(Q
i )and
r(Q
i
)=ro(Q
i )=s(Q
i
),i2f1;2g.
It remains to onsider the ase when n = 6and l(Q) = Q. Then ls 3
(Q) =
ls 3
l(Q)=s 3
(Q). Toprove(3) itthussuÆes toverifythat rs(Q) =s(Q) and
thatos 3
(s 1
(Q))=s 1
(Q),byLemma3.3. FromProposition3.2weobtainthat
rs(Q)=s 2
lsl(Q)=s 2
s 1
(Q)=s(Q)andthat os 2
(Q)=lsl(Q)=s 1
(Q).
Let us investigate more losely ase (2) of Proposition 3.7. It involves (a)
ommutative loops, (b) loops in whih the left inverse is ommutative and ()
loops in whih the right inverse is ommutative. Sine J
Q
= I
Q
when Q is
ommutative,thereis J
Q
=I
Q
inotherasesaswell.
Now,l(Q)isommutativeifandonlyifJ(x)ny=J(y)nxforallx;y2Q. The
latterlawanbeequivalently expressedas xny =J(y)nI(x) ory =J(xy)nI(x)
orJ(xy)y=I(x).
Proposition 3.8. LetQbealoop.
(i) If Qsatises forsome "; 2 f 1;1ga lawxny =J
"
(y)nJ
(x) oralaw
J
"
(xy)y =J
(x), then I =J, and Qsatises alleightthese laws. This
takesplaeifandonlyif l(Q)isaommutativeloop.
(ii) If Q satises forsome "; 2 f 1;1g alaw y=x=I
(x)=I
"
(y) oralaw
yI
"
(yx) = I
(x), then I =J, and Q satisesall eight these laws. This
takesplaeifandonlyif r(Q) isaommutativeloop.
Ifbothl(Q)andr(Q)areommutativeloops,thenQisaommutativeWIPloop
and l(Q)=r(Q). If l(Q) (or r(Q)) isommutativeand Q isnot ommutative,
then I(Q)
= D
12=d
and jI(Q)j =6=d, where d =2 if Q satises theAIP, and
d=1otherwise.
Proof: Ifxny =J
"
(y)nJ
(x), theny =J
"
(xy)nJ
(x)andJ
"
(xy)y=J
(x). If
J(xy)y=J(x),thenI(x)=J(xI(x))I(x)=J(x). IfJ(xy)y=I(x)orI(xy)y=
J(x),thenI(x)=J(x)anbeobtainedbysetting y=1. IfI(xy)y=I(x),then
I(y)y = 1, and soI(y) = J(y). We have already observed above that l(Q) is
ommutativeifandonly ifxny =J(y)nI(x) forallx;y 2Q. That provespoint
(i). Point(ii)followsbymirrorsymmetry.
Now,l=osandr=os 1
,byProposition3.2. Hene ol(Q)=l(Q),s(Q)=
2 1 1 2
Ifbothol(Q) =l(Q)and or(Q) =r(Q) aretrue, then s (Q) =s (Q) is om-
mutative,andheneQ
=s 3
(Q)isommutativeaswell,byLemma3.4. Insuha
aseQ=s 2
(Q),QisaommutativeWIPloopandweanuseProposition3.6.
If l(Q) is ommutative and Q is not ommutative, then no ase of Proposi-
tion3.6applies,andheneoneofasesofProposition3.7hastobesatised.
Loops that satisfy the equality J(xy)y = J(x) (i.e. loops in whih the left
inverse is ommutative) were introdued by Johnson and Sharma [13℄ and re-
entlystudied byGreer andKinyon[12℄. Theyareknownasweak ommutative
inverseproperty loops,orWCIPloops. Inthispaperweshallallthemleftross-
ommutativeloops. Loopsinwhihtherightinverseisommutativewillbealled
rightross-ommutative. BysayingthatQisross-ommutative wemeanthatit
isleftross-ommutativeorrightross-ommutative.
Situations that are not overed by Proposition 3.7 and Proposition 3.6 are
desribed in the following statement. The laims about the m-inversity follow
fromProposition3.5.
Proposition 3.9. Suppose that Q is neither WIPnor LIP norRIP norAAIP
loop,andthat itis neitherommutativenorross-ommutative. ThenI(Q)is a
regularpermutationgroupthatisisomorphieithertotheinnitedihedralgroup,
orto D
2n
, n 3. If n = 3k+", where " 2 f 1;1g, then Q is "k-inverse. If
n = 3k, then I k
2 AutQ (and Q is m-inverse for no m 2 Z). On the other
hand,if I(Q)isregularandnonommutative,thenQisneitherommutativenor
ross-ommutativenorWIPnorLIPnorRIPnorAAIPloop.
Lemma3.4 impliesthat I(Q) ontainsatmostsix isomorphismlasses. This
ispreisedin detailin Setion5.
4. Paratopismsand nulei
LetQbeaquasigroup. IsotopismsQ!Qarealled autotopisms. Theyform
a group that will be denoted by Atp (Q). An autotopism an be seen as a
paratopism(id ;):Q!Q,andvie versa.
Hene eah paratopism f = (;) : Q ! R yields an isomorphism f
:
Atp (Q)!Atp(R )that sends2Atp (Q)to()
1
2Atp (R ). Indeed,
(;)(id;)(;) 1
=(;)(
1
;( 1
)
1
)=(id;( 1
)
1
):
Foreveryi2f1;2;3gdenotebyAtp
i
(Q)thegroupofall(
1
;
2
;
3
)2Atp(Q)
with
i
=id
Q .
Lemma 4.1. Let f =(;) : Q ! R be a paratopism of quasigroups. Then
f
(Atp
i
(Q))=Atp
(i)
(R ) foreveryi2f1;2;3g.
Proof: If 2 Atp(Q), then 2 Atp
i
(Q) if and only if
i
= id
Q
. Now, the
(i)thomponentof ( 1
)
1
is equalto
i
i
1
i
. Clearly
i
i
1
i
=id
R if
andonlyif =id .
Weshallinlude awellknown fataboutnuleiof loops. Theproofissimple
enoughtowarrantomitting. Reallthat N
=N
(Q)=fa2Q;a(xy)=(ax)y
forall x;y 2Qgis knownastheleft nuleus,while themiddle and right nulei
N
andN
areobtainedbyshiftingtotherightthepositionofa.
Lemma 4.2. LetQ bea loop. Then Atp
1
(Q) equals f(id
Q
;R
a
;R
a
); a 2N
g,
Atp
2
(Q) equals f(L
a
;id
Q
;L
a
); a 2 N
g, and Atp
3
(Q) equals f(R 1
a
;L
a
;id
Q );
a2N
g.
TheonnetionmakesunderstandablewhyAtp
i
(Q)isalledanA
i
-nuleusby
someauthors. Lemma4.2makeslearthatforloopstheonstrutofAtp
i (Q)is
notneeded,unlessitanbeemployedwithadvantageinaproof. Thisisexatly
whatweshalldobelow. Tomaketheonnetiondiret,wedubN
(Q)asN
1 (Q),
N
(Q)asN
2
(Q) andN
(Q)asN
3 (Q).
Lemma4.3. Let(;) :Q!R be aparatopismof loopssuhthat
i (1)=1
foralli2f1;2;3g. Then
N
(i)
(R )=
j (N
i
(Q)) forall i;j2f1;2;3g suhthat i6=j:
Proof: Let i and j be as assumed. By Lemma 4.2, elements of N
i
(Q) are
exatlythose that anbeexpressed as
j
(1) for =(
1
;
2
;
3
) 2Atp
i (Q). If
2Atp
i
(Q),then( 1
)
1
2Atp
(i)
(R )byLemma4.1. ElementsofN
(i) (R )
an be expressed as
(j)
(1), where 2 Atp
(i)
(R ), by Lemma 4.2. If =
( 1
)
1
, 2Atp
i
(Q), then
(j)
=
j
j
1
j
. Thus
(j)
(1)=
j (
j (1))2
j (N
i
). We haveproved that
j (N
i
(Q)) N
(i)
(R ). By onsidering (;) 1
we get 1
j (N
(i)
(R )) N
i
(Q), and hene the required equality really takes
plae.
If (;) : Q ! R is an isostrophism, then N
(i)
(R ) = N
i
(Q) sine
j is a
powerofI. Table5showsthenuleioftheisostrophethatappearsintheseond
olumn(says 3k +1
(Q)). Thevalueof isin therstolumn ((123) fors 3k +1
),
and olumns 3-5 show the soures fornulei of the given loop in the order N
,
N
andN
. Forexampleappearsintheolumn3in therowofs 3k +1
(Q),and
thatmeansthatN
(s
3k +1
(Q))=N
(Q).
id s
3k
(Q)
(1 23) s 3k +1
(Q)
(1 32) s 3k +2
(Q)
(1 2) os 3k
(Q)
(2 3) ls 3k
(Q)
(1 3) rs 3k
(Q)
Table5. Isostrophiesandtheinterdependeneofnulei
ThefatthatisostrophismsswiththenuleiwasobservedalreadybyArtzy[3℄.
HealsonotedtheonsequenesforLIP,RIPandAAIPloops.
Inm-inverseloopss 3m+1
(Q) =Q,and soTable5shows thatall threenulei
haveto oinide. ThatwasprovedbyKarklinsand Karklin[14℄in adiretway.
Wereordtheseresultsinthenextstatement. Theproofanbederiveddiretly
fromTable5.
Proposition 4.4. If Qisanm-inverseloop,thenN
=N
=N
. If Qhasthe
LIP,thenN
=N
. If QhastheRIP,thenN
=N
. If QhastheAAIPoris
ommutative,thenN
=N
.
KarklinsandKarklin[14℄alsonote that N(Q)=Z(Q) ifQis2k-inverse. We
shall explain this phenomenon in Corollary 4.8. As a preparatory step let us
reordthefollowingeasyfats:
Lemma 4.5. Let Q be a loop. If a 2 N
\N
, then I(ax) = I(x)a 1
and
J(xa)=a 1
J(x). If a2N
,thenI(xa)=a 1
I(x)and J(ax)=J(x)a 1
.
Proof: Fullling I(ax)=I(x)a meansfullling 1=(ax)(I(x)a). That learly
holdsifa2N
\N
. Theotherasesan beprovedsimilarly.
Corollary4.6. LetQbealoop. Ifa2N(Q),x2Qandk2Z,then
I 2k
(ax)=aI 2k
(x); I 2k
(xa)=I 2k
(x)a;
I 2k +1
(ax)=I 2k +1
(x)a 1
and I 2k +1
(xa)=a 1
I 2k +1
(x):
Proof: ProeedbyindutionusingLemma 4.5.
Theorem 4.7. LetQ bealoop. Then I 2k +1
2 Aut (Q)for somek 2 Zifand
onlyif sisofanoddorderinI(Q). InsuhaaseN(Q)=Z(Q).
Proof: ByLemma3.4,s 3r
(Q)=QifandonlyifI r
2Aut(Q). Ifthisistruefor
anoddr, thenI r
(xa)=I r
(x)I r
(a)=I r
(x)a 1
foreveryx 2Q anda2N(Q).
However,byCorollary4.6wealsohaveI r
(xa)=a 1
I r
(x).
Corollary4.8(KarklinsandKarklin). If Qisa2h-inverseloopforsomeh2Z,
thenN(Q)=Z(Q).
Proof: IfQis2h-inverse,thenI
`
2Aut(Q)for`=6h+1,byProposition3.5.
5. Isomorphismsand the leftand rightinverses
Lemma5.1. Supposethat 2S
3
, andthat (;
i
)is aquasigroup paratopism
Q
i
!R
i
,i2f1;2g. ThenQ
1
isisotopitoQ
2
ifandonlyif R
1
isisotopitoR
2 .
Proof: AnisotopismR
1
!R
2
anbeobtainedfromanisotopism:Q
1
!Q
2
asaomposition(;
2
)(id;)(;
1 )
1
=(;
2 )(
1
;( 1
1 )
1
)=
(id ;
2 (
1
)
1
).
Proposition 5.2. Let Q
1 and Q
2
be loops and let f 2 I be an isostrophism.
ThenQ
1
=Q
2
ifand onlyif f(Q
1 )
=f(Q
2
). Furthermore,Q
1
is isotopitoQ
2
ifandonlyif f(Q
1
)isisotopitof(Q
2 ).
Proof: Thepartaboutisotopyfollowsfrom Lemma5.1. Fortheisomorphisms
just notethatif':Q
1
= Q
2
,then'(t(x;y))=t('(x);'(y))foranyt2F(x;y).
Proposition 5.3. Let Q be an m-inverse loop for some m 2 Z. Then every
elementof I(Q)isisomorphitoQortoQ op
. IfQisommutative,thenQ=Q op
.
If QisanAAIPloop,thenQ
= Q
op
aswell.
Proof: IfQ
1
;Q
2
2I(Q)areinthesameorbitofs,thenthereexistsk1suh
that s 3k
(Q
1 )=Q
2
sinethelengthof theorbitis j3m+1j. ByLemma3.4 this
settlestheaseofm-inverseloops. Therestisobvious.
Proposition5.4. LetQbealoopthatisnotanIPloop. If QisaLIPorRIPor
AAIPorommutativeloop,thenI(Q)ontainsexatlythreeisomorphismtypes.
Theyarerepresentedbys k
(Q),jkj1.
Proof: By Proposition 3.7 eah element of I(Q) an be expressed as s k
(Q),
k2Z. Henewegetallpossibleisomorphismtypesifkisrestritedto 1,0and
1,byLemma3.4. Weneedtoprovethatnotwoofthemmaybeisomorphi. For
ases(1)and(3)ofProposition3.7thisfollowsfromthefataloopisanIPloop
ifitsatisesatleasttwooftheLI,RIandAAIproperties. Suppose nowthatQ
isommutative. Thenoannotxs k
(Q)fork2f 1;1;2gsineQands 3
(Q)are
theonly pointsof I(Q) that arexed byo. If s 1
(Q)
=
s(Q), thenQ
= s
2
(Q)
byProposition5.2. However,theommutativeloopQannotbeisomorphitoa
nonommutativeloops k
(Q),k2f 1;1;2g.
ByProposition3.9,theonlyasesnotoveredbyPropositions5.3and5.4are
thosefor whih I(Q) isregularand, if nite, oforder 6k, k1. Forsuhloops
weanusethefollowinggeneralstatement:
Theorem 5.5 (Artzy). Let Qbea loop. Then I(Q) ontains1 or2or3 or6
isomorphismlasses.
Proof: ByProposition5.2isomorphiloopsyielduponI(Q)asetofonjugate
bloks. Consider the ation of I(Q) upon this set. The kernel of the ation
ontains s 3
, by Lemma 3.4. The image of the ation is hene equivalent to a
transitiveationofS
3
sineI=hs 3
i
= S
3
.
Intherestofthissetionweshalladdressthefollowingquestion: Startingfrom
Qiterativelyonstrutleftand rightinverses. WhendowegetfullI(Q)?
Note rst that by Proposition 3.2 the subgroup hl;ri I is of index 2, and
equalshl;s 2
i=hos;s 2
i=fs 2k
;os 2k +1
;k2Zg.Hene eitherhl;ri(Q)=I(Q),or
hl;ri halvesI(Q) into twodierent orbits. Intheformer aseweshallsay that
ItislearthatQisofoddtypeifitisommutativeorifitisaCIPloop. WIP
loops are those loops Q for whih l(Q) = r(Q), and so nonommutative WIP
loopsareofeventype,byProposition3.6.
ThenonregulargroupsofProposition3.7areofoddtypeinases(1)and(2),
andofeventypeinase(3).
SupposethatI(Q)isregularnonommutative(Proposition3.9). Ifitisinnite,
thenitisofeventype,andthatisalsotrueintheniteaseif4dividesjI(Qj=
jI(Q)j. Theremainingasesareofoddtype.
WethusknowwhenQisofoddoreventypeinallases. UsingProposition3.5
it is easy to verifythat ourresults an be formulatedin thefollowingompat
way:
Theorem5.6. AloopQisofoddtypeif I(Q)ontainsaommutativeloopor
ifthereexistsk2Zsuhthat I 2k +1
2Aut (Q).
Weanthusrestate Theorem 4.7as: If aloop is of odd type, then the entre
andthe nuleusoinide.
Lemma5.7. AloopQisofoddtypeifandonlyiftheationsof landrgenerate
thegroupI(Q).
Proof: IfI(Q)isregular,thenthereisnothingtoprove. SoitsuÆestoverify
thatlandrgenerateI(Q)inases(1)and(2)ofProposition3.7. Thatiseasy.
TheharaterizationofoddtypeloopsinTheorem5.6givesimmediately:
Corollary 5.8. A subloopor afatorloop of an oddtype loop is an oddtype
loop.
Proposition 5.9. Let Q be an oddtype loop. Suppose that V U Q are
subloopssuhthatVUandthatU=V isanIPloop. ThenU=V isommutative.
Proof: The loop U=V is of an odd type by Corollary 5.8. Hene I(U=V) is
generatedbylandr,byLemma5.7. Sineweareassumingl(U=V)=r(U=V)=
U=V,thesetI(U=V)hastoontainonlyoneelement. Thuso(U=V)=U=V.
6. Isostrophialvarieties
Sometimesitisusefulto denoteaquasigroupoperationbyaletterinsteadof
bya binaryoperator. If Q(A)is aquasigroup,then by A
we shalldenote the
parastrophe. (Thisisanadhonotationthatwillbeusedonlyintherstpart
of this setion.) Thus A= A
id
, and ifA(x;y) =xy, then A
(23)
(x;y) =xny,
A
(13)
(x;y)=x=yet. WehaveA
(a
1
;a
2 )=a
3 ,A(a
(1)
;a
(2) )=a
(3) ,whih
wereordintheform
A
(a
(1)
;a
(2) )=a
(3)
, A(a
(1)
;a
(2) )=a
(3) :
Lemma 6.1. Suppose that f = (;) : Q(A)! S(B) is a paratopism. Then
B(x;y)=
1
(3) (A
(
1
1
(1) (x);
1
1
(2)
(y))). If 2S
3 ,then
B
(x;y)=
1
1
(3) (A
(
1
1
1
(1) (x);
1
1
1
(2) (y))):
Proof: Thefatthatf isaparatopismanbeexpressedby
B
(
1
1
(1) (a
1
1
(1) );
1
1
(2) (a
1
(2) ))=
1
1
(3) (a
1
1
(3) )
as( 1
1
)=. Setx=
1
1
(1) (a
1
1
(1)
)andy=
1
1
(2) (a
1
1
(2) ).
Ourformulastatesthat B
(x;y)=
1
1
(3)
(z),where z =a
1
1
(3)
is equal
toA
(a
1
1
(1)
;a
1
1
1
(2)
). Bythehoieofx,a
1
1
(1)
= 1
1
1
(1) (x).
Theseondargumentdependsuponyinasimilarway,andthatgivestherequired
expressionofB
(x;y).
Theabovelemma is nothingelse, but aformalveriationthat if f =(;)
is a paratopism, then is an isotopism to the 1
parastrophe of the target
quasigroup |afat that hasbeenmentionedin Setion 2. Sine theoperation
B depends fully upon f and A, we an denote it by f(A). Note that Table 2
tabulatesf(A)fortheallpossiblevaluesof .
Lemma 6.2. Suppose that f : Q
1
! Q
2
and g : Q
2
! Q
3
are paratopisms.
Denotethequasigroupoperationof Q
1
byA. Thenthequasigroupoperationof
Q
3
anbeexpressedbothas g(f(A)) andas(gf)(A).
Proof: Sine gf is a paratopism Q
1
! Q
3
, the operation of Q
3
is equal to
(gf)(A). However, it is also equal to g(B), where B = f(A) is the operation
ofQ
2
.
Lemma6.3. ConsiderafreeloopF(X). Thenf(F(X))isalsoafreeloopwith
baseX,foreveryf 2I.
Proof: Every loop an be expressed as f(Q), for some loop Q. A mapping
':X !Q anbe extendedtoa(unique) loop homomorphism :F(X)!Q.
Bytermequivalene,amapping :F(X)!Qis ahomomorphismifandonly
ifitis ahomomorphismf(F(X))!f(Q).
SupposenowthatX =fx
1
;x
2
;:::g. ByLemma6.3thereexistsauniqueloop
homomorphismf
:F(X)!f(F(X))suhthat f
(x
i )=x
i
foreveryi1. To
omputef
(t) for aterm t use either Lemma 6.1 orTable2. Note that f
is a
mappingfrom F(X) to F(X), and heneit maps a reduedloop term upon a
reduedloopterm.
Lemma6.4. If f;g2I,theng
f
=(fg )
. Inpartiular,(f
) 1
=(f 1
)
.
Proof: DenotetheoperationofF(X)byA. Thenf
:F(X)(A)!F(X)(f(A))
andg
:F(X)(A)!F(X)(g (A))areloophomomorphisms. Hene
F(X)!F(X)(fg (A))isaloophomomomorphismaswell. This homomorphism
isidentialuponX,andheneithastoagreewith(fg )
.
Lemma 6.5. Suppose that Q is a loop, f 2 I, s;t = F(x
1
;:::;x
m
) and that
a
1
;:::;a
m
2Q. Thens(a
1
;:::;a
m
)isequaltot(a
1
;:::;a
m
)inf(Q) ifandonly
if (f
(s))(a
1
;:::;a
m
)isequalto(f
(t))(a
1
;:::;a
m )inQ.
Proof: PutF =F(x
1
;:::;x
m
)anddenoteby thehomomorphismF !f(Q)
that sendsx
i toa
i
. Furthermore,denoteby'the homomorphismf(F)!f(Q)
thatsendsx
i toa
i
. Thehomomorphisms and'f
agree upon x
1
;:::;x
m ,and
henetheyagreeeverywhere. Sine'anbealsointerpretedasahomomorphism
' : F ! Q wean write the equality (f
(s))(a
1
;:::;a
m ) = (f
(t))(a
1
;:::;a
m )
(whih is assumedto betrue in Q)as'(f
(s)) ='(f
(t)). This isthe sameas
(s)= (t),andthat meansthat s(a
1
;:::;a
m )=t(a
1
;:::;a
m
)in f(Q).
Corollary 6.6. Let V be a variety of loops and let f 2 I. Then the lass of
all f(Q), Q 2 V is also a variety of loops (we shall denote it by f(V)). A law
s(x
1
;:::;x
n
) = t(x
1
;:::;x
n
) holds in f(V) if and only if the law f
(s)= f
(t)
holdsinV.
VarietiesVandf
(V)aresaidtobeisostrophi. ByLemma3.4,Q
= s
3
(Q)for
anyloopQ. Henef(V)=fs 3k
(V). WeseethatS
3
atsuponvarietiesisostrophi
toV.
Corollary 6.7. There are 1or2 or3or 6varieties isostrophito avariety V.
EverysuhvarietyisequaltoV of l(V)orr(V),oritisavarietythatisopposite
tooneofthese threevarieties.
Todesribeisostrophivarietiesitthus suÆestobeableto expressthemul-
tipliationanddivisionsin o(Q),l(Q),r(Q) andr(Q). WedosoinTable6.
loop Q o(Q) l(Q) r(Q)
multipliation xy yx (1=x)ny x=(yn1)
leftdivision xny y=x (1=x)y 1=(ynx)
rightdivision x=y ynx (y=x)n1 x(yn1)
Table 6. Operationsin isostrophiloops
Proposition 6.8. Assume m 2 Z. Every variety isostrophi to the variety of
m-inverseloopsisequaltothatvariety.
Proof: A loopQis m-inverse ifitfulls J m+1
(x)J m
(yx) =J m
(y). Thelatter
lawisequivalenttoI m
(xy)I m+1
(x)=I m
(y)(f.Setion 3). Now,
o
(J m+1
(x)J m
(yx))=I m
(xy)I m+1
(x)ando
(J m
(y))=I m
(y). ThusQ op
isalso
anm-inverseloop. ThestatementthusfollowsfromProposition5.3.
Thevarietyofallloopsthatfulll alaws(x
1
;:::;x
m )=t(x
1
;:::;x
m
)will be
denotedbyEq[s(x
1
;:::;x
m )=t(x
1
;:::;x
m
)℄. InformulasweshalluseLIP,RIP
Lemma6.9. Supposethat";2f 1;1g. Then:
(i) LIP=Eq[I
"
(x)(xy)=y℄=Eq[(x=y)(y=x)=1℄;
(ii) RIP=Eq[(yx)I
"
(x)=y℄=Eq[(xny)(ynx)=1℄;and
(iii) AAIP=Eq[I
"
(x)I
(y)=I(yx)℄=Eq[J
"
(x)J
(y)=J(yx)℄.
Proof: InaLIP loopI(x)=J(x)by Lemma 3.3. If I(x)(xy)=y holds, then
y = 1 yields I(x)x = 1, and so I(x) = J(x) again. Now, (x=y)(y=x) = 1 is
equivalenttoy=(xy)=xn1,andthatis y=J(x)(xy).
InanAAIPloopI(x)=J(x)byLemma 3.3. Ifanyof"and isequalto 1,
thenwegetI =J bysettingx=1ory=1. Therestislear.
FromCorollary6.6 andTable6weseethatl
(I(x)(xy))=xn(J(x)ny). Sine
xn(J(x)ny) = y if and only if J(x)(xy) = y we see that l
(LIP) = LIP. Fur-
thermore,r
((x=y)(y=x))=(xI(y))=I(y(I(x)), andsor
(LIP)=Eq[J(x)I(y)=
I(yx)℄=AAIP. Inthiswayweobtainadiret proofforthefollowingstatement.
ThestatementanbealsoderivedfromProposition3.7. Wehavehosenadiret
prooftoillustratetheoneptofisostrophivarietiesuponawellknownandeasy
example.
Proposition 6.10. r
(AAIP) =LIP = o
(RIP), l
(AAIP) =RIP= o
(LIP),
and r
(LIP)=AAIP =l
(RIP). Furthermore,l
(LIP)=LIP, r
(RIP) =RIP,
ando
(AAIP)=AAIP.
Everyloop variety V ontainsasubvariety Itp(V) ofloopsQ suh that every
loopisotopeofQ isin V. Loopsofthis kindarealled isotopially invariant or
universal (withrespettoV). Notethat Itp(Itp (V))=Itp (V).
Proposition6.11. LetV andW beisostrophivarieties,withW=f(V),where
f 2I. ThenItp (W)=f(Itp (V)). Inpartiular,if Itp(V)=V,thenItp(W)=W.
Proof: This follows from the fat that f maps lasses of isotopesto lassesof
isotopes,byProposition5.2.
It is well known (and easy to prove) that ItpEq[xy = yx℄ is the variety of
abelian groups. If V is the variety of left ross-ommutativeloops(f. Proposi-
tion3.8),thenItp (V)isthevarietyofabeliangroupsagain,byProposition6.11.
Put lBol = Itp(LIP), mBol = Itp(AAIP) and rBol = Itp(RIP). Proposi-
tions6.10and6.11immediatelyyield:
Corollary 6.12. r
(mBol)=lBol=o
(rBol),l
(mBol)=rBol=o
(lBol),and
r
(lBol)=mBol=l
(rBol). Furthermore,l
(lBol)=lBol,r
(rBol)=rBol, and
o
(mBol)=mBol.
Lemma 6.13. Let V be a variety of loops suh that Q
= Q op
2 V for every
Q 2 V. Then Itp(V) onsists of all loops Q suh that the left isotope (x=e)y
belongsto V foreverye2Q.
Proof: SupposethataloopQfullstheonditionofthestatement. Weneedto
onsider an isomorphism ' : Q ! Q . We get '(x(eny)) =('(y)='(e))'(x).
Therightisotopeisheneisomorphitotheoppositeloopofaleftisotope. Sine
theleftisotopebelongstoV,theopposite loophastobelongtoV aswell.
Lemma6.14. LetQbealoopande2Q. DenotebyS theleftisotope(x=e)y.
ThenJ
S
(x)=(e=x)eand I
S
(x)=(x=e)ne.
Proof: Thisanbeveriedinadiretway.
The rst part of the next statement was formulated by Robinson as Theo-
rem 3.1 of [23℄. The seond part seems to have appeared for the rst time in
ChapterXIofBelousov'sbook [6℄.
Proposition 6.15. ThevarietylBol isequaltoEq[x(yxz)=(xyx)z℄,rBolis
equaltoEq[z(xyx)=(zxy)x℄. Furthermore,mBolisequaltoEq[(x=y)(znx)=
(x=(zy))x℄=Eq[(x=y)(znx)=x((zy)nx)℄.
Proof: ByCorollary6.12itsuÆesto proveonlytherstequalityin theeah
partofthestatementsineo(lBol)=rBol ando(mBol)=mBol.
Denoteby L
x
theleft translationy 7!xy. AloopQ hastheLIP ifandonly
ifL 1
y 2fL
x
; x 2Qg foreah y 2Q(i.e. theleft translationsare losed under
inverses).
Let Qbe aLIP loop. The left translations of aleft priniple isotope (x=e)y
arelosedunderinversesforanye2Q. Thelefttranslationsofarightpriniple
isotopex(fny)arelosedunderinversesifandonlyifforallx;f 2Qthereexists
z 2Q suh that (L
x L
1
f )
1
=L
z L
1
f
. Insuh aaseL
z
=L
f L
1
x L
f
. Thus if
Q2lBol,then forall x;y 2Qthere exists z 2Qsuhthat L
x L
y L
x
=L
z , and
sox(yxw) = (xyx)w for all x;y;x;w. By plugging y = 1=x we get the LI
property,andhenetheargumentanbereversed.
From Proposition6.10and Lemma 6.13itfollowsthat Q2mBol ifandonly
iftheloop (x=e)y hastheAAIP foreverye2Q. Fixeand denote theloopby
S. FromLemma6.14wegetJ
S
(y)=(e=y)eandI
S
(ze)=zne. FromLemma6.9
itfollowsthatS isanAAIPloopifandonlyif(e=y)(zne)=(e=(zy))e.
It isusual to allelementsof lBol,rBol and mBolleft, right and middle Bol
loops,respetively.
LetQbealeftBolloop. Thenxny =x 1
ysineitisaLIPloop. Theoperation
ofthemiddleBolloopr(Q)anbethusexpressedasx=y 1
. Gvaramija[11℄notes
thatthere isanotherexpression: y(y 1
xy). Thisfollowsfrom thefat thatthe
leftBolidentitygivesx=y=y 1
(yxy 1
).
Syrbu[25℄,[26℄givesfurthermiddleBolloopidentities. Theseidentitiesdier
onlybyrearrangingtherighthandside,whenweputu=zy andexpressx(unx)
(or(x=u)x)inanequivalentway. Weshallnishthissetionbyshowingthatthis
phenomenonanbeexplainedbythepropertiesof
ItpEq[1=x=xn1℄lBol[rBol:[mBol:
