New York Journal of Mathematics

New York J Math

Test Elements and the Retract Theorem in

Hyp erb olic Groups

John C ONeill and Edward C Turner

Abstract We prove that in many p erhaps all torsion free hyp erb olic groups

test elements are precisely those elements not contained in prop er retracts We

also show that all Fuchsian groups have this prop erty Finallyweshow that

all surface groups except Z Z have test elements

Contents

Intro duction

Torsion free hyp erb olic groups

The Retract Theorem in Fuchsian groups

Test elements in surface groups

References

 Intro duction

Denition If G is a then g G is a test element if for any endomorphism

G G g g implies that is an automorphism

The notion of test elements was rst considered in the context of free groups in

which they are called test words The rst example was provided in by Nielsen

N who showed that the basic commutator a b aba b is a test word in the

F a b Considerable progress has b een made recently in understanding

b oth test elements and test wordssee Tu for example and the references cited

there The following reformulation of the denition makes clear how test elements

are used to recognize automorphisms g is a test elementif g g for some

automorphism implies that is an automorphism Thus the issue of deciding

whether is an automorphism is replaced by that of deciding whether g andg

are equivalent under the action of the automorphism group AutG The classic

algorithm of J H C Whitehead W decides very eectively when two elements

of a free group F are equivalent under the action of AutF in a forthcoming

Received February

Mathematics Subject Classication E F

Key words and phrases test words test elements endomorphisms automorphisms hyp erb olic

groups

ISSN

John C ONeil l and Edward C Turner

pap er D J Collins and the second author will showhow to do the same thing in

a torsionfree hyp erb olic group

Pro ducing nontest elements is quite easy Supp ose for example that the sub

group R of G is a prop er retract ie the image of a nonsurjectivemap G G

called a retraction with the prop ertythat r r for all r R Then no element

of R can b e a test element In Tu the following theoremthe Retract Theorem

for freegroupswas proven characterizing test words in free groups

Theorem A word w in a free group F is a test word if and only if w is not in

any proper retract

We are interested in deciding whether the Retract Theorem is true for gen

eral hyp erb olic groups By hyp erb olic we mean word hyp erb olic in the sense of

Gromovsee eg GH In particular hyp erb olic groups are nitely generated

We succeed in proving it for torsion free hyp erb olic groups that are stably hyperbolic

in the following sense

Denition A hyp erb olic group is stably hyperbolic if for every endomorphism

n

G G there are arbitrarily large values of n so that Gishyp erb olic

Anyhyp erb olic group with the prop ertythat every nitely generated is

hyp erb olic hyp erb olic surface groups for example clearly satises this prop erty A

famous application of the Rips construction R shows that some hyp erb olic groups

We mo dify this construction have nitely generated nonhyp erb olic

in Section to pro duce an example in which such a subgroup is the image of

an endomorphism Its relatively easy to extend this to nd endomorphisms of

hyp erb olic groups that have arbitrarily many nonhyp erb olic forward images but

in all cases weve studied the images are eventually hyp erb olic It seems quite

p ossible that all hyp erb olic groups are actually stably hyp erb olic

Our sp ecic results are the following

Theorem If G is a torsion free stably and g G then g is a

test element if and only if g is not in any proper retract

uchsian group and h H then h is a Theorem If H is a nitely generated F

test element if and only if h is not in any proper retract This applies in particular

if H is a nite freeproduct of cyclic groups

Theorem If G is a surfacegroup other than Z Z then G has test elements

The situation for surface groups is interestingall surface groups except Z Z

have test elements explicit examples given in Section and all except the funda

mental group of the Klein b ottle ha b j aba b i satisfy the Retract Theorem

In V it was shown that b lies in no prop er retract but is nevertheless not a test

word since it is xed by aa bb

We will use the following terminology

Denition If G is a group and G G is an endomorphism then

n n

n

j G G and

n

G

j G G

G

T

n

where G

i

The group G is Hopan if it is not isomorphic to any of its prop er quotients and

is coHopan if it is not isomorphic to any of its prop er subgroups

Test Elements and the Retract Theorem in Hyperbolic Groups

 Torsion free hyp erb olic groups

This section is devoted to a pro of of Theorem Our pro of of Theorem is

mo deled on the pro of of the Retract Theorem in Tu the main technical to ol b eing

Prop osition whose pro of we defer to the end of the section

Prop osition Suppose that G G G G is a freeproduct of innite

m

cyclic and freely indecomposable coHopan groups If G G is a monomor

phism then G is a freeoffactor G

Prop osition If G is a torsionfree hyperbolic group and G G is an

n

endomorphism with the property that G is hyperbolic for arbitrarily large n

N

then G is a free factor of G for some N

Pro of It was proven by Sela Se that if G G is an endomorphism of a

n n

n

torsionfree hyp erb olic group G then j G G is a monomorphism

G

n

for large enough n Let n b e large enough so that j is a monomorphism and

G

n n

that G is hyp erb olic and consider the free pro duct decomp osition G

H H into freely indecomp osable hyp erb olic factors Sela Se has also

m

proven that every freely indecomp osable torsion free hyp erb olic group is either

coHopan or innite cyclic Prop osition now follows from Prop osition

Pro of of Theorem The fact that elements of prop er retracts are not test el

ements is trivially true in all groups since a prop er retraction is not an automor

phism To prove the converse we b egin with the following general observation

if G G is a monomorphism of a group G then is an automorphism of

G It is clear that is injective To see that is surjective let g G

n

then for every n there exists g Gsuchthat g g since is injective

n n

g g for all m n Hence g G and is surjective

m n

Now supp ose that G is a stably hyp erb olic group that g is not a test element

and that is a endomorphism which is not an automorphism so that g g we

show that g lies in a prop er retract byshowing that G is a prop er retract

n

For large enough n is a monomorphism by Se since is an

automorphism by the observation ab ove According to Prop osition Gisa

N N

free factor of G for some N cho ose suchanN and let G G

b e a free factor pro jection mapping Then

N N

G G

is a retraction mapping Thus is a retract

It may b e that the Retract Theorem is true for general hyp erb olic groups Theo

rem is a partial result in this direction It may also b e the case that all hyp erb olic

groups are stably hyp erb olic However the following example suggested byGA

Swarup shows that it may b e that Gisnothyp erb olic but in this case G

is trivial

Example Supp ose that

F F h x x x x j x x x x x x x x i

and let

i F F F F by x x for

i

John C ONeil l and Edward C Turner

In Dr it was shown that ker has rank and is not nitely presented and is

therefore not a hyp erb olic group

Nowlet G F F F F hy y i and G F hx y y i

r r r r r

by

x x

i i

y y

i i

ker Then P has rank and let P

Now the Rips construction R pro duces a hyp erb olic group H with rank r

that maps onto G by a map with kernel generated bytwo added generators a and

b Then H P isanonhyp erb olic subgroup of H of rank at most

P H

ha bi H G

F

r

Nowlet r F H b e a surjection and H H Then H

r

is hyp erb olic and is an endomorphism whose image H is not hyp erb olic

Pro of of Prop osition We begin by observing that it suces to show that

n

G is a free factor of G for some n in fact we will show that this is

n

true for all suciently large n For if G G G for some G then the

monomorphism

n n

G G

pulls this factorization backto G

n n

G G G

n

But it is straightforward to show that G G

The pro of is a geometric generalization of the Takahasi Theorem Ta and is

mo deled on the pro of of the Takahasi result outlined in problem on page of

MKS We will need the following slightvariant of the usual normal form measure

of length in a free pro duct which dep ends on the pro duct decomp osition

Denition If G G G G is a free pro duct of freely indecomp osable

m

factors then the length jg j is

G

t

X

jg j jg j

j

G

G

i

j

j i

form g g g g g G for some i where g has free pro duct normal

t j i j

j

th

jg j Z and jg j if G n if G Z and g is the n power of a

j G j G i i j

i i j j

j j generator

Test Elements and the Retract Theorem in Hyperbolic Groups

We order the factors of G so that G Z for i thus

i

G F G G

m

Let X b e the natural complex with fundamental group G whichisthewedge

of circles and complexes K K K G each of whichisjoinedto

m i i

the wedge p ointby a line segment

The nested sequence of subgroups

n

G G G G G

determines a sequence of coverings of X

p p

p p

k  k 



k

X X X X X

k k

k

where X G Let P p p p X X

k k k k

In the covering space X a connected comp onent K of P K will b e called

k i

i

k

K country otherwise an essential K country if K and an inessential

i i

i

Let M b e a contractable subspace of X which is the union of a maximal tree in

k k

X xed for the remainder of the argument with all the the inessential countries

k

in X Then M whichwe call a representing subspacedetermines a free pro duct

k k

k

representation for G as in the Kurosh Subgroup Theorem

We dene a unit path in X to b e a path in X that b egins and ends at a lift of

k k

the basep ointofX and never passes through

i an intermediary basep oint

ii an inessential country

iii an edge of the maximal tree

Clearly each path in X is uniquely a pro duct of unit paths and paths without unit

k

k

subpieces If g Gandf is the lift to X of a closed lo op f representing g

k

then relative to the free pro duct representation corresp onding to M jg j is

k

k

G

just the numb er of unit paths in f

k

For g G let kg k be the length of the shortest representation of g

k

G

k k

in G with resp ect to any free pro duct representation for G into freely

n o

N

indecomp osable factors Supp ose that min kg k t is attained in G

k

G

k N

G G G with G Z for i and supp ose that g g g g

t

m

i

where g G for some i m

j j

i

j

N N

Applying the Kurosh Theorem to G as a subgroup of G and using

N N

the fact that G Gwe get that

N

G GF G

m m

m m

N

where F F G for i m and G G is a monomorphism

i j

j i

j

for ji m

j

N

The factors of G can b e rearranged and iterated as often as needed so

that

N

G G G G G GF

m p p

N r

GF G G

p p

p p

G G

p p m m

p m m

p

John C ONeil l and Edward C Turner

where F F G G and G G are monomorphisms and

i k i j

i i i

k

k

N

are elements of Gforii p pk m

k

Since G is coHopan for i pwemay write

i

N r

GF G G

p

p p

G G

p p m m

p m m

p

We note that X isacovering space of X with the prop ertythatanyessential

N r N

country in X is a cover of a sub complex K for i p

N r

i

Since G is a subgroup of a nitely generated group it is countable and

we may list the elements Let g be the rst element in this list and denote by

I f kg the set of subscripts app earing in the normal form representation

g

for g The rst main step in the pro of of the Prop osition is the following claim

n o

Claim Suppose that kg k min jj g jj t and furthermore that

N k

G G

k

for i m Then G is a freefactor g g g g whereeach g G

j k I t j

g

i

k

j

of G

Pro of of Claim Supp ose that g G There are three cases to consider

i



according to the index i

Case Supp ose that i

n

h x i g x for some n Consider a lo op f representing x in X Since G

i i N

i i



and its lift f in X for kN If f is not a lo op then it is a contractible path in X

k k

t whichwe can extend to a representing space M Note that kg k jg j

k

k

G M

k

con tradicting the assumption that t was minimal Hence f is a lo op and therefore

k

G is a free factor of G for all k N

i

Case Supp ose that i p

Let K b e the sub complex in X such that K G Consider the cover

i N i

 

i



K of K in X which is adjacent to the basep oint A priori there are three

i i N r

 

p ossibilities for K

i



K Sub case a

i



K is a nontrivial subgroup of G Sub case b

i



i



G K Sub case c

i



i



In all cases let f be the loop in K representing g and let f b e the lift of f in

i



K

i



In Sub case a K and hence K is contractible weincludeK in an

i i i

  

representing space M for X Then kg k jg j t contradicting

k

N r N r

G M

N r

the assumption that t was minimal

In Sub case b w K e assume that H a nontrivial prop er subgroup of

i



Up to rearrangement of factors wemay assume that G

i



H G G G

p p pt

Test Elements and the Retract Theorem in Hyperbolic Groups

N r

and so that Therefore wemay rewrite Gas

p p pt

follows

N r

GF G G

p

p p

G G

p pt

p pt

G G

pt pt m m

pt m m

pt

where otherwise is not injective and if then G is not a

i j j j



subgroup of G

i



Wenow turn our attention to

N r

GF G G

p

p

G G

i i p q

 p  q

i i

 

G G

m m q q

m m q

q

t and consider the cover of K in X which is adjacenttothe where q p

i N r



basep oint say K If K were essential then either or for

i i i j

  

q j m The former contradicts injectivityof If the latter o ccurs then in

fact but we previously assumed that in this case G was not a subgroup

j j j

of G Hence K is inessential and we nd that kg k jg j t which

k

i i

 

G M

N r

is a contradiction

h do es not contradict the minimality Then Sub case c is the only p ossibilitywhic

N r

and G is a free factor of G Similarly K G of t Therefore

i i

 

i



N r k

is a free factor of G for all p N Hence G is a free factor of G G

i



i



for all k N and it is an easy exercise to prove then that G is a free factor of

i



G

Case Supp ose that pi m

n o

N

We note that if min jg j t and g g g is in G then g G

k

t i

j

G

k N

for pj mifK is the subspace of X with K G andK is the cover

j N j j

j

of K in X which is adjacent to the basep oint then K is inessential

j N r j

This shows that in all cases G is a free factor of G

i



We rep eat the argument for all g involved in g Each g is represented by a lo op

i i

k

in X which is either a lo op adjacent to the basep oint or f that lifts to a path f

k i

i

a noncontractible path that lies in a inessential country adjacent to the basep oint

These essential countries and lo ops are actually homeomorphic to those countries

they cover Thus G for i I is actually a free factor of X for all k nand

I i j j k

j j

hence a free factor of G as well This completes the pro of of Claim

N

Wenow reorder the factors of G in the following way

N

GG G G G G G

L L Q Q m

where

k

L G is a free factor of G for all k N and for each further for i

i

iterate of the corresp onding Kurosh conjugator

i

Z for Li Q G

i

for Qi m G is a coHopan group i

John C ONeil l and Edward C Turner

Now let g be the rst element in the ordering of G so that g is not

L L

contained in G If one cannot do this then G G is a free

i i i i

k

factor of G for all k N and the Prop osition is complete We dene jg j

k

Gj



to b e the minimal length of g with resp ect to all free pro duct representations of

k

Gwhichcontain as a free factor Let

n o

jg j s kg k min

k k

Gj Gj

 

k N

and supp ose that this minimum is attained in



N

GG G G G G G G

L L Q Q m

where G G if i L G Z if L i Q and G is coHopan if

i i i i

Q i m

Let g g g be the normal form of g with resp ect to this representation

s

All of the arguments p ertaining to the length of g in Part I of the where g G

j i

j

argument directly apply to g for each G containing g in the element g wehave

i j

j



a corresp onding subspace K in X that lifts to a homeomorphic copy of itself in

i N

j

X for k N If this fails to b e the case for some k N thenwe can nd some

k



k k for which jg j kg k which contradicts the assumption of

k k

Gj Gj

 

the minimality



L k

In this waywe obtain G which is a free factor of Gforallk N

i i

and is hence a free factor of G This completes the pro of of the Prop osition

since the pro cess must terminate this is since the free factors of G are also

M

free factors of G for large enough M and the numb er of these free factors is

b ounded by m

 The Retract Theorem in Fuchsian groups

In this section weshow that the Retract Theorem holds for a nitely generated

hsian group G generalizing results of Voce V and V See B for background Fuc

information on Fuchsian groups This theorem also holds by the same techniques

for groups of isometries of the hyp erb olic plane H that include orientation reversing

isometries We b egin with the sp ecial case of nite free pro ducts of cyclic groups

Lemma If G is a nite freeproduct of cyclic groups and is an endomorphism

of G then G is a retract The Retract Theorem therefore holds for G

Pro of We begin by showing that is a monomorphism for suciently large

N

then r ank Gr s By the Kurosh Theorem Z N If G F Z

k r k

s 

n n

the images G all have the same form as well and rank G r s

n n

is a nonincreasing function of n In fact the values of r are nonincreasing by

n

the following argument Abstractly is a surjection from F Z Z to

r k k

 s





F Z Z Following by pro jection onto F we get a map whichmust

r k r

k

 

s





be trivial on each factor Z inducing a surjection of F onto F so r r

k r r

i 

similarly r r The values of r are therefore eventually constant and so the

n n n

M

values of s are also eventually constant Now replace G with G

n M

for suciently large M here the values of r and s are constant and it suces

n n

to prove the Lemma in this case

Test Elements and the Retract Theorem in Hyperbolic Groups

Its not hard to see that k k k k Consider the ab elianization

s

s

L L

r r





Z This is true as well for all soeventually Z Z Z Z Z

k n k k

k

 s

s



this pro duct stops decreasing at whichpoint b ecomes a monomorphism

N

The Lemma nowfollows by the same arguments as in Prop osition and Theo

rem

Pro of of Theorem Supp ose that G is a nitely generated Fuchsian group with

fundamental p olygon P and orbifold H G If there are no cusps then G is torsion

free and the Retract Theorem holds by Theorem If G is not cocompact ie if

H G is not compact or if the genus n is then G is a pro duct of cyclic groups

and the Retract Theorem holds by Lemma Wemay therefore assume that P is

compact and that G has presentation

k

i

G a b a b c c j a b a b c c c i

n n t n n t i

with t and n By considering the ab elianization of G it is easy to see that

r ank Gn t The orbifold H G has genus n top ologically and has t cone

ts with cone angles Denote the total cone angle of H G by poin

k k

t

t

X

Cone

G

k

j

j

Then by B p P has area

t

X

r ank G Cone A n

G

k

j

j

As b efore it suces to show that for any endomorphism of G that G

is a retract Consider rst the case in which the index jG Gj of GinG is

innite namely jG Gj In HKS it is proven that a subgroup of innite

index in a Fuchsian group is a free pro duct of cyclic groupsreplacing G and

with Gand completes the argument in this case

Now assume that G has nite index in G and consider the family of subgroups

G G G G

each of whichisFuchsian If for any G is either not cocompact or has genus

orhasnoconepoints then the Retract Theorem holds for G and Gis

a retract of both G and of G So we can assume that for all G is co

compact has genus n has t conepoints has rank r n t

has total cone angle Cone and area A r Cone

The sequence r is nonincreasing so it eventually stabilizes The sequence of

indices j G Gj is also nonincreasing and eventually stabilizes at a value

q If at anypoint G G then G Gandwe are done

So by replacing G and by Gand j for suitably large wemayassume

G

that all the ranks and all the indices are equal

A qA since a fundamental p olygon for Gisq nonoverlapping Now

fundamental p olygons for G Thus

lim A lim Cone

This is a clear contradiction completing the argument

John C ONeil l and Edward C Turner

 Test elements in surface groups

In this section wewillprove the existence of test elements in the fundamental

groups of orientable surfaces other than the torus For the remainder of this section

we will consider the surface S of genus n with fundamental group

n

hx x x j x x x x x x i

n n n n

Theorem The group for n contains test elements In particular the

n

words

k k k

w x x x k

k

n

are test elements

Pro of The Retract Theorem holds for since it is Fuchsian so it suces to

n

show that for k w lies in no prop er retract of

k n

Supp ose that is a prop er retraction with H and that

n n n

w H In general a subgroup K of is a surface group and if K k

k n n

then the Euler characteristics and ranks are related by

K k r ank K k n n r ank

n n

Since the r ank H r ank H has innite index Let S be the covering space

n H

of S corresp onding to H Since S is a noncompact surface H is free

n H

Consider the standard CW complex structure on S with one cell n cells

n

and one cell and the map f S S that represents w F S

n n

k n

where S is the circle and S is the skeleton of S Since f w H f lifts

n

n k

e

to a map f as indicated



S

S

H

H



p

p

e

p

f

p

p

p

p

p

p

p

p

p

p

f

 

S

S

S n

n

homotopic in S to f so that the image Claim There isamapf S S

n

n

e e

f S of the lift f is contained in a topological retract V of S The subgroup

H

K F representedbyV is a retract of F

n n

S

H



p

p

p



e

p

f

p

p

p

p

p

p

p



p

p

f



S

S n

Pro of of Claim The noncompact surface S retracts onto a homotopy equiv

H

alent compact subsurface with b oundary T which contains f S Then T strong

deformation retracts onto a subset V of its skeleton T for example perform

the sequence of simple homotopies that push in on a free edge of any remaining

e e

cell This strong deformation retract will homotop the map f to a map f whose

image is again in the skeleton f is dened by pro jecting to S Since a graph

n

e

retracts onto any of its connected subgraphs f S is a retract of V whichisin

e

turn a retract of S thus f S is a retract of S

H H

Test Elements and the Retract Theorem in Hyperbolic Groups

The diagram of spaces on the left determines the diagram of groups on the right

below in which is the presentation map is the inverse of the

induced by the deformation retraction and i and i are inclusion maps

K H

 

S

V K H

H

i i

K H

 

S F

S

n n n

n

The retraction of F to K is This completes the pro of of the Claim

n

Now let v K F b e the element represented by f Since K is a prop er

k n

retract of F the retract index v of v is see Tu On the other hand

n k k

n

v w since the denition of dep ends only on the image in Z and v

k k k

n

Z But w isamultiple and w have the same image in F

n k k n n

of k Tu page This contradicts the existence of the retraction

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DFMS United States Air Force Academy CO

JohnOneillusafaafmil httpwwwusafaafmildfmsp eopleoneillhtm

Dept of Math University at Albany Albany NY

tedmathalbanyedu httpnyjmalbanyeduted

This pap er is available via httpnyjmalbanyedujhtml