Vol. 23, No. 9 (2000) 585–595 S0161171200002969 ©Hindawi Publishing Corp.
ON INVERTOR ELEMENTS AND FINITELY GENERATED
SUBGROUPS OF GROUPS ACTING ON TREES
WITH INVERSIONS
R. M. S. MAHMOOD and M. I. KHANFAR
(Received 9 February 1999)
Abstract.An element of a group acting on a graph is called invertor if it transfers an edge of the graph to its inverse. In this paper, we show that ifGis a group acting on a tree Xwith inversions such thatGdoes not fix any element ofX, then an elementgofGis invertor if and only ifgis not in any vertex stabilizer ofGandg2is in an edge stabilizer ofG. Moreover, ifHis a finitely generated subgroup ofG, thenHcontains an invertor element or some conjugate ofHcontains a cyclically reduced element of length at least one on whichHis not in any vertex stabilizer ofG, orHis in a vertex stabilizer ofG. Keywords and phrases. Groups acting on trees, invertor elements, finitely generated sub-groups.
2000 Mathematics Subject Classification. Primary 20F65, 20E06.
1. Introduction. Lyndon and Schupp [5, Lemma 6.8, page 212] proved that ifGis a nontrivial free product with amalgamation, then for any finitely generated subgroup HofG,His contained in a conjugate of a factor ofGor some conjugate ofHcontains a cyclically reduced element of length at least two. Similarly, ifGis anHNN group then, for any finitely generated subgroupH ofG,H is contained in a conjugate of the base or some conjugate ofH contains a cyclically reduced element of length at least two. These results can be easily generalized to the result that finitely generated subgroups of groups acting on nontrivial trees without inversions are contained in a vertex stabilizers or some of their conjugates contain cyclically reduced elements of length at least one. In this paper, we show that the situation of groups acting on trees with inversions is different in the sense that ifGis a group acting on a treeXwith inversions such thatGdoes not fix any element ofXand ifHis a finitely generated subgroup ofG, thenH is contained in a vertex stabilizer ofG or a conjugate ofH contains a cyclically reduced element of length at least one orHcontains an invertor element on whichH is not in any vertex stabilizer ofG. This paper is arranged as follows. In Section 2, we give preliminary definitions and results needed for the rest of sections. In Section 3, we discuss some properties of invertor elements. In Section 4, we discuss finitely generated subgroups of groups acting on trees. In Section 5, we apply the results of previous sections to new groups called quasiHNN groups that are generalization ofHNN groups.
nonempty, equipped with two mappingsE(X)→V (X)×V(X),y→(o(y),t(y))and, E(X)→E(X), y→y¯satisfying the conditions ¯¯y=yando(y)¯ =t(y)for ally∈E(X). The case ¯y=yis possible for somey∈E(X). Fory∈E(X),o(y)andt(y)are called the ends ofyand ¯yis called the inverse ofy. There are obvious definitions of trees, morphism of graphs and Aut(X), the set of all automorphisms of the graphXwhich is a group under the composition of morphisms. For more details we refer the readers to Mahmood [6, 7] or Serre [10]. We say that a groupGacts on a graphXif there is a group homomorphismφ:G→Aut(X). Ifx∈X(vertex or edge) andg∈G, we writeg(x)for (φ(g))(x). Thus, ifg∈Gandy∈E(X), theng(o(y))=o(g(y)), g(t(y))=t(g(y)), andg(y)¯ =g(y). The caseg(y)=y¯for someg∈Gandy∈E(X)may occur. That is,Gacts with inversions onX. We have the following notations related to the action of the groupGon the graphX.
(1) Ifx∈X (vertex or edge), defineG(x)to be the setG(x)= {g(x):g∈G}. This set is called the orbit ofXcontainingx.
(2) If x,y ∈X define G(x,y) to be the set G(x,y)= {g ∈G: g(x)=y}, and G(x,x)=Gx, the stabilizer ofx. Thus,G(x,y)≠ϕ if and only ifx and y are in the same orbit. Ify∈E(X)andu∈ {o(y),t(y)}, then it is clear thatGy¯=Gy and
Gy≤Gu.
(3) The set of elements ofX fixed byGis denoted byXG. That is,XG= {x∈X:
Gx= G}.
(4) IfXis connected, then a subtreeT ofX is called a tree of representatives for the action ofGonXifT contains exactly one vertex from each vertex orbit, and the subgraphY ofXcontaining a tree of representativesT, say, is called a fundamental domain for the action ofGonXif each edge ofY has at least one end inT, andY contains exactly one edgey, say, from each edge orbit such thatG(y,y)¯ =ϕ, and exactly one pairx,x¯from each edge orbit such thatG(x,x)¯ ≠ϕ. For the existence of T andY (see Khanfar and Mahmood [4]).
Henceforth,Gwill be a group acting on a treeX,T a tree of representatives for the action ofG onX, andY a fundamental domain for the action ofGonX such that XG=ϕandT⊆Y.
Definition2.1. For any vertexvofXdefinev∗to be the unique vertex ofTsuch thatG(v,v∗)≠ϕ. That is,vandv∗are in the same vertex orbit. It is clear that ifvis inT, thenv∗=v and in generalv∗∗=v∗. Moreover, ifuandvare two vertices of Xsuch thatG(u,v)≠ϕ, thenu∗=v∗.
Definition2.2. For each edgeyofY define the following.
(1) Define[y]to be an element ofG(t(y),t(y)∗). That is,[y]((t(y))∗)=t(y)to be chosen as follows:
(a) ifo(y)∈V (T ), then[y]=1 in casey∈E(T ), and[y](y)=y¯ifG(y,y)¯ ≠ϕ, (b) ifo(y)∈V (T ), then[y]=[y]¯ −1 ifG(y,y)¯ =ϕ, otherwise[y]=[y]¯ if
G(y,y)¯ ≠ϕ.
(3) Defineφyto be the mapφy:G−y→G+ygiven byφy(g)=[y]g[y]−1. It is clear thatφyis an isomorphism.
(4) Defineδy to be the elementδy=[y][y]¯ . It is clear thatδy=1 ifG(y,y)¯ =ϕ. Otherwiseδy=[y]2. Consequentlyδy∈Gy, δy¯=δy andφy(δy)=δy.
Proposition2.3. Gis generated by the generators ofGv and by the elements[y],
wherevruns overV (T )andyruns overE(Y ).
Proof. See Mahmood [6].
Definition2.4. By a wordwofGwe mean an expression of the form
w=g0·y1·g1·y2·g2···yn·gn, n≥0,
yi∈E(Y ), fori=1,2,...,n (2.1)
such that
(1) g0∈G(o(y1))∗;
(2) gi∈G(t(yi))∗, fori=1,2,...,n;
(3) (t(yi))∗=(o(yi+1))∗, fori=1,2,...,n−1. We defineo(w)=(o(y1))∗andt(w)=(t(y
n))∗.
Ifo(w)=t(w), thenwis called a closed word ofGof typev,v=o(w). We have the following definitions related to the wordwdefined above:
(i) nis called the length ofwand is denoted by|w| =n. (ii) wis called a trivial word ofGif|w| =0, orw=g0. (iii) The value ofwdenoted[w]is defined to be the element
[w]=g0[y1]g1[y2]g2···[yn]gn ofG. (2.2)
(iv) The inverse ofwdenotedw−1is defined to be the word
w−1=g−1
n ·y¯n·δ−1yngn−1−1···g−12 ·y2¯ ·δ−1y2g−11 ·y1¯ ·δ−1y1g0−1 ofG. (2.3) It is clear that[w−1]=[w]−1but(w−1)−1≠wifwcontains an edgey(say) such that G(y,y)¯ ≠ϕ. Otherwise(w−1)−1=w.
(v) Ifw=h0·x1·h1·x2·h2· ··· ·x
m·hmis a word ofGsuch thatt(w)=o(w), thenw·wis defined to be the word
w·w=g0·y1·g1·y2·g2· ··· ·y
n·gnh0·x1·h1·x2·h2· ··· ·xm·hm ofG. (2.4)
It is clear that[w·w]=[w][w], o(w·w)=o(w), andt(w·w)=t(w). (vi) wis called a reduced word ofGifwcontains no subword of the forms
(1) yi·gi·y¯iifgi∈G−yi fori=1,...,n,
(2) yi·gi·yiifgi∈Gyi andG(yi,y¯i)≠ϕfori=1,...,n. We take trivial words to be reduced.
Definition2.5. Bythe edge reductionon the word
w=g0·y1·g1·y2·g2· ··· ·yn·gn (2.5)
(1) replace a subword of the formyi·gi·y¯i, ifgi∈G−yi, byφyi(giδyi),
(2) replace a subword of the form yi·gi·yi, ifgi∈Gyi and G(yi,y¯i)≠ϕ, by φyi(giδyi).
We definew0to be the identity element 1 ofG.
Theorem2.6. Every element ofGis the value of a closed and reduced word ofG. Moreover, ifwa closed word ofGof value1, the identity element ofG, thenwis not reduced.
Proof. See Mahmood [7].
Now we generalize Theorem 2.6 as follows.
Lemma2.7. For every elementgofGand any two verticesuandvofTthere exists a reduced wordw ofGsuch thato(w)=u, t(w)=v, and[w]=g. Moreover, ifw is reduced, then for any wordw ofG such thato(w)=o(w), t(w)=t(w), and [w]=[w]=g, we have|w| = |w|if and only ifwis a reduced.
Proof. SinceG is generated by the generators of Gz and by the elements[y], wherezruns over the vertices ofTandyover the edges ofY, thereforegcan be writ-ten as the productg=g0[y1]g1[y2]g2···[yn]gn, wheregi∈Gui for some vertices u0,u1,...,unofT, and edgesy1,y2,...,ynofY. By taking the reduced paths inT be-tweenuandui, betweenuiand(o(yi))∗, and between(t(yi))∗andv, and the identity element 1 ofGwe may choose this product so thatw=g0·y1·g1·y2·g2···yn·gn is a word ofGsuch thato(w)=u, t(w)=v, and[w]=g. Ifwis not reduced then for somei, 1≤i≤n−1 we haveyi+1=y¯iandgi∈G−yi oryi+1=yi,gi∈Gyi, and G(yi,y¯i)≠ϕ. Ifyi+1=y¯iandgi∈G−yi, then we replaceyi·gi·y¯iby the element φyi(giδyi)inw. We get a new wordg0·y1·g1·y2·g2···yi−1·gi−1φyi(giδyi)·yi+1· ··· ·yn·gn.
Ifyi+1=yi,gi∈Gyi, andG(yi,y¯i)≠ϕ, then we replaceyi·gi·yi+1by the element φyi(giδyi)inw. We get a new word as above.
Continuing the above processes onw yields a reduced word ofG satisfying the required properties. In other words, the performance of the edge reductions on w yields a reduced word ofGof value[w].
Now letw=g0·y1·g1·y2·g2· ··· ·yn·gn, n≥0 and
w=h0·x1·h1·x2·h2· ··· ·x
m·hm
be two words ofGsuch thatw is reducedo(w)=o(w), t(w)=t(w)and [w]= [w]=g. Assume thatwis reduced. We need to show thatn=m. Since[w][w]−1= 1, the identity ofG, therefore by Theorem 2.6, the word
w0=h0·x1·h1·x2·h2· ··· ·xm·hmgn−1·¯yn·δ−1yngn−1−1· ··· ·g2−1·y2¯ ·δ−1y2g−11 ·y1¯ ·δ−1y1g0−1 (2.6)
is not reduced. Sincew andware reduced, thereforey
xm·hmgn−1·y¯nor(xm·hmgn−1·yn)byφyn(hmgn−1δyn)inw0. We get a new word
w1=h0·x1·h1·x2·h2· ··· ·xm−1·hm−1Aδ−1yngn−1−1·y¯n−1···g−12 ·y2¯ ·δ−1y2g1−1·¯y1·δ−1y1g−10 , (2.7)
whereA=φyn(hmgn−1δyn). Thenw1is not reduced. Similarly, we havexm−1=y¯n−1, andhm−1φyn(hmg−1n δyn)δ−1yng−1n−1∈Gyn−1orxm−1=yn−1ifG(yn−1,y¯n−1)≠ϕ.
Now continuing above processes yieldsx1=y1orx1=y1¯ ifG(y1,y1)¯ ≠ϕ. This implies that|w| = |w|.
Conversely, assume thatwis reduced and|w| = |w|. We need to show thatwis reduced. For, ifwis not reduced then by applying edge reductions on w yields a reduced wordw ofG such thato(w)=o(w), t(w)=t(w), [w]=[w]and |w|<|w|. Contradiction. Hencewis reduced. This completes the proof.
Definition2.8. For each elementgofGand each vertexvofV(T )define|g|v to be the length of a reduced word ofGof typevand valueg. In view of Theorem 2.6 and Lemma 2.7, this concept is clear.|g|v is called the length ofgwith respect tov. Proposition2.9. Letgbe an element ofGandva vertex ofV(T )such that|g|v
is even. Then for any vertexuofV(T ),|g|uis even.
Proof. Letw1 andw2be two reduced and closed words ofG of typesv andu
respectively such thatw1andw2are of valuegand|w1|is even. We need to show that|w2|is even. Letw0be the word
w0=1·y1·1·y2·1· ··· ·yn·1·w1·1·y¯n·1· ··· ·1·y2¯ ·1·y1¯ ·1, (2.8)
wherey1,y2,...,yn is the reduced path inT joininguandv. It is clear thatw0is a word ofGof typeuand of value[w2], andw0is of even length. By applying edge reductions onw0yields a reduced word ofGof typeu, value[w2]and of even length. Then Lemma 2.7 implies that|w2|is even. This completes the proof.
Proposition2.10. Ifgis an element ofGv for some vertexvofV (X), then|g|uis
even for any vertexuofV (T ).
Proof. g=aba−1wherea∈G, andb∈Gv∗. Letube any vertex ofV (T )and wa reduced word ofGsuch thato(w)=u, t(u)=v∗, and[w]=a. Then the word w0=w·b·w−1is closed of typeu, valueg, and is of even length. Now applying edge reductions onw0yields a reduced word ofGof typeu, valueg, and of even length. This implies that|g|uis even. This completes the proof.
Definition2.11. Letwbe a closed and reduced word ofG. We say thatw is a cyclically reduced word ofGifw2is reduced.
It is clear that every word of length zero is a cyclically reduced. The proof of the following proposition is clear.
Proposition2.12. Letw=g0·y1·g1·y2·g2· ··· ·yn·gn,n >0be a closed and
(3) gn−1·yn·gng0·y1·g1is a reduced word ofG, (4) |wm| = |m||w|for every integerm,
(5) |wm|<|wm+1|for every integerm, m≥0.
Proposition2.13. Letw1andw2be two closed and reduced words ofGsuch that w1andw2are of the same type and of the same value, andw1be cyclically reduced. Thenw2is cyclically reduced. Moreover, if |w1| ≥1, then[w1] is not in any vertex stabilizer ofG.
Proof. By Lemma 2.7,|w1| = |w2|. If|w1| =0, then by definitionw2is cyclically reduced. Letw1=g0·y1·g1·y2·g2· ··· ·yn·gn, n >0 and
w2=h0·x1·h1·x2·h2· ··· ·xn·hn. (2.9)
We need to show that 1·xn·hnh0·x1·1 is reduced. Sincew1is cyclically reduced, therefore gng0∈G−yn. By Lemma 2.7,hnh0∈G−yn. So 1·xn·hnh0·x1·1 is re-duced. Thereforew2is cyclically reduced. Now assume that|w1| ≥1. If[w1]∈Gvfor
v∈V(X), then[w1]=aba−1, wherea∈G, andb∈Gv∗. By Lemma 2.7, there exists a reduced wordw ofGsuch thato(w)=o(w1), t(w)=v∗, anda=[w]. Then the wordsw1andw·b·w−1are of the same type and of the same value. Now applying edge reductions onw·b·w−1yields a reduced word ofGof length zero which con-tradicts Lemma 2.7, or yields a reduced wordwofG such thatw is not cyclically reduced. This contradicts the first part of the proposition. Hence[w1]is not in any vertex stabilizer ofG. This completes the proof.
Definition2.14. An elementgofGis called cyclically reduced ifgis the value of a cyclically reduced word ofG. In view of Proposition 2.13, this concept is well defined.
In the next section, we show that some elements ofG are conjugate to cyclically reduced elements ofGand some not.
3. On invertor elements of groups acting on trees. Throughout this section,G will be a group acting on a treeX,Ta tree of representatives for the action ofGonX andY a fundamental domain for the action ofGonXsuch thatXG=ϕandT⊆Y. We have the following definition.
Definition3.1. Letgbe an element ofGandxbe an edge ofX. Ifg(x)=x¯, then gis called an invertor element ofGandxis called inverted edge underg. It is clear that ifxis an inverted edge underg, then ¯xis an inverted edge undergandg2∈G
x. The main result of this section is the following theorem.
Theorem3.2. Letgbe an element ofG. Then the following are equivalent: (i) gis an invertor element ofG;
(ii) gis conjugate to an element of the form[y]a, whereyis an edge ofY such thatG(y,y)¯ ≠ϕ, anda∈Gy;
(iii) g2is in an edge stabilizer ofG, andgis not in any vertex stabilizer ofG; (iv) gis not conjugate to any cyclically reduced element ofG;
Proof. (i)⇒(ii). There exists an edgexofXsuch thatg(x)=x¯. The structure ofY implies that there exist an edgeyofY and an elementfofGsuch thatx=f (y)and G(y,y)¯ ≠ϕ. Theng(x)=x¯ implies thatgf (y)=f [y](y). Henceg=f [y]a f−1, wherea∈Gy.
(ii)⇒(iii). If g=f [y]a f−1, wherey is an edge of Y such that G(y,y)¯ ≠ϕ and a∈Gy, theng2∈Gx, wherexis the edgef (y). Now we show thatgis not in any vertex stabilizer ofG. For, ifgis in a vertex stabilizer ofG, then[y]a∈Gv, where
vis a vertex ofX. Then[y]a=bcb−1, whereb∈Gandc∈G
v∗. Letwbe the word w=1·y·a. It is clear thatwis a cyclically reduced word ofGof type(o(y))∗, value [y]a, and of length one. By Lemma 2.7, there exists a reduced word w such that o(w)=(o(y))∗, t(w)=v∗, and[w]=b.
Letw0be the wordw0=w·c·w−1. Now applying edge reductions onw0yields a reduced word ofGof type(o(y))∗, value[w], and of even length. Sincewis reduced of length 1, this contradicts Lemma 2.7. Hencegis not in any vertex stabilizer ofG.
(iii)⇒(iv). Let w be a cyclically reduced word of G of type u, f ∈ G, and v = f (u)such thatg=f [w]f−1. If |w| =0, then[w]∈Gu andg∈Gv. This contra-dicts the assumption that g is not in any vertex stabilizer of G. If |w| ≥1, then w2 is a cyclically reduced word of G of type u, and |w2| ≥ 2. This implies that g2=f [w2]f−1. Then[w2]=f−1g2f. Sinceg2∈Gv, therefore[w2]∈Gu. This con-tradicts Proposition 2.13.
(iv)⇒(v). Letwbe a closed and reduced word ofGof typevand valueg. Sinceg is not conjugate to any cyclically reduced elements ofG, thereforewcan be written asw=w0·a·y·b·w−1
0 , wherew0 is a reduced word ofG such thato(w0)=v, t(w0)=(o(y))∗, andy is an edge ofY such thatG(y,y)¯ ≠ϕanda,b∈G
y. Then
g2is the value of the wordw0·c·w−1
0 , wherec=a[y]ba[y]b. Now applying edge reductions onwyields a reduced word ofGof odd length. Similarly, by applying edge reductions onw0·c·w−1
0 yields a reduced word ofGof even length. This implies that|g|vis odd and|g2|v<|g|v.
(v)⇒(i).gis the value of the closed and reduced wordw=w0·y·w1ofGof type avertexvofT, wherew0andw1are reduced words ofGsuch that|w0| = |w1|, and yis an edge ofY. Theng2is the value ofw2. Since|g2|v<|g|v, thereforew2is not reduced. This implies that[w1]=[w0]−1, andG(y,y)¯ ≠ϕ. Sog=[w0][y][w0]−1. Letxbe the edgex=[w0](y). Then
g(x)=[w0][y][w0]−1[w0](y)=[w0][y](y) =[w0](y),¯ becauseG(y,y)¯ ≠ϕ
=[w0](y)=x.¯
(3.1)
This completes the proof.
Remark3.3. We note that the conditionXG=ϕis essential. For instance, ifXG≠
ϕ, then there exists a vertexvofXsuch thatG=Gv. Then every invertor element of
Gis inGv. This contradicts Theorem 3.2(iii).
Corollary3.5. Letgbe an element ofG,vbe a vertex ofΓ, andnbe an integer. Then
(i) |g2n|v= |g2n+1|v ifg∈Gv, (ii) |g2n|
v<|g2n+1|v ifgis an invertor element ofG, (iii) |g2n|
v>|g2n+1|v ifgis noninvertor element ofG, g∉Gv.
4. Finitely generated subgroups of groups acting on trees. The main result of this section is the following theorem.
Theorem4.1. LetGbe a group acting on a treeX,T be a tree of representatives for the action ofGonXandYbe a fundamental domain for the action ofGonXsuch thatXG=ϕandT⊆Y. LetHbe a finitely generated subgroup ofG. ThenHcontains
an invertor element or a conjugate ofHcontains a cyclically reduced element of length at least one on whichHis not in any vertex stabilizer ofG, orHis in a vertex stabilizer ofG. IfHis contained in the vertex stabilizersGuandGv of the verticesuandvofX,
thenGu=Gv, orHlies in any edge stabilizerGx, wherexis any edge in the reduced
path inXjoininguandv.
Proof. IfHcontains an invertor element ofG, then by Theorem 3.2(iii),His not in any vertex stabilizer ofG. Now assume thatHcontains no invertor element ofG. SinceHis finitely generated, thereforeH= h1,...,hn. Letw1,...,wnbe closed and reduced words ofGof the same type such that[wi]=hi fori=1,...,n. The proof is by induction on the sumr of the lengths of the wordsw1,...,wn. Ifr=0, then it is clear thatH is contained inGv, wherevis the type of the wordsw1,...,wn. Now assume that finitely generated subgroups ofGhaving generators in which the sum of the lengths of closed and reduced words of G of the same type and values the generators of these subgroups is less thanr are contained in the vertex stabilizers of some vertices ofX, or have cyclically reduced elements of length at least one. If one of the wordsw1,...,wnis cyclically reduced of length at least one we are done. Assume that this is not the case.
Leth∈ {h1,...,hn}andw∈ {w1,...,wn}such that[w]=hand
w=g0·y1·g1···ym·gm, m≥1, (4.1)
his not cyclically reduced; for, ifhis cyclically reduced, thenh∈His cyclically re-duced of length at least one. This contradicts the assumption thatHhas no cyclically reduced element of length at least one. If h∈ {h1,...,h
n}and w∈ {w1,...,wn} such that[w]=h and w=f0·x1·f1· ··· ·x
s·fs, thenhh is not cyclically re-duced. Otherwisehhwill be cyclically reduced of length at least one. This implies that[ym]gm=a(g0[y1])−1, wherea∈G+ym, and
[xs]fs=(f0[x1])−1=g0[y1]a−1c, wherec=[ym]gmf0[x1]. (4.2)
Leth
i=(g0[y1])−1hig0[y1]fori=1,...,n. Then
H=(g0[y1])−1Hg0[y1]= h
1,...,hn. (4.3) Then the elementsh1,...,h
nare values of closed and reduced wordsw1,...,w nofGof type(t(y1))∗and[w
is less than the sum of the lengths of the wordsw1,...,wn. Then by hypothesis as-sumption onr, eitherH is contained inG
z for somez∈V(X), or H contains a cyclically reduced element of length at least one. IfHis contained inGz, then it is clear thatHis contained inGv, wherev=g0[y1](z). OtherwiseHis a conjugate of
H andHcontains a cyclically reduced element of length at least one. Now assume thatHis contained in the vertex stabilizersGuandGv of the verticesuandvofX such thatGu≠Gv. Then it is clear thatu≠v, and there exists a unique reduced path
x1,...,xninXjoininguandv. ConsequentlyGu∩Gv⊆ni=1Gxi, andHlies in any edge stabilizerGx, wherex∈ {x1,...,xn}.
This completes the proof.
Corollary4.2. IfHis a finite subgroup ofG, thenHcontains an invertor element, orHis in a vertex stabilizer ofG.
Proof. IfHcontains an invertor element, thenHis not in any vertex stabilizer of G. If a conjugate ofHcontains a cyclically reduced elementgof length at least one, thengnis a cyclically reduced element of length at least one for any integern. Then by Proposition 2.13,gis not in any vertex stabilizer ofG. Thusgis of infinite order. HenceHis in a vertex stabilizer ofG. This completes the proof.
Corollary4.3. IfHis a cyclic subgroup ofGgenerated by the elementgsuch that gis not in any vertex stabilizer ofG, andH∩Gxis trivial for any edge ofX, thenH
is cyclic of order 2 ifgis an invertor element, orHis infinite cyclic ifgis not invertor element.
Proof. Ifgis an invertor element, then by Theorem 3.2(iii),g2is in an edge sta-bilizer ofG. By assumptiong2=1, the identity element ofG. HenceHis a cyclic of order 2. Ifgis not invertor element, then from abovegis of infinite order. HenceH is infinite cyclic. This completes the proof.
5. Applications. This section is an application of Theorems 3.2 and 4.1. Groups acting on trees can be divided into two parts. The first part is of action fixing some vertices, and the other not. For groups acting on trees without fixing any vertices are some of actions without inversions, and the other with inversions. Free groups, free product of groups, free product of groups with amalgamation subgroup, tree product of groups, andHNN groups are examples of groups acting on trees without inversions. A new class of groups called quasiHNN groups are examples of groups acting on trees with inversions. In fact, free product of groups, free product of groups with amalgamation subgroup are special cases of tree product of groups, and, free groups andHNNgroups are special cases of quasiHNNgroups as we will see below. We start with tree product of groups. For more details of tree product of groups we refer the readers to Fisher [1] or Karrass and Solitar [2].
Proposition5.1. LetG=∗i∈I(Ai,Ujk=Ukj)be a nontrivial tree product of the
groupsAi, i∈I, andHbe a finitely generated subgroup ofG. Then a conjugate ofH
Proof. By Mahmood [8], there exists a treeXon whichGacts without inversions such that any tree of representatives for the action ofGonXequals the corresponding fundamental domain for the action ofGonX,XG=ϕ, and for every vertexvofX, and every edgexofX,Guis isomorphic toAi, i∈I, andGxis isomorphic toUik, for somei,k∈I. Moreover,Gcontains no invertor elements. Therefore by Theorem 4.1, the proof of Proposition 5.1 follows.
Corollary5.2. If G is a free product with amalgamation group, then for any
finitely generated subgroupHofGa conjugate ofHcontains a cyclically reduced ele-ment of length at least one orHis contained in a conjugate of a factor ofG. Moreover ifHis finite, thenHis contained in a conjugate of factor ofG.
Before we state our next proposition we introduce the concept of quasiHNNgroups. Khanfar and Mahmood [3] extended the class ofHNNgroups to a new class of groups called quasiHNN groups defined as follows.
LetGbe a group,{Ai:i∈I},{Bi:i∈I}, and{Cj:j∈J}be families of subgroups ofG. For eachi∈I, letφi:Ai→Bibe an onto isomorphism, and for eachj∈J let
αj:Cj→Cjbe an outer automorphism of order 2 such that the inner automorphism
α2
jis determined by ¯cj∈Cjfixed byαj. That is,αj∈Out(Cj)andα2j∈Inn(Cj)such thatαj(c¯j)=c¯jandα2j(c)=c¯jcc¯−1j for allc∈Cj.
LetG∗be the group of the presentation
G∗=G,t
i,tj|relG, tiAiti−1=Bi, tjCjtj−1=Cj, tj2=c¯j, i∈I, j∈J (5.1)
wheretiAit−1i =Bistands for the set of relationstiaiti−1=φi(ai)for allai∈Ai, and
tjCjtj−1=Cjstands for the set of relationstjcjtj−1=αj(cj)for allcj∈Cj. The group
G∗ is called a quasiHNN group of baseGand associated pairs(A
i,Bi), and(Cj,Cj),
i∈I,j∈Jof subgroups ofG. Moreover,Gis embedded inG∗. We note that ifJ= ∅, thenG∗is anHNN group of baseGand associated pairs(Ai,Bi),i∈Iof subgroups ofG.
Proposition5.3. LetG∗ be the quasi HNN group of baseGand associated pairs (Ai,Bi), and(Cj,Cj),i∈I,j∈Jof subgroups ofGdefined above. LetHbe a finitely
generated subgroup ofG∗. ThenH contains an element conjugate to an elementtj, j∈JorHcontains a cyclically reduced element of length at least one orHis contained in a conjugate ofG. Moreover, ifH is finite containing no element conjugate to the elementtj, j∈J, thenHis contained in a conjugate ofG.
Proof. By Mahmood and Khanfar [9], there exists a treeXon whichG∗acts with inversions such thatG∗is transitive on the setV(X), XG∗
=ϕ, and for every vertexv ofX, and every edgexofX,G∗
vis isomorphic toG, andGx∗is isomorphic toAi, i∈I, or is isomorphic toCj, j∈J. Moreover,G∗contains the invertor elements conjugate to an elementtj, j∈J. Therefore by Theorem 4.1, the proof of Proposition 5.3 follows.
Corollary5.4. For any finitely generated subgroupH of an HNN group G∗ of
orHis contained in a conjugate ofG. Moreover ifHis finite, thenHis contained in a conjugate ofG.
Proof. By takingJ= ∅, we see thatG∗of Proposition 5.3 above is anHNNgroup containing no invertor elements. Therefore the proof of Corollary 5.4 follows from Proposition 5.3.
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Mahmood: Ajman University of Science and Technology, Abu Dhabi, United Arab Emirates
E-mail address:[email protected]
