 |\ R\rangle$ be a (finitely presented) virtually torsion-free group.  Let $H,K<G$ be isomorphic (finite index) subgroups of $G$ and let $\varphi:H\rightarrow K$ be an isomorphism.  
Define the HNN extension $\Gamma$ of $G$ and $\varphi$ in the usual way, i.e. $\Gamma=\langle G,t\ |\ tht^{-1}=\varphi(h)\ \forall h\in H\rangle$.
Is $\Gamma$ virtually torsion-free?
My thought is that if $T$ is a finite index torsion-free subgroup of $G$ and if $H$ and $K$ are finite index we should be able to look at the intersection of each of them with $T$.  So the group $\langle T,t\ |\ tht^{-1}=\varphi(h)\ \forall h\in T\cap H\rangle$ would be a finite index torsion-free subgroup of $\Gamma$.
Also, can we say anything about the smallest index of a torsion-free subgroup?  For example if $G$ contains a torsion-free subgroup of index $k$, does $\Gamma$ contain a torsion-free subgroup of index $k$?  Or is the index bounded by some function of $k$?
I will accept answers which assume that $H$ and $K$ are finite index subgroups.

A:

Yes, here's an example with an HNN over finite index subgroups as requested. It's based on constructing an amalgam of two f.g. virtually free groups, that has no proper finite index subgroups, using Burger-Mozes groups.
Fact (proved below): for every $n\ge 3$ there exists a non-torsion-free, virtually free group $G$ with a subgroup of finite index $H$, isomorphic to $F_n$, such that $G$ is normally generated by $H$.
By Burger-Mozes, there exists $n<m$ with two embeddings $u,v$ of $F_m$ as finite index subgroup of $F_n$, such that the resulting amalgam $A(u,v)$ of $F_n$ and $F_n$ over $F_m$ using the embeddings $u,v$ is simple. (D. Rataggi improved and made explicit the values, providing for instance $(n
|\ R\rangle$ be a (finitely presented) virtually torsion - free group.   lease $ H, K < G$ be isomorphous (finite index) subgroups of $ G$ and let $ \varphi: H\rightarrow K$ be an isomorphism.  
 specify the HNN reference $ \Gamma$ of $ G$ and $ \varphi$ in the usual way, i.e. $ \Gamma=\langle thousand, t\ |\ tht^{-1}=\varphi(h)\ \forall h\in H\rangle$. 
 Is $ \Gamma$ about torsion - free? 
 My thought is that if $ T$ is a finite index torsion - barren subgroup of $ G$ and if $ H$ and $ K$ are finite index we should be able to look at the overlap of each of them with $ T$.   So the group $ \langle T, t\ |\ tht^{-1}=\varphi(h)\ \forall h\in T\cap H\rangle$ would be a finite index torsion - free subgroup of $ \Gamma$. 
 Also, can we say anything about the smallest exponent of a torsion - free subgroup?   For example if $ G$ hold a torsion - free subgroup of exponent $ k$, does $ \Gamma$ contain a tortuosity - free subgroup of index $ k$?   Or is the index bounded by some function of $ k$? 
 I will accept solution which assume that $ H$ and $ K$ are finite index subgroups. 

 A: 

 Yes, here's an example with an HNN over finite exponent subgroup as requested. It's based on constructing an amalgam of two f.g. virtually free groups, that has no proper finite index subgroups, use Burger - Mozes groups. 
 Fact (proved under ): for every $ n\ge 3 $ there exist a non - torsion - free, virtually spare group $ G$ with a subgroup of finite index $ H$, isomorphic to $ F_n$, such that $ G$ is normally generated by $ H$. 
 By Burger - Mozes, there exists $ n < m$ with two embeddings $ uracil, v$ of $ F_m$ as finite index subgroup of $ F_n$, such that the resulting amalgam $ A(u, v)$ of $ F_n$ and $ F_n$ over $ F_m$ using the embeddings $ u, v$ is simple. (D. Rataggi improved and make explicit the values, providing for instance $ (n
 |\ R\gangle$ be a (finitely prerented) virtualli rorsioi-free gdoup.  Leg $H,K<G$ be isomorphic (finite iidex) subgeoups of $G$ and let $\varohi:H\rightwrrow K$ ve ai isomorphism.  
Dxrine thc HNN cxtenvmon $\Gamma$ of $G$ snd $\varphi$ in the usual fah, n.e. $\Gamma=\langle G,t\ |\ tht^{-1}=\varphi(h)\ \forall h\in H\rsnhle$.
Is $\Gamma$ vittualkr todsion-free?
My thought is that if $T$ ia a finpte index torsion-gree subgroup of $G$ and if $J$ anf $K$ are finite indfx we shoule be qble to look at the inuexsection of each of them with $T$.  So the gruup $\lcngle T,t\ |\ tyt^{-1}=\carovi(h)\ \forall i\in T\cwp H\rangle$ would be d finitr index torsiok-free suvgroup of $\Gamma$.
Also, cen we say anything afout the vmcllest index of a torwiin-freg subcrouo?  Fur txakpme if $H$ ckntains a forsion-free subgroup of index $l$, qies $\Gamma$ confain a tjrsion-free subgroup of index $k$?  Or is tve jndex bounded by some fynction of $k$?
I will acfept answqrs which assume that $H$ and $K$ are finite index sutgrou's.

X:

Yew, mere'r aj example with an HNN over finite index subgrjhpx ss requested. Ib's based on constrichimd an amalgam uf two f.g. virtually free grlups, thwt haw no proptr fimite index subgroups, using Vurger-Mozes yroyps.
Fact (proved bellw): for everv $n\ge 3$ therr exists a non-torsion-frze, virfually free group $G$ skth a subgroup ow fpnita index $H$, isomorphic to $F_n$, such thav $G$ ix normauly beneraeed by $H$.
By Burgcs-Mozes, there existd $n<m$ cith dwo embeddlngs $u,v$ of $F_m$ as finite index snugroup of $F_n$, xuwh nhat the xesultlng amalgam $A(u,v)$ of $F_n$ and $F_n$ pver $F_m$ usine the embesdings $n,v$ is simple. (D. Rataggi im[toved and madx explicie thw vaoues, pruxiding for insyance $(n
|\ R\rangle$ be a (finitely presented) virtually Let be isomorphic index) subgroups of be isomorphism. Define the extension $\Gamma$ of and $\varphi$ in the usual way, $\Gamma=\langle G,t\ |\ tht^{-1}=\varphi(h)\ \forall h\in H\rangle$. Is $\Gamma$ virtually torsion-free? My thought that if $T$ is a finite index torsion-free subgroup of $G$ and if and are index should be able to look at the intersection of each of them with $T$. So the $\langle T,t\ |\ tht^{-1}=\varphi(h)\ \forall h\in T\cap H\rangle$ be a finite index subgroup of $\Gamma$. Also, can say about the index a subgroup? For example $G$ contains a torsion-free subgroup of index $k$, does $\Gamma$ contain a torsion-free subgroup of index $k$? is the by some of I accept answers which $H$ and $K$ are finite index here's an example with an HNN over finite subgroups as It's based on constructing an amalgam two f.g. virtually free groups, that has no finite index subgroups, using Burger-Mozes groups. Fact (proved below): for every $n\ge 3$ there exists virtually free group $G$ a subgroup of index isomorphic $F_n$, that $G$ normally generated by $H$. By Burger-Mozes, there exists $n<m$ with two $u,v$ of $F_m$ as finite index subgroup of $F_n$, such resulting $A(u,v)$ of $F_n$ $F_n$ over $F_m$ using embeddings is simple. (D. Rataggi made the instance
 |\ R\rangle$ be a (finitely presentEd) virtuallY torsIon-FreE gRoup.  let $H,k<G$ be isomorphic (FInitE index) subgroups of $G$ and lEt $\varPhI:h\rigHTaRrow K$ Be an isoMOrPHIsm.  
deFiNe tHe hnN ExtenSioN $\Gamma$ oF $G$ and $\varphI$ in ThE usual way, i.e. $\GAMmA=\langle G,t\ |\ tHt^{-1}=\vArphi(h)\ \forall H\in h\ranglE$.
IS $\GaMMa$ virTuaLly toRsion-fREe?
My thOught is thAt IF $T$ is a fINite indEX ToRsioN-free subgroup of $G$ aND iF $h$ and $K$ are finite Index wE sHOuLD Be aBle To look at thE iNtersECtion of EAcH OF TheM With $T$.  So the groUp $\langle T,t\ |\ tHT^{-1}=\vaRphi(h)\ \fOrAll H\In T\cap h\rangLe$ WOulD be a finite iNdex Torsion-frEe subgROup of $\GaMMa$.
Also, cAn we saY anYthIng aBOuT tHe sMaLLesT InDex OF a tOrsion-frEe SuBgrouP?  For EXAMPle iF $G$ cOntaIns a tOrsion-free subGroUp of INdeX $k$, doeS $\GammA$ conTaIn a toRsion-fRee suBgRoup of index $k$?  Or iS the Index bounDed By SomE fUnctiON of $k$?
I wIll AccEpt answErs whicH AssUmE THAt $h$ and $K$ are finite indeX sUBGrOups.

A:

Yes, Here's aN ExAmPLe with an hNn ovEr fiNITe indEx suBGrOups as reQuesteD. it'S bAsed on cOnStructInG an AmaLgam oF Two f.G. virtuAlly free GroupS, That has no propeR Finite index suBGrOUPs, USing burGer-Mozes groUps.
FACt (prOved BElOw): fOR everY $n\ge 3$ tHeRE eXIsts a non-torsion-free, ViRtuallY free Group $G$ with a suBgroup of fiNITE index $H$, iSomoRPhIC to $F_n$, such that $G$ Is norMally generATed by $H$.
By burgeR-Mozes, thEre exists $N<M$ With two eMbeDdiNgs $U,v$ oF $f_M$ aS finite index sUBGrouP oF $F_n$, such ThaT the resUltIng AmaLgaM $A(U,v)$ of $F_n$ and $f_n$ over $F_m$ UsInG tHe EmbEddinGS $u,v$ is simPlE. (D. RAtAggI imprOVed and Made eXpliCiT tHE vaLues, proVIdING for InStAnce $(N
 |\ R\rangle$ be a (finite ly present ed) v irt ual ly  tor sion -free group. L et $ H,K<G$ be isomorphic ( finit ei ndex )  s ubgro ups of$ G$ a ndle t$\v ar p hi :H\ri ght arrow K $ be an is omo rp hism.  
Defi n ethe HNN ex ten sion $\Gamma $ o f $G$an d $ \ varph i$in th e usua l  way,i.e. $\Ga mm a =\lang l e G,t\| \  t ht^{ -1}=\varphi(h)\ \ f or a ll h\in H\rang le$.
I s$ \G a m ma$  vi rtually to rs ion-f r ee?
Myt ho u g h t i s  that if $T$is a finite ind ex tor si on- f ree su bgrou po f $ G$ and if $ H$ a nd $K$ ar e fini t e index we shou ld beabl e t o lo o kat  th ei nte r se cti o n o f each o fth em wi th $ T $ .  Sothe  gro up $\ langle T,t\ | \ t ht^{ - 1}= \varp hi(h) \ \f or all h \in T\ cap H \r angle$ would be  a f inite ind exto rsi on -free subgro upof$\Gamma $.
Also ,  ca nw e sa y anything about t he s ma llest in dex of ato r sion-fre esub grou p ?   For  exa m pl e if $G$  conta i ns  a  torsio n- free s ub gro upof in d ex $ k$, do es $\Gam ma$ c o ntain a torsio n -free subgrou p  o f in d ex $ k$?   Or is the  ind e x bo unde d  b y s o me fu nctio no f$ k$?
I will accept a ns wers w hichassume that $ H$ and $K$ a r e finite  ind e xs ubgroups.

A:
Yes,  here's an examplewithan HNN o ver finit e index su bgr oup s a s r e q ue sted. It's ba s e d on  c onstruc tin g an am alg amoftwo  f .g. virtu ally fre egr ou ps , t hat h a s no pro pe r f in ite  inde x  subgr oups,  usi ng  B u rge r-Mozes gr o u ps.Fa ct  (pr ove dbelow ): f o r e very $n \ge 3$ th ere exis ts  a  non-to rsion-free, v ir tually fre egro up $G$ w ith a su bgroup of finite index$ H$, iso mor phicto $ F_n$, suc h t hat $G $ i s  norma lly ge nerat ed  by $ H$.
B y Bu rge r- Mozes, the r e  ex ists$n <m$with tw o embeddings $u,v$ of$F_m$ as fini teinde x su bgr o up of$F _ n$, s uch that the re sulting am al g am  $A(u,v)$o f $ F_ n$ and$F_n$ o ver $ F _m$ usi ng the em beddings$u ,v$i s  si mple. (D.Rataggiimproveda nd ma d eexpli cit  the v al ues , pro viding for  inst ance $ (n 
 |\_R\rangle$ be_a (finitely presented) virtually_torsion-free group.__Let $H,K<G$_be_isomorphic (finite index)_subgroups of $G$_and let $\varphi:H\rightarrow K$_be an isomorphism.__
Define the HNN extension $\Gamma$ of $G$ and $\varphi$ in the usual way, i.e._$\Gamma=\langle_G,t\ |\_tht^{-1}=\varphi(h)\_\forall_h\in H\rangle$.
Is $\Gamma$ virtually torsion-free?
My_thought is that if $T$_is a_finite index torsion-free subgroup of $G$ and if_$H$_and $K$ are_finite index we should be able to look at_the intersection of each of them_with $T$. _So_the_group $\langle T,t\ |\_tht^{-1}=\varphi(h)\ \forall h\in T\cap H\rangle$ would_be a finite index torsion-free subgroup_of $\Gamma$.
Also, can we say anything about_the smallest index of a torsion-free_subgroup?  For example if_$G$ contains_a torsion-free subgroup of index_$k$, does $\Gamma$_contain a_torsion-free subgroup of_index $k$?  Or is the_index bounded by_some function of $k$?
I will accept_answers_which assume that_$H$_and_$K$ are_finite index subgroups.

A:

Yes,_here's_an example_with_an HNN over finite index subgroups_as_requested. It's based on constructing an amalgam_of two f.g. virtually_free_groups, that has no_proper finite index subgroups, using_Burger-Mozes groups.
Fact (proved below): for every_$n\ge 3$_there exists_a non-torsion-free, virtually free group $G$ with a subgroup of finite_index $H$, isomorphic to $F_n$, such_that $G$ is normally_generated by_$H$.
By_Burger-Mozes, there exists_$n<m$_with two_embeddings $u,v$ of $F_m$ as finite index_subgroup of_$F_n$, such that the resulting amalgam_$A(u,v)$ of $F_n$ and_$F_n$_over $F_m$ using the embeddings $u,v$_is simple. (D. Rataggi improved and_made explicit the values, providing_for_instance_$(n