Is there an algorithm to decide if a word is in a finitely generated subgroup of a free group?












12












$begingroup$



Let $S$ be a finite set and $F$ is the free group on that set. Is there an algorithm which takes as input a sequence of $w,w_1,ldots,w_kin F$ and decides whether $win langle w_1,ldots,w_krangle$?




This question keeps appearing in some of my work. My intuition is that this has been solved somewhere. It seems very related to the Nielsen-Schreier theorem and, to my understanding, Nielsen's proof of this theorem gave an algorithm for finding a free generating set for any finitely generated subgroup of a free group - which is very closely related to this problem. I also have found various literature referring to this as a "generalized word problem" and various undecidability results relating to the problem in general - but, even though nothing suggests that this is undecidable for a free group, I've not come across any algorithm for deciding it.






gr.group-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you can decide this then you can decide whether $F/langle w_1,...,w_krangle$ is trivial, just check for each $w$ being an element of $S$, right? Thus it's undecideable...
    $endgroup$
    – Dima Pasechnik
    Dec 30 '18 at 4:21








  • 1




    $begingroup$
    @DimaPasechnik: What you write down is not a group since the subgroup you are trying to take the quotient by is not normal.
    $endgroup$
    – Andy Putman
    Dec 30 '18 at 4:23










  • $begingroup$
    oops, right. sorry for noise.
    $endgroup$
    – Dima Pasechnik
    Dec 30 '18 at 4:27






  • 3




    $begingroup$
    This is known as "solvable uniform (subgroup) membership problem".
    $endgroup$
    – YCor
    Dec 30 '18 at 5:31










  • $begingroup$
    I saw this in the hot questions list and went, "Huh! We worked on something similar at Hampshire but didn't get far.". Lo and behold :D
    $endgroup$
    – TreFox
    Dec 30 '18 at 16:06
















12












$begingroup$



Let $S$ be a finite set and $F$ is the free group on that set. Is there an algorithm which takes as input a sequence of $w,w_1,ldots,w_kin F$ and decides whether $win langle w_1,ldots,w_krangle$?




This question keeps appearing in some of my work. My intuition is that this has been solved somewhere. It seems very related to the Nielsen-Schreier theorem and, to my understanding, Nielsen's proof of this theorem gave an algorithm for finding a free generating set for any finitely generated subgroup of a free group - which is very closely related to this problem. I also have found various literature referring to this as a "generalized word problem" and various undecidability results relating to the problem in general - but, even though nothing suggests that this is undecidable for a free group, I've not come across any algorithm for deciding it.






gr.group-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you can decide this then you can decide whether $F/langle w_1,...,w_krangle$ is trivial, just check for each $w$ being an element of $S$, right? Thus it's undecideable...
    $endgroup$
    – Dima Pasechnik
    Dec 30 '18 at 4:21








  • 1




    $begingroup$
    @DimaPasechnik: What you write down is not a group since the subgroup you are trying to take the quotient by is not normal.
    $endgroup$
    – Andy Putman
    Dec 30 '18 at 4:23










  • $begingroup$
    oops, right. sorry for noise.
    $endgroup$
    – Dima Pasechnik
    Dec 30 '18 at 4:27






  • 3




    $begingroup$
    This is known as "solvable uniform (subgroup) membership problem".
    $endgroup$
    – YCor
    Dec 30 '18 at 5:31










  • $begingroup$
    I saw this in the hot questions list and went, "Huh! We worked on something similar at Hampshire but didn't get far.". Lo and behold :D
    $endgroup$
    – TreFox
    Dec 30 '18 at 16:06














12












12








12


2



$begingroup$



Let $S$ be a finite set and $F$ is the free group on that set. Is there an algorithm which takes as input a sequence of $w,w_1,ldots,w_kin F$ and decides whether $win langle w_1,ldots,w_krangle$?




This question keeps appearing in some of my work. My intuition is that this has been solved somewhere. It seems very related to the Nielsen-Schreier theorem and, to my understanding, Nielsen's proof of this theorem gave an algorithm for finding a free generating set for any finitely generated subgroup of a free group - which is very closely related to this problem. I also have found various literature referring to this as a "generalized word problem" and various undecidability results relating to the problem in general - but, even though nothing suggests that this is undecidable for a free group, I've not come across any algorithm for deciding it.






gr.group-theory







share|cite|improve this question











$endgroup$





Let $S$ be a finite set and $F$ is the free group on that set. Is there an algorithm which takes as input a sequence of $w,w_1,ldots,w_kin F$ and decides whether $win langle w_1,ldots,w_krangle$?




This question keeps appearing in some of my work. My intuition is that this has been solved somewhere. It seems very related to the Nielsen-Schreier theorem and, to my understanding, Nielsen's proof of this theorem gave an algorithm for finding a free generating set for any finitely generated subgroup of a free group - which is very closely related to this problem. I also have found various literature referring to this as a "generalized word problem" and various undecidability results relating to the problem in general - but, even though nothing suggests that this is undecidable for a free group, I've not come across any algorithm for deciding it.






gr.group-theory




gr.group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 5:25









YCor

27.4k481132




27.4k481132










asked Dec 30 '18 at 3:39









Milo BrandtMilo Brandt

228110




228110












  • $begingroup$
    If you can decide this then you can decide whether $F/langle w_1,...,w_krangle$ is trivial, just check for each $w$ being an element of $S$, right? Thus it's undecideable...
    $endgroup$
    – Dima Pasechnik
    Dec 30 '18 at 4:21








  • 1




    $begingroup$
    @DimaPasechnik: What you write down is not a group since the subgroup you are trying to take the quotient by is not normal.
    $endgroup$
    – Andy Putman
    Dec 30 '18 at 4:23










  • $begingroup$
    oops, right. sorry for noise.
    $endgroup$
    – Dima Pasechnik
    Dec 30 '18 at 4:27






  • 3




    $begingroup$
    This is known as "solvable uniform (subgroup) membership problem".
    $endgroup$
    – YCor
    Dec 30 '18 at 5:31










  • $begingroup$
    I saw this in the hot questions list and went, "Huh! We worked on something similar at Hampshire but didn't get far.". Lo and behold :D
    $endgroup$
    – TreFox
    Dec 30 '18 at 16:06


















  • $begingroup$
    If you can decide this then you can decide whether $F/langle w_1,...,w_krangle$ is trivial, just check for each $w$ being an element of $S$, right? Thus it's undecideable...
    $endgroup$
    – Dima Pasechnik
    Dec 30 '18 at 4:21








  • 1




    $begingroup$
    @DimaPasechnik: What you write down is not a group since the subgroup you are trying to take the quotient by is not normal.
    $endgroup$
    – Andy Putman
    Dec 30 '18 at 4:23










  • $begingroup$
    oops, right. sorry for noise.
    $endgroup$
    – Dima Pasechnik
    Dec 30 '18 at 4:27






  • 3




    $begingroup$
    This is known as "solvable uniform (subgroup) membership problem".
    $endgroup$
    – YCor
    Dec 30 '18 at 5:31










  • $begingroup$
    I saw this in the hot questions list and went, "Huh! We worked on something similar at Hampshire but didn't get far.". Lo and behold :D
    $endgroup$
    – TreFox
    Dec 30 '18 at 16:06
















$begingroup$
If you can decide this then you can decide whether $F/langle w_1,...,w_krangle$ is trivial, just check for each $w$ being an element of $S$, right? Thus it's undecideable...
$endgroup$
– Dima Pasechnik
Dec 30 '18 at 4:21






$begingroup$
If you can decide this then you can decide whether $F/langle w_1,...,w_krangle$ is trivial, just check for each $w$ being an element of $S$, right? Thus it's undecideable...
$endgroup$
– Dima Pasechnik
Dec 30 '18 at 4:21






1




1




$begingroup$
@DimaPasechnik: What you write down is not a group since the subgroup you are trying to take the quotient by is not normal.
$endgroup$
– Andy Putman
Dec 30 '18 at 4:23




$begingroup$
@DimaPasechnik: What you write down is not a group since the subgroup you are trying to take the quotient by is not normal.
$endgroup$
– Andy Putman
Dec 30 '18 at 4:23












$begingroup$
oops, right. sorry for noise.
$endgroup$
– Dima Pasechnik
Dec 30 '18 at 4:27




$begingroup$
oops, right. sorry for noise.
$endgroup$
– Dima Pasechnik
Dec 30 '18 at 4:27




3




3




$begingroup$
This is known as "solvable uniform (subgroup) membership problem".
$endgroup$
– YCor
Dec 30 '18 at 5:31




$begingroup$
This is known as "solvable uniform (subgroup) membership problem".
$endgroup$
– YCor
Dec 30 '18 at 5:31












$begingroup$
I saw this in the hot questions list and went, "Huh! We worked on something similar at Hampshire but didn't get far.". Lo and behold :D
$endgroup$
– TreFox
Dec 30 '18 at 16:06




$begingroup$
I saw this in the hot questions list and went, "Huh! We worked on something similar at Hampshire but didn't get far.". Lo and behold :D
$endgroup$
– TreFox
Dec 30 '18 at 16:06










1 Answer
1






active

oldest

votes


















17












$begingroup$

Let $T$ be a finite subset of the free group on a set $S$. Nielsen's original proof (described nicely in the beginning of Lyndon and Schupp's book) gives an algorithmic process to find a free generating set $T'$ for the subgroup generated by $T$ with the following very nice property: for a word $u$ in $T'$, the $T'$-length $|u|_{T'}$ of $u$ is at least the $S$-length $|u|_S$ of $u$. To recognize if a word $w$ in $S$ lies in the subgroup generated by $T$, it is thus enough to check whether it equals any of the finitely many words of length at most $|w|_S$ in $T'$.



But of course there are much faster and better ways to do this. The nicest algorithm (which runs very fast) is based on Stallings folding and can be found in



Stallings, John R.
Topology of finite graphs.
Invent. Math. 71 (1983), no. 3, 551–565.



I don't have the paper handy right now, so I'm not sure if the algorithm is made explicit in it, but if you understand this paper then it should be clear how to do what you want.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In fact the algorithm can be thought of as a special case of Todd-Coxeter coset enumeration.
    $endgroup$
    – Derek Holt
    Dec 30 '18 at 10:51






  • 1




    $begingroup$
    @DerekHolt: Conversely, having "grown up" with Stallings' algorithm, I've never needed to learn the Todd--Coxeter algorithm explicilty, since you can recover it in any group by pulling back to a free group. But I guess Todd--Coxeter comes with no guarantees that it will successfully distinguish cosets in general.
    $endgroup$
    – HJRW
    Dec 30 '18 at 15:11






  • 1




    $begingroup$
    Yes, in general if two cosets are equal, then Todd-Coxeter will eventually establish their equality, but you cannot predict how long it will take (which is inevitable since the question is undecidable). But for free groups, where there are no relations, there is a bound on the total time.
    $endgroup$
    – Derek Holt
    Dec 30 '18 at 20:36






  • 2




    $begingroup$
    The algorithm is made pretty explicit there.
    $endgroup$
    – Benjamin Steinberg
    Dec 30 '18 at 21:44











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









17












$begingroup$

Let $T$ be a finite subset of the free group on a set $S$. Nielsen's original proof (described nicely in the beginning of Lyndon and Schupp's book) gives an algorithmic process to find a free generating set $T'$ for the subgroup generated by $T$ with the following very nice property: for a word $u$ in $T'$, the $T'$-length $|u|_{T'}$ of $u$ is at least the $S$-length $|u|_S$ of $u$. To recognize if a word $w$ in $S$ lies in the subgroup generated by $T$, it is thus enough to check whether it equals any of the finitely many words of length at most $|w|_S$ in $T'$.



But of course there are much faster and better ways to do this. The nicest algorithm (which runs very fast) is based on Stallings folding and can be found in



Stallings, John R.
Topology of finite graphs.
Invent. Math. 71 (1983), no. 3, 551–565.



I don't have the paper handy right now, so I'm not sure if the algorithm is made explicit in it, but if you understand this paper then it should be clear how to do what you want.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In fact the algorithm can be thought of as a special case of Todd-Coxeter coset enumeration.
    $endgroup$
    – Derek Holt
    Dec 30 '18 at 10:51






  • 1




    $begingroup$
    @DerekHolt: Conversely, having "grown up" with Stallings' algorithm, I've never needed to learn the Todd--Coxeter algorithm explicilty, since you can recover it in any group by pulling back to a free group. But I guess Todd--Coxeter comes with no guarantees that it will successfully distinguish cosets in general.
    $endgroup$
    – HJRW
    Dec 30 '18 at 15:11






  • 1




    $begingroup$
    Yes, in general if two cosets are equal, then Todd-Coxeter will eventually establish their equality, but you cannot predict how long it will take (which is inevitable since the question is undecidable). But for free groups, where there are no relations, there is a bound on the total time.
    $endgroup$
    – Derek Holt
    Dec 30 '18 at 20:36






  • 2




    $begingroup$
    The algorithm is made pretty explicit there.
    $endgroup$
    – Benjamin Steinberg
    Dec 30 '18 at 21:44
















17












$begingroup$

Let $T$ be a finite subset of the free group on a set $S$. Nielsen's original proof (described nicely in the beginning of Lyndon and Schupp's book) gives an algorithmic process to find a free generating set $T'$ for the subgroup generated by $T$ with the following very nice property: for a word $u$ in $T'$, the $T'$-length $|u|_{T'}$ of $u$ is at least the $S$-length $|u|_S$ of $u$. To recognize if a word $w$ in $S$ lies in the subgroup generated by $T$, it is thus enough to check whether it equals any of the finitely many words of length at most $|w|_S$ in $T'$.



But of course there are much faster and better ways to do this. The nicest algorithm (which runs very fast) is based on Stallings folding and can be found in



Stallings, John R.
Topology of finite graphs.
Invent. Math. 71 (1983), no. 3, 551–565.



I don't have the paper handy right now, so I'm not sure if the algorithm is made explicit in it, but if you understand this paper then it should be clear how to do what you want.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In fact the algorithm can be thought of as a special case of Todd-Coxeter coset enumeration.
    $endgroup$
    – Derek Holt
    Dec 30 '18 at 10:51






  • 1




    $begingroup$
    @DerekHolt: Conversely, having "grown up" with Stallings' algorithm, I've never needed to learn the Todd--Coxeter algorithm explicilty, since you can recover it in any group by pulling back to a free group. But I guess Todd--Coxeter comes with no guarantees that it will successfully distinguish cosets in general.
    $endgroup$
    – HJRW
    Dec 30 '18 at 15:11






  • 1




    $begingroup$
    Yes, in general if two cosets are equal, then Todd-Coxeter will eventually establish their equality, but you cannot predict how long it will take (which is inevitable since the question is undecidable). But for free groups, where there are no relations, there is a bound on the total time.
    $endgroup$
    – Derek Holt
    Dec 30 '18 at 20:36






  • 2




    $begingroup$
    The algorithm is made pretty explicit there.
    $endgroup$
    – Benjamin Steinberg
    Dec 30 '18 at 21:44














17












17








17





$begingroup$

Let $T$ be a finite subset of the free group on a set $S$. Nielsen's original proof (described nicely in the beginning of Lyndon and Schupp's book) gives an algorithmic process to find a free generating set $T'$ for the subgroup generated by $T$ with the following very nice property: for a word $u$ in $T'$, the $T'$-length $|u|_{T'}$ of $u$ is at least the $S$-length $|u|_S$ of $u$. To recognize if a word $w$ in $S$ lies in the subgroup generated by $T$, it is thus enough to check whether it equals any of the finitely many words of length at most $|w|_S$ in $T'$.



But of course there are much faster and better ways to do this. The nicest algorithm (which runs very fast) is based on Stallings folding and can be found in



Stallings, John R.
Topology of finite graphs.
Invent. Math. 71 (1983), no. 3, 551–565.



I don't have the paper handy right now, so I'm not sure if the algorithm is made explicit in it, but if you understand this paper then it should be clear how to do what you want.






share|cite|improve this answer









$endgroup$



Let $T$ be a finite subset of the free group on a set $S$. Nielsen's original proof (described nicely in the beginning of Lyndon and Schupp's book) gives an algorithmic process to find a free generating set $T'$ for the subgroup generated by $T$ with the following very nice property: for a word $u$ in $T'$, the $T'$-length $|u|_{T'}$ of $u$ is at least the $S$-length $|u|_S$ of $u$. To recognize if a word $w$ in $S$ lies in the subgroup generated by $T$, it is thus enough to check whether it equals any of the finitely many words of length at most $|w|_S$ in $T'$.



But of course there are much faster and better ways to do this. The nicest algorithm (which runs very fast) is based on Stallings folding and can be found in



Stallings, John R.
Topology of finite graphs.
Invent. Math. 71 (1983), no. 3, 551–565.



I don't have the paper handy right now, so I'm not sure if the algorithm is made explicit in it, but if you understand this paper then it should be clear how to do what you want.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 30 '18 at 4:32









Andy PutmanAndy Putman

31.6k6134214




31.6k6134214












  • $begingroup$
    In fact the algorithm can be thought of as a special case of Todd-Coxeter coset enumeration.
    $endgroup$
    – Derek Holt
    Dec 30 '18 at 10:51






  • 1




    $begingroup$
    @DerekHolt: Conversely, having "grown up" with Stallings' algorithm, I've never needed to learn the Todd--Coxeter algorithm explicilty, since you can recover it in any group by pulling back to a free group. But I guess Todd--Coxeter comes with no guarantees that it will successfully distinguish cosets in general.
    $endgroup$
    – HJRW
    Dec 30 '18 at 15:11






  • 1




    $begingroup$
    Yes, in general if two cosets are equal, then Todd-Coxeter will eventually establish their equality, but you cannot predict how long it will take (which is inevitable since the question is undecidable). But for free groups, where there are no relations, there is a bound on the total time.
    $endgroup$
    – Derek Holt
    Dec 30 '18 at 20:36






  • 2




    $begingroup$
    The algorithm is made pretty explicit there.
    $endgroup$
    – Benjamin Steinberg
    Dec 30 '18 at 21:44


















  • $begingroup$
    In fact the algorithm can be thought of as a special case of Todd-Coxeter coset enumeration.
    $endgroup$
    – Derek Holt
    Dec 30 '18 at 10:51






  • 1




    $begingroup$
    @DerekHolt: Conversely, having "grown up" with Stallings' algorithm, I've never needed to learn the Todd--Coxeter algorithm explicilty, since you can recover it in any group by pulling back to a free group. But I guess Todd--Coxeter comes with no guarantees that it will successfully distinguish cosets in general.
    $endgroup$
    – HJRW
    Dec 30 '18 at 15:11






  • 1




    $begingroup$
    Yes, in general if two cosets are equal, then Todd-Coxeter will eventually establish their equality, but you cannot predict how long it will take (which is inevitable since the question is undecidable). But for free groups, where there are no relations, there is a bound on the total time.
    $endgroup$
    – Derek Holt
    Dec 30 '18 at 20:36






  • 2




    $begingroup$
    The algorithm is made pretty explicit there.
    $endgroup$
    – Benjamin Steinberg
    Dec 30 '18 at 21:44
















$begingroup$
In fact the algorithm can be thought of as a special case of Todd-Coxeter coset enumeration.
$endgroup$
– Derek Holt
Dec 30 '18 at 10:51




$begingroup$
In fact the algorithm can be thought of as a special case of Todd-Coxeter coset enumeration.
$endgroup$
– Derek Holt
Dec 30 '18 at 10:51




1




1




$begingroup$
@DerekHolt: Conversely, having "grown up" with Stallings' algorithm, I've never needed to learn the Todd--Coxeter algorithm explicilty, since you can recover it in any group by pulling back to a free group. But I guess Todd--Coxeter comes with no guarantees that it will successfully distinguish cosets in general.
$endgroup$
– HJRW
Dec 30 '18 at 15:11




$begingroup$
@DerekHolt: Conversely, having "grown up" with Stallings' algorithm, I've never needed to learn the Todd--Coxeter algorithm explicilty, since you can recover it in any group by pulling back to a free group. But I guess Todd--Coxeter comes with no guarantees that it will successfully distinguish cosets in general.
$endgroup$
– HJRW
Dec 30 '18 at 15:11




1




1




$begingroup$
Yes, in general if two cosets are equal, then Todd-Coxeter will eventually establish their equality, but you cannot predict how long it will take (which is inevitable since the question is undecidable). But for free groups, where there are no relations, there is a bound on the total time.
$endgroup$
– Derek Holt
Dec 30 '18 at 20:36




$begingroup$
Yes, in general if two cosets are equal, then Todd-Coxeter will eventually establish their equality, but you cannot predict how long it will take (which is inevitable since the question is undecidable). But for free groups, where there are no relations, there is a bound on the total time.
$endgroup$
– Derek Holt
Dec 30 '18 at 20:36




2




2




$begingroup$
The algorithm is made pretty explicit there.
$endgroup$
– Benjamin Steinberg
Dec 30 '18 at 21:44




$begingroup$
The algorithm is made pretty explicit there.
$endgroup$
– Benjamin Steinberg
Dec 30 '18 at 21:44


















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Monofisismo

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Monofisismo (do grego: monos - "único, singular" e physis - "natureza") é o ponto de vista cristológico que defende que, depois da união do divino e do humano na encarnação histórica, Jesus Cristo, como encarnação do Filho ou Verbo ( Logos ) de Deus, teria apenas uma única "natureza", a divina, e não uma síntese de ambas. O monofisismo é contraposto pelo diofisismo (ou "diafisismo"), que defende que Jesus preservou em si as duas naturezas. Historicamente, o monofisismo se refere primordialmente à posição dos que (especialmente no Egito e, em menor grau, na Síria) rejeitaram as decisões do Concílio de Calcedônia em 451 (o quarto concílio ecumênico). Os membros mais moderados entre eles, porém, defendem a teologia "miafisita", que se tornou a oficial para as Igrejas Ortodoxas Orientais. Muitos ortodoxos orientais, porém, rejeitam essa classificação, mesmo como um termo genérico, mas ele é amplamente utilizado na literatura históri...
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Angular Downloading a file using contenturl with Basic Authentication

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0 I have a contentUrl when user clicks on that,file should be downloaded. If I try to browse the contentUrl in the browser it asks for Userid and password.I believe this is Basic Authentication.I dont want user to enter UserId and Password so I am passing it Authorization(Please see below code). Url looks some thing like this: http://XXX.XXX.XX.XXX:8080/flowable-rest/service/runtime/tasks/90875/attachments/90555/content let username : string = 'admin'; let pwd : string = 'test'; let headers = new Headers(); headers.set('Accept', 'text/json'); headers.append('Content-Type', 'application/json'); headers.set('Access-Control-Allow-Origin', '*'); headers.set('Access-Control-Allow-Methods','POST, GET, OPTIONS, PUT, DELETE'); head...
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Olmecas

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Monumento 1, uma das quatro cabeças colossais encontradas em La Venta, com altura aproximada de 3 metros Olmecas é a designação do povo e da civilização que estiveram na origem da antiga cultura pré-colombiana da Mesoamérica e que se desenvolveram nas regiões tropicais do centro-sul do atual México durante o pré-clássico, próximo de onde hoje estão localizados os estados mexicanos de Veracruz e Tabasco, no Istmo de Tehuantepec, numa zona designada área nuclear olmeca. A cultura olmeca floresceu nesta região aproximadamente entre 1500 e 400 a.C., [ 1 ] e crê-se que tenha sido a civilização-mãe de todas as civilizações mesoamericanas que se desenvolveram posteriormente. [ 2 ] No entanto, desconhece-se a sua exacta filiação étnica, ainda que existam numerosas hipóteses colocadas para tentar resolver esta questão. O etnónimo olmeca foi cunhado pelos arqueólogos do século XX, e não devem confundir-se com os muito posteriores olmecas-xicalancas que ocuparam vários locais do México...
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