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












1















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-Schrier 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







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  • 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...
    – Dima Pasechnik
    52 mins ago












  • @DimaPasechnik: What you write down is not a group since the subgroup you are trying to take the quotient by is not normal.
    – Andy Putman
    50 mins ago










  • oops, right. sorry for noise.
    – Dima Pasechnik
    46 mins ago
















1















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-Schrier 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
























  • 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...
    – Dima Pasechnik
    52 mins ago












  • @DimaPasechnik: What you write down is not a group since the subgroup you are trying to take the quotient by is not normal.
    – Andy Putman
    50 mins ago










  • oops, right. sorry for noise.
    – Dima Pasechnik
    46 mins ago














1












1








1








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-Schrier 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
















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-Schrier 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






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edited 1 hour ago

























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Milo Brandt

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  • 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...
    – Dima Pasechnik
    52 mins ago












  • @DimaPasechnik: What you write down is not a group since the subgroup you are trying to take the quotient by is not normal.
    – Andy Putman
    50 mins ago










  • oops, right. sorry for noise.
    – Dima Pasechnik
    46 mins ago


















  • 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...
    – Dima Pasechnik
    52 mins ago












  • @DimaPasechnik: What you write down is not a group since the subgroup you are trying to take the quotient by is not normal.
    – Andy Putman
    50 mins ago










  • oops, right. sorry for noise.
    – Dima Pasechnik
    46 mins ago
















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...
– Dima Pasechnik
52 mins ago






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...
– Dima Pasechnik
52 mins ago














@DimaPasechnik: What you write down is not a group since the subgroup you are trying to take the quotient by is not normal.
– Andy Putman
50 mins ago




@DimaPasechnik: What you write down is not a group since the subgroup you are trying to take the quotient by is not normal.
– Andy Putman
50 mins ago












oops, right. sorry for noise.
– Dima Pasechnik
46 mins ago




oops, right. sorry for noise.
– Dima Pasechnik
46 mins ago










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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.






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    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


























      3














      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
























        3












        3








        3






        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












        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 41 mins ago









        Andy Putman

        31.2k5132212




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