Is there an algorithm to decide if a word is in a finitely generated subgroup of a free group?
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
edited 33 mins ago
YCor
27.1k380132
asked 2 hours ago
Milo Brandt
1789
add a comment |
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
edited 33 mins ago
YCor
27.1k380132
asked 2 hours ago
Milo Brandt
1789
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
1 hour 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
1 hour ago
oops, right. sorry for noise.
– Dima Pasechnik
1 hour ago
This is known as "solvable uniform (subgroup) membership problem".
– YCor
27 mins ago
add a comment |
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
edited 33 mins ago
YCor
27.1k380132
asked 2 hours ago
Milo Brandt
1789
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
edited 33 mins ago
YCor
27.1k380132
asked 2 hours ago
Milo Brandt
1789
edited 33 mins ago
YCor
27.1k380132
asked 2 hours ago
Milo Brandt
1789
edited 33 mins ago
YCor
27.1k380132
edited 33 mins ago
YCor
27.1k380132
edited 33 mins ago
YCor
27.1k380132
27.1k380132
asked 2 hours ago
Milo Brandt
1789
asked 2 hours ago
Milo Brandt
1789
asked 2 hours ago
Milo Brandt
1789
1789
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
1 hour 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
1 hour ago
oops, right. sorry for noise.
– Dima Pasechnik
1 hour ago
This is known as "solvable uniform (subgroup) membership problem".
– YCor
27 mins ago
add a comment |
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
1 hour 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
1 hour ago
oops, right. sorry for noise.
– Dima Pasechnik
1 hour ago
This is known as "solvable uniform (subgroup) membership problem".
– YCor
27 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
oops, right. sorry for noise.
– Dima Pasechnik
1 hour ago
oops, right. sorry for noise.
– Dima Pasechnik
1 hour ago
This is known as "solvable uniform (subgroup) membership problem".
– YCor
27 mins ago
This is known as "solvable uniform (subgroup) membership problem".
– YCor
27 mins ago
add a comment |
1 Answer
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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.
answered 1 hour ago
Andy Putman
31.2k5132212
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
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votes
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.
answered 1 hour ago
Andy Putman
31.2k5132212
add a comment |
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.
answered 1 hour ago
Andy Putman
31.2k5132212
add a comment |
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.
answered 1 hour ago
Andy Putman
31.2k5132212
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.
answered 1 hour ago
Andy Putman
31.2k5132212
answered 1 hour ago
Andy Putman
31.2k5132212
answered 1 hour ago
Andy Putman
31.2k5132212
answered 1 hour ago
Andy Putman
31.2k5132212
31.2k5132212
add a comment |
add a comment |
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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
1 hour 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
1 hour ago
oops, right. sorry for noise.
– Dima Pasechnik
1 hour ago
This is known as "solvable uniform (subgroup) membership problem".
– YCor
27 mins ago