How can I work out if a certain group presentation implies a certain relation?
$begingroup$
I thought that maybe it would be possible to answer this question using the concept of a group presentation.
Let $x_1,x_2,ldots,x_k$ be k different elements of a group G and $kgeq4$.
If we know that $x_i$ commutes with $x_{i+1}$ and $x_k$ commutes with $x_1$, can we say that all $x_i$ commute with each other ?
Thinking in terms of group presentations, we can ask: does the group
$$langle x_1, ldots, x_k mid (forall i<k) x_ix_{i+1}=x_{i+1}x_i, x_kx_1=x_1x_k rangle$$
satisfy $x_ix_j$ for all $i,j$ ? In other words, does the normal subgroup generated by $[x_i, x_{i+1}]$ for each $i<k$ and $[x_k, x_1]$ contain $[x_i, x_j]$ for each $i, j$ ?
But how could we prove or disprove that? Are there any methods that can be used to work out if a certain element is in a certain generated normal subgroup?
group-theory group-presentation combinatorial-group-theory
edited Nov 30 '18 at 3:11
Shaun
8,893113681
asked Jan 17 '15 at 23:12
Jack MJack M
18.6k33880
$endgroup$
add a comment |
$begingroup$
I thought that maybe it would be possible to answer this question using the concept of a group presentation.
Let $x_1,x_2,ldots,x_k$ be k different elements of a group G and $kgeq4$.
If we know that $x_i$ commutes with $x_{i+1}$ and $x_k$ commutes with $x_1$, can we say that all $x_i$ commute with each other ?
Thinking in terms of group presentations, we can ask: does the group
$$langle x_1, ldots, x_k mid (forall i<k) x_ix_{i+1}=x_{i+1}x_i, x_kx_1=x_1x_k rangle$$
satisfy $x_ix_j$ for all $i,j$ ? In other words, does the normal subgroup generated by $[x_i, x_{i+1}]$ for each $i<k$ and $[x_k, x_1]$ contain $[x_i, x_j]$ for each $i, j$ ?
But how could we prove or disprove that? Are there any methods that can be used to work out if a certain element is in a certain generated normal subgroup?
group-theory group-presentation combinatorial-group-theory
edited Nov 30 '18 at 3:11
Shaun
8,893113681
asked Jan 17 '15 at 23:12
Jack MJack M
18.6k33880
$endgroup$
1
$begingroup$
I believe in this particular case the answer is 'no' and the easiest way to show it is to find a group that satisfies those relations but $x_1x_3neq x_3x_1$ like the automorphism group of a tiling of the hyperbolic plane with orthogonal polygons. For some level of generality, this is a very difficult, in fact undecidable problem. But this particular presentation has nice properties: I think the group is hyperbolic and hence automatic. See the book 'Word processing in groups' by Epstein et. al. (coauthored by @DerekHolt btw).
$endgroup$
– Myself
Jan 17 '15 at 23:24
$begingroup$
@Myself Providing counterexamples was the type of answer provided on the original question - I'm specifically interested in "direct" approaches with the presentation.
$endgroup$
– Jack M
Jan 17 '15 at 23:27
$begingroup$
That won't work I'm afraid, at least not in general. The situation is entirely comparable to the relation between proof theory and model theory (maybe even an instance of it). You cannot know from the proof theory if no proof exists because you have not sought hard enough or because the statement is undecidable. Maybe some logician could explain you precisely why that is. But sometimes, like here, things are easier for instance because your group is automatic, or your (in model theory) the theory admits quantifier elimination etc.
$endgroup$
– Myself
Jan 17 '15 at 23:39
$begingroup$
@JackM: I do not really understand your question, but the group you consider is just the right-angled Artin group associated to a cycle of length $k$.
$endgroup$
– Seirios
Jan 19 '15 at 8:31
add a comment |
$begingroup$
I thought that maybe it would be possible to answer this question using the concept of a group presentation.
Let $x_1,x_2,ldots,x_k$ be k different elements of a group G and $kgeq4$.
If we know that $x_i$ commutes with $x_{i+1}$ and $x_k$ commutes with $x_1$, can we say that all $x_i$ commute with each other ?
Thinking in terms of group presentations, we can ask: does the group
$$langle x_1, ldots, x_k mid (forall i<k) x_ix_{i+1}=x_{i+1}x_i, x_kx_1=x_1x_k rangle$$
satisfy $x_ix_j$ for all $i,j$ ? In other words, does the normal subgroup generated by $[x_i, x_{i+1}]$ for each $i<k$ and $[x_k, x_1]$ contain $[x_i, x_j]$ for each $i, j$ ?
But how could we prove or disprove that? Are there any methods that can be used to work out if a certain element is in a certain generated normal subgroup?
group-theory group-presentation combinatorial-group-theory
edited Nov 30 '18 at 3:11
Shaun
8,893113681
asked Jan 17 '15 at 23:12
Jack MJack M
18.6k33880
$endgroup$
I thought that maybe it would be possible to answer this question using the concept of a group presentation.
Let $x_1,x_2,ldots,x_k$ be k different elements of a group G and $kgeq4$.
If we know that $x_i$ commutes with $x_{i+1}$ and $x_k$ commutes with $x_1$, can we say that all $x_i$ commute with each other ?
Thinking in terms of group presentations, we can ask: does the group
$$langle x_1, ldots, x_k mid (forall i<k) x_ix_{i+1}=x_{i+1}x_i, x_kx_1=x_1x_k rangle$$
satisfy $x_ix_j$ for all $i,j$ ? In other words, does the normal subgroup generated by $[x_i, x_{i+1}]$ for each $i<k$ and $[x_k, x_1]$ contain $[x_i, x_j]$ for each $i, j$ ?
But how could we prove or disprove that? Are there any methods that can be used to work out if a certain element is in a certain generated normal subgroup?
group-theory group-presentation combinatorial-group-theory
group-theory group-presentation combinatorial-group-theory
edited Nov 30 '18 at 3:11
Shaun
8,893113681
asked Jan 17 '15 at 23:12
Jack MJack M
18.6k33880
edited Nov 30 '18 at 3:11
Shaun
8,893113681
asked Jan 17 '15 at 23:12
Jack MJack M
18.6k33880
edited Nov 30 '18 at 3:11
Shaun
8,893113681
edited Nov 30 '18 at 3:11
Shaun
8,893113681
edited Nov 30 '18 at 3:11
Shaun
8,893113681
8,893113681
asked Jan 17 '15 at 23:12
Jack MJack M
18.6k33880
asked Jan 17 '15 at 23:12
Jack MJack M
18.6k33880
asked Jan 17 '15 at 23:12
Jack MJack M
18.6k33880
18.6k33880
1
$begingroup$
I believe in this particular case the answer is 'no' and the easiest way to show it is to find a group that satisfies those relations but $x_1x_3neq x_3x_1$ like the automorphism group of a tiling of the hyperbolic plane with orthogonal polygons. For some level of generality, this is a very difficult, in fact undecidable problem. But this particular presentation has nice properties: I think the group is hyperbolic and hence automatic. See the book 'Word processing in groups' by Epstein et. al. (coauthored by @DerekHolt btw).
$endgroup$
– Myself
Jan 17 '15 at 23:24
$begingroup$
@Myself Providing counterexamples was the type of answer provided on the original question - I'm specifically interested in "direct" approaches with the presentation.
$endgroup$
– Jack M
Jan 17 '15 at 23:27
$begingroup$
That won't work I'm afraid, at least not in general. The situation is entirely comparable to the relation between proof theory and model theory (maybe even an instance of it). You cannot know from the proof theory if no proof exists because you have not sought hard enough or because the statement is undecidable. Maybe some logician could explain you precisely why that is. But sometimes, like here, things are easier for instance because your group is automatic, or your (in model theory) the theory admits quantifier elimination etc.
$endgroup$
– Myself
Jan 17 '15 at 23:39
$begingroup$
@JackM: I do not really understand your question, but the group you consider is just the right-angled Artin group associated to a cycle of length $k$.
$endgroup$
– Seirios
Jan 19 '15 at 8:31
add a comment |
1
$begingroup$
I believe in this particular case the answer is 'no' and the easiest way to show it is to find a group that satisfies those relations but $x_1x_3neq x_3x_1$ like the automorphism group of a tiling of the hyperbolic plane with orthogonal polygons. For some level of generality, this is a very difficult, in fact undecidable problem. But this particular presentation has nice properties: I think the group is hyperbolic and hence automatic. See the book 'Word processing in groups' by Epstein et. al. (coauthored by @DerekHolt btw).
$endgroup$
– Myself
Jan 17 '15 at 23:24
$begingroup$
@Myself Providing counterexamples was the type of answer provided on the original question - I'm specifically interested in "direct" approaches with the presentation.
$endgroup$
– Jack M
Jan 17 '15 at 23:27
$begingroup$
That won't work I'm afraid, at least not in general. The situation is entirely comparable to the relation between proof theory and model theory (maybe even an instance of it). You cannot know from the proof theory if no proof exists because you have not sought hard enough or because the statement is undecidable. Maybe some logician could explain you precisely why that is. But sometimes, like here, things are easier for instance because your group is automatic, or your (in model theory) the theory admits quantifier elimination etc.
$endgroup$
– Myself
Jan 17 '15 at 23:39
$begingroup$
@JackM: I do not really understand your question, but the group you consider is just the right-angled Artin group associated to a cycle of length $k$.
$endgroup$
– Seirios
Jan 19 '15 at 8:31
1
1
$begingroup$
I believe in this particular case the answer is 'no' and the easiest way to show it is to find a group that satisfies those relations but $x_1x_3neq x_3x_1$ like the automorphism group of a tiling of the hyperbolic plane with orthogonal polygons. For some level of generality, this is a very difficult, in fact undecidable problem. But this particular presentation has nice properties: I think the group is hyperbolic and hence automatic. See the book 'Word processing in groups' by Epstein et. al. (coauthored by @DerekHolt btw).
$endgroup$
– Myself
Jan 17 '15 at 23:24
$begingroup$
I believe in this particular case the answer is 'no' and the easiest way to show it is to find a group that satisfies those relations but $x_1x_3neq x_3x_1$ like the automorphism group of a tiling of the hyperbolic plane with orthogonal polygons. For some level of generality, this is a very difficult, in fact undecidable problem. But this particular presentation has nice properties: I think the group is hyperbolic and hence automatic. See the book 'Word processing in groups' by Epstein et. al. (coauthored by @DerekHolt btw).
$endgroup$
– Myself
Jan 17 '15 at 23:24
$begingroup$
@Myself Providing counterexamples was the type of answer provided on the original question - I'm specifically interested in "direct" approaches with the presentation.
$endgroup$
– Jack M
Jan 17 '15 at 23:27
$begingroup$
@Myself Providing counterexamples was the type of answer provided on the original question - I'm specifically interested in "direct" approaches with the presentation.
$endgroup$
– Jack M
Jan 17 '15 at 23:27
$begingroup$
That won't work I'm afraid, at least not in general. The situation is entirely comparable to the relation between proof theory and model theory (maybe even an instance of it). You cannot know from the proof theory if no proof exists because you have not sought hard enough or because the statement is undecidable. Maybe some logician could explain you precisely why that is. But sometimes, like here, things are easier for instance because your group is automatic, or your (in model theory) the theory admits quantifier elimination etc.
$endgroup$
– Myself
Jan 17 '15 at 23:39
$begingroup$
That won't work I'm afraid, at least not in general. The situation is entirely comparable to the relation between proof theory and model theory (maybe even an instance of it). You cannot know from the proof theory if no proof exists because you have not sought hard enough or because the statement is undecidable. Maybe some logician could explain you precisely why that is. But sometimes, like here, things are easier for instance because your group is automatic, or your (in model theory) the theory admits quantifier elimination etc.
$endgroup$
– Myself
Jan 17 '15 at 23:39
$begingroup$
@JackM: I do not really understand your question, but the group you consider is just the right-angled Artin group associated to a cycle of length $k$.
$endgroup$
– Seirios
Jan 19 '15 at 8:31
$begingroup$
@JackM: I do not really understand your question, but the group you consider is just the right-angled Artin group associated to a cycle of length $k$.
$endgroup$
– Seirios
Jan 19 '15 at 8:31
add a comment |
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$begingroup$
I believe in this particular case the answer is 'no' and the easiest way to show it is to find a group that satisfies those relations but $x_1x_3neq x_3x_1$ like the automorphism group of a tiling of the hyperbolic plane with orthogonal polygons. For some level of generality, this is a very difficult, in fact undecidable problem. But this particular presentation has nice properties: I think the group is hyperbolic and hence automatic. See the book 'Word processing in groups' by Epstein et. al. (coauthored by @DerekHolt btw).
$endgroup$
– Myself
Jan 17 '15 at 23:24
$begingroup$
@Myself Providing counterexamples was the type of answer provided on the original question - I'm specifically interested in "direct" approaches with the presentation.
$endgroup$
– Jack M
Jan 17 '15 at 23:27
$begingroup$
That won't work I'm afraid, at least not in general. The situation is entirely comparable to the relation between proof theory and model theory (maybe even an instance of it). You cannot know from the proof theory if no proof exists because you have not sought hard enough or because the statement is undecidable. Maybe some logician could explain you precisely why that is. But sometimes, like here, things are easier for instance because your group is automatic, or your (in model theory) the theory admits quantifier elimination etc.
$endgroup$
– Myself
Jan 17 '15 at 23:39
$begingroup$
@JackM: I do not really understand your question, but the group you consider is just the right-angled Artin group associated to a cycle of length $k$.
$endgroup$
– Seirios
Jan 19 '15 at 8:31