Solvability of group presentations with 2 “almost disjoint” relations
I am interested in a certain type of $2$-relator group presentations arising in algebraic topology which have two relators that only contain a single generator in common. Specifically, suppose I have group presentation of the form
$$
G cong langle{S_1} cup {S_2} cup {b} mid P_1 = b^{n_1}, P_2 = b^{n_2}rangle
$$
Where $S_1 = {x_1, dots , x_n}$, $S_2 = {y_1, dots, y_m}$ are sets of generators (one or both possibly empty). In the first relation, $P_1$ is a word which is either a product of commutators of the form $P_1 = [x_1,x_2][x_3,x_4]dots [x_{n-1},x_n]$, OR is of the form $P_1 = x_1^2 x_2^2 dots x_n^2$. Similarly, $P_2$ takes on one of those two forms but in the $y_i$ generators.
Finally, the $n_{1,2}$ may be any non-zero integers (in particular, they may be positive or negative).
We therefore have that $P_1$ is a word only in the generators of $S_1$ and $P_2$ is a word only in the generators of $S_2$.
My Question:
I am aware that in general, two relator group presentation word/conjugacy/isomorphism problems are not tractable. However, given the restricted form that the group relations have (they are formed from disjoint sets of generators, except for $b$), are there any theorems or methods which provide additional power to work with these groups?
Observations:
If $d = gcd(n_1, n_2) = 1$, then the $b$ generator can be eliminated.
If both of $S_{1,2}$ are empty, then the group is cyclic (and trivial if $d=1$).
If one of $S_{1,2}$ is empty, there are certain simplifications that can be made, but everything seems to be case by case depending on d and the form of $P_1$, $P_2$.
Are there any results which would allow me to exactly classify when this group presentation can be reduced to a one relatiion group and when two relations is the minimum number possible?
Playing around with different special cases, it intuitively seems obvious that this should be the case (but intuition may be incorrect), but I am constantly getting stuck trying to prove that there could not possibly exist any other isomorphic presentations which have a single relation.
Any strategies, tactics, useful lemmas/theorems would be greatly appreciated!
group-theory algebraic-topology algebraic-groups group-presentation
edited Dec 3 '18 at 0:37
Shaun
8,810113680
asked Nov 9 '16 at 16:04
user334137
515210
|
show 2 more comments
I am interested in a certain type of $2$-relator group presentations arising in algebraic topology which have two relators that only contain a single generator in common. Specifically, suppose I have group presentation of the form
$$
G cong langle{S_1} cup {S_2} cup {b} mid P_1 = b^{n_1}, P_2 = b^{n_2}rangle
$$
Where $S_1 = {x_1, dots , x_n}$, $S_2 = {y_1, dots, y_m}$ are sets of generators (one or both possibly empty). In the first relation, $P_1$ is a word which is either a product of commutators of the form $P_1 = [x_1,x_2][x_3,x_4]dots [x_{n-1},x_n]$, OR is of the form $P_1 = x_1^2 x_2^2 dots x_n^2$. Similarly, $P_2$ takes on one of those two forms but in the $y_i$ generators.
Finally, the $n_{1,2}$ may be any non-zero integers (in particular, they may be positive or negative).
We therefore have that $P_1$ is a word only in the generators of $S_1$ and $P_2$ is a word only in the generators of $S_2$.
My Question:
I am aware that in general, two relator group presentation word/conjugacy/isomorphism problems are not tractable. However, given the restricted form that the group relations have (they are formed from disjoint sets of generators, except for $b$), are there any theorems or methods which provide additional power to work with these groups?
Observations:
If $d = gcd(n_1, n_2) = 1$, then the $b$ generator can be eliminated.
If both of $S_{1,2}$ are empty, then the group is cyclic (and trivial if $d=1$).
If one of $S_{1,2}$ is empty, there are certain simplifications that can be made, but everything seems to be case by case depending on d and the form of $P_1$, $P_2$.
Are there any results which would allow me to exactly classify when this group presentation can be reduced to a one relatiion group and when two relations is the minimum number possible?
Playing around with different special cases, it intuitively seems obvious that this should be the case (but intuition may be incorrect), but I am constantly getting stuck trying to prove that there could not possibly exist any other isomorphic presentations which have a single relation.
Any strategies, tactics, useful lemmas/theorems would be greatly appreciated!
group-theory algebraic-topology algebraic-groups group-presentation
edited Dec 3 '18 at 0:37
Shaun
8,810113680
asked Nov 9 '16 at 16:04
user334137
515210
2
It looks like a free product of two groups with the subgroup $langle b rangle$ amalgamated. I don't know if that helps!
– Derek Holt
Nov 9 '16 at 16:48
In fact, that is exactly what it is! This is the fundamental group of a class of spaces. However, I am new to the field. Are there any standard tools/results from the theory of amalgamations which would address the types of problems I mentioned?
– user334137
Nov 9 '16 at 19:57
Small cancellation theory might help.
– Shaun
Dec 3 '18 at 0:33
I'm not sure, though. I wonder what @DerekHolt thinks :)
– Shaun
Dec 3 '18 at 0:37
@Shaun It's been a couple years since I posted this--I forgot it had never been answered. This question was a peripheral one to the primary question I was interested in at the time, and I used small cancellation theory to address my primary question in the end, bypassing this one. However, I would still be interested to know the result here out of curiosity. Small cancellation theory together with Sela's solution to the isomorphism problem for hyperbolic groups would definitely answer this question in some instances, but wouldn't say anything if there were too many $b$'s in the relations.
– user334137
Dec 3 '18 at 1:42
|
show 2 more comments
I am interested in a certain type of $2$-relator group presentations arising in algebraic topology which have two relators that only contain a single generator in common. Specifically, suppose I have group presentation of the form
$$
G cong langle{S_1} cup {S_2} cup {b} mid P_1 = b^{n_1}, P_2 = b^{n_2}rangle
$$
Where $S_1 = {x_1, dots , x_n}$, $S_2 = {y_1, dots, y_m}$ are sets of generators (one or both possibly empty). In the first relation, $P_1$ is a word which is either a product of commutators of the form $P_1 = [x_1,x_2][x_3,x_4]dots [x_{n-1},x_n]$, OR is of the form $P_1 = x_1^2 x_2^2 dots x_n^2$. Similarly, $P_2$ takes on one of those two forms but in the $y_i$ generators.
Finally, the $n_{1,2}$ may be any non-zero integers (in particular, they may be positive or negative).
We therefore have that $P_1$ is a word only in the generators of $S_1$ and $P_2$ is a word only in the generators of $S_2$.
My Question:
I am aware that in general, two relator group presentation word/conjugacy/isomorphism problems are not tractable. However, given the restricted form that the group relations have (they are formed from disjoint sets of generators, except for $b$), are there any theorems or methods which provide additional power to work with these groups?
Observations:
If $d = gcd(n_1, n_2) = 1$, then the $b$ generator can be eliminated.
If both of $S_{1,2}$ are empty, then the group is cyclic (and trivial if $d=1$).
If one of $S_{1,2}$ is empty, there are certain simplifications that can be made, but everything seems to be case by case depending on d and the form of $P_1$, $P_2$.
Are there any results which would allow me to exactly classify when this group presentation can be reduced to a one relatiion group and when two relations is the minimum number possible?
Playing around with different special cases, it intuitively seems obvious that this should be the case (but intuition may be incorrect), but I am constantly getting stuck trying to prove that there could not possibly exist any other isomorphic presentations which have a single relation.
Any strategies, tactics, useful lemmas/theorems would be greatly appreciated!
group-theory algebraic-topology algebraic-groups group-presentation
edited Dec 3 '18 at 0:37
Shaun
8,810113680
asked Nov 9 '16 at 16:04
user334137
515210
I am interested in a certain type of $2$-relator group presentations arising in algebraic topology which have two relators that only contain a single generator in common. Specifically, suppose I have group presentation of the form
$$
G cong langle{S_1} cup {S_2} cup {b} mid P_1 = b^{n_1}, P_2 = b^{n_2}rangle
$$
Where $S_1 = {x_1, dots , x_n}$, $S_2 = {y_1, dots, y_m}$ are sets of generators (one or both possibly empty). In the first relation, $P_1$ is a word which is either a product of commutators of the form $P_1 = [x_1,x_2][x_3,x_4]dots [x_{n-1},x_n]$, OR is of the form $P_1 = x_1^2 x_2^2 dots x_n^2$. Similarly, $P_2$ takes on one of those two forms but in the $y_i$ generators.
Finally, the $n_{1,2}$ may be any non-zero integers (in particular, they may be positive or negative).
We therefore have that $P_1$ is a word only in the generators of $S_1$ and $P_2$ is a word only in the generators of $S_2$.
My Question:
I am aware that in general, two relator group presentation word/conjugacy/isomorphism problems are not tractable. However, given the restricted form that the group relations have (they are formed from disjoint sets of generators, except for $b$), are there any theorems or methods which provide additional power to work with these groups?
Observations:
If $d = gcd(n_1, n_2) = 1$, then the $b$ generator can be eliminated.
If both of $S_{1,2}$ are empty, then the group is cyclic (and trivial if $d=1$).
If one of $S_{1,2}$ is empty, there are certain simplifications that can be made, but everything seems to be case by case depending on d and the form of $P_1$, $P_2$.
Are there any results which would allow me to exactly classify when this group presentation can be reduced to a one relatiion group and when two relations is the minimum number possible?
Playing around with different special cases, it intuitively seems obvious that this should be the case (but intuition may be incorrect), but I am constantly getting stuck trying to prove that there could not possibly exist any other isomorphic presentations which have a single relation.
Any strategies, tactics, useful lemmas/theorems would be greatly appreciated!
group-theory algebraic-topology algebraic-groups group-presentation
group-theory algebraic-topology algebraic-groups group-presentation
edited Dec 3 '18 at 0:37
Shaun
8,810113680
asked Nov 9 '16 at 16:04
user334137
515210
edited Dec 3 '18 at 0:37
Shaun
8,810113680
asked Nov 9 '16 at 16:04
user334137
515210
edited Dec 3 '18 at 0:37
Shaun
8,810113680
edited Dec 3 '18 at 0:37
Shaun
8,810113680
edited Dec 3 '18 at 0:37
Shaun
8,810113680
8,810113680
asked Nov 9 '16 at 16:04
user334137
515210
asked Nov 9 '16 at 16:04
user334137
515210
asked Nov 9 '16 at 16:04
user334137
515210
515210
2
It looks like a free product of two groups with the subgroup $langle b rangle$ amalgamated. I don't know if that helps!
– Derek Holt
Nov 9 '16 at 16:48
In fact, that is exactly what it is! This is the fundamental group of a class of spaces. However, I am new to the field. Are there any standard tools/results from the theory of amalgamations which would address the types of problems I mentioned?
– user334137
Nov 9 '16 at 19:57
Small cancellation theory might help.
– Shaun
Dec 3 '18 at 0:33
I'm not sure, though. I wonder what @DerekHolt thinks :)
– Shaun
Dec 3 '18 at 0:37
@Shaun It's been a couple years since I posted this--I forgot it had never been answered. This question was a peripheral one to the primary question I was interested in at the time, and I used small cancellation theory to address my primary question in the end, bypassing this one. However, I would still be interested to know the result here out of curiosity. Small cancellation theory together with Sela's solution to the isomorphism problem for hyperbolic groups would definitely answer this question in some instances, but wouldn't say anything if there were too many $b$'s in the relations.
– user334137
Dec 3 '18 at 1:42
|
show 2 more comments
2
It looks like a free product of two groups with the subgroup $langle b rangle$ amalgamated. I don't know if that helps!
– Derek Holt
Nov 9 '16 at 16:48
In fact, that is exactly what it is! This is the fundamental group of a class of spaces. However, I am new to the field. Are there any standard tools/results from the theory of amalgamations which would address the types of problems I mentioned?
– user334137
Nov 9 '16 at 19:57
Small cancellation theory might help.
– Shaun
Dec 3 '18 at 0:33
I'm not sure, though. I wonder what @DerekHolt thinks :)
– Shaun
Dec 3 '18 at 0:37
@Shaun It's been a couple years since I posted this--I forgot it had never been answered. This question was a peripheral one to the primary question I was interested in at the time, and I used small cancellation theory to address my primary question in the end, bypassing this one. However, I would still be interested to know the result here out of curiosity. Small cancellation theory together with Sela's solution to the isomorphism problem for hyperbolic groups would definitely answer this question in some instances, but wouldn't say anything if there were too many $b$'s in the relations.
– user334137
Dec 3 '18 at 1:42
2
2
It looks like a free product of two groups with the subgroup $langle b rangle$ amalgamated. I don't know if that helps!
– Derek Holt
Nov 9 '16 at 16:48
It looks like a free product of two groups with the subgroup $langle b rangle$ amalgamated. I don't know if that helps!
– Derek Holt
Nov 9 '16 at 16:48
In fact, that is exactly what it is! This is the fundamental group of a class of spaces. However, I am new to the field. Are there any standard tools/results from the theory of amalgamations which would address the types of problems I mentioned?
– user334137
Nov 9 '16 at 19:57
In fact, that is exactly what it is! This is the fundamental group of a class of spaces. However, I am new to the field. Are there any standard tools/results from the theory of amalgamations which would address the types of problems I mentioned?
– user334137
Nov 9 '16 at 19:57
Small cancellation theory might help.
– Shaun
Dec 3 '18 at 0:33
Small cancellation theory might help.
– Shaun
Dec 3 '18 at 0:33
I'm not sure, though. I wonder what @DerekHolt thinks :)
– Shaun
Dec 3 '18 at 0:37
I'm not sure, though. I wonder what @DerekHolt thinks :)
– Shaun
Dec 3 '18 at 0:37
@Shaun It's been a couple years since I posted this--I forgot it had never been answered. This question was a peripheral one to the primary question I was interested in at the time, and I used small cancellation theory to address my primary question in the end, bypassing this one. However, I would still be interested to know the result here out of curiosity. Small cancellation theory together with Sela's solution to the isomorphism problem for hyperbolic groups would definitely answer this question in some instances, but wouldn't say anything if there were too many $b$'s in the relations.
– user334137
Dec 3 '18 at 1:42
@Shaun It's been a couple years since I posted this--I forgot it had never been answered. This question was a peripheral one to the primary question I was interested in at the time, and I used small cancellation theory to address my primary question in the end, bypassing this one. However, I would still be interested to know the result here out of curiosity. Small cancellation theory together with Sela's solution to the isomorphism problem for hyperbolic groups would definitely answer this question in some instances, but wouldn't say anything if there were too many $b$'s in the relations.
– user334137
Dec 3 '18 at 1:42
|
show 2 more comments
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2006744%2fsolvability-of-group-presentations-with-2-almost-disjoint-relations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2006744%2fsolvability-of-group-presentations-with-2-almost-disjoint-relations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
It looks like a free product of two groups with the subgroup $langle b rangle$ amalgamated. I don't know if that helps!
– Derek Holt
Nov 9 '16 at 16:48
In fact, that is exactly what it is! This is the fundamental group of a class of spaces. However, I am new to the field. Are there any standard tools/results from the theory of amalgamations which would address the types of problems I mentioned?
– user334137
Nov 9 '16 at 19:57
Small cancellation theory might help.
– Shaun
Dec 3 '18 at 0:33
I'm not sure, though. I wonder what @DerekHolt thinks :)
– Shaun
Dec 3 '18 at 0:37
@Shaun It's been a couple years since I posted this--I forgot it had never been answered. This question was a peripheral one to the primary question I was interested in at the time, and I used small cancellation theory to address my primary question in the end, bypassing this one. However, I would still be interested to know the result here out of curiosity. Small cancellation theory together with Sela's solution to the isomorphism problem for hyperbolic groups would definitely answer this question in some instances, but wouldn't say anything if there were too many $b$'s in the relations.
– user334137
Dec 3 '18 at 1:42