Showing $langle a,bmid abab^{-1}rangle$ and $ langle c,d mid c^2d^2rangle$ are isomorphic.
$begingroup$
I computed the fundamental group of the Klein bottle in two different ways and obtained two seemingly different answers:
$$
langle a,b mid abab^{-1}rangle
$$
and
$$
langle c,d mid c^2d^2rangle.
$$
I then tried to show that these two groups are in fact the same:
begin{align*}
langle a,b mid abab^{-1}rangle
=langle ab,b^{-1} mid abab^{-1}rangle
&=langle ab,b^{-1} mid (ab)(ab)b^{-1}b^{-1}rangle\
&=langle c,d mid c^2d^2rangle
end{align*}
Is my method for doing so valid?
group-theory proof-verification algebraic-topology fundamental-groups group-presentation
edited Dec 10 '18 at 6:04
Shaun
8,939113681
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
998
$endgroup$
add a comment |
$begingroup$
I computed the fundamental group of the Klein bottle in two different ways and obtained two seemingly different answers:
$$
langle a,b mid abab^{-1}rangle
$$
and
$$
langle c,d mid c^2d^2rangle.
$$
I then tried to show that these two groups are in fact the same:
begin{align*}
langle a,b mid abab^{-1}rangle
=langle ab,b^{-1} mid abab^{-1}rangle
&=langle ab,b^{-1} mid (ab)(ab)b^{-1}b^{-1}rangle\
&=langle c,d mid c^2d^2rangle
end{align*}
Is my method for doing so valid?
group-theory proof-verification algebraic-topology fundamental-groups group-presentation
edited Dec 10 '18 at 6:04
Shaun
8,939113681
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
998
$endgroup$
2
$begingroup$
Take a look at en.wikipedia.org/wiki/Tietze_transformations
$endgroup$
– Kyle Miller
Dec 6 '18 at 0:45
add a comment |
$begingroup$
I computed the fundamental group of the Klein bottle in two different ways and obtained two seemingly different answers:
$$
langle a,b mid abab^{-1}rangle
$$
and
$$
langle c,d mid c^2d^2rangle.
$$
I then tried to show that these two groups are in fact the same:
begin{align*}
langle a,b mid abab^{-1}rangle
=langle ab,b^{-1} mid abab^{-1}rangle
&=langle ab,b^{-1} mid (ab)(ab)b^{-1}b^{-1}rangle\
&=langle c,d mid c^2d^2rangle
end{align*}
Is my method for doing so valid?
group-theory proof-verification algebraic-topology fundamental-groups group-presentation
edited Dec 10 '18 at 6:04
Shaun
8,939113681
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
998
$endgroup$
I computed the fundamental group of the Klein bottle in two different ways and obtained two seemingly different answers:
$$
langle a,b mid abab^{-1}rangle
$$
and
$$
langle c,d mid c^2d^2rangle.
$$
I then tried to show that these two groups are in fact the same:
begin{align*}
langle a,b mid abab^{-1}rangle
=langle ab,b^{-1} mid abab^{-1}rangle
&=langle ab,b^{-1} mid (ab)(ab)b^{-1}b^{-1}rangle\
&=langle c,d mid c^2d^2rangle
end{align*}
Is my method for doing so valid?
group-theory proof-verification algebraic-topology fundamental-groups group-presentation
group-theory proof-verification algebraic-topology fundamental-groups group-presentation
edited Dec 10 '18 at 6:04
Shaun
8,939113681
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
998
edited Dec 10 '18 at 6:04
Shaun
8,939113681
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
998
edited Dec 10 '18 at 6:04
Shaun
8,939113681
edited Dec 10 '18 at 6:04
Shaun
8,939113681
edited Dec 10 '18 at 6:04
Shaun
8,939113681
8,939113681
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
998
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
998
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
998
998
2
$begingroup$
Take a look at en.wikipedia.org/wiki/Tietze_transformations
$endgroup$
– Kyle Miller
Dec 6 '18 at 0:45
add a comment |
2
$begingroup$
Take a look at en.wikipedia.org/wiki/Tietze_transformations
$endgroup$
– Kyle Miller
Dec 6 '18 at 0:45
2
2
$begingroup$
Take a look at en.wikipedia.org/wiki/Tietze_transformations
$endgroup$
– Kyle Miller
Dec 6 '18 at 0:45
$begingroup$
Take a look at en.wikipedia.org/wiki/Tietze_transformations
$endgroup$
– Kyle Miller
Dec 6 '18 at 0:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Sure. The second step doesn't really make sense as a presentation, but if you skip it then everything is fine. It might be clearest that these groups are isomorphic if you added the intermediate group $$langle a,b,c,d| abab^{-1},ab=c,d=b^{-1},c^2d^2rangle$$
which admits obvious maps from each of the two desired groups which one readily proves to be surjective (since $c$ and $d$ are in the subgroup generated by $a$ and $b$) and injective (since the first and last relations are each implied by the other three.)
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
32.8k23372
$endgroup$
add a comment |
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%2f3027704%2fshowing-langle-a-b-mid-abab-1-rangle-and-langle-c-d-mid-c2d2-rangle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Sure. The second step doesn't really make sense as a presentation, but if you skip it then everything is fine. It might be clearest that these groups are isomorphic if you added the intermediate group $$langle a,b,c,d| abab^{-1},ab=c,d=b^{-1},c^2d^2rangle$$
which admits obvious maps from each of the two desired groups which one readily proves to be surjective (since $c$ and $d$ are in the subgroup generated by $a$ and $b$) and injective (since the first and last relations are each implied by the other three.)
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
32.8k23372
$endgroup$
add a comment |
$begingroup$
Sure. The second step doesn't really make sense as a presentation, but if you skip it then everything is fine. It might be clearest that these groups are isomorphic if you added the intermediate group $$langle a,b,c,d| abab^{-1},ab=c,d=b^{-1},c^2d^2rangle$$
which admits obvious maps from each of the two desired groups which one readily proves to be surjective (since $c$ and $d$ are in the subgroup generated by $a$ and $b$) and injective (since the first and last relations are each implied by the other three.)
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
32.8k23372
$endgroup$
add a comment |
$begingroup$
Sure. The second step doesn't really make sense as a presentation, but if you skip it then everything is fine. It might be clearest that these groups are isomorphic if you added the intermediate group $$langle a,b,c,d| abab^{-1},ab=c,d=b^{-1},c^2d^2rangle$$
which admits obvious maps from each of the two desired groups which one readily proves to be surjective (since $c$ and $d$ are in the subgroup generated by $a$ and $b$) and injective (since the first and last relations are each implied by the other three.)
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
32.8k23372
$endgroup$
Sure. The second step doesn't really make sense as a presentation, but if you skip it then everything is fine. It might be clearest that these groups are isomorphic if you added the intermediate group $$langle a,b,c,d| abab^{-1},ab=c,d=b^{-1},c^2d^2rangle$$
which admits obvious maps from each of the two desired groups which one readily proves to be surjective (since $c$ and $d$ are in the subgroup generated by $a$ and $b$) and injective (since the first and last relations are each implied by the other three.)
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
32.8k23372
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
32.8k23372
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
32.8k23372
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
32.8k23372
32.8k23372
add a comment |
add a comment |
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.
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%2f3027704%2fshowing-langle-a-b-mid-abab-1-rangle-and-langle-c-d-mid-c2d2-rangle%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
$begingroup$
Take a look at en.wikipedia.org/wiki/Tietze_transformations
$endgroup$
– Kyle Miller
Dec 6 '18 at 0:45