Solving the equation $abc=cba$ in the free group.
$begingroup$
It is known that if two words $a,b$ commute in the free group $F$, then they are powers of the same word, i.e. $a=c^r$ and $b=c^s$, where $cin F$ and $r,s in mathbb Z$.
What happens if there are three words $a,b,c in F$ such that $a b c=c b a$?
Is there a similar property as above? Or, if not, is there anything that follows from the equation, any information about $a,b$ or $c$? And why?
My goal is to find all solutions for the equation $a b c=c b a$ in the free group of rank two.
group-theory free-groups
asked Jan 6 at 13:20
BluBlu
466
$endgroup$
add a comment |
$begingroup$
It is known that if two words $a,b$ commute in the free group $F$, then they are powers of the same word, i.e. $a=c^r$ and $b=c^s$, where $cin F$ and $r,s in mathbb Z$.
What happens if there are three words $a,b,c in F$ such that $a b c=c b a$?
Is there a similar property as above? Or, if not, is there anything that follows from the equation, any information about $a,b$ or $c$? And why?
My goal is to find all solutions for the equation $a b c=c b a$ in the free group of rank two.
group-theory free-groups
asked Jan 6 at 13:20
BluBlu
466
$endgroup$
$begingroup$
An obvious family of solutions is to take $a=c$ with $b$ arbitrary. So perhaps we should fix $b$ and inquire about solutions where $aneq c$.
$endgroup$
– hardmath
Jan 6 at 14:02
add a comment |
$begingroup$
It is known that if two words $a,b$ commute in the free group $F$, then they are powers of the same word, i.e. $a=c^r$ and $b=c^s$, where $cin F$ and $r,s in mathbb Z$.
What happens if there are three words $a,b,c in F$ such that $a b c=c b a$?
Is there a similar property as above? Or, if not, is there anything that follows from the equation, any information about $a,b$ or $c$? And why?
My goal is to find all solutions for the equation $a b c=c b a$ in the free group of rank two.
group-theory free-groups
asked Jan 6 at 13:20
BluBlu
466
$endgroup$
It is known that if two words $a,b$ commute in the free group $F$, then they are powers of the same word, i.e. $a=c^r$ and $b=c^s$, where $cin F$ and $r,s in mathbb Z$.
What happens if there are three words $a,b,c in F$ such that $a b c=c b a$?
Is there a similar property as above? Or, if not, is there anything that follows from the equation, any information about $a,b$ or $c$? And why?
My goal is to find all solutions for the equation $a b c=c b a$ in the free group of rank two.
group-theory free-groups
group-theory free-groups
asked Jan 6 at 13:20
BluBlu
466
asked Jan 6 at 13:20
BluBlu
466
edited Jan 6 at 13:28
Blu
asked Jan 6 at 13:20
BluBlu
466
asked Jan 6 at 13:20
BluBlu
466
asked Jan 6 at 13:20
BluBlu
466
466
$begingroup$
An obvious family of solutions is to take $a=c$ with $b$ arbitrary. So perhaps we should fix $b$ and inquire about solutions where $aneq c$.
$endgroup$
– hardmath
Jan 6 at 14:02
add a comment |
$begingroup$
An obvious family of solutions is to take $a=c$ with $b$ arbitrary. So perhaps we should fix $b$ and inquire about solutions where $aneq c$.
$endgroup$
– hardmath
Jan 6 at 14:02
$begingroup$
An obvious family of solutions is to take $a=c$ with $b$ arbitrary. So perhaps we should fix $b$ and inquire about solutions where $aneq c$.
$endgroup$
– hardmath
Jan 6 at 14:02
$begingroup$
An obvious family of solutions is to take $a=c$ with $b$ arbitrary. So perhaps we should fix $b$ and inquire about solutions where $aneq c$.
$endgroup$
– hardmath
Jan 6 at 14:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your identity is equivalent to
$$
a b = c (b a) c^ {-1}.
$$
If you choose $a, b$ arbitrarily, you are thus looking for all $c$ that conjugate $b a$ to $a b$. One of the solutions will be $c = a$. All the solutions will thus be of the form $c = a z$, where $z (b a) = (b a) z$, which gets you back to your first statement.
answered Jan 6 at 13:59
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
$begingroup$
Yes, these are solutions. Thanks! I just don’t see why these are all, could you explain it? Thank you!
$endgroup$
– Blu
Jan 6 at 16:58
$begingroup$
Let $c$ be a solution of the equation $a b = c (b a) c^{-1}$, for $a, b$ given. Then $(a^{-1} c) (b a) (a^{-1} c)^{-1} = a^{-1} (c (b a) c^{-1}) a = a^{-1} (a b) a = b a$, so that $a^{-1} c = z$ centralises $b a$, and $c = a z$.
$endgroup$
– Andreas Caranti
Jan 6 at 17:22
add a comment |
$begingroup$
Another way to solve this equation is to replace the product $bc$ with a new variable, $g$ say, and replace $c^{-1}a$ another new variable, $h$ say*. Then your equation becomes
$$
begin{align*}
abc&=cba\
c^{-1}abc&=bcc^{-1}a\
hg&=gh
end{align*}
$$
Therefore, solutions over the free group $F$ are assignments $(g, h)rightarrow (w^i, w^j)$ with $i, jinmathbb{Z}$ and $win F$ such that there is no element $uin F$ such that $u^k=w$, $k>1$. Substituting in our replacements of $gleftrightarrow bc$ and $hleftrightarrow c^{-1}a$, we have all solutions are of the form $$(a, b, c)rightarrow (vw^i, w^jv^{-1}, v)$$ where $w$ is as above and $vin F$ is arbitrary.
*This second assignment isn't initially obvious.
answered Jan 7 at 10:39
user1729user1729
17.1k64193
$endgroup$
$begingroup$
But the equation is $abc=cba$ and not $abc=bca$, so setting $d=bc$ will only give $ad=cdc^{-1}a$.
$endgroup$
– Blu
Jan 8 at 1:18
$begingroup$
@Blu I've fixed my solution now.
$endgroup$
– user1729
Jan 8 at 11:31
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%2f3063847%2fsolving-the-equation-abc-cba-in-the-free-group%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your identity is equivalent to
$$
a b = c (b a) c^ {-1}.
$$
If you choose $a, b$ arbitrarily, you are thus looking for all $c$ that conjugate $b a$ to $a b$. One of the solutions will be $c = a$. All the solutions will thus be of the form $c = a z$, where $z (b a) = (b a) z$, which gets you back to your first statement.
answered Jan 6 at 13:59
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
$begingroup$
Yes, these are solutions. Thanks! I just don’t see why these are all, could you explain it? Thank you!
$endgroup$
– Blu
Jan 6 at 16:58
$begingroup$
Let $c$ be a solution of the equation $a b = c (b a) c^{-1}$, for $a, b$ given. Then $(a^{-1} c) (b a) (a^{-1} c)^{-1} = a^{-1} (c (b a) c^{-1}) a = a^{-1} (a b) a = b a$, so that $a^{-1} c = z$ centralises $b a$, and $c = a z$.
$endgroup$
– Andreas Caranti
Jan 6 at 17:22
add a comment |
$begingroup$
Your identity is equivalent to
$$
a b = c (b a) c^ {-1}.
$$
If you choose $a, b$ arbitrarily, you are thus looking for all $c$ that conjugate $b a$ to $a b$. One of the solutions will be $c = a$. All the solutions will thus be of the form $c = a z$, where $z (b a) = (b a) z$, which gets you back to your first statement.
answered Jan 6 at 13:59
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
$begingroup$
Yes, these are solutions. Thanks! I just don’t see why these are all, could you explain it? Thank you!
$endgroup$
– Blu
Jan 6 at 16:58
$begingroup$
Let $c$ be a solution of the equation $a b = c (b a) c^{-1}$, for $a, b$ given. Then $(a^{-1} c) (b a) (a^{-1} c)^{-1} = a^{-1} (c (b a) c^{-1}) a = a^{-1} (a b) a = b a$, so that $a^{-1} c = z$ centralises $b a$, and $c = a z$.
$endgroup$
– Andreas Caranti
Jan 6 at 17:22
add a comment |
$begingroup$
Your identity is equivalent to
$$
a b = c (b a) c^ {-1}.
$$
If you choose $a, b$ arbitrarily, you are thus looking for all $c$ that conjugate $b a$ to $a b$. One of the solutions will be $c = a$. All the solutions will thus be of the form $c = a z$, where $z (b a) = (b a) z$, which gets you back to your first statement.
answered Jan 6 at 13:59
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
Your identity is equivalent to
$$
a b = c (b a) c^ {-1}.
$$
If you choose $a, b$ arbitrarily, you are thus looking for all $c$ that conjugate $b a$ to $a b$. One of the solutions will be $c = a$. All the solutions will thus be of the form $c = a z$, where $z (b a) = (b a) z$, which gets you back to your first statement.
answered Jan 6 at 13:59
Andreas CarantiAndreas Caranti
56.6k34395
answered Jan 6 at 13:59
Andreas CarantiAndreas Caranti
56.6k34395
answered Jan 6 at 13:59
Andreas CarantiAndreas Caranti
56.6k34395
answered Jan 6 at 13:59
Andreas CarantiAndreas Caranti
56.6k34395
56.6k34395
$begingroup$
Yes, these are solutions. Thanks! I just don’t see why these are all, could you explain it? Thank you!
$endgroup$
– Blu
Jan 6 at 16:58
$begingroup$
Let $c$ be a solution of the equation $a b = c (b a) c^{-1}$, for $a, b$ given. Then $(a^{-1} c) (b a) (a^{-1} c)^{-1} = a^{-1} (c (b a) c^{-1}) a = a^{-1} (a b) a = b a$, so that $a^{-1} c = z$ centralises $b a$, and $c = a z$.
$endgroup$
– Andreas Caranti
Jan 6 at 17:22
add a comment |
$begingroup$
Yes, these are solutions. Thanks! I just don’t see why these are all, could you explain it? Thank you!
$endgroup$
– Blu
Jan 6 at 16:58
$begingroup$
Let $c$ be a solution of the equation $a b = c (b a) c^{-1}$, for $a, b$ given. Then $(a^{-1} c) (b a) (a^{-1} c)^{-1} = a^{-1} (c (b a) c^{-1}) a = a^{-1} (a b) a = b a$, so that $a^{-1} c = z$ centralises $b a$, and $c = a z$.
$endgroup$
– Andreas Caranti
Jan 6 at 17:22
$begingroup$
Yes, these are solutions. Thanks! I just don’t see why these are all, could you explain it? Thank you!
$endgroup$
– Blu
Jan 6 at 16:58
$begingroup$
Yes, these are solutions. Thanks! I just don’t see why these are all, could you explain it? Thank you!
$endgroup$
– Blu
Jan 6 at 16:58
$begingroup$
Let $c$ be a solution of the equation $a b = c (b a) c^{-1}$, for $a, b$ given. Then $(a^{-1} c) (b a) (a^{-1} c)^{-1} = a^{-1} (c (b a) c^{-1}) a = a^{-1} (a b) a = b a$, so that $a^{-1} c = z$ centralises $b a$, and $c = a z$.
$endgroup$
– Andreas Caranti
Jan 6 at 17:22
$begingroup$
Let $c$ be a solution of the equation $a b = c (b a) c^{-1}$, for $a, b$ given. Then $(a^{-1} c) (b a) (a^{-1} c)^{-1} = a^{-1} (c (b a) c^{-1}) a = a^{-1} (a b) a = b a$, so that $a^{-1} c = z$ centralises $b a$, and $c = a z$.
$endgroup$
– Andreas Caranti
Jan 6 at 17:22
add a comment |
$begingroup$
Another way to solve this equation is to replace the product $bc$ with a new variable, $g$ say, and replace $c^{-1}a$ another new variable, $h$ say*. Then your equation becomes
$$
begin{align*}
abc&=cba\
c^{-1}abc&=bcc^{-1}a\
hg&=gh
end{align*}
$$
Therefore, solutions over the free group $F$ are assignments $(g, h)rightarrow (w^i, w^j)$ with $i, jinmathbb{Z}$ and $win F$ such that there is no element $uin F$ such that $u^k=w$, $k>1$. Substituting in our replacements of $gleftrightarrow bc$ and $hleftrightarrow c^{-1}a$, we have all solutions are of the form $$(a, b, c)rightarrow (vw^i, w^jv^{-1}, v)$$ where $w$ is as above and $vin F$ is arbitrary.
*This second assignment isn't initially obvious.
answered Jan 7 at 10:39
user1729user1729
17.1k64193
$endgroup$
$begingroup$
But the equation is $abc=cba$ and not $abc=bca$, so setting $d=bc$ will only give $ad=cdc^{-1}a$.
$endgroup$
– Blu
Jan 8 at 1:18
$begingroup$
@Blu I've fixed my solution now.
$endgroup$
– user1729
Jan 8 at 11:31
add a comment |
$begingroup$
Another way to solve this equation is to replace the product $bc$ with a new variable, $g$ say, and replace $c^{-1}a$ another new variable, $h$ say*. Then your equation becomes
$$
begin{align*}
abc&=cba\
c^{-1}abc&=bcc^{-1}a\
hg&=gh
end{align*}
$$
Therefore, solutions over the free group $F$ are assignments $(g, h)rightarrow (w^i, w^j)$ with $i, jinmathbb{Z}$ and $win F$ such that there is no element $uin F$ such that $u^k=w$, $k>1$. Substituting in our replacements of $gleftrightarrow bc$ and $hleftrightarrow c^{-1}a$, we have all solutions are of the form $$(a, b, c)rightarrow (vw^i, w^jv^{-1}, v)$$ where $w$ is as above and $vin F$ is arbitrary.
*This second assignment isn't initially obvious.
answered Jan 7 at 10:39
user1729user1729
17.1k64193
$endgroup$
$begingroup$
But the equation is $abc=cba$ and not $abc=bca$, so setting $d=bc$ will only give $ad=cdc^{-1}a$.
$endgroup$
– Blu
Jan 8 at 1:18
$begingroup$
@Blu I've fixed my solution now.
$endgroup$
– user1729
Jan 8 at 11:31
add a comment |
$begingroup$
Another way to solve this equation is to replace the product $bc$ with a new variable, $g$ say, and replace $c^{-1}a$ another new variable, $h$ say*. Then your equation becomes
$$
begin{align*}
abc&=cba\
c^{-1}abc&=bcc^{-1}a\
hg&=gh
end{align*}
$$
Therefore, solutions over the free group $F$ are assignments $(g, h)rightarrow (w^i, w^j)$ with $i, jinmathbb{Z}$ and $win F$ such that there is no element $uin F$ such that $u^k=w$, $k>1$. Substituting in our replacements of $gleftrightarrow bc$ and $hleftrightarrow c^{-1}a$, we have all solutions are of the form $$(a, b, c)rightarrow (vw^i, w^jv^{-1}, v)$$ where $w$ is as above and $vin F$ is arbitrary.
*This second assignment isn't initially obvious.
answered Jan 7 at 10:39
user1729user1729
17.1k64193
$endgroup$
Another way to solve this equation is to replace the product $bc$ with a new variable, $g$ say, and replace $c^{-1}a$ another new variable, $h$ say*. Then your equation becomes
$$
begin{align*}
abc&=cba\
c^{-1}abc&=bcc^{-1}a\
hg&=gh
end{align*}
$$
Therefore, solutions over the free group $F$ are assignments $(g, h)rightarrow (w^i, w^j)$ with $i, jinmathbb{Z}$ and $win F$ such that there is no element $uin F$ such that $u^k=w$, $k>1$. Substituting in our replacements of $gleftrightarrow bc$ and $hleftrightarrow c^{-1}a$, we have all solutions are of the form $$(a, b, c)rightarrow (vw^i, w^jv^{-1}, v)$$ where $w$ is as above and $vin F$ is arbitrary.
*This second assignment isn't initially obvious.
answered Jan 7 at 10:39
user1729user1729
17.1k64193
edited Jan 8 at 11:30
answered Jan 7 at 10:39
user1729user1729
17.1k64193
answered Jan 7 at 10:39
user1729user1729
17.1k64193
answered Jan 7 at 10:39
user1729user1729
17.1k64193
17.1k64193
$begingroup$
But the equation is $abc=cba$ and not $abc=bca$, so setting $d=bc$ will only give $ad=cdc^{-1}a$.
$endgroup$
– Blu
Jan 8 at 1:18
$begingroup$
@Blu I've fixed my solution now.
$endgroup$
– user1729
Jan 8 at 11:31
add a comment |
$begingroup$
But the equation is $abc=cba$ and not $abc=bca$, so setting $d=bc$ will only give $ad=cdc^{-1}a$.
$endgroup$
– Blu
Jan 8 at 1:18
$begingroup$
@Blu I've fixed my solution now.
$endgroup$
– user1729
Jan 8 at 11:31
$begingroup$
But the equation is $abc=cba$ and not $abc=bca$, so setting $d=bc$ will only give $ad=cdc^{-1}a$.
$endgroup$
– Blu
Jan 8 at 1:18
$begingroup$
But the equation is $abc=cba$ and not $abc=bca$, so setting $d=bc$ will only give $ad=cdc^{-1}a$.
$endgroup$
– Blu
Jan 8 at 1:18
$begingroup$
@Blu I've fixed my solution now.
$endgroup$
– user1729
Jan 8 at 11:31
$begingroup$
@Blu I've fixed my solution now.
$endgroup$
– user1729
Jan 8 at 11:31
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%2f3063847%2fsolving-the-equation-abc-cba-in-the-free-group%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
$begingroup$
An obvious family of solutions is to take $a=c$ with $b$ arbitrary. So perhaps we should fix $b$ and inquire about solutions where $aneq c$.
$endgroup$
– hardmath
Jan 6 at 14:02