Solving the equation $abc=cba$ in the free group.












2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 13:28







Blu

















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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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





















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















6












6








6





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 6 at 13:59









Andreas CarantiAndreas Caranti

56.4k34395




56.4k34395












  • $begingroup$
    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










  • $begingroup$
    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













0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 11:30

























answered Jan 7 at 10:39









user1729user1729

17.3k64085




17.3k64085












  • $begingroup$
    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










  • $begingroup$
    @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


















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