Algorithm to find the n-th root in the free group












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Given a word $w$ in the free group $F$, is there an algorithm to find it’s n-th root, if it exists?
Thanks!






group-theory algorithms roots free-groups







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  • $begingroup$
    What is your definition of a free group? What did you try t solve this problem?
    $endgroup$
    – Moishe Cohen
    Jan 6 at 16:49
















-1












$begingroup$


Given a word $w$ in the free group $F$, is there an algorithm to find it’s n-th root, if it exists?
Thanks!






group-theory algorithms roots free-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is your definition of a free group? What did you try t solve this problem?
    $endgroup$
    – Moishe Cohen
    Jan 6 at 16:49














-1












-1








-1





$begingroup$


Given a word $w$ in the free group $F$, is there an algorithm to find it’s n-th root, if it exists?
Thanks!






group-theory algorithms roots free-groups







share|cite|improve this question











$endgroup$




Given a word $w$ in the free group $F$, is there an algorithm to find it’s n-th root, if it exists?
Thanks!






group-theory algorithms roots free-groups




group-theory algorithms roots free-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Jan 6 at 13:55







Blu

















asked Jan 6 at 13:47









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  • $begingroup$
    What is your definition of a free group? What did you try t solve this problem?
    $endgroup$
    – Moishe Cohen
    Jan 6 at 16:49


















  • $begingroup$
    What is your definition of a free group? What did you try t solve this problem?
    $endgroup$
    – Moishe Cohen
    Jan 6 at 16:49
















$begingroup$
What is your definition of a free group? What did you try t solve this problem?
$endgroup$
– Moishe Cohen
Jan 6 at 16:49




$begingroup$
What is your definition of a free group? What did you try t solve this problem?
$endgroup$
– Moishe Cohen
Jan 6 at 16:49










1 Answer
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$begingroup$

If $w$ has an $n$th root, then so does every conjugate of it.
Among all pairs $(x,u)$ such that $w=u^{-1}x^nu$, pick one that minimizes first the length $|x|$ and then $|u|$.



If the first and last letter of $x$ are inverse of each other, i.e., $x=aya^{-1}$
for some letter $a$ and word $y$ with $|y|<|x|$, then $w=v^{-1}y^nv$ with $v=au$, contradicting minimality of $|x|$.
Hence $x^n$ is really just concatenation (i.e., without any cancellation among the factors).



If the first letters of $u$ and $x$ are the same, i.e., $u=av$, $x=ay$ for some letter $a$ and words $v,y$ with $|v|<|u|$, $|y|<|x|$, then
$w=v^{-1}a^{-1}(ay)^nav=v^{-1}(ya)^nv$, contradicting minimality (as $v|<|u|$ and $|ya|le |x|$).



Similarly, if the first letter of $u$ and the last letter of $x$ are inverses, i.e., $u=av$, $x=ya^{-1}$ for some letter $a$ and words $v,y$ with $|v|<|u|$, $|y|<|x|$, then $w=v^{-1}a^{-1}(ya^{-1})^nav=v^{-1}(a^{-1}y)^nv$, contradicting minimality.



We conclude that no cancellation occurs when multiplying the factors $u^{-1},x,x,ldots, x,u$.



This suggests the following algorithm:




  1. Let $uleftarrowepsilon$, $vleftarrow w$.

  2. If $|v|$ is a multiple of $n$ and is the simple concatenation of $n$ copies of its length $|v|/n$ prefix $x$, then output "$uxu^{-1} $" and terminate. (Note that this condition also triggers when $v=epsilon$, hence we may assume $vneepsilon$ in the next step)

  3. Let $aleftarrow operatorname{first}(v)$, $bleftarrow operatorname{last}(v)$.

  4. If $ane b^{-1}$ output "NOT AN $n$TH POWER" and terminate.


  5. (As $a=b^{-1}$, we have $ane b$, hence $|v|ge 2$ and so $v=av'b$ for some shorter word $v'$) Let $vleftarrow v'$, $uleftarrow ua$, and go back to step 2.






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    $begingroup$

    If $w$ has an $n$th root, then so does every conjugate of it.
    Among all pairs $(x,u)$ such that $w=u^{-1}x^nu$, pick one that minimizes first the length $|x|$ and then $|u|$.



    If the first and last letter of $x$ are inverse of each other, i.e., $x=aya^{-1}$
    for some letter $a$ and word $y$ with $|y|<|x|$, then $w=v^{-1}y^nv$ with $v=au$, contradicting minimality of $|x|$.
    Hence $x^n$ is really just concatenation (i.e., without any cancellation among the factors).



    If the first letters of $u$ and $x$ are the same, i.e., $u=av$, $x=ay$ for some letter $a$ and words $v,y$ with $|v|<|u|$, $|y|<|x|$, then
    $w=v^{-1}a^{-1}(ay)^nav=v^{-1}(ya)^nv$, contradicting minimality (as $v|<|u|$ and $|ya|le |x|$).



    Similarly, if the first letter of $u$ and the last letter of $x$ are inverses, i.e., $u=av$, $x=ya^{-1}$ for some letter $a$ and words $v,y$ with $|v|<|u|$, $|y|<|x|$, then $w=v^{-1}a^{-1}(ya^{-1})^nav=v^{-1}(a^{-1}y)^nv$, contradicting minimality.



    We conclude that no cancellation occurs when multiplying the factors $u^{-1},x,x,ldots, x,u$.



    This suggests the following algorithm:




    1. Let $uleftarrowepsilon$, $vleftarrow w$.

    2. If $|v|$ is a multiple of $n$ and is the simple concatenation of $n$ copies of its length $|v|/n$ prefix $x$, then output "$uxu^{-1} $" and terminate. (Note that this condition also triggers when $v=epsilon$, hence we may assume $vneepsilon$ in the next step)

    3. Let $aleftarrow operatorname{first}(v)$, $bleftarrow operatorname{last}(v)$.

    4. If $ane b^{-1}$ output "NOT AN $n$TH POWER" and terminate.


    5. (As $a=b^{-1}$, we have $ane b$, hence $|v|ge 2$ and so $v=av'b$ for some shorter word $v'$) Let $vleftarrow v'$, $uleftarrow ua$, and go back to step 2.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If $w$ has an $n$th root, then so does every conjugate of it.
      Among all pairs $(x,u)$ such that $w=u^{-1}x^nu$, pick one that minimizes first the length $|x|$ and then $|u|$.



      If the first and last letter of $x$ are inverse of each other, i.e., $x=aya^{-1}$
      for some letter $a$ and word $y$ with $|y|<|x|$, then $w=v^{-1}y^nv$ with $v=au$, contradicting minimality of $|x|$.
      Hence $x^n$ is really just concatenation (i.e., without any cancellation among the factors).



      If the first letters of $u$ and $x$ are the same, i.e., $u=av$, $x=ay$ for some letter $a$ and words $v,y$ with $|v|<|u|$, $|y|<|x|$, then
      $w=v^{-1}a^{-1}(ay)^nav=v^{-1}(ya)^nv$, contradicting minimality (as $v|<|u|$ and $|ya|le |x|$).



      Similarly, if the first letter of $u$ and the last letter of $x$ are inverses, i.e., $u=av$, $x=ya^{-1}$ for some letter $a$ and words $v,y$ with $|v|<|u|$, $|y|<|x|$, then $w=v^{-1}a^{-1}(ya^{-1})^nav=v^{-1}(a^{-1}y)^nv$, contradicting minimality.



      We conclude that no cancellation occurs when multiplying the factors $u^{-1},x,x,ldots, x,u$.



      This suggests the following algorithm:




      1. Let $uleftarrowepsilon$, $vleftarrow w$.

      2. If $|v|$ is a multiple of $n$ and is the simple concatenation of $n$ copies of its length $|v|/n$ prefix $x$, then output "$uxu^{-1} $" and terminate. (Note that this condition also triggers when $v=epsilon$, hence we may assume $vneepsilon$ in the next step)

      3. Let $aleftarrow operatorname{first}(v)$, $bleftarrow operatorname{last}(v)$.

      4. If $ane b^{-1}$ output "NOT AN $n$TH POWER" and terminate.


      5. (As $a=b^{-1}$, we have $ane b$, hence $|v|ge 2$ and so $v=av'b$ for some shorter word $v'$) Let $vleftarrow v'$, $uleftarrow ua$, and go back to step 2.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If $w$ has an $n$th root, then so does every conjugate of it.
        Among all pairs $(x,u)$ such that $w=u^{-1}x^nu$, pick one that minimizes first the length $|x|$ and then $|u|$.



        If the first and last letter of $x$ are inverse of each other, i.e., $x=aya^{-1}$
        for some letter $a$ and word $y$ with $|y|<|x|$, then $w=v^{-1}y^nv$ with $v=au$, contradicting minimality of $|x|$.
        Hence $x^n$ is really just concatenation (i.e., without any cancellation among the factors).



        If the first letters of $u$ and $x$ are the same, i.e., $u=av$, $x=ay$ for some letter $a$ and words $v,y$ with $|v|<|u|$, $|y|<|x|$, then
        $w=v^{-1}a^{-1}(ay)^nav=v^{-1}(ya)^nv$, contradicting minimality (as $v|<|u|$ and $|ya|le |x|$).



        Similarly, if the first letter of $u$ and the last letter of $x$ are inverses, i.e., $u=av$, $x=ya^{-1}$ for some letter $a$ and words $v,y$ with $|v|<|u|$, $|y|<|x|$, then $w=v^{-1}a^{-1}(ya^{-1})^nav=v^{-1}(a^{-1}y)^nv$, contradicting minimality.



        We conclude that no cancellation occurs when multiplying the factors $u^{-1},x,x,ldots, x,u$.



        This suggests the following algorithm:




        1. Let $uleftarrowepsilon$, $vleftarrow w$.

        2. If $|v|$ is a multiple of $n$ and is the simple concatenation of $n$ copies of its length $|v|/n$ prefix $x$, then output "$uxu^{-1} $" and terminate. (Note that this condition also triggers when $v=epsilon$, hence we may assume $vneepsilon$ in the next step)

        3. Let $aleftarrow operatorname{first}(v)$, $bleftarrow operatorname{last}(v)$.

        4. If $ane b^{-1}$ output "NOT AN $n$TH POWER" and terminate.


        5. (As $a=b^{-1}$, we have $ane b$, hence $|v|ge 2$ and so $v=av'b$ for some shorter word $v'$) Let $vleftarrow v'$, $uleftarrow ua$, and go back to step 2.






        share|cite|improve this answer









        $endgroup$



        If $w$ has an $n$th root, then so does every conjugate of it.
        Among all pairs $(x,u)$ such that $w=u^{-1}x^nu$, pick one that minimizes first the length $|x|$ and then $|u|$.



        If the first and last letter of $x$ are inverse of each other, i.e., $x=aya^{-1}$
        for some letter $a$ and word $y$ with $|y|<|x|$, then $w=v^{-1}y^nv$ with $v=au$, contradicting minimality of $|x|$.
        Hence $x^n$ is really just concatenation (i.e., without any cancellation among the factors).



        If the first letters of $u$ and $x$ are the same, i.e., $u=av$, $x=ay$ for some letter $a$ and words $v,y$ with $|v|<|u|$, $|y|<|x|$, then
        $w=v^{-1}a^{-1}(ay)^nav=v^{-1}(ya)^nv$, contradicting minimality (as $v|<|u|$ and $|ya|le |x|$).



        Similarly, if the first letter of $u$ and the last letter of $x$ are inverses, i.e., $u=av$, $x=ya^{-1}$ for some letter $a$ and words $v,y$ with $|v|<|u|$, $|y|<|x|$, then $w=v^{-1}a^{-1}(ya^{-1})^nav=v^{-1}(a^{-1}y)^nv$, contradicting minimality.



        We conclude that no cancellation occurs when multiplying the factors $u^{-1},x,x,ldots, x,u$.



        This suggests the following algorithm:




        1. Let $uleftarrowepsilon$, $vleftarrow w$.

        2. If $|v|$ is a multiple of $n$ and is the simple concatenation of $n$ copies of its length $|v|/n$ prefix $x$, then output "$uxu^{-1} $" and terminate. (Note that this condition also triggers when $v=epsilon$, hence we may assume $vneepsilon$ in the next step)

        3. Let $aleftarrow operatorname{first}(v)$, $bleftarrow operatorname{last}(v)$.

        4. If $ane b^{-1}$ output "NOT AN $n$TH POWER" and terminate.


        5. (As $a=b^{-1}$, we have $ane b$, hence $|v|ge 2$ and so $v=av'b$ for some shorter word $v'$) Let $vleftarrow v'$, $uleftarrow ua$, and go back to step 2.







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        answered Jan 6 at 17:15









        Hagen von EitzenHagen von Eitzen

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