Conjugation of permutations: left-to-right versus right-to-left.












4












$begingroup$


In the group $S_n$ I usually use the fact that if $(a_1 a_2 dots a_r) in S_n$ is an r-cycle and $sigma in S_n$ then $sigma (a_1 a_2 dots a_r)sigma^{-1} = (sigma(a_1)sigma(a_2) dots sigma(a_r))$. My question is: does this use the convention that permutations act from left-to-right, or right-to-left?



Here's a proof for this identity:



Let $rho in S_n$ be such that $rho (a_i) = rho(a_{i+1 text{ (mod } r)})$, in other words $rho = (sigma(a_1) sigma(a_2) dots sigma(a_r))$.



Then $a_i overset{sigma}{longmapsto} sigma(a_i) overset{rho}{longmapsto} sigma(a_{i+1}) overset{sigma^{-1}}{longmapsto} a_{i+1} implies sigma^{-1}rhosigma=(a_1 a_2 dots a_r)$ and $sigma(a_1 a_2 dots a_r)sigma^{-1} = (sigma(a_1) sigma(a_2) dots sigma(a_r)) = rho$.



Here I have used the convention that permutations act from right-to-left. However, in most other situations I prefer to read permutations from left to right - this is probably the most common convention among group theorists (see here).



For consistency, it seems I should be using $sigma(a_1 a_2 dots a_r)sigma^{-1} = (sigma^{-1}(a_1) sigma^{-1}(a_2) dots sigma^{-1}(a_r))$ instead of the other version. Am I correct, or have I made a misunderstanding somewhere?






group-theory permutations symmetric-groups convention







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$endgroup$












  • $begingroup$
    If you apply (and compose) permutations from the left, then it is clear that $sigma(a_{1}ldots a_{r})sigma^{-1}[sigma(a_{i})] = sigma(a_{i+1})$ where the subscript $i+1$ is read $(mod r)$.
    $endgroup$
    – Geoff Robinson
    May 30 '15 at 14:32










  • $begingroup$
    I've always let the rightmost permutation go first, so that $(1 2)(1 3) = (1 3 2)$. For conjugation, I would always apply $sigma^{-1}$ on the left; i.e., $sigma^{-1}(a_1 a_2ldots a_r)sigma$. I believe this is the most consistent "functions act from the left" notation with $f(x)$ and $(f circ g)(x) = f(g(x))$ rather than $(x)f$. But, I constantly just need to remind myself how it all works.
    $endgroup$
    – pjs36
    May 30 '15 at 16:10
















4












$begingroup$


In the group $S_n$ I usually use the fact that if $(a_1 a_2 dots a_r) in S_n$ is an r-cycle and $sigma in S_n$ then $sigma (a_1 a_2 dots a_r)sigma^{-1} = (sigma(a_1)sigma(a_2) dots sigma(a_r))$. My question is: does this use the convention that permutations act from left-to-right, or right-to-left?



Here's a proof for this identity:



Let $rho in S_n$ be such that $rho (a_i) = rho(a_{i+1 text{ (mod } r)})$, in other words $rho = (sigma(a_1) sigma(a_2) dots sigma(a_r))$.



Then $a_i overset{sigma}{longmapsto} sigma(a_i) overset{rho}{longmapsto} sigma(a_{i+1}) overset{sigma^{-1}}{longmapsto} a_{i+1} implies sigma^{-1}rhosigma=(a_1 a_2 dots a_r)$ and $sigma(a_1 a_2 dots a_r)sigma^{-1} = (sigma(a_1) sigma(a_2) dots sigma(a_r)) = rho$.



Here I have used the convention that permutations act from right-to-left. However, in most other situations I prefer to read permutations from left to right - this is probably the most common convention among group theorists (see here).



For consistency, it seems I should be using $sigma(a_1 a_2 dots a_r)sigma^{-1} = (sigma^{-1}(a_1) sigma^{-1}(a_2) dots sigma^{-1}(a_r))$ instead of the other version. Am I correct, or have I made a misunderstanding somewhere?






group-theory permutations symmetric-groups convention







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you apply (and compose) permutations from the left, then it is clear that $sigma(a_{1}ldots a_{r})sigma^{-1}[sigma(a_{i})] = sigma(a_{i+1})$ where the subscript $i+1$ is read $(mod r)$.
    $endgroup$
    – Geoff Robinson
    May 30 '15 at 14:32










  • $begingroup$
    I've always let the rightmost permutation go first, so that $(1 2)(1 3) = (1 3 2)$. For conjugation, I would always apply $sigma^{-1}$ on the left; i.e., $sigma^{-1}(a_1 a_2ldots a_r)sigma$. I believe this is the most consistent "functions act from the left" notation with $f(x)$ and $(f circ g)(x) = f(g(x))$ rather than $(x)f$. But, I constantly just need to remind myself how it all works.
    $endgroup$
    – pjs36
    May 30 '15 at 16:10














4












4








4


2



$begingroup$


In the group $S_n$ I usually use the fact that if $(a_1 a_2 dots a_r) in S_n$ is an r-cycle and $sigma in S_n$ then $sigma (a_1 a_2 dots a_r)sigma^{-1} = (sigma(a_1)sigma(a_2) dots sigma(a_r))$. My question is: does this use the convention that permutations act from left-to-right, or right-to-left?



Here's a proof for this identity:



Let $rho in S_n$ be such that $rho (a_i) = rho(a_{i+1 text{ (mod } r)})$, in other words $rho = (sigma(a_1) sigma(a_2) dots sigma(a_r))$.



Then $a_i overset{sigma}{longmapsto} sigma(a_i) overset{rho}{longmapsto} sigma(a_{i+1}) overset{sigma^{-1}}{longmapsto} a_{i+1} implies sigma^{-1}rhosigma=(a_1 a_2 dots a_r)$ and $sigma(a_1 a_2 dots a_r)sigma^{-1} = (sigma(a_1) sigma(a_2) dots sigma(a_r)) = rho$.



Here I have used the convention that permutations act from right-to-left. However, in most other situations I prefer to read permutations from left to right - this is probably the most common convention among group theorists (see here).



For consistency, it seems I should be using $sigma(a_1 a_2 dots a_r)sigma^{-1} = (sigma^{-1}(a_1) sigma^{-1}(a_2) dots sigma^{-1}(a_r))$ instead of the other version. Am I correct, or have I made a misunderstanding somewhere?






group-theory permutations symmetric-groups convention







share|cite|improve this question











$endgroup$




In the group $S_n$ I usually use the fact that if $(a_1 a_2 dots a_r) in S_n$ is an r-cycle and $sigma in S_n$ then $sigma (a_1 a_2 dots a_r)sigma^{-1} = (sigma(a_1)sigma(a_2) dots sigma(a_r))$. My question is: does this use the convention that permutations act from left-to-right, or right-to-left?



Here's a proof for this identity:



Let $rho in S_n$ be such that $rho (a_i) = rho(a_{i+1 text{ (mod } r)})$, in other words $rho = (sigma(a_1) sigma(a_2) dots sigma(a_r))$.



Then $a_i overset{sigma}{longmapsto} sigma(a_i) overset{rho}{longmapsto} sigma(a_{i+1}) overset{sigma^{-1}}{longmapsto} a_{i+1} implies sigma^{-1}rhosigma=(a_1 a_2 dots a_r)$ and $sigma(a_1 a_2 dots a_r)sigma^{-1} = (sigma(a_1) sigma(a_2) dots sigma(a_r)) = rho$.



Here I have used the convention that permutations act from right-to-left. However, in most other situations I prefer to read permutations from left to right - this is probably the most common convention among group theorists (see here).



For consistency, it seems I should be using $sigma(a_1 a_2 dots a_r)sigma^{-1} = (sigma^{-1}(a_1) sigma^{-1}(a_2) dots sigma^{-1}(a_r))$ instead of the other version. Am I correct, or have I made a misunderstanding somewhere?






group-theory permutations symmetric-groups convention




group-theory permutations symmetric-groups convention






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 22 '18 at 2:46









Shaun

10.5k113687




10.5k113687










asked May 30 '15 at 14:12









StanleyStanley

2,0391021




2,0391021












  • $begingroup$
    If you apply (and compose) permutations from the left, then it is clear that $sigma(a_{1}ldots a_{r})sigma^{-1}[sigma(a_{i})] = sigma(a_{i+1})$ where the subscript $i+1$ is read $(mod r)$.
    $endgroup$
    – Geoff Robinson
    May 30 '15 at 14:32










  • $begingroup$
    I've always let the rightmost permutation go first, so that $(1 2)(1 3) = (1 3 2)$. For conjugation, I would always apply $sigma^{-1}$ on the left; i.e., $sigma^{-1}(a_1 a_2ldots a_r)sigma$. I believe this is the most consistent "functions act from the left" notation with $f(x)$ and $(f circ g)(x) = f(g(x))$ rather than $(x)f$. But, I constantly just need to remind myself how it all works.
    $endgroup$
    – pjs36
    May 30 '15 at 16:10


















  • $begingroup$
    If you apply (and compose) permutations from the left, then it is clear that $sigma(a_{1}ldots a_{r})sigma^{-1}[sigma(a_{i})] = sigma(a_{i+1})$ where the subscript $i+1$ is read $(mod r)$.
    $endgroup$
    – Geoff Robinson
    May 30 '15 at 14:32










  • $begingroup$
    I've always let the rightmost permutation go first, so that $(1 2)(1 3) = (1 3 2)$. For conjugation, I would always apply $sigma^{-1}$ on the left; i.e., $sigma^{-1}(a_1 a_2ldots a_r)sigma$. I believe this is the most consistent "functions act from the left" notation with $f(x)$ and $(f circ g)(x) = f(g(x))$ rather than $(x)f$. But, I constantly just need to remind myself how it all works.
    $endgroup$
    – pjs36
    May 30 '15 at 16:10
















$begingroup$
If you apply (and compose) permutations from the left, then it is clear that $sigma(a_{1}ldots a_{r})sigma^{-1}[sigma(a_{i})] = sigma(a_{i+1})$ where the subscript $i+1$ is read $(mod r)$.
$endgroup$
– Geoff Robinson
May 30 '15 at 14:32




$begingroup$
If you apply (and compose) permutations from the left, then it is clear that $sigma(a_{1}ldots a_{r})sigma^{-1}[sigma(a_{i})] = sigma(a_{i+1})$ where the subscript $i+1$ is read $(mod r)$.
$endgroup$
– Geoff Robinson
May 30 '15 at 14:32












$begingroup$
I've always let the rightmost permutation go first, so that $(1 2)(1 3) = (1 3 2)$. For conjugation, I would always apply $sigma^{-1}$ on the left; i.e., $sigma^{-1}(a_1 a_2ldots a_r)sigma$. I believe this is the most consistent "functions act from the left" notation with $f(x)$ and $(f circ g)(x) = f(g(x))$ rather than $(x)f$. But, I constantly just need to remind myself how it all works.
$endgroup$
– pjs36
May 30 '15 at 16:10




$begingroup$
I've always let the rightmost permutation go first, so that $(1 2)(1 3) = (1 3 2)$. For conjugation, I would always apply $sigma^{-1}$ on the left; i.e., $sigma^{-1}(a_1 a_2ldots a_r)sigma$. I believe this is the most consistent "functions act from the left" notation with $f(x)$ and $(f circ g)(x) = f(g(x))$ rather than $(x)f$. But, I constantly just need to remind myself how it all works.
$endgroup$
– pjs36
May 30 '15 at 16:10










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Yes, you are correct; the convention used is often clear from the context. It can be tricky sometimes though, like when group theory (which tends to compose maps right-to-left) gets used in semigroup theory (which tends to compose maps left-to-right).






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    $begingroup$

    Yes, you are correct; the convention used is often clear from the context. It can be tricky sometimes though, like when group theory (which tends to compose maps right-to-left) gets used in semigroup theory (which tends to compose maps left-to-right).






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Yes, you are correct; the convention used is often clear from the context. It can be tricky sometimes though, like when group theory (which tends to compose maps right-to-left) gets used in semigroup theory (which tends to compose maps left-to-right).






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Yes, you are correct; the convention used is often clear from the context. It can be tricky sometimes though, like when group theory (which tends to compose maps right-to-left) gets used in semigroup theory (which tends to compose maps left-to-right).






        share|cite|improve this answer









        $endgroup$



        Yes, you are correct; the convention used is often clear from the context. It can be tricky sometimes though, like when group theory (which tends to compose maps right-to-left) gets used in semigroup theory (which tends to compose maps left-to-right).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 2:48









        ShaunShaun

        10.5k113687




        10.5k113687






























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