Conjugacy in $S_n$ with composing permutations left to right vs. right to left












5












$begingroup$


I realize there are two conventions for composing permutations.



Left to right: $(1 2)(1 3) = (1 2 3)$



Right to left: $(1 2)(1 3) = (1 3 2)$



Among others, Dummit and Foote and Contemporary Abstract Algebra (Gallian) use the right to left convention, while Handbook of Computational Group Theory (Holt), Sage, and GAP use the left to right convention.



Now, given permutations $sigma, tau in S_n$, where $sigma = (sigma_1 sigma_2 ldots sigma_n) ldots$ (in cycle notation), in the right to left convention we have the convenient fact that $tau sigma tau^{-1} = (tau(sigma_1) tau(sigma_2) ldots tau(sigma_n) ) ldots$.



Proof: Observe that if $sigma(i) = j$, then $tau sigma tau^{-1} (tau(i)) = tau(j)$. Therefore if the ordered pair $i, j$ appears in the cycle decomposition of $sigma$, then the ordered pair $tau(i), tau(j)$ appears in the cycle decomposition of $tau sigma tau^{-1}$.



Now if you believe that, you can use a similar proof to obtain the ugly fact that in the left to right notation $tau sigma tau^{-1} = (tau^{-1}(sigma_1) tau^{-1}(sigma_2) ldots tau^{-1}(sigma_n))ldots$



To me this makes the left to right convention inferior, because conjugation follows a less natural rule. Am I missing something? Are there other reasons to compose permutations from left to right that result in simpler algebraic laws? And if not, I am curious why we don't exclusively use right to left notation.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Notationally I understand why both conventions exist, as you explain. However, algebraically are there reasons for left to right composition? I just showed an algebraic reason to prefer right to left.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 13:32
















5












$begingroup$


I realize there are two conventions for composing permutations.



Left to right: $(1 2)(1 3) = (1 2 3)$



Right to left: $(1 2)(1 3) = (1 3 2)$



Among others, Dummit and Foote and Contemporary Abstract Algebra (Gallian) use the right to left convention, while Handbook of Computational Group Theory (Holt), Sage, and GAP use the left to right convention.



Now, given permutations $sigma, tau in S_n$, where $sigma = (sigma_1 sigma_2 ldots sigma_n) ldots$ (in cycle notation), in the right to left convention we have the convenient fact that $tau sigma tau^{-1} = (tau(sigma_1) tau(sigma_2) ldots tau(sigma_n) ) ldots$.



Proof: Observe that if $sigma(i) = j$, then $tau sigma tau^{-1} (tau(i)) = tau(j)$. Therefore if the ordered pair $i, j$ appears in the cycle decomposition of $sigma$, then the ordered pair $tau(i), tau(j)$ appears in the cycle decomposition of $tau sigma tau^{-1}$.



Now if you believe that, you can use a similar proof to obtain the ugly fact that in the left to right notation $tau sigma tau^{-1} = (tau^{-1}(sigma_1) tau^{-1}(sigma_2) ldots tau^{-1}(sigma_n))ldots$



To me this makes the left to right convention inferior, because conjugation follows a less natural rule. Am I missing something? Are there other reasons to compose permutations from left to right that result in simpler algebraic laws? And if not, I am curious why we don't exclusively use right to left notation.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Notationally I understand why both conventions exist, as you explain. However, algebraically are there reasons for left to right composition? I just showed an algebraic reason to prefer right to left.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 13:32














5












5








5





$begingroup$


I realize there are two conventions for composing permutations.



Left to right: $(1 2)(1 3) = (1 2 3)$



Right to left: $(1 2)(1 3) = (1 3 2)$



Among others, Dummit and Foote and Contemporary Abstract Algebra (Gallian) use the right to left convention, while Handbook of Computational Group Theory (Holt), Sage, and GAP use the left to right convention.



Now, given permutations $sigma, tau in S_n$, where $sigma = (sigma_1 sigma_2 ldots sigma_n) ldots$ (in cycle notation), in the right to left convention we have the convenient fact that $tau sigma tau^{-1} = (tau(sigma_1) tau(sigma_2) ldots tau(sigma_n) ) ldots$.



Proof: Observe that if $sigma(i) = j$, then $tau sigma tau^{-1} (tau(i)) = tau(j)$. Therefore if the ordered pair $i, j$ appears in the cycle decomposition of $sigma$, then the ordered pair $tau(i), tau(j)$ appears in the cycle decomposition of $tau sigma tau^{-1}$.



Now if you believe that, you can use a similar proof to obtain the ugly fact that in the left to right notation $tau sigma tau^{-1} = (tau^{-1}(sigma_1) tau^{-1}(sigma_2) ldots tau^{-1}(sigma_n))ldots$



To me this makes the left to right convention inferior, because conjugation follows a less natural rule. Am I missing something? Are there other reasons to compose permutations from left to right that result in simpler algebraic laws? And if not, I am curious why we don't exclusively use right to left notation.










share|cite|improve this question











$endgroup$




I realize there are two conventions for composing permutations.



Left to right: $(1 2)(1 3) = (1 2 3)$



Right to left: $(1 2)(1 3) = (1 3 2)$



Among others, Dummit and Foote and Contemporary Abstract Algebra (Gallian) use the right to left convention, while Handbook of Computational Group Theory (Holt), Sage, and GAP use the left to right convention.



Now, given permutations $sigma, tau in S_n$, where $sigma = (sigma_1 sigma_2 ldots sigma_n) ldots$ (in cycle notation), in the right to left convention we have the convenient fact that $tau sigma tau^{-1} = (tau(sigma_1) tau(sigma_2) ldots tau(sigma_n) ) ldots$.



Proof: Observe that if $sigma(i) = j$, then $tau sigma tau^{-1} (tau(i)) = tau(j)$. Therefore if the ordered pair $i, j$ appears in the cycle decomposition of $sigma$, then the ordered pair $tau(i), tau(j)$ appears in the cycle decomposition of $tau sigma tau^{-1}$.



Now if you believe that, you can use a similar proof to obtain the ugly fact that in the left to right notation $tau sigma tau^{-1} = (tau^{-1}(sigma_1) tau^{-1}(sigma_2) ldots tau^{-1}(sigma_n))ldots$



To me this makes the left to right convention inferior, because conjugation follows a less natural rule. Am I missing something? Are there other reasons to compose permutations from left to right that result in simpler algebraic laws? And if not, I am curious why we don't exclusively use right to left notation.







abstract-algebra group-theory permutations gap convention






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 22:37









Alexander Konovalov

5,24222057




5,24222057










asked Aug 12 '15 at 13:26









Bryce SandlundBryce Sandlund

1926




1926












  • $begingroup$
    Notationally I understand why both conventions exist, as you explain. However, algebraically are there reasons for left to right composition? I just showed an algebraic reason to prefer right to left.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 13:32


















  • $begingroup$
    Notationally I understand why both conventions exist, as you explain. However, algebraically are there reasons for left to right composition? I just showed an algebraic reason to prefer right to left.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 13:32
















$begingroup$
Notationally I understand why both conventions exist, as you explain. However, algebraically are there reasons for left to right composition? I just showed an algebraic reason to prefer right to left.
$endgroup$
– Bryce Sandlund
Aug 12 '15 at 13:32




$begingroup$
Notationally I understand why both conventions exist, as you explain. However, algebraically are there reasons for left to right composition? I just showed an algebraic reason to prefer right to left.
$endgroup$
– Bryce Sandlund
Aug 12 '15 at 13:32










2 Answers
2






active

oldest

votes


















2












$begingroup$

Left-to-right is basically a by-product of western languages being scanned this way. The convention has groups acting (group actions are the raison-d-être for group theory) on the right, and the permutation product is a consequence.



If you have a group $G$ acting on a set $Omega$, then $omega^{gh}=(omega^g)^h$ in left-to-right action, while when acting on the left we have at best that $(gh)(omega)=g(h(omega))$ (or in some convention even the perverse $h(g(omega))$). For many group theorists (the use seems to differ between group theory and other areas) the first version is easier to write down, especially if the product gets longer.



A free bonus is that right action corresponds to row vectors, which are easier to typeset than column vectors.



In my (biased) view the main reasons for the right-to-left convention are the historical use in calculus (one writes $sin(a)$, not $a^{sin}$, even though the second use would be on many pocket calculators nowadays), as well as in Linear Algebra textbooks that tend to write equation systems universally as $Ax=b$, not $xA=b$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Appreciate the insight. It makes sense when you use group actions to have $omega^{gh} = (omega^{g})^h$.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 19:15



















3












$begingroup$

The "ugly" formula $tau sigma tau^{-1} = (tau^{-1}(sigma_1) tau^{-1}(sigma_2) ldots tau^{-1}(sigma_n))$ can be easily rewritten
in the form $tau^{-1} sigma tau = (tau(sigma_1) tau(sigma_2) ldots tau(sigma_n))$, using the bijection $tau^{-1}mapsto tau$. So I would not say that the left to right convention is "inferior". Each of the conventions has certain advantages which depend on the context. In an abstract group $G$ the words are usually given by $w=a_1^{e_1}cdots a_r^{e_r}$ from left to right, but if the elements are maps, it seems sometimes better to use the right to left convention in order to have $(fg)(x)=f(g(x))$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the comment. The problem I noticed is that $tau^{-1} sigma tau$ is not $sigma$ conjugated by $tau$, whereas $tau sigma tau^{-1}$ is. I do agree that both notations have advantages at times though.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 19:14






  • 1




    $begingroup$
    I do not think so. $sigma$ is conjugated by $tau$ either means $tau^{-1}sigmatau$, or $tausigmatau^{-1}$, depending on the conventions "right to left", or "left to right". Both things are equivalent, of course.
    $endgroup$
    – Dietrich Burde
    Aug 12 '15 at 19:22












  • $begingroup$
    You're right. My understanding was false.
    $endgroup$
    – Bryce Sandlund
    Jan 29 '16 at 1:47












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2 Answers
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2 Answers
2






active

oldest

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active

oldest

votes






active

oldest

votes









2












$begingroup$

Left-to-right is basically a by-product of western languages being scanned this way. The convention has groups acting (group actions are the raison-d-être for group theory) on the right, and the permutation product is a consequence.



If you have a group $G$ acting on a set $Omega$, then $omega^{gh}=(omega^g)^h$ in left-to-right action, while when acting on the left we have at best that $(gh)(omega)=g(h(omega))$ (or in some convention even the perverse $h(g(omega))$). For many group theorists (the use seems to differ between group theory and other areas) the first version is easier to write down, especially if the product gets longer.



A free bonus is that right action corresponds to row vectors, which are easier to typeset than column vectors.



In my (biased) view the main reasons for the right-to-left convention are the historical use in calculus (one writes $sin(a)$, not $a^{sin}$, even though the second use would be on many pocket calculators nowadays), as well as in Linear Algebra textbooks that tend to write equation systems universally as $Ax=b$, not $xA=b$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Appreciate the insight. It makes sense when you use group actions to have $omega^{gh} = (omega^{g})^h$.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 19:15
















2












$begingroup$

Left-to-right is basically a by-product of western languages being scanned this way. The convention has groups acting (group actions are the raison-d-être for group theory) on the right, and the permutation product is a consequence.



If you have a group $G$ acting on a set $Omega$, then $omega^{gh}=(omega^g)^h$ in left-to-right action, while when acting on the left we have at best that $(gh)(omega)=g(h(omega))$ (or in some convention even the perverse $h(g(omega))$). For many group theorists (the use seems to differ between group theory and other areas) the first version is easier to write down, especially if the product gets longer.



A free bonus is that right action corresponds to row vectors, which are easier to typeset than column vectors.



In my (biased) view the main reasons for the right-to-left convention are the historical use in calculus (one writes $sin(a)$, not $a^{sin}$, even though the second use would be on many pocket calculators nowadays), as well as in Linear Algebra textbooks that tend to write equation systems universally as $Ax=b$, not $xA=b$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Appreciate the insight. It makes sense when you use group actions to have $omega^{gh} = (omega^{g})^h$.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 19:15














2












2








2





$begingroup$

Left-to-right is basically a by-product of western languages being scanned this way. The convention has groups acting (group actions are the raison-d-être for group theory) on the right, and the permutation product is a consequence.



If you have a group $G$ acting on a set $Omega$, then $omega^{gh}=(omega^g)^h$ in left-to-right action, while when acting on the left we have at best that $(gh)(omega)=g(h(omega))$ (or in some convention even the perverse $h(g(omega))$). For many group theorists (the use seems to differ between group theory and other areas) the first version is easier to write down, especially if the product gets longer.



A free bonus is that right action corresponds to row vectors, which are easier to typeset than column vectors.



In my (biased) view the main reasons for the right-to-left convention are the historical use in calculus (one writes $sin(a)$, not $a^{sin}$, even though the second use would be on many pocket calculators nowadays), as well as in Linear Algebra textbooks that tend to write equation systems universally as $Ax=b$, not $xA=b$.






share|cite|improve this answer











$endgroup$



Left-to-right is basically a by-product of western languages being scanned this way. The convention has groups acting (group actions are the raison-d-être for group theory) on the right, and the permutation product is a consequence.



If you have a group $G$ acting on a set $Omega$, then $omega^{gh}=(omega^g)^h$ in left-to-right action, while when acting on the left we have at best that $(gh)(omega)=g(h(omega))$ (or in some convention even the perverse $h(g(omega))$). For many group theorists (the use seems to differ between group theory and other areas) the first version is easier to write down, especially if the product gets longer.



A free bonus is that right action corresponds to row vectors, which are easier to typeset than column vectors.



In my (biased) view the main reasons for the right-to-left convention are the historical use in calculus (one writes $sin(a)$, not $a^{sin}$, even though the second use would be on many pocket calculators nowadays), as well as in Linear Algebra textbooks that tend to write equation systems universally as $Ax=b$, not $xA=b$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 15 at 21:03









darij grinberg

11.4k33167




11.4k33167










answered Aug 12 '15 at 14:09









ahulpkeahulpke

7,2021026




7,2021026












  • $begingroup$
    Appreciate the insight. It makes sense when you use group actions to have $omega^{gh} = (omega^{g})^h$.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 19:15


















  • $begingroup$
    Appreciate the insight. It makes sense when you use group actions to have $omega^{gh} = (omega^{g})^h$.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 19:15
















$begingroup$
Appreciate the insight. It makes sense when you use group actions to have $omega^{gh} = (omega^{g})^h$.
$endgroup$
– Bryce Sandlund
Aug 12 '15 at 19:15




$begingroup$
Appreciate the insight. It makes sense when you use group actions to have $omega^{gh} = (omega^{g})^h$.
$endgroup$
– Bryce Sandlund
Aug 12 '15 at 19:15











3












$begingroup$

The "ugly" formula $tau sigma tau^{-1} = (tau^{-1}(sigma_1) tau^{-1}(sigma_2) ldots tau^{-1}(sigma_n))$ can be easily rewritten
in the form $tau^{-1} sigma tau = (tau(sigma_1) tau(sigma_2) ldots tau(sigma_n))$, using the bijection $tau^{-1}mapsto tau$. So I would not say that the left to right convention is "inferior". Each of the conventions has certain advantages which depend on the context. In an abstract group $G$ the words are usually given by $w=a_1^{e_1}cdots a_r^{e_r}$ from left to right, but if the elements are maps, it seems sometimes better to use the right to left convention in order to have $(fg)(x)=f(g(x))$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the comment. The problem I noticed is that $tau^{-1} sigma tau$ is not $sigma$ conjugated by $tau$, whereas $tau sigma tau^{-1}$ is. I do agree that both notations have advantages at times though.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 19:14






  • 1




    $begingroup$
    I do not think so. $sigma$ is conjugated by $tau$ either means $tau^{-1}sigmatau$, or $tausigmatau^{-1}$, depending on the conventions "right to left", or "left to right". Both things are equivalent, of course.
    $endgroup$
    – Dietrich Burde
    Aug 12 '15 at 19:22












  • $begingroup$
    You're right. My understanding was false.
    $endgroup$
    – Bryce Sandlund
    Jan 29 '16 at 1:47
















3












$begingroup$

The "ugly" formula $tau sigma tau^{-1} = (tau^{-1}(sigma_1) tau^{-1}(sigma_2) ldots tau^{-1}(sigma_n))$ can be easily rewritten
in the form $tau^{-1} sigma tau = (tau(sigma_1) tau(sigma_2) ldots tau(sigma_n))$, using the bijection $tau^{-1}mapsto tau$. So I would not say that the left to right convention is "inferior". Each of the conventions has certain advantages which depend on the context. In an abstract group $G$ the words are usually given by $w=a_1^{e_1}cdots a_r^{e_r}$ from left to right, but if the elements are maps, it seems sometimes better to use the right to left convention in order to have $(fg)(x)=f(g(x))$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the comment. The problem I noticed is that $tau^{-1} sigma tau$ is not $sigma$ conjugated by $tau$, whereas $tau sigma tau^{-1}$ is. I do agree that both notations have advantages at times though.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 19:14






  • 1




    $begingroup$
    I do not think so. $sigma$ is conjugated by $tau$ either means $tau^{-1}sigmatau$, or $tausigmatau^{-1}$, depending on the conventions "right to left", or "left to right". Both things are equivalent, of course.
    $endgroup$
    – Dietrich Burde
    Aug 12 '15 at 19:22












  • $begingroup$
    You're right. My understanding was false.
    $endgroup$
    – Bryce Sandlund
    Jan 29 '16 at 1:47














3












3








3





$begingroup$

The "ugly" formula $tau sigma tau^{-1} = (tau^{-1}(sigma_1) tau^{-1}(sigma_2) ldots tau^{-1}(sigma_n))$ can be easily rewritten
in the form $tau^{-1} sigma tau = (tau(sigma_1) tau(sigma_2) ldots tau(sigma_n))$, using the bijection $tau^{-1}mapsto tau$. So I would not say that the left to right convention is "inferior". Each of the conventions has certain advantages which depend on the context. In an abstract group $G$ the words are usually given by $w=a_1^{e_1}cdots a_r^{e_r}$ from left to right, but if the elements are maps, it seems sometimes better to use the right to left convention in order to have $(fg)(x)=f(g(x))$.






share|cite|improve this answer









$endgroup$



The "ugly" formula $tau sigma tau^{-1} = (tau^{-1}(sigma_1) tau^{-1}(sigma_2) ldots tau^{-1}(sigma_n))$ can be easily rewritten
in the form $tau^{-1} sigma tau = (tau(sigma_1) tau(sigma_2) ldots tau(sigma_n))$, using the bijection $tau^{-1}mapsto tau$. So I would not say that the left to right convention is "inferior". Each of the conventions has certain advantages which depend on the context. In an abstract group $G$ the words are usually given by $w=a_1^{e_1}cdots a_r^{e_r}$ from left to right, but if the elements are maps, it seems sometimes better to use the right to left convention in order to have $(fg)(x)=f(g(x))$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 12 '15 at 13:52









Dietrich BurdeDietrich Burde

81.5k648106




81.5k648106












  • $begingroup$
    Thanks for the comment. The problem I noticed is that $tau^{-1} sigma tau$ is not $sigma$ conjugated by $tau$, whereas $tau sigma tau^{-1}$ is. I do agree that both notations have advantages at times though.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 19:14






  • 1




    $begingroup$
    I do not think so. $sigma$ is conjugated by $tau$ either means $tau^{-1}sigmatau$, or $tausigmatau^{-1}$, depending on the conventions "right to left", or "left to right". Both things are equivalent, of course.
    $endgroup$
    – Dietrich Burde
    Aug 12 '15 at 19:22












  • $begingroup$
    You're right. My understanding was false.
    $endgroup$
    – Bryce Sandlund
    Jan 29 '16 at 1:47


















  • $begingroup$
    Thanks for the comment. The problem I noticed is that $tau^{-1} sigma tau$ is not $sigma$ conjugated by $tau$, whereas $tau sigma tau^{-1}$ is. I do agree that both notations have advantages at times though.
    $endgroup$
    – Bryce Sandlund
    Aug 12 '15 at 19:14






  • 1




    $begingroup$
    I do not think so. $sigma$ is conjugated by $tau$ either means $tau^{-1}sigmatau$, or $tausigmatau^{-1}$, depending on the conventions "right to left", or "left to right". Both things are equivalent, of course.
    $endgroup$
    – Dietrich Burde
    Aug 12 '15 at 19:22












  • $begingroup$
    You're right. My understanding was false.
    $endgroup$
    – Bryce Sandlund
    Jan 29 '16 at 1:47
















$begingroup$
Thanks for the comment. The problem I noticed is that $tau^{-1} sigma tau$ is not $sigma$ conjugated by $tau$, whereas $tau sigma tau^{-1}$ is. I do agree that both notations have advantages at times though.
$endgroup$
– Bryce Sandlund
Aug 12 '15 at 19:14




$begingroup$
Thanks for the comment. The problem I noticed is that $tau^{-1} sigma tau$ is not $sigma$ conjugated by $tau$, whereas $tau sigma tau^{-1}$ is. I do agree that both notations have advantages at times though.
$endgroup$
– Bryce Sandlund
Aug 12 '15 at 19:14




1




1




$begingroup$
I do not think so. $sigma$ is conjugated by $tau$ either means $tau^{-1}sigmatau$, or $tausigmatau^{-1}$, depending on the conventions "right to left", or "left to right". Both things are equivalent, of course.
$endgroup$
– Dietrich Burde
Aug 12 '15 at 19:22






$begingroup$
I do not think so. $sigma$ is conjugated by $tau$ either means $tau^{-1}sigmatau$, or $tausigmatau^{-1}$, depending on the conventions "right to left", or "left to right". Both things are equivalent, of course.
$endgroup$
– Dietrich Burde
Aug 12 '15 at 19:22














$begingroup$
You're right. My understanding was false.
$endgroup$
– Bryce Sandlund
Jan 29 '16 at 1:47




$begingroup$
You're right. My understanding was false.
$endgroup$
– Bryce Sandlund
Jan 29 '16 at 1:47


















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