Conjugation of permutations: left-to-right versus right-to-left.
$begingroup$
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
edited Dec 22 '18 at 2:46
Shaun
9,804113684
asked May 30 '15 at 14:12
StanleyStanley
2,0411021
$endgroup$
add a comment |
$begingroup$
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
edited Dec 22 '18 at 2:46
Shaun
9,804113684
asked May 30 '15 at 14:12
StanleyStanley
2,0411021
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 22 '18 at 2:46
Shaun
9,804113684
asked May 30 '15 at 14:12
StanleyStanley
2,0411021
$endgroup$
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
edited Dec 22 '18 at 2:46
Shaun
9,804113684
asked May 30 '15 at 14:12
StanleyStanley
2,0411021
edited Dec 22 '18 at 2:46
Shaun
9,804113684
asked May 30 '15 at 14:12
StanleyStanley
2,0411021
edited Dec 22 '18 at 2:46
Shaun
9,804113684
edited Dec 22 '18 at 2:46
Shaun
9,804113684
edited Dec 22 '18 at 2:46
Shaun
9,804113684
9,804113684
asked May 30 '15 at 14:12
StanleyStanley
2,0411021
asked May 30 '15 at 14:12
StanleyStanley
2,0411021
asked May 30 '15 at 14:12
StanleyStanley
2,0411021
2,0411021
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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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).
answered Dec 22 '18 at 2:48
ShaunShaun
9,804113684
$endgroup$
add a comment |
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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).
answered Dec 22 '18 at 2:48
ShaunShaun
9,804113684
$endgroup$
add a comment |
$begingroup$
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).
answered Dec 22 '18 at 2:48
ShaunShaun
9,804113684
$endgroup$
add a comment |
$begingroup$
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).
answered Dec 22 '18 at 2:48
ShaunShaun
9,804113684
$endgroup$
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).
answered Dec 22 '18 at 2:48
ShaunShaun
9,804113684
answered Dec 22 '18 at 2:48
ShaunShaun
9,804113684
answered Dec 22 '18 at 2:48
ShaunShaun
9,804113684
answered Dec 22 '18 at 2:48
ShaunShaun
9,804113684
9,804113684
add a comment |
add a comment |
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$begingroup$
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