How does a permutation act on a string?Why Composition and Dihedral Group have reverse order of operation?Permutation of a groupFinding a permutation, and number of, from powers of the permutationShowing that there is a permutation $rho$ that fixes a number that $sigma$ moves when $rho sigma rho^-1=sigma^-1$Permutation of composition factors?Blocks in permutation group theory (D&F)Establishing a bijection between permutations and permutation matrices.Finding the smallest exponent $k$ for a non-cyclic permutation $sigma$, so that $sigma^k = id$.What is the following way of indexing permutations called?Mth position of the Nth Ordered permutation of an ordrered set all elements taken at onceSymmetries of the Tetrahedron - Geometric description and isomorphic correlations
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How does a permutation act on a string?
Why Composition and Dihedral Group have reverse order of operation?Permutation of a groupFinding a permutation, and number of, from powers of the permutationShowing that there is a permutation $rho$ that fixes a number that $sigma$ moves when $rho sigma rho^-1=sigma^-1$Permutation of composition factors?Blocks in permutation group theory (D&F)Establishing a bijection between permutations and permutation matrices.Finding the smallest exponent $k$ for a non-cyclic permutation $sigma$, so that $sigma^k = id$.What is the following way of indexing permutations called?Mth position of the Nth Ordered permutation of an ordrered set all elements taken at onceSymmetries of the Tetrahedron - Geometric description and isomorphic correlations
$begingroup$
Is there a conventional way to have a permutation act on a list of objects? It seems like there are two possible ways, one being the inverse of the other.
Suppose I have a permutation $sigma in S_4$ which is concretely specified as a function from the set $S = 1,2,3,4$ to itself. Specifically,
$$beginarrayc
i & 1 & 2 & 3 & 4 \ hline
sigma(i) & 4 & 3 & 1 & 2
endarray$$
Say I want to permute the string "STAR" by $sigma$. One way to do it would be to send the letter at position $i$ to position $sigma(i)$ in the result, giving "ARTS". Another way to do it would be to populate the $i^textth$ entry of the result using the $sigma(i)^textth$ entry of the original. That would give "RAST".
The first one seems more correct, but the second is more appealing because the string "1234" permutes to "4312", which you read directly off the table.
EDIT: I realize this is equivalent to asking if a permutation matrix should have ones in entries $a_i,sigma(i)$ or $a_sigma(i),i$.
permutations
asked May 14 at 23:41
orlandpmorlandpm
4,80922038
$endgroup$
add a comment |
$begingroup$
Is there a conventional way to have a permutation act on a list of objects? It seems like there are two possible ways, one being the inverse of the other.
Suppose I have a permutation $sigma in S_4$ which is concretely specified as a function from the set $S = 1,2,3,4$ to itself. Specifically,
$$beginarrayc
i & 1 & 2 & 3 & 4 \ hline
sigma(i) & 4 & 3 & 1 & 2
endarray$$
Say I want to permute the string "STAR" by $sigma$. One way to do it would be to send the letter at position $i$ to position $sigma(i)$ in the result, giving "ARTS". Another way to do it would be to populate the $i^textth$ entry of the result using the $sigma(i)^textth$ entry of the original. That would give "RAST".
The first one seems more correct, but the second is more appealing because the string "1234" permutes to "4312", which you read directly off the table.
EDIT: I realize this is equivalent to asking if a permutation matrix should have ones in entries $a_i,sigma(i)$ or $a_sigma(i),i$.
permutations
asked May 14 at 23:41
orlandpmorlandpm
4,80922038
$endgroup$
$begingroup$
For each $iin S$, $i$ should go to $sigma(i)$. Since the characters in the string are usually mapped to their indices, a permutation of the string is a permutation on its index set. Therefore, I would say your first way is correct.
$endgroup$
– John Douma
May 14 at 23:55
3
$begingroup$
Both are correct: one is a left action and the other is a right action.
$endgroup$
– Catalin Zara
May 15 at 0:17
$begingroup$
@CatalinZara could you expand this into an answer?
$endgroup$
– Q the Platypus
May 15 at 1:46
add a comment |
$begingroup$
Is there a conventional way to have a permutation act on a list of objects? It seems like there are two possible ways, one being the inverse of the other.
Suppose I have a permutation $sigma in S_4$ which is concretely specified as a function from the set $S = 1,2,3,4$ to itself. Specifically,
$$beginarrayc
i & 1 & 2 & 3 & 4 \ hline
sigma(i) & 4 & 3 & 1 & 2
endarray$$
Say I want to permute the string "STAR" by $sigma$. One way to do it would be to send the letter at position $i$ to position $sigma(i)$ in the result, giving "ARTS". Another way to do it would be to populate the $i^textth$ entry of the result using the $sigma(i)^textth$ entry of the original. That would give "RAST".
The first one seems more correct, but the second is more appealing because the string "1234" permutes to "4312", which you read directly off the table.
EDIT: I realize this is equivalent to asking if a permutation matrix should have ones in entries $a_i,sigma(i)$ or $a_sigma(i),i$.
permutations
asked May 14 at 23:41
orlandpmorlandpm
4,80922038
$endgroup$
Is there a conventional way to have a permutation act on a list of objects? It seems like there are two possible ways, one being the inverse of the other.
Suppose I have a permutation $sigma in S_4$ which is concretely specified as a function from the set $S = 1,2,3,4$ to itself. Specifically,
$$beginarrayc
i & 1 & 2 & 3 & 4 \ hline
sigma(i) & 4 & 3 & 1 & 2
endarray$$
Say I want to permute the string "STAR" by $sigma$. One way to do it would be to send the letter at position $i$ to position $sigma(i)$ in the result, giving "ARTS". Another way to do it would be to populate the $i^textth$ entry of the result using the $sigma(i)^textth$ entry of the original. That would give "RAST".
The first one seems more correct, but the second is more appealing because the string "1234" permutes to "4312", which you read directly off the table.
EDIT: I realize this is equivalent to asking if a permutation matrix should have ones in entries $a_i,sigma(i)$ or $a_sigma(i),i$.
permutations
permutations
asked May 14 at 23:41
orlandpmorlandpm
4,80922038
asked May 14 at 23:41
orlandpmorlandpm
4,80922038
edited May 14 at 23:56
orlandpm
asked May 14 at 23:41
orlandpmorlandpm
4,80922038
asked May 14 at 23:41
orlandpmorlandpm
4,80922038
asked May 14 at 23:41
orlandpmorlandpm
4,80922038
4,80922038
$begingroup$
For each $iin S$, $i$ should go to $sigma(i)$. Since the characters in the string are usually mapped to their indices, a permutation of the string is a permutation on its index set. Therefore, I would say your first way is correct.
$endgroup$
– John Douma
May 14 at 23:55
3
$begingroup$
Both are correct: one is a left action and the other is a right action.
$endgroup$
– Catalin Zara
May 15 at 0:17
$begingroup$
@CatalinZara could you expand this into an answer?
$endgroup$
– Q the Platypus
May 15 at 1:46
add a comment |
$begingroup$
For each $iin S$, $i$ should go to $sigma(i)$. Since the characters in the string are usually mapped to their indices, a permutation of the string is a permutation on its index set. Therefore, I would say your first way is correct.
$endgroup$
– John Douma
May 14 at 23:55
3
$begingroup$
Both are correct: one is a left action and the other is a right action.
$endgroup$
– Catalin Zara
May 15 at 0:17
$begingroup$
@CatalinZara could you expand this into an answer?
$endgroup$
– Q the Platypus
May 15 at 1:46
$begingroup$
For each $iin S$, $i$ should go to $sigma(i)$. Since the characters in the string are usually mapped to their indices, a permutation of the string is a permutation on its index set. Therefore, I would say your first way is correct.
$endgroup$
– John Douma
May 14 at 23:55
$begingroup$
For each $iin S$, $i$ should go to $sigma(i)$. Since the characters in the string are usually mapped to their indices, a permutation of the string is a permutation on its index set. Therefore, I would say your first way is correct.
$endgroup$
– John Douma
May 14 at 23:55
3
3
$begingroup$
Both are correct: one is a left action and the other is a right action.
$endgroup$
– Catalin Zara
May 15 at 0:17
$begingroup$
Both are correct: one is a left action and the other is a right action.
$endgroup$
– Catalin Zara
May 15 at 0:17
$begingroup$
@CatalinZara could you expand this into an answer?
$endgroup$
– Q the Platypus
May 15 at 1:46
$begingroup$
@CatalinZara could you expand this into an answer?
$endgroup$
– Q the Platypus
May 15 at 1:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Both actions are correct: one is a left action and the other is a right action.
[See https://en.wikipedia.org/wiki/Group_action_(mathematics)]
For the first action: to a permutation $sigma$ and a string $x$, we associate a string $sigma cdot x$, defined by $(sigma cdot x)_sigma(i) = x_i$ ("letter at position $i$ is sent to position $sigma(i)$"), for all indices $i$ or, equivalently, $(sigma cdot x)_j = x_sigma^-1(j)$ for all indices $j$. That is a left action, since for two permutations $sigma$ and $tau$, we have
$$[sigmacdot (taucdot x))]_i = (taucdot x)_sigma^-1(i) = x_tau^-1(sigma^-1(i))= x_(sigmatau)^-1(i) = [(sigma tau)cdot x]_i,$$
hence
$$sigmacdot (taucdot x) = (sigma tau)cdot x.$$
Applying (i.e. "multiplying" by) $tau$ and then $sigma$ is the same as applying $sigmatau$. That is how multiplication to the left works, hence the term ''left action.''
For the second action: to a permutation $sigma$ and a string $x$, we associate a string $xstar sigma$, defined by $(xstar sigma)_i = x_sigma(i)$ ("$i^th$ entry of the result is the $sigma(i)^th$ entry of the original."). That is a right action, since for two permutations $sigma$ and $tau$, we have
$$[(xstar sigma)star tau]_i = [xstarsigma]_tau(i)= x_sigma(tau(i)) = x_(sigmatau)(i) = [xstar (sigmatau)]_i,$$
hence
$$(xstar sigma)star tau = xstar(sigmatau).$$
Applying (i.e. "multiplying" by) $sigma$ and then $tau$ is the same as applying $sigmatau$. That is how multiplication to the right works, hence the term ''right action.''
The two actions are indeed related by
$$(sigma cdot x)star sigma = x = sigma cdot (xstar sigma),$$
because
$$xstar sigma = sigma^-1cdot x.$$
answered May 15 at 2:22
Catalin ZaraCatalin Zara
3,982515
$endgroup$
$begingroup$
Very well explained. Failing to recognize that the two possible conventions lead to actions on opposite sides can result in much confusion, as in this older question: math.stackexchange.com/questions/1373136
$endgroup$
– Ted
May 15 at 3:07
add a comment |
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$begingroup$
Both actions are correct: one is a left action and the other is a right action.
[See https://en.wikipedia.org/wiki/Group_action_(mathematics)]
For the first action: to a permutation $sigma$ and a string $x$, we associate a string $sigma cdot x$, defined by $(sigma cdot x)_sigma(i) = x_i$ ("letter at position $i$ is sent to position $sigma(i)$"), for all indices $i$ or, equivalently, $(sigma cdot x)_j = x_sigma^-1(j)$ for all indices $j$. That is a left action, since for two permutations $sigma$ and $tau$, we have
$$[sigmacdot (taucdot x))]_i = (taucdot x)_sigma^-1(i) = x_tau^-1(sigma^-1(i))= x_(sigmatau)^-1(i) = [(sigma tau)cdot x]_i,$$
hence
$$sigmacdot (taucdot x) = (sigma tau)cdot x.$$
Applying (i.e. "multiplying" by) $tau$ and then $sigma$ is the same as applying $sigmatau$. That is how multiplication to the left works, hence the term ''left action.''
For the second action: to a permutation $sigma$ and a string $x$, we associate a string $xstar sigma$, defined by $(xstar sigma)_i = x_sigma(i)$ ("$i^th$ entry of the result is the $sigma(i)^th$ entry of the original."). That is a right action, since for two permutations $sigma$ and $tau$, we have
$$[(xstar sigma)star tau]_i = [xstarsigma]_tau(i)= x_sigma(tau(i)) = x_(sigmatau)(i) = [xstar (sigmatau)]_i,$$
hence
$$(xstar sigma)star tau = xstar(sigmatau).$$
Applying (i.e. "multiplying" by) $sigma$ and then $tau$ is the same as applying $sigmatau$. That is how multiplication to the right works, hence the term ''right action.''
The two actions are indeed related by
$$(sigma cdot x)star sigma = x = sigma cdot (xstar sigma),$$
because
$$xstar sigma = sigma^-1cdot x.$$
answered May 15 at 2:22
Catalin ZaraCatalin Zara
3,982515
$endgroup$
$begingroup$
Very well explained. Failing to recognize that the two possible conventions lead to actions on opposite sides can result in much confusion, as in this older question: math.stackexchange.com/questions/1373136
$endgroup$
– Ted
May 15 at 3:07
add a comment |
$begingroup$
Both actions are correct: one is a left action and the other is a right action.
[See https://en.wikipedia.org/wiki/Group_action_(mathematics)]
For the first action: to a permutation $sigma$ and a string $x$, we associate a string $sigma cdot x$, defined by $(sigma cdot x)_sigma(i) = x_i$ ("letter at position $i$ is sent to position $sigma(i)$"), for all indices $i$ or, equivalently, $(sigma cdot x)_j = x_sigma^-1(j)$ for all indices $j$. That is a left action, since for two permutations $sigma$ and $tau$, we have
$$[sigmacdot (taucdot x))]_i = (taucdot x)_sigma^-1(i) = x_tau^-1(sigma^-1(i))= x_(sigmatau)^-1(i) = [(sigma tau)cdot x]_i,$$
hence
$$sigmacdot (taucdot x) = (sigma tau)cdot x.$$
Applying (i.e. "multiplying" by) $tau$ and then $sigma$ is the same as applying $sigmatau$. That is how multiplication to the left works, hence the term ''left action.''
For the second action: to a permutation $sigma$ and a string $x$, we associate a string $xstar sigma$, defined by $(xstar sigma)_i = x_sigma(i)$ ("$i^th$ entry of the result is the $sigma(i)^th$ entry of the original."). That is a right action, since for two permutations $sigma$ and $tau$, we have
$$[(xstar sigma)star tau]_i = [xstarsigma]_tau(i)= x_sigma(tau(i)) = x_(sigmatau)(i) = [xstar (sigmatau)]_i,$$
hence
$$(xstar sigma)star tau = xstar(sigmatau).$$
Applying (i.e. "multiplying" by) $sigma$ and then $tau$ is the same as applying $sigmatau$. That is how multiplication to the right works, hence the term ''right action.''
The two actions are indeed related by
$$(sigma cdot x)star sigma = x = sigma cdot (xstar sigma),$$
because
$$xstar sigma = sigma^-1cdot x.$$
answered May 15 at 2:22
Catalin ZaraCatalin Zara
3,982515
$endgroup$
$begingroup$
Very well explained. Failing to recognize that the two possible conventions lead to actions on opposite sides can result in much confusion, as in this older question: math.stackexchange.com/questions/1373136
$endgroup$
– Ted
May 15 at 3:07
add a comment |
$begingroup$
Both actions are correct: one is a left action and the other is a right action.
[See https://en.wikipedia.org/wiki/Group_action_(mathematics)]
For the first action: to a permutation $sigma$ and a string $x$, we associate a string $sigma cdot x$, defined by $(sigma cdot x)_sigma(i) = x_i$ ("letter at position $i$ is sent to position $sigma(i)$"), for all indices $i$ or, equivalently, $(sigma cdot x)_j = x_sigma^-1(j)$ for all indices $j$. That is a left action, since for two permutations $sigma$ and $tau$, we have
$$[sigmacdot (taucdot x))]_i = (taucdot x)_sigma^-1(i) = x_tau^-1(sigma^-1(i))= x_(sigmatau)^-1(i) = [(sigma tau)cdot x]_i,$$
hence
$$sigmacdot (taucdot x) = (sigma tau)cdot x.$$
Applying (i.e. "multiplying" by) $tau$ and then $sigma$ is the same as applying $sigmatau$. That is how multiplication to the left works, hence the term ''left action.''
For the second action: to a permutation $sigma$ and a string $x$, we associate a string $xstar sigma$, defined by $(xstar sigma)_i = x_sigma(i)$ ("$i^th$ entry of the result is the $sigma(i)^th$ entry of the original."). That is a right action, since for two permutations $sigma$ and $tau$, we have
$$[(xstar sigma)star tau]_i = [xstarsigma]_tau(i)= x_sigma(tau(i)) = x_(sigmatau)(i) = [xstar (sigmatau)]_i,$$
hence
$$(xstar sigma)star tau = xstar(sigmatau).$$
Applying (i.e. "multiplying" by) $sigma$ and then $tau$ is the same as applying $sigmatau$. That is how multiplication to the right works, hence the term ''right action.''
The two actions are indeed related by
$$(sigma cdot x)star sigma = x = sigma cdot (xstar sigma),$$
because
$$xstar sigma = sigma^-1cdot x.$$
answered May 15 at 2:22
Catalin ZaraCatalin Zara
3,982515
$endgroup$
Both actions are correct: one is a left action and the other is a right action.
[See https://en.wikipedia.org/wiki/Group_action_(mathematics)]
For the first action: to a permutation $sigma$ and a string $x$, we associate a string $sigma cdot x$, defined by $(sigma cdot x)_sigma(i) = x_i$ ("letter at position $i$ is sent to position $sigma(i)$"), for all indices $i$ or, equivalently, $(sigma cdot x)_j = x_sigma^-1(j)$ for all indices $j$. That is a left action, since for two permutations $sigma$ and $tau$, we have
$$[sigmacdot (taucdot x))]_i = (taucdot x)_sigma^-1(i) = x_tau^-1(sigma^-1(i))= x_(sigmatau)^-1(i) = [(sigma tau)cdot x]_i,$$
hence
$$sigmacdot (taucdot x) = (sigma tau)cdot x.$$
Applying (i.e. "multiplying" by) $tau$ and then $sigma$ is the same as applying $sigmatau$. That is how multiplication to the left works, hence the term ''left action.''
For the second action: to a permutation $sigma$ and a string $x$, we associate a string $xstar sigma$, defined by $(xstar sigma)_i = x_sigma(i)$ ("$i^th$ entry of the result is the $sigma(i)^th$ entry of the original."). That is a right action, since for two permutations $sigma$ and $tau$, we have
$$[(xstar sigma)star tau]_i = [xstarsigma]_tau(i)= x_sigma(tau(i)) = x_(sigmatau)(i) = [xstar (sigmatau)]_i,$$
hence
$$(xstar sigma)star tau = xstar(sigmatau).$$
Applying (i.e. "multiplying" by) $sigma$ and then $tau$ is the same as applying $sigmatau$. That is how multiplication to the right works, hence the term ''right action.''
The two actions are indeed related by
$$(sigma cdot x)star sigma = x = sigma cdot (xstar sigma),$$
because
$$xstar sigma = sigma^-1cdot x.$$
answered May 15 at 2:22
Catalin ZaraCatalin Zara
3,982515
edited May 15 at 3:01
answered May 15 at 2:22
Catalin ZaraCatalin Zara
3,982515
answered May 15 at 2:22
Catalin ZaraCatalin Zara
3,982515
answered May 15 at 2:22
Catalin ZaraCatalin Zara
3,982515
3,982515
$begingroup$
Very well explained. Failing to recognize that the two possible conventions lead to actions on opposite sides can result in much confusion, as in this older question: math.stackexchange.com/questions/1373136
$endgroup$
– Ted
May 15 at 3:07
add a comment |
$begingroup$
Very well explained. Failing to recognize that the two possible conventions lead to actions on opposite sides can result in much confusion, as in this older question: math.stackexchange.com/questions/1373136
$endgroup$
– Ted
May 15 at 3:07
$begingroup$
Very well explained. Failing to recognize that the two possible conventions lead to actions on opposite sides can result in much confusion, as in this older question: math.stackexchange.com/questions/1373136
$endgroup$
– Ted
May 15 at 3:07
$begingroup$
Very well explained. Failing to recognize that the two possible conventions lead to actions on opposite sides can result in much confusion, as in this older question: math.stackexchange.com/questions/1373136
$endgroup$
– Ted
May 15 at 3:07
add a comment |
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$begingroup$
For each $iin S$, $i$ should go to $sigma(i)$. Since the characters in the string are usually mapped to their indices, a permutation of the string is a permutation on its index set. Therefore, I would say your first way is correct.
$endgroup$
– John Douma
May 14 at 23:55
3
$begingroup$
Both are correct: one is a left action and the other is a right action.
$endgroup$
– Catalin Zara
May 15 at 0:17
$begingroup$
@CatalinZara could you expand this into an answer?
$endgroup$
– Q the Platypus
May 15 at 1:46