How to represent a product of cycles in matrix form?

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I have a permutation a in a product of disjoint cycles form as follows

$a = (1,9,3,7)(2,11,6)(4,8,5,10)$

I want to represent it in a matrix form A such that

$A = beginpmatrix1&2&3&4&5&6&7&8&9&10&11\9&11&7&8&10&2&1&5&3&4&6
endpmatrix$

I believe a can be defined in Mathematica as

a = Cycles[1, 9, 3, 7, 2, 11, 6, 4, 8, 5, 10]

How do I convert a to A?

asked 51 mins ago

Heisenberg

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up vote
3
down vote

favorite

I have a permutation a in a product of disjoint cycles form as follows

$a = (1,9,3,7)(2,11,6)(4,8,5,10)$

I want to represent it in a matrix form A such that

$A = beginpmatrix1&2&3&4&5&6&7&8&9&10&11\9&11&7&8&10&2&1&5&3&4&6
endpmatrix$

I believe a can be defined in Mathematica as

a = Cycles[1, 9, 3, 7, 2, 11, 6, 4, 8, 5, 10]

How do I convert a to A?

asked 51 mins ago

Heisenberg

1162

New contributor

add a commentÂ |Â

up vote
3
down vote

favorite

I have a permutation a in a product of disjoint cycles form as follows

$a = (1,9,3,7)(2,11,6)(4,8,5,10)$

I want to represent it in a matrix form A such that

$A = beginpmatrix1&2&3&4&5&6&7&8&9&10&11\9&11&7&8&10&2&1&5&3&4&6
endpmatrix$

I believe a can be defined in Mathematica as

a = Cycles[1, 9, 3, 7, 2, 11, 6, 4, 8, 5, 10]

How do I convert a to A?

asked 51 mins ago

Heisenberg

1162

New contributor

I have a permutation a in a product of disjoint cycles form as follows

$a = (1,9,3,7)(2,11,6)(4,8,5,10)$

I want to represent it in a matrix form A such that

$A = beginpmatrix1&2&3&4&5&6&7&8&9&10&11\9&11&7&8&10&2&1&5&3&4&6
endpmatrix$

I believe a can be defined in Mathematica as

a = Cycles[1, 9, 3, 7, 2, 11, 6, 4, 8, 5, 10]

How do I convert a to A?

permutation group-theory products

asked 51 mins ago

Heisenberg

1162

New contributor

asked 51 mins ago

Heisenberg

1162

New contributor

asked 51 mins ago

Heisenberg

1162

New contributor

asked 51 mins ago

Heisenberg

1162

asked 51 mins ago

Heisenberg

1162

New contributor

Heisenberg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2 Answers
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up vote
3
down vote

mat = Sort @ #, # & @ PermutationList[a];
MatrixForm @ mat // TeXForm

$ left(
beginarrayccccccccccc
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \
9 & 11 & 7 & 8 & 10 & 2 & 1 & 5 & 3 & 4 & 6 \
endarray
right)$

answered 44 mins ago

kglr

162k8185385

add a commentÂ |Â

up vote
2
down vote

One idea is to overload MatrixForm so that it does this for you automatically:

Unprotect[MatrixForm];
MatrixForm /: MakeBoxes[MatrixForm[cyc_Cycles], StandardForm] := With[
 list=PermutationList[cyc],
 ToBoxes[MatrixForm[Range@Length@list, list], StandardForm]
]
Protect[MatrixForm];

Then:

Cycles[1, 9, 3, 7, 2, 11, 6, 4, 8, 5, 10] //MatrixForm

enter image description here

answered 32 mins ago

Carl Woll

58.4k275150

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
3
down vote

mat = Sort @ #, # & @ PermutationList[a];
MatrixForm @ mat // TeXForm

$ left(
beginarrayccccccccccc
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \
9 & 11 & 7 & 8 & 10 & 2 & 1 & 5 & 3 & 4 & 6 \
endarray
right)$

answered 44 mins ago

kglr

162k8185385

add a commentÂ |Â

up vote
3
down vote

mat = Sort @ #, # & @ PermutationList[a];
MatrixForm @ mat // TeXForm

$ left(
beginarrayccccccccccc
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \
9 & 11 & 7 & 8 & 10 & 2 & 1 & 5 & 3 & 4 & 6 \
endarray
right)$

answered 44 mins ago

kglr

162k8185385

add a commentÂ |Â

up vote
3
down vote

mat = Sort @ #, # & @ PermutationList[a];
MatrixForm @ mat // TeXForm

$ left(
beginarrayccccccccccc
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \
9 & 11 & 7 & 8 & 10 & 2 & 1 & 5 & 3 & 4 & 6 \
endarray
right)$

answered 44 mins ago

kglr

162k8185385

mat = Sort @ #, # & @ PermutationList[a];
MatrixForm @ mat // TeXForm

$ left(
beginarrayccccccccccc
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \
9 & 11 & 7 & 8 & 10 & 2 & 1 & 5 & 3 & 4 & 6 \
endarray
right)$

answered 44 mins ago

kglr

162k8185385

answered 44 mins ago

kglr

162k8185385

answered 44 mins ago

kglr

162k8185385

answered 44 mins ago

kglr

162k8185385

add a commentÂ |Â

up vote
2
down vote

One idea is to overload MatrixForm so that it does this for you automatically:

Unprotect[MatrixForm];
MatrixForm /: MakeBoxes[MatrixForm[cyc_Cycles], StandardForm] := With[
 list=PermutationList[cyc],
 ToBoxes[MatrixForm[Range@Length@list, list], StandardForm]
]
Protect[MatrixForm];

Then:

Cycles[1, 9, 3, 7, 2, 11, 6, 4, 8, 5, 10] //MatrixForm

enter image description here

answered 32 mins ago

Carl Woll

58.4k275150

add a commentÂ |Â

up vote
2
down vote

One idea is to overload MatrixForm so that it does this for you automatically:

Unprotect[MatrixForm];
MatrixForm /: MakeBoxes[MatrixForm[cyc_Cycles], StandardForm] := With[
 list=PermutationList[cyc],
 ToBoxes[MatrixForm[Range@Length@list, list], StandardForm]
]
Protect[MatrixForm];

Then:

Cycles[1, 9, 3, 7, 2, 11, 6, 4, 8, 5, 10] //MatrixForm

enter image description here

answered 32 mins ago

Carl Woll

58.4k275150

add a commentÂ |Â

up vote
2
down vote

One idea is to overload MatrixForm so that it does this for you automatically:

Unprotect[MatrixForm];
MatrixForm /: MakeBoxes[MatrixForm[cyc_Cycles], StandardForm] := With[
 list=PermutationList[cyc],
 ToBoxes[MatrixForm[Range@Length@list, list], StandardForm]
]
Protect[MatrixForm];

Then:

Cycles[1, 9, 3, 7, 2, 11, 6, 4, 8, 5, 10] //MatrixForm

enter image description here

answered 32 mins ago

Carl Woll

58.4k275150

One idea is to overload MatrixForm so that it does this for you automatically:

Unprotect[MatrixForm];
MatrixForm /: MakeBoxes[MatrixForm[cyc_Cycles], StandardForm] := With[
 list=PermutationList[cyc],
 ToBoxes[MatrixForm[Range@Length@list, list], StandardForm]
]
Protect[MatrixForm];

Then:

Cycles[1, 9, 3, 7, 2, 11, 6, 4, 8, 5, 10] //MatrixForm

enter image description here

answered 32 mins ago

Carl Woll

58.4k275150

answered 32 mins ago

Carl Woll

58.4k275150

answered 32 mins ago

Carl Woll

58.4k275150

answered 32 mins ago

Carl Woll

58.4k275150

add a commentÂ |Â

Heisenberg is a new contributor. Be nice, and check out our Code of Conduct.

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