Proving the sign map is a homomorphism
$begingroup$
Definition: A transposition is a 2-cycle permutation.
Definition: A permutation $sigma$ is called even if its decomposition into transpositions has even number of transpositions; analogously for odd permutations.
Now, coming to the question, how do I prove using this definition that the sign map $operatorname{sgn}colon S_nto{pm 1}$ assigning a permutation in the symmetric group $S_n$ to its sign (+1 if even, -1 if odd) is a homomorphism?
I know how to prove it using the definition of the sign map being the determinant of the $ntimes n$ permutation matrix associated to a permutation (in which case it is obvious since $operatorname{det}$ is multiplicative).
group-theory permutation-cycles
asked Dec 26 '18 at 17:21
learnerlearner
689416
$endgroup$
add a comment |
$begingroup$
Definition: A transposition is a 2-cycle permutation.
Definition: A permutation $sigma$ is called even if its decomposition into transpositions has even number of transpositions; analogously for odd permutations.
Now, coming to the question, how do I prove using this definition that the sign map $operatorname{sgn}colon S_nto{pm 1}$ assigning a permutation in the symmetric group $S_n$ to its sign (+1 if even, -1 if odd) is a homomorphism?
I know how to prove it using the definition of the sign map being the determinant of the $ntimes n$ permutation matrix associated to a permutation (in which case it is obvious since $operatorname{det}$ is multiplicative).
group-theory permutation-cycles
asked Dec 26 '18 at 17:21
learnerlearner
689416
$endgroup$
add a comment |
$begingroup$
Definition: A transposition is a 2-cycle permutation.
Definition: A permutation $sigma$ is called even if its decomposition into transpositions has even number of transpositions; analogously for odd permutations.
Now, coming to the question, how do I prove using this definition that the sign map $operatorname{sgn}colon S_nto{pm 1}$ assigning a permutation in the symmetric group $S_n$ to its sign (+1 if even, -1 if odd) is a homomorphism?
I know how to prove it using the definition of the sign map being the determinant of the $ntimes n$ permutation matrix associated to a permutation (in which case it is obvious since $operatorname{det}$ is multiplicative).
group-theory permutation-cycles
asked Dec 26 '18 at 17:21
learnerlearner
689416
$endgroup$
Definition: A transposition is a 2-cycle permutation.
Definition: A permutation $sigma$ is called even if its decomposition into transpositions has even number of transpositions; analogously for odd permutations.
Now, coming to the question, how do I prove using this definition that the sign map $operatorname{sgn}colon S_nto{pm 1}$ assigning a permutation in the symmetric group $S_n$ to its sign (+1 if even, -1 if odd) is a homomorphism?
I know how to prove it using the definition of the sign map being the determinant of the $ntimes n$ permutation matrix associated to a permutation (in which case it is obvious since $operatorname{det}$ is multiplicative).
group-theory permutation-cycles
group-theory permutation-cycles
asked Dec 26 '18 at 17:21
learnerlearner
689416
asked Dec 26 '18 at 17:21
learnerlearner
689416
asked Dec 26 '18 at 17:21
learnerlearner
689416
asked Dec 26 '18 at 17:21
learnerlearner
689416
asked Dec 26 '18 at 17:21
learnerlearner
689416
689416
add a comment |
add a comment |
1 Answer
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$begingroup$
If it enough to know that the parity of the number of transpositions which appear in a given decomposition does not depend of the specific decomposition.
Then if $sigma=t_1...t_n, sign(sigma)=(-1)^n$ this is well defined since the parity of $n$ does not depend of the composition. If $sigma'=t'_1....t'_{n'}, sign(sigma')=(-1)^{n'}$ and as $sigmasigma'=t_1....t_nt'_1...t'_{n'}, sign(sigmasigma')=(-1)^{n+n'}$.
answered Dec 26 '18 at 17:26
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
$begingroup$
I know that; I don't see how that answers my question.
$endgroup$
– learner
Dec 26 '18 at 17:28
$begingroup$
you see now how it answer ?
$endgroup$
– Tsemo Aristide
Dec 26 '18 at 17:30
$begingroup$
Yes, I do. And I feel quite silly that I didn't think of it myself. Thanks.
$endgroup$
– learner
Dec 26 '18 at 17:32
add a comment |
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1 Answer
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active
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1 Answer
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active
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active
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$begingroup$
If it enough to know that the parity of the number of transpositions which appear in a given decomposition does not depend of the specific decomposition.
Then if $sigma=t_1...t_n, sign(sigma)=(-1)^n$ this is well defined since the parity of $n$ does not depend of the composition. If $sigma'=t'_1....t'_{n'}, sign(sigma')=(-1)^{n'}$ and as $sigmasigma'=t_1....t_nt'_1...t'_{n'}, sign(sigmasigma')=(-1)^{n+n'}$.
answered Dec 26 '18 at 17:26
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
$begingroup$
I know that; I don't see how that answers my question.
$endgroup$
– learner
Dec 26 '18 at 17:28
$begingroup$
you see now how it answer ?
$endgroup$
– Tsemo Aristide
Dec 26 '18 at 17:30
$begingroup$
Yes, I do. And I feel quite silly that I didn't think of it myself. Thanks.
$endgroup$
– learner
Dec 26 '18 at 17:32
add a comment |
$begingroup$
If it enough to know that the parity of the number of transpositions which appear in a given decomposition does not depend of the specific decomposition.
Then if $sigma=t_1...t_n, sign(sigma)=(-1)^n$ this is well defined since the parity of $n$ does not depend of the composition. If $sigma'=t'_1....t'_{n'}, sign(sigma')=(-1)^{n'}$ and as $sigmasigma'=t_1....t_nt'_1...t'_{n'}, sign(sigmasigma')=(-1)^{n+n'}$.
answered Dec 26 '18 at 17:26
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
$begingroup$
I know that; I don't see how that answers my question.
$endgroup$
– learner
Dec 26 '18 at 17:28
$begingroup$
you see now how it answer ?
$endgroup$
– Tsemo Aristide
Dec 26 '18 at 17:30
$begingroup$
Yes, I do. And I feel quite silly that I didn't think of it myself. Thanks.
$endgroup$
– learner
Dec 26 '18 at 17:32
add a comment |
$begingroup$
If it enough to know that the parity of the number of transpositions which appear in a given decomposition does not depend of the specific decomposition.
Then if $sigma=t_1...t_n, sign(sigma)=(-1)^n$ this is well defined since the parity of $n$ does not depend of the composition. If $sigma'=t'_1....t'_{n'}, sign(sigma')=(-1)^{n'}$ and as $sigmasigma'=t_1....t_nt'_1...t'_{n'}, sign(sigmasigma')=(-1)^{n+n'}$.
answered Dec 26 '18 at 17:26
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
If it enough to know that the parity of the number of transpositions which appear in a given decomposition does not depend of the specific decomposition.
Then if $sigma=t_1...t_n, sign(sigma)=(-1)^n$ this is well defined since the parity of $n$ does not depend of the composition. If $sigma'=t'_1....t'_{n'}, sign(sigma')=(-1)^{n'}$ and as $sigmasigma'=t_1....t_nt'_1...t'_{n'}, sign(sigmasigma')=(-1)^{n+n'}$.
answered Dec 26 '18 at 17:26
Tsemo AristideTsemo Aristide
58.1k11445
edited Dec 26 '18 at 17:29
answered Dec 26 '18 at 17:26
Tsemo AristideTsemo Aristide
58.1k11445
answered Dec 26 '18 at 17:26
Tsemo AristideTsemo Aristide
58.1k11445
answered Dec 26 '18 at 17:26
Tsemo AristideTsemo Aristide
58.1k11445
58.1k11445
$begingroup$
I know that; I don't see how that answers my question.
$endgroup$
– learner
Dec 26 '18 at 17:28
$begingroup$
you see now how it answer ?
$endgroup$
– Tsemo Aristide
Dec 26 '18 at 17:30
$begingroup$
Yes, I do. And I feel quite silly that I didn't think of it myself. Thanks.
$endgroup$
– learner
Dec 26 '18 at 17:32
add a comment |
$begingroup$
I know that; I don't see how that answers my question.
$endgroup$
– learner
Dec 26 '18 at 17:28
$begingroup$
you see now how it answer ?
$endgroup$
– Tsemo Aristide
Dec 26 '18 at 17:30
$begingroup$
Yes, I do. And I feel quite silly that I didn't think of it myself. Thanks.
$endgroup$
– learner
Dec 26 '18 at 17:32
$begingroup$
I know that; I don't see how that answers my question.
$endgroup$
– learner
Dec 26 '18 at 17:28
$begingroup$
I know that; I don't see how that answers my question.
$endgroup$
– learner
Dec 26 '18 at 17:28
$begingroup$
you see now how it answer ?
$endgroup$
– Tsemo Aristide
Dec 26 '18 at 17:30
$begingroup$
you see now how it answer ?
$endgroup$
– Tsemo Aristide
Dec 26 '18 at 17:30
$begingroup$
Yes, I do. And I feel quite silly that I didn't think of it myself. Thanks.
$endgroup$
– learner
Dec 26 '18 at 17:32
$begingroup$
Yes, I do. And I feel quite silly that I didn't think of it myself. Thanks.
$endgroup$
– learner
Dec 26 '18 at 17:32
add a comment |
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