commutativity of cycle permutations [closed]
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How is the following true, (given two cycle permutations a and b) $(a)^{-1} (b)^{-1} = ((b)(a))^{-1}$ where b and a contain one of the same elements. isn't it only disjoint cycles that are commutative?
abstract-algebra group-theory permutation-cycles
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closed as off-topic by GNUSupporter 8964民主女神 地下教會, Derek Holt, Dietrich Burde, Namaste, Adrian Keister Dec 17 '18 at 15:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How is the following true, (given two cycle permutations a and b) $(a)^{-1} (b)^{-1} = ((b)(a))^{-1}$ where b and a contain one of the same elements. isn't it only disjoint cycles that are commutative?
abstract-algebra group-theory permutation-cycles
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closed as off-topic by GNUSupporter 8964民主女神 地下教會, Derek Holt, Dietrich Burde, Namaste, Adrian Keister Dec 17 '18 at 15:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, Derek Holt, Dietrich Burde, Namaste, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
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This is not about commutativity as commutativity is $(a)(b)=(b)(a)$ whithout the inverse signs. Here we have :$(b)(a)(a)^{-1}(b)^{-1}=(b)((a)(a)^{-1})(b)^{-1}=(b)(b)^{-1}=e$ also, $(a)^{-1}(b)^{-1}(b)(a)=(a)^{-1}((b)^{-1}(b))(a)=(a)^{-1}a=(a)=e$?
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– nafhgood
Dec 17 '18 at 13:51
add a comment |
$begingroup$
How is the following true, (given two cycle permutations a and b) $(a)^{-1} (b)^{-1} = ((b)(a))^{-1}$ where b and a contain one of the same elements. isn't it only disjoint cycles that are commutative?
abstract-algebra group-theory permutation-cycles
$endgroup$
How is the following true, (given two cycle permutations a and b) $(a)^{-1} (b)^{-1} = ((b)(a))^{-1}$ where b and a contain one of the same elements. isn't it only disjoint cycles that are commutative?
abstract-algebra group-theory permutation-cycles
abstract-algebra group-theory permutation-cycles
asked Dec 17 '18 at 13:37
user571032
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Derek Holt, Dietrich Burde, Namaste, Adrian Keister Dec 17 '18 at 15:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, Derek Holt, Dietrich Burde, Namaste, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Derek Holt, Dietrich Burde, Namaste, Adrian Keister Dec 17 '18 at 15:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, Derek Holt, Dietrich Burde, Namaste, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
This is not about commutativity as commutativity is $(a)(b)=(b)(a)$ whithout the inverse signs. Here we have :$(b)(a)(a)^{-1}(b)^{-1}=(b)((a)(a)^{-1})(b)^{-1}=(b)(b)^{-1}=e$ also, $(a)^{-1}(b)^{-1}(b)(a)=(a)^{-1}((b)^{-1}(b))(a)=(a)^{-1}a=(a)=e$?
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– nafhgood
Dec 17 '18 at 13:51
add a comment |
$begingroup$
This is not about commutativity as commutativity is $(a)(b)=(b)(a)$ whithout the inverse signs. Here we have :$(b)(a)(a)^{-1}(b)^{-1}=(b)((a)(a)^{-1})(b)^{-1}=(b)(b)^{-1}=e$ also, $(a)^{-1}(b)^{-1}(b)(a)=(a)^{-1}((b)^{-1}(b))(a)=(a)^{-1}a=(a)=e$?
$endgroup$
– nafhgood
Dec 17 '18 at 13:51
$begingroup$
This is not about commutativity as commutativity is $(a)(b)=(b)(a)$ whithout the inverse signs. Here we have :$(b)(a)(a)^{-1}(b)^{-1}=(b)((a)(a)^{-1})(b)^{-1}=(b)(b)^{-1}=e$ also, $(a)^{-1}(b)^{-1}(b)(a)=(a)^{-1}((b)^{-1}(b))(a)=(a)^{-1}a=(a)=e$?
$endgroup$
– nafhgood
Dec 17 '18 at 13:51
$begingroup$
This is not about commutativity as commutativity is $(a)(b)=(b)(a)$ whithout the inverse signs. Here we have :$(b)(a)(a)^{-1}(b)^{-1}=(b)((a)(a)^{-1})(b)^{-1}=(b)(b)^{-1}=e$ also, $(a)^{-1}(b)^{-1}(b)(a)=(a)^{-1}((b)^{-1}(b))(a)=(a)^{-1}a=(a)=e$?
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– nafhgood
Dec 17 '18 at 13:51
add a comment |
1 Answer
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Two things to note:
(1) $a^{-1}b^{-1}=(ba)^{-1}$ is not showing commutativity of two elements.
(Commutativity would be $a^{-1}b^{-1}=b^{-1}a^{-1}$.)
(2) Sometimes nondisjoint cycles commute. For instance $(12345)(13524)=(13524)(12345)$
answered Dec 17 '18 at 13:51
paw88789paw88789
29.4k12349
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Two things to note:
(1) $a^{-1}b^{-1}=(ba)^{-1}$ is not showing commutativity of two elements.
(Commutativity would be $a^{-1}b^{-1}=b^{-1}a^{-1}$.)
(2) Sometimes nondisjoint cycles commute. For instance $(12345)(13524)=(13524)(12345)$
answered Dec 17 '18 at 13:51
paw88789paw88789
29.4k12349
$endgroup$
add a comment |
$begingroup$
Two things to note:
(1) $a^{-1}b^{-1}=(ba)^{-1}$ is not showing commutativity of two elements.
(Commutativity would be $a^{-1}b^{-1}=b^{-1}a^{-1}$.)
(2) Sometimes nondisjoint cycles commute. For instance $(12345)(13524)=(13524)(12345)$
answered Dec 17 '18 at 13:51
paw88789paw88789
29.4k12349
$endgroup$
add a comment |
$begingroup$
Two things to note:
(1) $a^{-1}b^{-1}=(ba)^{-1}$ is not showing commutativity of two elements.
(Commutativity would be $a^{-1}b^{-1}=b^{-1}a^{-1}$.)
(2) Sometimes nondisjoint cycles commute. For instance $(12345)(13524)=(13524)(12345)$
answered Dec 17 '18 at 13:51
paw88789paw88789
29.4k12349
$endgroup$
Two things to note:
(1) $a^{-1}b^{-1}=(ba)^{-1}$ is not showing commutativity of two elements.
(Commutativity would be $a^{-1}b^{-1}=b^{-1}a^{-1}$.)
(2) Sometimes nondisjoint cycles commute. For instance $(12345)(13524)=(13524)(12345)$
answered Dec 17 '18 at 13:51
paw88789paw88789
29.4k12349
answered Dec 17 '18 at 13:51
paw88789paw88789
29.4k12349
answered Dec 17 '18 at 13:51
paw88789paw88789
29.4k12349
answered Dec 17 '18 at 13:51
paw88789paw88789
29.4k12349
29.4k12349
add a comment |
add a comment |
$begingroup$
This is not about commutativity as commutativity is $(a)(b)=(b)(a)$ whithout the inverse signs. Here we have :$(b)(a)(a)^{-1}(b)^{-1}=(b)((a)(a)^{-1})(b)^{-1}=(b)(b)^{-1}=e$ also, $(a)^{-1}(b)^{-1}(b)(a)=(a)^{-1}((b)^{-1}(b))(a)=(a)^{-1}a=(a)=e$?
$endgroup$
– nafhgood
Dec 17 '18 at 13:51