Mixing
When $2$-cycles (transpositions) are not disjoint, they don't necessarily commute?
$begingroup$
Consider a permutation that can be written as a product of $2$-cycles,
for instance:
$$
(1632)(457)=(12)(13)(16)(47)(45)
$$
The $2$-cycles on the right-hand side of the identity above don't necessarily commute, correct? Since they are not disjoint.
abstract-algebra group-theory permutations
$endgroup$
add a comment |
$begingroup$
Consider a permutation that can be written as a product of $2$-cycles,
for instance:
$$
(1632)(457)=(12)(13)(16)(47)(45)
$$
The $2$-cycles on the right-hand side of the identity above don't necessarily commute, correct? Since they are not disjoint.
abstract-algebra group-theory permutations
$endgroup$
6
$begingroup$
Try it out. Compute $(12)(23)$ and $(23)(12)$.
$endgroup$
– Joe Johnson 126
Feb 16 '18 at 23:16
1
$begingroup$
Non-disjoint two-cycles will never commute.
$endgroup$
– saulspatz
Feb 16 '18 at 23:22
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@saulspatz Yes, I understand the computation, but is there an example where they might commute, and if they can't how could you prove it in general?
$endgroup$
– user462561
Feb 16 '18 at 23:24
1
$begingroup$
No, they'll never commute. Look at Bernard's answer. Note that if $a=b$ then you don't have two two-cycles.
$endgroup$
– saulspatz
Feb 16 '18 at 23:31
add a comment |
$begingroup$
Consider a permutation that can be written as a product of $2$-cycles,
for instance:
$$
(1632)(457)=(12)(13)(16)(47)(45)
$$
The $2$-cycles on the right-hand side of the identity above don't necessarily commute, correct? Since they are not disjoint.
abstract-algebra group-theory permutations
$endgroup$
Consider a permutation that can be written as a product of $2$-cycles,
for instance:
$$
(1632)(457)=(12)(13)(16)(47)(45)
$$
The $2$-cycles on the right-hand side of the identity above don't necessarily commute, correct? Since they are not disjoint.
abstract-algebra group-theory permutations
abstract-algebra group-theory permutations
edited Feb 3 at 4:07
asked Feb 16 '18 at 23:15
user462561
6
$begingroup$
Try it out. Compute $(12)(23)$ and $(23)(12)$.
$endgroup$
– Joe Johnson 126
Feb 16 '18 at 23:16
1
$begingroup$
Non-disjoint two-cycles will never commute.
$endgroup$
– saulspatz
Feb 16 '18 at 23:22
$begingroup$
@saulspatz Yes, I understand the computation, but is there an example where they might commute, and if they can't how could you prove it in general?
$endgroup$
– user462561
Feb 16 '18 at 23:24
1
$begingroup$
No, they'll never commute. Look at Bernard's answer. Note that if $a=b$ then you don't have two two-cycles.
$endgroup$
– saulspatz
Feb 16 '18 at 23:31
add a comment |
6
$begingroup$
Try it out. Compute $(12)(23)$ and $(23)(12)$.
$endgroup$
– Joe Johnson 126
Feb 16 '18 at 23:16
1
$begingroup$
Non-disjoint two-cycles will never commute.
$endgroup$
– saulspatz
Feb 16 '18 at 23:22
$begingroup$
@saulspatz Yes, I understand the computation, but is there an example where they might commute, and if they can't how could you prove it in general?
$endgroup$
– user462561
Feb 16 '18 at 23:24
1
$begingroup$
No, they'll never commute. Look at Bernard's answer. Note that if $a=b$ then you don't have two two-cycles.
$endgroup$
– saulspatz
Feb 16 '18 at 23:31
6
6
$begingroup$
Try it out. Compute $(12)(23)$ and $(23)(12)$.
$endgroup$
– Joe Johnson 126
Feb 16 '18 at 23:16
$begingroup$
Try it out. Compute $(12)(23)$ and $(23)(12)$.
$endgroup$
– Joe Johnson 126
Feb 16 '18 at 23:16
1
1
$begingroup$
Non-disjoint two-cycles will never commute.
$endgroup$
– saulspatz
Feb 16 '18 at 23:22
$begingroup$
Non-disjoint two-cycles will never commute.
$endgroup$
– saulspatz
Feb 16 '18 at 23:22
$begingroup$
@saulspatz Yes, I understand the computation, but is there an example where they might commute, and if they can't how could you prove it in general?
$endgroup$
– user462561
Feb 16 '18 at 23:24
$begingroup$
@saulspatz Yes, I understand the computation, but is there an example where they might commute, and if they can't how could you prove it in general?
$endgroup$
– user462561
Feb 16 '18 at 23:24
1
1
$begingroup$
No, they'll never commute. Look at Bernard's answer. Note that if $a=b$ then you don't have two two-cycles.
$endgroup$
– saulspatz
Feb 16 '18 at 23:31
$begingroup$
No, they'll never commute. Look at Bernard's answer. Note that if $a=b$ then you don't have two two-cycles.
$endgroup$
– saulspatz
Feb 16 '18 at 23:31
add a comment |
2 Answers
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$begingroup$
$(a,b)(b,c)=(a,b,c),qquad (b,c)(a,b)=(a,c,b)=(a,b,c)^{-1}$.
answered Feb 16 '18 at 23:27
BernardBernard
124k742117
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add a comment |
$begingroup$
No, you are correct, non-disjoint two-cycles do not necessarily commute.
One example: $$(12)(23) = (123)$$
$$(23)(12) = (132)$$
We see $$(123) neq (132)$$
Hence $$(12)(23) neq (23)(12)$$
answered Feb 16 '18 at 23:27
NamasteNamaste
1
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$begingroup$
$(a,b)(b,c)=(a,b,c),qquad (b,c)(a,b)=(a,c,b)=(a,b,c)^{-1}$.
answered Feb 16 '18 at 23:27
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
$(a,b)(b,c)=(a,b,c),qquad (b,c)(a,b)=(a,c,b)=(a,b,c)^{-1}$.
answered Feb 16 '18 at 23:27
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
$(a,b)(b,c)=(a,b,c),qquad (b,c)(a,b)=(a,c,b)=(a,b,c)^{-1}$.
answered Feb 16 '18 at 23:27
BernardBernard
124k742117
$endgroup$
$(a,b)(b,c)=(a,b,c),qquad (b,c)(a,b)=(a,c,b)=(a,b,c)^{-1}$.
answered Feb 16 '18 at 23:27
BernardBernard
124k742117
answered Feb 16 '18 at 23:27
BernardBernard
124k742117
answered Feb 16 '18 at 23:27
BernardBernard
124k742117
answered Feb 16 '18 at 23:27
BernardBernard
124k742117
124k742117
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$begingroup$
No, you are correct, non-disjoint two-cycles do not necessarily commute.
One example: $$(12)(23) = (123)$$
$$(23)(12) = (132)$$
We see $$(123) neq (132)$$
Hence $$(12)(23) neq (23)(12)$$
answered Feb 16 '18 at 23:27
NamasteNamaste
1
$endgroup$
add a comment |
$begingroup$
No, you are correct, non-disjoint two-cycles do not necessarily commute.
One example: $$(12)(23) = (123)$$
$$(23)(12) = (132)$$
We see $$(123) neq (132)$$
Hence $$(12)(23) neq (23)(12)$$
answered Feb 16 '18 at 23:27
NamasteNamaste
1
$endgroup$
add a comment |
$begingroup$
No, you are correct, non-disjoint two-cycles do not necessarily commute.
One example: $$(12)(23) = (123)$$
$$(23)(12) = (132)$$
We see $$(123) neq (132)$$
Hence $$(12)(23) neq (23)(12)$$
answered Feb 16 '18 at 23:27
NamasteNamaste
1
$endgroup$
No, you are correct, non-disjoint two-cycles do not necessarily commute.
One example: $$(12)(23) = (123)$$
$$(23)(12) = (132)$$
We see $$(123) neq (132)$$
Hence $$(12)(23) neq (23)(12)$$
answered Feb 16 '18 at 23:27
NamasteNamaste
1
edited Aug 3 '18 at 20:49
answered Feb 16 '18 at 23:27
NamasteNamaste
1
answered Feb 16 '18 at 23:27
NamasteNamaste
1
answered Feb 16 '18 at 23:27
NamasteNamaste
1
1
add a comment |
add a comment |
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6
$begingroup$
Try it out. Compute $(12)(23)$ and $(23)(12)$.
$endgroup$
– Joe Johnson 126
Feb 16 '18 at 23:16
1
$begingroup$
Non-disjoint two-cycles will never commute.
$endgroup$
– saulspatz
Feb 16 '18 at 23:22
$begingroup$
@saulspatz Yes, I understand the computation, but is there an example where they might commute, and if they can't how could you prove it in general?
$endgroup$
– user462561
Feb 16 '18 at 23:24
1
$begingroup$
No, they'll never commute. Look at Bernard's answer. Note that if $a=b$ then you don't have two two-cycles.
$endgroup$
– saulspatz
Feb 16 '18 at 23:31