Elements of prime order in symmetric groups
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I have been assigned the following task:
“Explain briefly why there can be no odd elements of prime order except for order $2$ in any symmetric group $S_n$.”
Unless I am largely mistaken in my understanding of the notions of order and the sign of a permutation, I have come up with the following:
$[3,3,3,1]$ is an element of $S_{10}$ with order $3$, and is even.
$[7, 1, 1]$ is an element of $S_9$ with order $7$, and is odd.
Am I mistaken in thinking that this shows that not only do there exist odd elements of prime order $p≠2$, but also that there exist even elements of order $2$ (the former contradicting the statement in the question)?
abstract-algebra group-theory
asked Nov 29 '18 at 18:50
TomHanneyTomHanney
174
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add a comment |
$begingroup$
I have been assigned the following task:
“Explain briefly why there can be no odd elements of prime order except for order $2$ in any symmetric group $S_n$.”
Unless I am largely mistaken in my understanding of the notions of order and the sign of a permutation, I have come up with the following:
$[3,3,3,1]$ is an element of $S_{10}$ with order $3$, and is even.
$[7, 1, 1]$ is an element of $S_9$ with order $7$, and is odd.
Am I mistaken in thinking that this shows that not only do there exist odd elements of prime order $p≠2$, but also that there exist even elements of order $2$ (the former contradicting the statement in the question)?
abstract-algebra group-theory
asked Nov 29 '18 at 18:50
TomHanneyTomHanney
174
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What is this notation: $[3,3,3,1]$?
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– Berci
Nov 29 '18 at 19:04
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Anyway, it's odd, but a cycle of odd length is an even permutation..
$endgroup$
– Berci
Nov 29 '18 at 19:07
1
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@Berci It's the lengths of the disjoint cycles.
$endgroup$
– verret
Nov 29 '18 at 20:11
add a comment |
$begingroup$
I have been assigned the following task:
“Explain briefly why there can be no odd elements of prime order except for order $2$ in any symmetric group $S_n$.”
Unless I am largely mistaken in my understanding of the notions of order and the sign of a permutation, I have come up with the following:
$[3,3,3,1]$ is an element of $S_{10}$ with order $3$, and is even.
$[7, 1, 1]$ is an element of $S_9$ with order $7$, and is odd.
Am I mistaken in thinking that this shows that not only do there exist odd elements of prime order $p≠2$, but also that there exist even elements of order $2$ (the former contradicting the statement in the question)?
abstract-algebra group-theory
asked Nov 29 '18 at 18:50
TomHanneyTomHanney
174
$endgroup$
I have been assigned the following task:
“Explain briefly why there can be no odd elements of prime order except for order $2$ in any symmetric group $S_n$.”
Unless I am largely mistaken in my understanding of the notions of order and the sign of a permutation, I have come up with the following:
$[3,3,3,1]$ is an element of $S_{10}$ with order $3$, and is even.
$[7, 1, 1]$ is an element of $S_9$ with order $7$, and is odd.
Am I mistaken in thinking that this shows that not only do there exist odd elements of prime order $p≠2$, but also that there exist even elements of order $2$ (the former contradicting the statement in the question)?
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 29 '18 at 18:50
TomHanneyTomHanney
174
asked Nov 29 '18 at 18:50
TomHanneyTomHanney
174
asked Nov 29 '18 at 18:50
TomHanneyTomHanney
174
asked Nov 29 '18 at 18:50
TomHanneyTomHanney
174
asked Nov 29 '18 at 18:50
TomHanneyTomHanney
174
174
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What is this notation: $[3,3,3,1]$?
$endgroup$
– Berci
Nov 29 '18 at 19:04
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Anyway, it's odd, but a cycle of odd length is an even permutation..
$endgroup$
– Berci
Nov 29 '18 at 19:07
1
$begingroup$
@Berci It's the lengths of the disjoint cycles.
$endgroup$
– verret
Nov 29 '18 at 20:11
add a comment |
$begingroup$
What is this notation: $[3,3,3,1]$?
$endgroup$
– Berci
Nov 29 '18 at 19:04
$begingroup$
Anyway, it's odd, but a cycle of odd length is an even permutation..
$endgroup$
– Berci
Nov 29 '18 at 19:07
1
$begingroup$
@Berci It's the lengths of the disjoint cycles.
$endgroup$
– verret
Nov 29 '18 at 20:11
$begingroup$
What is this notation: $[3,3,3,1]$?
$endgroup$
– Berci
Nov 29 '18 at 19:04
$begingroup$
What is this notation: $[3,3,3,1]$?
$endgroup$
– Berci
Nov 29 '18 at 19:04
$begingroup$
Anyway, it's odd, but a cycle of odd length is an even permutation..
$endgroup$
– Berci
Nov 29 '18 at 19:07
$begingroup$
Anyway, it's odd, but a cycle of odd length is an even permutation..
$endgroup$
– Berci
Nov 29 '18 at 19:07
1
1
$begingroup$
@Berci It's the lengths of the disjoint cycles.
$endgroup$
– verret
Nov 29 '18 at 20:11
$begingroup$
@Berci It's the lengths of the disjoint cycles.
$endgroup$
– verret
Nov 29 '18 at 20:11
add a comment |
1 Answer
1
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A $7$-cycle is even because we can break it into transpositions $$(a_1,a_7)(a_1,a_6)(a_1,a_5)(a_1,a_4)(a_1,a_3)(a_1,a_2)$$
And as you can see there are six transpositions, so this permutation is even.
answered Nov 29 '18 at 20:39
CyclotomicFieldCyclotomicField
2,4081314
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Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
$endgroup$
– TomHanney
Nov 29 '18 at 22:09
add a comment |
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$begingroup$
A $7$-cycle is even because we can break it into transpositions $$(a_1,a_7)(a_1,a_6)(a_1,a_5)(a_1,a_4)(a_1,a_3)(a_1,a_2)$$
And as you can see there are six transpositions, so this permutation is even.
answered Nov 29 '18 at 20:39
CyclotomicFieldCyclotomicField
2,4081314
$endgroup$
$begingroup$
Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
$endgroup$
– TomHanney
Nov 29 '18 at 22:09
add a comment |
$begingroup$
A $7$-cycle is even because we can break it into transpositions $$(a_1,a_7)(a_1,a_6)(a_1,a_5)(a_1,a_4)(a_1,a_3)(a_1,a_2)$$
And as you can see there are six transpositions, so this permutation is even.
answered Nov 29 '18 at 20:39
CyclotomicFieldCyclotomicField
2,4081314
$endgroup$
$begingroup$
Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
$endgroup$
– TomHanney
Nov 29 '18 at 22:09
add a comment |
$begingroup$
A $7$-cycle is even because we can break it into transpositions $$(a_1,a_7)(a_1,a_6)(a_1,a_5)(a_1,a_4)(a_1,a_3)(a_1,a_2)$$
And as you can see there are six transpositions, so this permutation is even.
answered Nov 29 '18 at 20:39
CyclotomicFieldCyclotomicField
2,4081314
$endgroup$
A $7$-cycle is even because we can break it into transpositions $$(a_1,a_7)(a_1,a_6)(a_1,a_5)(a_1,a_4)(a_1,a_3)(a_1,a_2)$$
And as you can see there are six transpositions, so this permutation is even.
answered Nov 29 '18 at 20:39
CyclotomicFieldCyclotomicField
2,4081314
answered Nov 29 '18 at 20:39
CyclotomicFieldCyclotomicField
2,4081314
answered Nov 29 '18 at 20:39
CyclotomicFieldCyclotomicField
2,4081314
answered Nov 29 '18 at 20:39
CyclotomicFieldCyclotomicField
2,4081314
2,4081314
$begingroup$
Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
$endgroup$
– TomHanney
Nov 29 '18 at 22:09
add a comment |
$begingroup$
Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
$endgroup$
– TomHanney
Nov 29 '18 at 22:09
$begingroup$
Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
$endgroup$
– TomHanney
Nov 29 '18 at 22:09
$begingroup$
Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
$endgroup$
– TomHanney
Nov 29 '18 at 22:09
add a comment |
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$begingroup$
What is this notation: $[3,3,3,1]$?
$endgroup$
– Berci
Nov 29 '18 at 19:04
$begingroup$
Anyway, it's odd, but a cycle of odd length is an even permutation..
$endgroup$
– Berci
Nov 29 '18 at 19:07
1
$begingroup$
@Berci It's the lengths of the disjoint cycles.
$endgroup$
– verret
Nov 29 '18 at 20:11