Mixing
Finding element of order in the symmetric group
$begingroup$
What is the proper and technical way to find if there exists an element in the symmetric group $S_n$ (Wikipedia article on symmetric groups.). I do know how to disprove if there does not exists such elements, but how should I find the element which does exists of this order?
For example, in the $S_{12}$, there is not such element of order $13$ because The only elements of order $13$ in $S_n$ are unions of disjoint $13$-cycles, since $13$ is prime. This would require $S_{12}$ to contain at least $13$ symbols, which it does not. But I think that there is an element of order $35$ in $S_{12}$. How should I find it?
EDIT: I do understand that I have to find a $7$-cycle and a $5$-cycle, but how?
group-theory permutations
edited Jan 1 at 15:09
Shaun
8,820113681
asked Jan 1 at 14:59
vesiivesii
886
$endgroup$
add a comment |
$begingroup$
What is the proper and technical way to find if there exists an element in the symmetric group $S_n$ (Wikipedia article on symmetric groups.). I do know how to disprove if there does not exists such elements, but how should I find the element which does exists of this order?
For example, in the $S_{12}$, there is not such element of order $13$ because The only elements of order $13$ in $S_n$ are unions of disjoint $13$-cycles, since $13$ is prime. This would require $S_{12}$ to contain at least $13$ symbols, which it does not. But I think that there is an element of order $35$ in $S_{12}$. How should I find it?
EDIT: I do understand that I have to find a $7$-cycle and a $5$-cycle, but how?
group-theory permutations
edited Jan 1 at 15:09
Shaun
8,820113681
asked Jan 1 at 14:59
vesiivesii
886
$endgroup$
$begingroup$
$35=5times 7$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:02
$begingroup$
First, prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. Hence, by the answer provided by @Ross, you can prove your claim.
$endgroup$
– toric_actions
Jan 1 at 15:16
add a comment |
$begingroup$
What is the proper and technical way to find if there exists an element in the symmetric group $S_n$ (Wikipedia article on symmetric groups.). I do know how to disprove if there does not exists such elements, but how should I find the element which does exists of this order?
For example, in the $S_{12}$, there is not such element of order $13$ because The only elements of order $13$ in $S_n$ are unions of disjoint $13$-cycles, since $13$ is prime. This would require $S_{12}$ to contain at least $13$ symbols, which it does not. But I think that there is an element of order $35$ in $S_{12}$. How should I find it?
EDIT: I do understand that I have to find a $7$-cycle and a $5$-cycle, but how?
group-theory permutations
edited Jan 1 at 15:09
Shaun
8,820113681
asked Jan 1 at 14:59
vesiivesii
886
$endgroup$
What is the proper and technical way to find if there exists an element in the symmetric group $S_n$ (Wikipedia article on symmetric groups.). I do know how to disprove if there does not exists such elements, but how should I find the element which does exists of this order?
For example, in the $S_{12}$, there is not such element of order $13$ because The only elements of order $13$ in $S_n$ are unions of disjoint $13$-cycles, since $13$ is prime. This would require $S_{12}$ to contain at least $13$ symbols, which it does not. But I think that there is an element of order $35$ in $S_{12}$. How should I find it?
EDIT: I do understand that I have to find a $7$-cycle and a $5$-cycle, but how?
group-theory permutations
group-theory permutations
edited Jan 1 at 15:09
Shaun
8,820113681
asked Jan 1 at 14:59
vesiivesii
886
edited Jan 1 at 15:09
Shaun
8,820113681
asked Jan 1 at 14:59
vesiivesii
886
edited Jan 1 at 15:09
Shaun
8,820113681
edited Jan 1 at 15:09
Shaun
8,820113681
edited Jan 1 at 15:09
Shaun
8,820113681
8,820113681
asked Jan 1 at 14:59
vesiivesii
886
asked Jan 1 at 14:59
vesiivesii
886
asked Jan 1 at 14:59
vesiivesii
886
886
$begingroup$
$35=5times 7$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:02
$begingroup$
First, prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. Hence, by the answer provided by @Ross, you can prove your claim.
$endgroup$
– toric_actions
Jan 1 at 15:16
add a comment |
$begingroup$
$35=5times 7$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:02
$begingroup$
First, prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. Hence, by the answer provided by @Ross, you can prove your claim.
$endgroup$
– toric_actions
Jan 1 at 15:16
$begingroup$
$35=5times 7$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:02
$begingroup$
$35=5times 7$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:02
$begingroup$
First, prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. Hence, by the answer provided by @Ross, you can prove your claim.
$endgroup$
– toric_actions
Jan 1 at 15:16
$begingroup$
First, prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. Hence, by the answer provided by @Ross, you can prove your claim.
$endgroup$
– toric_actions
Jan 1 at 15:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you want an element of order $k$, you can factor $k$ and make some cycles whose $operatorname {LCM}$ is $k$. In your example of $35$ we note that $35=5cdot 7$, so if you make a $5-$cycle and a $7-$cycle you will have an element of order $35$. As $5+7=12$ you have enough room to do this in $S_{12}$
answered Jan 1 at 15:04
Ross MillikanRoss Millikan
293k23197371
$endgroup$
$begingroup$
Thanks I get it now. How can I find the maximum order possible?
$endgroup$
– vesii
Jan 1 at 15:15
1
$begingroup$
$35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
$endgroup$
– Ross Millikan
Jan 1 at 15:26
add a comment |
$begingroup$
Hint: $35 = 5 cdot 7$ and $5+7=12$.
answered Jan 1 at 15:03
lhflhf
163k10168388
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
If you want an element of order $k$, you can factor $k$ and make some cycles whose $operatorname {LCM}$ is $k$. In your example of $35$ we note that $35=5cdot 7$, so if you make a $5-$cycle and a $7-$cycle you will have an element of order $35$. As $5+7=12$ you have enough room to do this in $S_{12}$
answered Jan 1 at 15:04
Ross MillikanRoss Millikan
293k23197371
$endgroup$
$begingroup$
Thanks I get it now. How can I find the maximum order possible?
$endgroup$
– vesii
Jan 1 at 15:15
1
$begingroup$
$35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
$endgroup$
– Ross Millikan
Jan 1 at 15:26
add a comment |
$begingroup$
If you want an element of order $k$, you can factor $k$ and make some cycles whose $operatorname {LCM}$ is $k$. In your example of $35$ we note that $35=5cdot 7$, so if you make a $5-$cycle and a $7-$cycle you will have an element of order $35$. As $5+7=12$ you have enough room to do this in $S_{12}$
answered Jan 1 at 15:04
Ross MillikanRoss Millikan
293k23197371
$endgroup$
$begingroup$
Thanks I get it now. How can I find the maximum order possible?
$endgroup$
– vesii
Jan 1 at 15:15
1
$begingroup$
$35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
$endgroup$
– Ross Millikan
Jan 1 at 15:26
add a comment |
$begingroup$
If you want an element of order $k$, you can factor $k$ and make some cycles whose $operatorname {LCM}$ is $k$. In your example of $35$ we note that $35=5cdot 7$, so if you make a $5-$cycle and a $7-$cycle you will have an element of order $35$. As $5+7=12$ you have enough room to do this in $S_{12}$
answered Jan 1 at 15:04
Ross MillikanRoss Millikan
293k23197371
$endgroup$
If you want an element of order $k$, you can factor $k$ and make some cycles whose $operatorname {LCM}$ is $k$. In your example of $35$ we note that $35=5cdot 7$, so if you make a $5-$cycle and a $7-$cycle you will have an element of order $35$. As $5+7=12$ you have enough room to do this in $S_{12}$
answered Jan 1 at 15:04
Ross MillikanRoss Millikan
293k23197371
answered Jan 1 at 15:04
Ross MillikanRoss Millikan
293k23197371
answered Jan 1 at 15:04
Ross MillikanRoss Millikan
293k23197371
answered Jan 1 at 15:04
Ross MillikanRoss Millikan
293k23197371
293k23197371
$begingroup$
Thanks I get it now. How can I find the maximum order possible?
$endgroup$
– vesii
Jan 1 at 15:15
1
$begingroup$
$35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
$endgroup$
– Ross Millikan
Jan 1 at 15:26
add a comment |
$begingroup$
Thanks I get it now. How can I find the maximum order possible?
$endgroup$
– vesii
Jan 1 at 15:15
1
$begingroup$
$35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
$endgroup$
– Ross Millikan
Jan 1 at 15:26
$begingroup$
Thanks I get it now. How can I find the maximum order possible?
$endgroup$
– vesii
Jan 1 at 15:15
$begingroup$
Thanks I get it now. How can I find the maximum order possible?
$endgroup$
– vesii
Jan 1 at 15:15
1
1
$begingroup$
$35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
$endgroup$
– Ross Millikan
Jan 1 at 15:26
$begingroup$
$35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in oeis.org/A000793
$endgroup$
– Ross Millikan
Jan 1 at 15:26
add a comment |
$begingroup$
Hint: $35 = 5 cdot 7$ and $5+7=12$.
answered Jan 1 at 15:03
lhflhf
163k10168388
$endgroup$
add a comment |
$begingroup$
Hint: $35 = 5 cdot 7$ and $5+7=12$.
answered Jan 1 at 15:03
lhflhf
163k10168388
$endgroup$
add a comment |
$begingroup$
Hint: $35 = 5 cdot 7$ and $5+7=12$.
answered Jan 1 at 15:03
lhflhf
163k10168388
$endgroup$
Hint: $35 = 5 cdot 7$ and $5+7=12$.
answered Jan 1 at 15:03
lhflhf
163k10168388
answered Jan 1 at 15:03
lhflhf
163k10168388
answered Jan 1 at 15:03
lhflhf
163k10168388
answered Jan 1 at 15:03
lhflhf
163k10168388
163k10168388
add a comment |
add a comment |
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$begingroup$
$35=5times 7$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:02
$begingroup$
First, prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. Hence, by the answer provided by @Ross, you can prove your claim.
$endgroup$
– toric_actions
Jan 1 at 15:16