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What does order of element mean in the Symmetric Group?
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I have seen several references to "order" of an element in the Symmetric Group. Specifically, that the order of a cycle is the least common multiple of the lengths of the cycles in its decomposition.
But the Symmetric Group is not cyclic, and I'm only familiar with the concept of "order" for cyclic groups. So what does it mean in this context?
abstract-algebra group-theory symmetric-groups
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Clotex
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up vote
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I have seen several references to "order" of an element in the Symmetric Group. Specifically, that the order of a cycle is the least common multiple of the lengths of the cycles in its decomposition.
But the Symmetric Group is not cyclic, and I'm only familiar with the concept of "order" for cyclic groups. So what does it mean in this context?
abstract-algebra group-theory symmetric-groups
asked 1 hour ago
Clotex
111
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Clotex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have seen several references to "order" of an element in the Symmetric Group. Specifically, that the order of a cycle is the least common multiple of the lengths of the cycles in its decomposition.
But the Symmetric Group is not cyclic, and I'm only familiar with the concept of "order" for cyclic groups. So what does it mean in this context?
abstract-algebra group-theory symmetric-groups
asked 1 hour ago
Clotex
111
New contributor
Clotex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have seen several references to "order" of an element in the Symmetric Group. Specifically, that the order of a cycle is the least common multiple of the lengths of the cycles in its decomposition.
But the Symmetric Group is not cyclic, and I'm only familiar with the concept of "order" for cyclic groups. So what does it mean in this context?
abstract-algebra group-theory symmetric-groups
abstract-algebra group-theory symmetric-groups
asked 1 hour ago
Clotex
111
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Clotex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 1 hour ago
Clotex
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asked 1 hour ago
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Clotex
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asked 1 hour ago
Clotex
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2 Answers
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The order of an element $g$ in a finite group $G$ is the smallest integer $n$ such that $g^n = e$ (the neutral element of the group). This is well-defined for every finite group $G$, so in particular for the Symmetric group as well.
The Lagrange theorem implies indeed, that for any finite group $G$ with $p$ elements and any $g in G$, then $g^p = e$, which shows that every $g$ has finite order, smaller than $p$.
answered 1 hour ago
seamp
1965
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Since you know about cyclic groups.
Think of order of an element in the symmetric group $S_n$ as the size of the cyclic group generated by it.
Example 1: Let us take $S_3$. Take $g=(123)$. Consider the cyclic group generated by $g$, that is $<g>$. $<g>=(123),(132),(1)$. So order of $g$ is 3 since cardinality of $<g>$ is 3.
Example 2: Consider $S_4$ and take $g=(12)(34)$ Then $<g>=(12)(34),(1)$. Since cardinality of $<g>$ is 2, order of $g$ is $2$.
Example 3(Take it as an exercise): Consider $S_5$ and take $g=(123)(45)$. Find $<g>$ by explicitly writing each element and then calculate the order from $g$ from there
answered 1 hour ago
StammeringMathematician
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
The order of an element $g$ in a finite group $G$ is the smallest integer $n$ such that $g^n = e$ (the neutral element of the group). This is well-defined for every finite group $G$, so in particular for the Symmetric group as well.
The Lagrange theorem implies indeed, that for any finite group $G$ with $p$ elements and any $g in G$, then $g^p = e$, which shows that every $g$ has finite order, smaller than $p$.
answered 1 hour ago
seamp
1965
New contributor
seamp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
4
down vote
The order of an element $g$ in a finite group $G$ is the smallest integer $n$ such that $g^n = e$ (the neutral element of the group). This is well-defined for every finite group $G$, so in particular for the Symmetric group as well.
The Lagrange theorem implies indeed, that for any finite group $G$ with $p$ elements and any $g in G$, then $g^p = e$, which shows that every $g$ has finite order, smaller than $p$.
answered 1 hour ago
seamp
1965
New contributor
seamp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The order of an element $g$ in a finite group $G$ is the smallest integer $n$ such that $g^n = e$ (the neutral element of the group). This is well-defined for every finite group $G$, so in particular for the Symmetric group as well.
The Lagrange theorem implies indeed, that for any finite group $G$ with $p$ elements and any $g in G$, then $g^p = e$, which shows that every $g$ has finite order, smaller than $p$.
answered 1 hour ago
seamp
1965
New contributor
seamp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The order of an element $g$ in a finite group $G$ is the smallest integer $n$ such that $g^n = e$ (the neutral element of the group). This is well-defined for every finite group $G$, so in particular for the Symmetric group as well.
The Lagrange theorem implies indeed, that for any finite group $G$ with $p$ elements and any $g in G$, then $g^p = e$, which shows that every $g$ has finite order, smaller than $p$.
answered 1 hour ago
seamp
1965
New contributor
seamp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
seamp
1965
New contributor
seamp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
seamp
1965
answered 1 hour ago
seamp
1965
1965
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seamp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
seamp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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seamp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
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up vote
3
down vote
Since you know about cyclic groups.
Think of order of an element in the symmetric group $S_n$ as the size of the cyclic group generated by it.
Example 1: Let us take $S_3$. Take $g=(123)$. Consider the cyclic group generated by $g$, that is $<g>$. $<g>=(123),(132),(1)$. So order of $g$ is 3 since cardinality of $<g>$ is 3.
Example 2: Consider $S_4$ and take $g=(12)(34)$ Then $<g>=(12)(34),(1)$. Since cardinality of $<g>$ is 2, order of $g$ is $2$.
Example 3(Take it as an exercise): Consider $S_5$ and take $g=(123)(45)$. Find $<g>$ by explicitly writing each element and then calculate the order from $g$ from there
answered 1 hour ago
StammeringMathematician
2,0591222
add a comment |Â
up vote
3
down vote
Since you know about cyclic groups.
Think of order of an element in the symmetric group $S_n$ as the size of the cyclic group generated by it.
Example 1: Let us take $S_3$. Take $g=(123)$. Consider the cyclic group generated by $g$, that is $<g>$. $<g>=(123),(132),(1)$. So order of $g$ is 3 since cardinality of $<g>$ is 3.
Example 2: Consider $S_4$ and take $g=(12)(34)$ Then $<g>=(12)(34),(1)$. Since cardinality of $<g>$ is 2, order of $g$ is $2$.
Example 3(Take it as an exercise): Consider $S_5$ and take $g=(123)(45)$. Find $<g>$ by explicitly writing each element and then calculate the order from $g$ from there
answered 1 hour ago
StammeringMathematician
2,0591222
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Since you know about cyclic groups.
Think of order of an element in the symmetric group $S_n$ as the size of the cyclic group generated by it.
Example 1: Let us take $S_3$. Take $g=(123)$. Consider the cyclic group generated by $g$, that is $<g>$. $<g>=(123),(132),(1)$. So order of $g$ is 3 since cardinality of $<g>$ is 3.
Example 2: Consider $S_4$ and take $g=(12)(34)$ Then $<g>=(12)(34),(1)$. Since cardinality of $<g>$ is 2, order of $g$ is $2$.
Example 3(Take it as an exercise): Consider $S_5$ and take $g=(123)(45)$. Find $<g>$ by explicitly writing each element and then calculate the order from $g$ from there
answered 1 hour ago
StammeringMathematician
2,0591222
Since you know about cyclic groups.
Think of order of an element in the symmetric group $S_n$ as the size of the cyclic group generated by it.
Example 1: Let us take $S_3$. Take $g=(123)$. Consider the cyclic group generated by $g$, that is $<g>$. $<g>=(123),(132),(1)$. So order of $g$ is 3 since cardinality of $<g>$ is 3.
Example 2: Consider $S_4$ and take $g=(12)(34)$ Then $<g>=(12)(34),(1)$. Since cardinality of $<g>$ is 2, order of $g$ is $2$.
Example 3(Take it as an exercise): Consider $S_5$ and take $g=(123)(45)$. Find $<g>$ by explicitly writing each element and then calculate the order from $g$ from there
answered 1 hour ago
StammeringMathematician
2,0591222
edited 1 hour ago
answered 1 hour ago
StammeringMathematician
2,0591222
answered 1 hour ago
StammeringMathematician
2,0591222
answered 1 hour ago
StammeringMathematician
2,0591222
2,0591222
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