About the Proof for “the Order of a Permutation $sigma$ is the $lcm$ of the Orders of its Disjoint...
$begingroup$
About the lemma below:
The order of a permutation $ sigma in S_n $ is the least common multiple of the orders of its disjoint cycles.
I didn't get the proof which consists of the following reasoning:
Let $ sigma = sigma_1 ... sigma_r $ with $ r le n $ and $ { sigma_i }_{1 le i le r} $ disjoint cycles in question. Since every $ sigma_i $ commute in this case, we have: $ sigma^m = sigma_{1}^{m} ... sigma_{r}^{m} $, for all $ m in mathbb{Z} $, and $ sigma^m = (1) $ iff $ sigma_{i}^{m} = (1) $ for all $ i $. Therefore, $ sigma^m = (1) $ iff $ | sigma_i |$ divides $m$ for all $i$ (for $sigma$ is cyclic). Since $ | sigma | $ is the least such $m$, the conclusion follows.
I have big trouble to understand why we needed to use this $m$-th power with, furthermore, considering the case about $sigma^m$ being the identity permutation.
Thank you,
group-theory permutations proof-explanation symmetric-groups cyclic-groups
edited Dec 27 '18 at 20:07
Shaun
9,444113684
asked Dec 27 '18 at 19:28
freehumoristfreehumorist
351214
$endgroup$
add a comment |
$begingroup$
About the lemma below:
The order of a permutation $ sigma in S_n $ is the least common multiple of the orders of its disjoint cycles.
I didn't get the proof which consists of the following reasoning:
Let $ sigma = sigma_1 ... sigma_r $ with $ r le n $ and $ { sigma_i }_{1 le i le r} $ disjoint cycles in question. Since every $ sigma_i $ commute in this case, we have: $ sigma^m = sigma_{1}^{m} ... sigma_{r}^{m} $, for all $ m in mathbb{Z} $, and $ sigma^m = (1) $ iff $ sigma_{i}^{m} = (1) $ for all $ i $. Therefore, $ sigma^m = (1) $ iff $ | sigma_i |$ divides $m$ for all $i$ (for $sigma$ is cyclic). Since $ | sigma | $ is the least such $m$, the conclusion follows.
I have big trouble to understand why we needed to use this $m$-th power with, furthermore, considering the case about $sigma^m$ being the identity permutation.
Thank you,
group-theory permutations proof-explanation symmetric-groups cyclic-groups
edited Dec 27 '18 at 20:07
Shaun
9,444113684
asked Dec 27 '18 at 19:28
freehumoristfreehumorist
351214
$endgroup$
4
$begingroup$
I'm not sure where your confusion lies. The $m$-th power is needed because one is seeking to find the order of $sigma$ in $S_n$.
$endgroup$
– user458276
Dec 27 '18 at 19:32
2
$begingroup$
What do you think that “order of $sigma$” means?
$endgroup$
– José Carlos Santos
Dec 27 '18 at 19:43
$begingroup$
I'm totally ok with your comments. I hadn't considered enough the question before posting it. But, I wonder if we wouldn't have here a slice of reasoning missing still? If not, I get it.
$endgroup$
– freehumorist
Dec 27 '18 at 20:14
add a comment |
$begingroup$
About the lemma below:
The order of a permutation $ sigma in S_n $ is the least common multiple of the orders of its disjoint cycles.
I didn't get the proof which consists of the following reasoning:
Let $ sigma = sigma_1 ... sigma_r $ with $ r le n $ and $ { sigma_i }_{1 le i le r} $ disjoint cycles in question. Since every $ sigma_i $ commute in this case, we have: $ sigma^m = sigma_{1}^{m} ... sigma_{r}^{m} $, for all $ m in mathbb{Z} $, and $ sigma^m = (1) $ iff $ sigma_{i}^{m} = (1) $ for all $ i $. Therefore, $ sigma^m = (1) $ iff $ | sigma_i |$ divides $m$ for all $i$ (for $sigma$ is cyclic). Since $ | sigma | $ is the least such $m$, the conclusion follows.
I have big trouble to understand why we needed to use this $m$-th power with, furthermore, considering the case about $sigma^m$ being the identity permutation.
Thank you,
group-theory permutations proof-explanation symmetric-groups cyclic-groups
edited Dec 27 '18 at 20:07
Shaun
9,444113684
asked Dec 27 '18 at 19:28
freehumoristfreehumorist
351214
$endgroup$
About the lemma below:
The order of a permutation $ sigma in S_n $ is the least common multiple of the orders of its disjoint cycles.
I didn't get the proof which consists of the following reasoning:
Let $ sigma = sigma_1 ... sigma_r $ with $ r le n $ and $ { sigma_i }_{1 le i le r} $ disjoint cycles in question. Since every $ sigma_i $ commute in this case, we have: $ sigma^m = sigma_{1}^{m} ... sigma_{r}^{m} $, for all $ m in mathbb{Z} $, and $ sigma^m = (1) $ iff $ sigma_{i}^{m} = (1) $ for all $ i $. Therefore, $ sigma^m = (1) $ iff $ | sigma_i |$ divides $m$ for all $i$ (for $sigma$ is cyclic). Since $ | sigma | $ is the least such $m$, the conclusion follows.
I have big trouble to understand why we needed to use this $m$-th power with, furthermore, considering the case about $sigma^m$ being the identity permutation.
Thank you,
group-theory permutations proof-explanation symmetric-groups cyclic-groups
group-theory permutations proof-explanation symmetric-groups cyclic-groups
edited Dec 27 '18 at 20:07
Shaun
9,444113684
asked Dec 27 '18 at 19:28
freehumoristfreehumorist
351214
edited Dec 27 '18 at 20:07
Shaun
9,444113684
asked Dec 27 '18 at 19:28
freehumoristfreehumorist
351214
edited Dec 27 '18 at 20:07
Shaun
9,444113684
edited Dec 27 '18 at 20:07
Shaun
9,444113684
edited Dec 27 '18 at 20:07
Shaun
9,444113684
9,444113684
asked Dec 27 '18 at 19:28
freehumoristfreehumorist
351214
asked Dec 27 '18 at 19:28
freehumoristfreehumorist
351214
asked Dec 27 '18 at 19:28
freehumoristfreehumorist
351214
351214
4
$begingroup$
I'm not sure where your confusion lies. The $m$-th power is needed because one is seeking to find the order of $sigma$ in $S_n$.
$endgroup$
– user458276
Dec 27 '18 at 19:32
2
$begingroup$
What do you think that “order of $sigma$” means?
$endgroup$
– José Carlos Santos
Dec 27 '18 at 19:43
$begingroup$
I'm totally ok with your comments. I hadn't considered enough the question before posting it. But, I wonder if we wouldn't have here a slice of reasoning missing still? If not, I get it.
$endgroup$
– freehumorist
Dec 27 '18 at 20:14
add a comment |
4
$begingroup$
I'm not sure where your confusion lies. The $m$-th power is needed because one is seeking to find the order of $sigma$ in $S_n$.
$endgroup$
– user458276
Dec 27 '18 at 19:32
2
$begingroup$
What do you think that “order of $sigma$” means?
$endgroup$
– José Carlos Santos
Dec 27 '18 at 19:43
$begingroup$
I'm totally ok with your comments. I hadn't considered enough the question before posting it. But, I wonder if we wouldn't have here a slice of reasoning missing still? If not, I get it.
$endgroup$
– freehumorist
Dec 27 '18 at 20:14
4
4
$begingroup$
I'm not sure where your confusion lies. The $m$-th power is needed because one is seeking to find the order of $sigma$ in $S_n$.
$endgroup$
– user458276
Dec 27 '18 at 19:32
$begingroup$
I'm not sure where your confusion lies. The $m$-th power is needed because one is seeking to find the order of $sigma$ in $S_n$.
$endgroup$
– user458276
Dec 27 '18 at 19:32
2
2
$begingroup$
What do you think that “order of $sigma$” means?
$endgroup$
– José Carlos Santos
Dec 27 '18 at 19:43
$begingroup$
What do you think that “order of $sigma$” means?
$endgroup$
– José Carlos Santos
Dec 27 '18 at 19:43
$begingroup$
I'm totally ok with your comments. I hadn't considered enough the question before posting it. But, I wonder if we wouldn't have here a slice of reasoning missing still? If not, I get it.
$endgroup$
– freehumorist
Dec 27 '18 at 20:14
$begingroup$
I'm totally ok with your comments. I hadn't considered enough the question before posting it. But, I wonder if we wouldn't have here a slice of reasoning missing still? If not, I get it.
$endgroup$
– freehumorist
Dec 27 '18 at 20:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As for all groups, the order of a permutation $sigmain S_n$ is, by definition, the least positive integer $m$ such that $sigma^n=operatorname{id}$.
In other words, it is the positive generator of the group homomorphism
begin{align}
(bf Z,+)&longrightarrow (S_n,circ)\
k &longmapsto sigma^k=underbrace{sigmacircsigmacircdotscircsigma}_{k :text{ factors}}end{align}
answered Dec 27 '18 at 20:17
BernardBernard
123k741116
$endgroup$
add a comment |
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1 Answer
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$begingroup$
As for all groups, the order of a permutation $sigmain S_n$ is, by definition, the least positive integer $m$ such that $sigma^n=operatorname{id}$.
In other words, it is the positive generator of the group homomorphism
begin{align}
(bf Z,+)&longrightarrow (S_n,circ)\
k &longmapsto sigma^k=underbrace{sigmacircsigmacircdotscircsigma}_{k :text{ factors}}end{align}
answered Dec 27 '18 at 20:17
BernardBernard
123k741116
$endgroup$
add a comment |
$begingroup$
As for all groups, the order of a permutation $sigmain S_n$ is, by definition, the least positive integer $m$ such that $sigma^n=operatorname{id}$.
In other words, it is the positive generator of the group homomorphism
begin{align}
(bf Z,+)&longrightarrow (S_n,circ)\
k &longmapsto sigma^k=underbrace{sigmacircsigmacircdotscircsigma}_{k :text{ factors}}end{align}
answered Dec 27 '18 at 20:17
BernardBernard
123k741116
$endgroup$
add a comment |
$begingroup$
As for all groups, the order of a permutation $sigmain S_n$ is, by definition, the least positive integer $m$ such that $sigma^n=operatorname{id}$.
In other words, it is the positive generator of the group homomorphism
begin{align}
(bf Z,+)&longrightarrow (S_n,circ)\
k &longmapsto sigma^k=underbrace{sigmacircsigmacircdotscircsigma}_{k :text{ factors}}end{align}
answered Dec 27 '18 at 20:17
BernardBernard
123k741116
$endgroup$
As for all groups, the order of a permutation $sigmain S_n$ is, by definition, the least positive integer $m$ such that $sigma^n=operatorname{id}$.
In other words, it is the positive generator of the group homomorphism
begin{align}
(bf Z,+)&longrightarrow (S_n,circ)\
k &longmapsto sigma^k=underbrace{sigmacircsigmacircdotscircsigma}_{k :text{ factors}}end{align}
answered Dec 27 '18 at 20:17
BernardBernard
123k741116
answered Dec 27 '18 at 20:17
BernardBernard
123k741116
answered Dec 27 '18 at 20:17
BernardBernard
123k741116
answered Dec 27 '18 at 20:17
BernardBernard
123k741116
123k741116
add a comment |
add a comment |
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4
$begingroup$
I'm not sure where your confusion lies. The $m$-th power is needed because one is seeking to find the order of $sigma$ in $S_n$.
$endgroup$
– user458276
Dec 27 '18 at 19:32
2
$begingroup$
What do you think that “order of $sigma$” means?
$endgroup$
– José Carlos Santos
Dec 27 '18 at 19:43
$begingroup$
I'm totally ok with your comments. I hadn't considered enough the question before posting it. But, I wonder if we wouldn't have here a slice of reasoning missing still? If not, I get it.
$endgroup$
– freehumorist
Dec 27 '18 at 20:14