Mixing
Compositions & Transpositions of permutations
$begingroup$
Consider the set of all permutations $S_n$.
Fix an element $tauin S_n$.
Then the sets ${sigmacirctaumid sigmain S_n}= {tau circsigmamid sigmain S_n}$ have exactly $n!$ elements.
I am confused what the above theorem stands for? How can ${sigmacirctau mid sigmain S_n}={tau circsigma mid sigma in S_n} = S_n$?
(This has been stated as an equivalence statement of the above theorem)? What is the logic behind? Does the term element above refer to a complete permutation? See link.
group-theory permutation-cycles
edited Jan 16 at 23:38
jordan_glen
1
asked Jan 16 at 20:33
Onkar SinghOnkar Singh
297
$endgroup$
add a comment |
$begingroup$
Consider the set of all permutations $S_n$.
Fix an element $tauin S_n$.
Then the sets ${sigmacirctaumid sigmain S_n}= {tau circsigmamid sigmain S_n}$ have exactly $n!$ elements.
I am confused what the above theorem stands for? How can ${sigmacirctau mid sigmain S_n}={tau circsigma mid sigma in S_n} = S_n$?
(This has been stated as an equivalence statement of the above theorem)? What is the logic behind? Does the term element above refer to a complete permutation? See link.
group-theory permutation-cycles
edited Jan 16 at 23:38
jordan_glen
1
asked Jan 16 at 20:33
Onkar SinghOnkar Singh
297
$endgroup$
add a comment |
$begingroup$
Consider the set of all permutations $S_n$.
Fix an element $tauin S_n$.
Then the sets ${sigmacirctaumid sigmain S_n}= {tau circsigmamid sigmain S_n}$ have exactly $n!$ elements.
I am confused what the above theorem stands for? How can ${sigmacirctau mid sigmain S_n}={tau circsigma mid sigma in S_n} = S_n$?
(This has been stated as an equivalence statement of the above theorem)? What is the logic behind? Does the term element above refer to a complete permutation? See link.
group-theory permutation-cycles
edited Jan 16 at 23:38
jordan_glen
1
asked Jan 16 at 20:33
Onkar SinghOnkar Singh
297
$endgroup$
Consider the set of all permutations $S_n$.
Fix an element $tauin S_n$.
Then the sets ${sigmacirctaumid sigmain S_n}= {tau circsigmamid sigmain S_n}$ have exactly $n!$ elements.
I am confused what the above theorem stands for? How can ${sigmacirctau mid sigmain S_n}={tau circsigma mid sigma in S_n} = S_n$?
(This has been stated as an equivalence statement of the above theorem)? What is the logic behind? Does the term element above refer to a complete permutation? See link.
group-theory permutation-cycles
group-theory permutation-cycles
edited Jan 16 at 23:38
jordan_glen
1
asked Jan 16 at 20:33
Onkar SinghOnkar Singh
297
edited Jan 16 at 23:38
jordan_glen
1
asked Jan 16 at 20:33
Onkar SinghOnkar Singh
297
edited Jan 16 at 23:38
jordan_glen
1
edited Jan 16 at 23:38
jordan_glen
1
edited Jan 16 at 23:38
jordan_glen
1
1
asked Jan 16 at 20:33
Onkar SinghOnkar Singh
297
asked Jan 16 at 20:33
Onkar SinghOnkar Singh
297
asked Jan 16 at 20:33
Onkar SinghOnkar Singh
297
297
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $sigma in S_n$ then $sigma circ tau^{-1} in S_n$, and $tau^{-1} circ sigma in S_n$ so multiplying by $sigma$ on either side is a surjective map $S_n rightarrow S_n$.
answered Jan 16 at 20:48
Anthony TerAnthony Ter
36116
$endgroup$
add a comment |
$begingroup$
The first thing you should note is that if $tau, sigma in S_n$, then $tau circ sigma in S_n$. This means that if you have two permutations, then their product is also a permutation of the same permutation group.
You also know that $S_n$ has exactly $n!$ elements.
Now imagine I give you a set $A$ where all of its elements are permutations from $S_n$. Mathematically speaking this means that $A subset S_n$.
Now let's think about how many elements $A$ can have.
If $A = S_n$, then $A$ contains all permutations from $S_n$. So how many elements does $A$ have? Exactly $n!$, since there are exactly $n!$ permutations in $S_n$.
What if $A$ contains exactly $n!$ permutations from $S_n$? Then $A$ must contain all permutations from $S_n$, since there are only $n!$ elements in $S_n$. So $A = S_n$
What I have shown is that any set $A$ that only contains permutations from $S_n$ has exactly $n!$ elements if and only if $A = S_n$.
The key point here is that $A$ only has permutations from $S_n$ as its elements. And if it has $n!$ permutations (that is, all permutations) as its elements, then it must be equal to $S_n$ (and vice versa).
This is essentially all the proposition says.
The set $left{sigma circ tau : sigma in S_nright}$ is a subset of $S_n$, that is it only contains permutations from $S_n$. Why? See my first statement at the top of this post.
So $left{sigma circ tau : sigma in S_nright} = S_n$ is equivalent to saying that $left{sigma circ tau : sigma in S_nright}$ has exactly $n!$ elements.
The same holds for $left{tau circ sigma: sigma in S_nright} = S_n$
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
If $sigma in S_n$ then $sigma circ tau^{-1} in S_n$, and $tau^{-1} circ sigma in S_n$ so multiplying by $sigma$ on either side is a surjective map $S_n rightarrow S_n$.
answered Jan 16 at 20:48
Anthony TerAnthony Ter
36116
$endgroup$
add a comment |
$begingroup$
If $sigma in S_n$ then $sigma circ tau^{-1} in S_n$, and $tau^{-1} circ sigma in S_n$ so multiplying by $sigma$ on either side is a surjective map $S_n rightarrow S_n$.
answered Jan 16 at 20:48
Anthony TerAnthony Ter
36116
$endgroup$
add a comment |
$begingroup$
If $sigma in S_n$ then $sigma circ tau^{-1} in S_n$, and $tau^{-1} circ sigma in S_n$ so multiplying by $sigma$ on either side is a surjective map $S_n rightarrow S_n$.
answered Jan 16 at 20:48
Anthony TerAnthony Ter
36116
$endgroup$
If $sigma in S_n$ then $sigma circ tau^{-1} in S_n$, and $tau^{-1} circ sigma in S_n$ so multiplying by $sigma$ on either side is a surjective map $S_n rightarrow S_n$.
answered Jan 16 at 20:48
Anthony TerAnthony Ter
36116
answered Jan 16 at 20:48
Anthony TerAnthony Ter
36116
answered Jan 16 at 20:48
Anthony TerAnthony Ter
36116
answered Jan 16 at 20:48
Anthony TerAnthony Ter
36116
36116
add a comment |
add a comment |
$begingroup$
The first thing you should note is that if $tau, sigma in S_n$, then $tau circ sigma in S_n$. This means that if you have two permutations, then their product is also a permutation of the same permutation group.
You also know that $S_n$ has exactly $n!$ elements.
Now imagine I give you a set $A$ where all of its elements are permutations from $S_n$. Mathematically speaking this means that $A subset S_n$.
Now let's think about how many elements $A$ can have.
If $A = S_n$, then $A$ contains all permutations from $S_n$. So how many elements does $A$ have? Exactly $n!$, since there are exactly $n!$ permutations in $S_n$.
What if $A$ contains exactly $n!$ permutations from $S_n$? Then $A$ must contain all permutations from $S_n$, since there are only $n!$ elements in $S_n$. So $A = S_n$
What I have shown is that any set $A$ that only contains permutations from $S_n$ has exactly $n!$ elements if and only if $A = S_n$.
The key point here is that $A$ only has permutations from $S_n$ as its elements. And if it has $n!$ permutations (that is, all permutations) as its elements, then it must be equal to $S_n$ (and vice versa).
This is essentially all the proposition says.
The set $left{sigma circ tau : sigma in S_nright}$ is a subset of $S_n$, that is it only contains permutations from $S_n$. Why? See my first statement at the top of this post.
So $left{sigma circ tau : sigma in S_nright} = S_n$ is equivalent to saying that $left{sigma circ tau : sigma in S_nright}$ has exactly $n!$ elements.
The same holds for $left{tau circ sigma: sigma in S_nright} = S_n$
$endgroup$
add a comment |
$begingroup$
The first thing you should note is that if $tau, sigma in S_n$, then $tau circ sigma in S_n$. This means that if you have two permutations, then their product is also a permutation of the same permutation group.
You also know that $S_n$ has exactly $n!$ elements.
Now imagine I give you a set $A$ where all of its elements are permutations from $S_n$. Mathematically speaking this means that $A subset S_n$.
Now let's think about how many elements $A$ can have.
If $A = S_n$, then $A$ contains all permutations from $S_n$. So how many elements does $A$ have? Exactly $n!$, since there are exactly $n!$ permutations in $S_n$.
What if $A$ contains exactly $n!$ permutations from $S_n$? Then $A$ must contain all permutations from $S_n$, since there are only $n!$ elements in $S_n$. So $A = S_n$
What I have shown is that any set $A$ that only contains permutations from $S_n$ has exactly $n!$ elements if and only if $A = S_n$.
The key point here is that $A$ only has permutations from $S_n$ as its elements. And if it has $n!$ permutations (that is, all permutations) as its elements, then it must be equal to $S_n$ (and vice versa).
This is essentially all the proposition says.
The set $left{sigma circ tau : sigma in S_nright}$ is a subset of $S_n$, that is it only contains permutations from $S_n$. Why? See my first statement at the top of this post.
So $left{sigma circ tau : sigma in S_nright} = S_n$ is equivalent to saying that $left{sigma circ tau : sigma in S_nright}$ has exactly $n!$ elements.
The same holds for $left{tau circ sigma: sigma in S_nright} = S_n$
$endgroup$
add a comment |
$begingroup$
The first thing you should note is that if $tau, sigma in S_n$, then $tau circ sigma in S_n$. This means that if you have two permutations, then their product is also a permutation of the same permutation group.
You also know that $S_n$ has exactly $n!$ elements.
Now imagine I give you a set $A$ where all of its elements are permutations from $S_n$. Mathematically speaking this means that $A subset S_n$.
Now let's think about how many elements $A$ can have.
If $A = S_n$, then $A$ contains all permutations from $S_n$. So how many elements does $A$ have? Exactly $n!$, since there are exactly $n!$ permutations in $S_n$.
What if $A$ contains exactly $n!$ permutations from $S_n$? Then $A$ must contain all permutations from $S_n$, since there are only $n!$ elements in $S_n$. So $A = S_n$
What I have shown is that any set $A$ that only contains permutations from $S_n$ has exactly $n!$ elements if and only if $A = S_n$.
The key point here is that $A$ only has permutations from $S_n$ as its elements. And if it has $n!$ permutations (that is, all permutations) as its elements, then it must be equal to $S_n$ (and vice versa).
This is essentially all the proposition says.
The set $left{sigma circ tau : sigma in S_nright}$ is a subset of $S_n$, that is it only contains permutations from $S_n$. Why? See my first statement at the top of this post.
So $left{sigma circ tau : sigma in S_nright} = S_n$ is equivalent to saying that $left{sigma circ tau : sigma in S_nright}$ has exactly $n!$ elements.
The same holds for $left{tau circ sigma: sigma in S_nright} = S_n$
$endgroup$
The first thing you should note is that if $tau, sigma in S_n$, then $tau circ sigma in S_n$. This means that if you have two permutations, then their product is also a permutation of the same permutation group.
You also know that $S_n$ has exactly $n!$ elements.
Now imagine I give you a set $A$ where all of its elements are permutations from $S_n$. Mathematically speaking this means that $A subset S_n$.
Now let's think about how many elements $A$ can have.
If $A = S_n$, then $A$ contains all permutations from $S_n$. So how many elements does $A$ have? Exactly $n!$, since there are exactly $n!$ permutations in $S_n$.
What if $A$ contains exactly $n!$ permutations from $S_n$? Then $A$ must contain all permutations from $S_n$, since there are only $n!$ elements in $S_n$. So $A = S_n$
What I have shown is that any set $A$ that only contains permutations from $S_n$ has exactly $n!$ elements if and only if $A = S_n$.
The key point here is that $A$ only has permutations from $S_n$ as its elements. And if it has $n!$ permutations (that is, all permutations) as its elements, then it must be equal to $S_n$ (and vice versa).
This is essentially all the proposition says.
The set $left{sigma circ tau : sigma in S_nright}$ is a subset of $S_n$, that is it only contains permutations from $S_n$. Why? See my first statement at the top of this post.
So $left{sigma circ tau : sigma in S_nright} = S_n$ is equivalent to saying that $left{sigma circ tau : sigma in S_nright}$ has exactly $n!$ elements.
The same holds for $left{tau circ sigma: sigma in S_nright} = S_n$
answered Jan 16 at 23:05
user635162
add a comment |
add a comment |
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