Compositions & Transpositions of permutations












0












$begingroup$


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







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$endgroup$

















    0












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 16 at 23:38









      jordan_glen

      1




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      asked Jan 16 at 20:33









      Onkar SinghOnkar Singh

      317




      317






















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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 $S_n rightarrow S_n$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            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 $left{tau circ sigma: sigma in S_nright} = S_n$






            share|cite|improve this answer









            $endgroup$














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              0












              $begingroup$

              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 $S_n rightarrow S_n$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                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 $S_n rightarrow S_n$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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 $S_n rightarrow S_n$.






                  share|cite|improve this answer









                  $endgroup$



                  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 $S_n rightarrow S_n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 16 at 20:48









                  Anthony TerAnthony Ter

                  37116




                  37116























                      0












                      $begingroup$

                      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 $left{tau circ sigma: sigma in S_nright} = S_n$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        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 $left{tau circ sigma: sigma in S_nright} = S_n$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          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 $left{tau circ sigma: sigma in S_nright} = S_n$






                          share|cite|improve this answer









                          $endgroup$



                          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 $left{tau circ sigma: sigma in S_nright} = S_n$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 16 at 23:05







                          user635162





































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