Verifying the calculation of the number of conjugate elements












2












$begingroup$


I would like to verify my understanding:



Consider action of $S_7$ on itself by conjugation. I'm trying to compute:




  1. $Stab_{S_7}((1 2))$

  2. $Stab_{S_7}((1 2 3 4 5 6 7))$

  3. $Stab_{S_7}((1 2 3)(4 5 6))$


I'm following this rule: two elements are conjugate if and only if they have the same cycle type.



For the first one it should be $(x_1x_2)$ so I choose 2 elements from 7 and get the orbit: $binom{7}{2}$ meaning $Stab_{S_7}((1 2))=frac{7!}{binom{7}{2}}$



For the second one it should be cycle type $(7)$ so there is only one cohse meaning $Stab_{S_7}((1 2 3 4 5 6 7))=7!$



For the third one it should be cycle type $(3,3)$ so we choose 3 elements from 7 for the first and organize in circle with $binom{7}{3}cdot 2!$ possibilities.
Then we choose $3$ from $4$ for the second one and organize in circle with $binom{4}{3}cdot 2!$. All that we should divide by $2$, so we get:



$$frac{binom{7}{3}binom{4}{3}cdot 4}{2}=binom{7}{3}binom{4}{3}cdot 2$$



And then: $Stab_{S_7}((1 2 3)(4 5 6)) = binom{7}{3}binom{4}{3}cdot 2$.



I feel like it isn't correct. If so, how to solve it?






abstract-algebra group-theory symmetric-groups group-actions







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I would like to verify my understanding:



    Consider action of $S_7$ on itself by conjugation. I'm trying to compute:




    1. $Stab_{S_7}((1 2))$

    2. $Stab_{S_7}((1 2 3 4 5 6 7))$

    3. $Stab_{S_7}((1 2 3)(4 5 6))$


    I'm following this rule: two elements are conjugate if and only if they have the same cycle type.



    For the first one it should be $(x_1x_2)$ so I choose 2 elements from 7 and get the orbit: $binom{7}{2}$ meaning $Stab_{S_7}((1 2))=frac{7!}{binom{7}{2}}$



    For the second one it should be cycle type $(7)$ so there is only one cohse meaning $Stab_{S_7}((1 2 3 4 5 6 7))=7!$



    For the third one it should be cycle type $(3,3)$ so we choose 3 elements from 7 for the first and organize in circle with $binom{7}{3}cdot 2!$ possibilities.
    Then we choose $3$ from $4$ for the second one and organize in circle with $binom{4}{3}cdot 2!$. All that we should divide by $2$, so we get:



    $$frac{binom{7}{3}binom{4}{3}cdot 4}{2}=binom{7}{3}binom{4}{3}cdot 2$$



    And then: $Stab_{S_7}((1 2 3)(4 5 6)) = binom{7}{3}binom{4}{3}cdot 2$.



    I feel like it isn't correct. If so, how to solve it?






    abstract-algebra group-theory symmetric-groups group-actions







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I would like to verify my understanding:



      Consider action of $S_7$ on itself by conjugation. I'm trying to compute:




      1. $Stab_{S_7}((1 2))$

      2. $Stab_{S_7}((1 2 3 4 5 6 7))$

      3. $Stab_{S_7}((1 2 3)(4 5 6))$


      I'm following this rule: two elements are conjugate if and only if they have the same cycle type.



      For the first one it should be $(x_1x_2)$ so I choose 2 elements from 7 and get the orbit: $binom{7}{2}$ meaning $Stab_{S_7}((1 2))=frac{7!}{binom{7}{2}}$



      For the second one it should be cycle type $(7)$ so there is only one cohse meaning $Stab_{S_7}((1 2 3 4 5 6 7))=7!$



      For the third one it should be cycle type $(3,3)$ so we choose 3 elements from 7 for the first and organize in circle with $binom{7}{3}cdot 2!$ possibilities.
      Then we choose $3$ from $4$ for the second one and organize in circle with $binom{4}{3}cdot 2!$. All that we should divide by $2$, so we get:



      $$frac{binom{7}{3}binom{4}{3}cdot 4}{2}=binom{7}{3}binom{4}{3}cdot 2$$



      And then: $Stab_{S_7}((1 2 3)(4 5 6)) = binom{7}{3}binom{4}{3}cdot 2$.



      I feel like it isn't correct. If so, how to solve it?






      abstract-algebra group-theory symmetric-groups group-actions







      share|cite|improve this question











      $endgroup$




      I would like to verify my understanding:



      Consider action of $S_7$ on itself by conjugation. I'm trying to compute:




      1. $Stab_{S_7}((1 2))$

      2. $Stab_{S_7}((1 2 3 4 5 6 7))$

      3. $Stab_{S_7}((1 2 3)(4 5 6))$


      I'm following this rule: two elements are conjugate if and only if they have the same cycle type.



      For the first one it should be $(x_1x_2)$ so I choose 2 elements from 7 and get the orbit: $binom{7}{2}$ meaning $Stab_{S_7}((1 2))=frac{7!}{binom{7}{2}}$



      For the second one it should be cycle type $(7)$ so there is only one cohse meaning $Stab_{S_7}((1 2 3 4 5 6 7))=7!$



      For the third one it should be cycle type $(3,3)$ so we choose 3 elements from 7 for the first and organize in circle with $binom{7}{3}cdot 2!$ possibilities.
      Then we choose $3$ from $4$ for the second one and organize in circle with $binom{4}{3}cdot 2!$. All that we should divide by $2$, so we get:



      $$frac{binom{7}{3}binom{4}{3}cdot 4}{2}=binom{7}{3}binom{4}{3}cdot 2$$



      And then: $Stab_{S_7}((1 2 3)(4 5 6)) = binom{7}{3}binom{4}{3}cdot 2$.



      I feel like it isn't correct. If so, how to solve it?






      abstract-algebra group-theory symmetric-groups group-actions




      abstract-algebra group-theory symmetric-groups group-actions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 20 at 20:42









      user26857

      39.3k124183




      39.3k124183










      asked Jan 20 at 18:00









      abuka123abuka123

      344




      344






















          2 Answers
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          $begingroup$

          The first one is correct. For the second one, there are many cycles of $7$ elements. Thinking of it as a permutation $f$, you have $6$ choices for $f(1)$, $5$ choices for $f(f(1))$, $4$ choices for $f(f(f(1)))$, and so on until the last thing must map to $1$, so there are $6!$ cycles of length $7$, hence $mathrm{Stab}_{S_7}((1234567))=7!/6!=7$.



          For the third one, you are correct in counting the number of permutations of cycle type $(3,3)$, but that is not the same as $mathrm{Stab}_{S_7}((123)(456))$. As you do in the other ones, you have to take that value and divide $7!$ by it.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            I will attempt to do the 3rd one for you:
            According to the type formula, we know that the conjugate of (123)(456) is some sort of double 3 cycle.
            Note that the stabilizer here is heavily determined:
            If i send 1 to 5, for example, then i must send 2 to 6 and 3 to 4. I then have a choice of whether i send 4 to 1,2 or 3. By using this property, we can reason as follows:
            1. I can send 1 to 6 possible things(excluding 7). Each choice also determines where i send 2 and 3.
            2. After having sent 1, i now have 3 choices on where to send 3(once again cant send to 7).
            So we have in total 18 choices, and each one corresponds to a different permutation of $$mathrm{S_7)$$






            share|cite|improve this answer









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              2 Answers
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              $begingroup$

              The first one is correct. For the second one, there are many cycles of $7$ elements. Thinking of it as a permutation $f$, you have $6$ choices for $f(1)$, $5$ choices for $f(f(1))$, $4$ choices for $f(f(f(1)))$, and so on until the last thing must map to $1$, so there are $6!$ cycles of length $7$, hence $mathrm{Stab}_{S_7}((1234567))=7!/6!=7$.



              For the third one, you are correct in counting the number of permutations of cycle type $(3,3)$, but that is not the same as $mathrm{Stab}_{S_7}((123)(456))$. As you do in the other ones, you have to take that value and divide $7!$ by it.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                The first one is correct. For the second one, there are many cycles of $7$ elements. Thinking of it as a permutation $f$, you have $6$ choices for $f(1)$, $5$ choices for $f(f(1))$, $4$ choices for $f(f(f(1)))$, and so on until the last thing must map to $1$, so there are $6!$ cycles of length $7$, hence $mathrm{Stab}_{S_7}((1234567))=7!/6!=7$.



                For the third one, you are correct in counting the number of permutations of cycle type $(3,3)$, but that is not the same as $mathrm{Stab}_{S_7}((123)(456))$. As you do in the other ones, you have to take that value and divide $7!$ by it.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The first one is correct. For the second one, there are many cycles of $7$ elements. Thinking of it as a permutation $f$, you have $6$ choices for $f(1)$, $5$ choices for $f(f(1))$, $4$ choices for $f(f(f(1)))$, and so on until the last thing must map to $1$, so there are $6!$ cycles of length $7$, hence $mathrm{Stab}_{S_7}((1234567))=7!/6!=7$.



                  For the third one, you are correct in counting the number of permutations of cycle type $(3,3)$, but that is not the same as $mathrm{Stab}_{S_7}((123)(456))$. As you do in the other ones, you have to take that value and divide $7!$ by it.






                  share|cite|improve this answer









                  $endgroup$



                  The first one is correct. For the second one, there are many cycles of $7$ elements. Thinking of it as a permutation $f$, you have $6$ choices for $f(1)$, $5$ choices for $f(f(1))$, $4$ choices for $f(f(f(1)))$, and so on until the last thing must map to $1$, so there are $6!$ cycles of length $7$, hence $mathrm{Stab}_{S_7}((1234567))=7!/6!=7$.



                  For the third one, you are correct in counting the number of permutations of cycle type $(3,3)$, but that is not the same as $mathrm{Stab}_{S_7}((123)(456))$. As you do in the other ones, you have to take that value and divide $7!$ by it.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 20 at 19:56









                  Kevin LongKevin Long

                  3,57121431




                  3,57121431























                      0












                      $begingroup$

                      I will attempt to do the 3rd one for you:
                      According to the type formula, we know that the conjugate of (123)(456) is some sort of double 3 cycle.
                      Note that the stabilizer here is heavily determined:
                      If i send 1 to 5, for example, then i must send 2 to 6 and 3 to 4. I then have a choice of whether i send 4 to 1,2 or 3. By using this property, we can reason as follows:
                      1. I can send 1 to 6 possible things(excluding 7). Each choice also determines where i send 2 and 3.
                      2. After having sent 1, i now have 3 choices on where to send 3(once again cant send to 7).
                      So we have in total 18 choices, and each one corresponds to a different permutation of $$mathrm{S_7)$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        I will attempt to do the 3rd one for you:
                        According to the type formula, we know that the conjugate of (123)(456) is some sort of double 3 cycle.
                        Note that the stabilizer here is heavily determined:
                        If i send 1 to 5, for example, then i must send 2 to 6 and 3 to 4. I then have a choice of whether i send 4 to 1,2 or 3. By using this property, we can reason as follows:
                        1. I can send 1 to 6 possible things(excluding 7). Each choice also determines where i send 2 and 3.
                        2. After having sent 1, i now have 3 choices on where to send 3(once again cant send to 7).
                        So we have in total 18 choices, and each one corresponds to a different permutation of $$mathrm{S_7)$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          I will attempt to do the 3rd one for you:
                          According to the type formula, we know that the conjugate of (123)(456) is some sort of double 3 cycle.
                          Note that the stabilizer here is heavily determined:
                          If i send 1 to 5, for example, then i must send 2 to 6 and 3 to 4. I then have a choice of whether i send 4 to 1,2 or 3. By using this property, we can reason as follows:
                          1. I can send 1 to 6 possible things(excluding 7). Each choice also determines where i send 2 and 3.
                          2. After having sent 1, i now have 3 choices on where to send 3(once again cant send to 7).
                          So we have in total 18 choices, and each one corresponds to a different permutation of $$mathrm{S_7)$$






                          share|cite|improve this answer









                          $endgroup$



                          I will attempt to do the 3rd one for you:
                          According to the type formula, we know that the conjugate of (123)(456) is some sort of double 3 cycle.
                          Note that the stabilizer here is heavily determined:
                          If i send 1 to 5, for example, then i must send 2 to 6 and 3 to 4. I then have a choice of whether i send 4 to 1,2 or 3. By using this property, we can reason as follows:
                          1. I can send 1 to 6 possible things(excluding 7). Each choice also determines where i send 2 and 3.
                          2. After having sent 1, i now have 3 choices on where to send 3(once again cant send to 7).
                          So we have in total 18 choices, and each one corresponds to a different permutation of $$mathrm{S_7)$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 20 at 20:31









                          Marat AlievMarat Aliev

                          1312




                          1312






























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