Proving $(a b)(c d)=(a b c)(b c d)$ [closed]












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$begingroup$


I was reading a book in Group theory and they said that $(a, b)(c, d)=(a, b, c)(b, c, d),$ but they did not prove it, for some reason (maybe it's too obvious but i can't seem to see it).




How to formally prove that $$(a, b)(c, d)=(a, b, c)(b, c, d),$$ when $(a, b),(c, d),(a, b, c),(b, c, d)in S_n$ so $ngeq 3$.







group-theory permutations symmetric-groups







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closed as off-topic by amWhy, Paul Frost, Chris Godsil, KReiser, M. Vinay Jan 2 at 7:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, Chris Godsil, KReiser, M. Vinay

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    Please specify which book you are reading, using an edit.
    $endgroup$
    – Shaun
    Jan 1 at 19:24






  • 2




    $begingroup$
    it is obvious if one knows the rules to do it: form right to left. THat's all.
    $endgroup$
    – DonAntonio
    Jan 1 at 19:26
















0












$begingroup$


I was reading a book in Group theory and they said that $(a, b)(c, d)=(a, b, c)(b, c, d),$ but they did not prove it, for some reason (maybe it's too obvious but i can't seem to see it).




How to formally prove that $$(a, b)(c, d)=(a, b, c)(b, c, d),$$ when $(a, b),(c, d),(a, b, c),(b, c, d)in S_n$ so $ngeq 3$.







group-theory permutations symmetric-groups







share|cite|improve this question











$endgroup$



closed as off-topic by amWhy, Paul Frost, Chris Godsil, KReiser, M. Vinay Jan 2 at 7:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, Chris Godsil, KReiser, M. Vinay

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    Please specify which book you are reading, using an edit.
    $endgroup$
    – Shaun
    Jan 1 at 19:24






  • 2




    $begingroup$
    it is obvious if one knows the rules to do it: form right to left. THat's all.
    $endgroup$
    – DonAntonio
    Jan 1 at 19:26














0












0








0





$begingroup$


I was reading a book in Group theory and they said that $(a, b)(c, d)=(a, b, c)(b, c, d),$ but they did not prove it, for some reason (maybe it's too obvious but i can't seem to see it).




How to formally prove that $$(a, b)(c, d)=(a, b, c)(b, c, d),$$ when $(a, b),(c, d),(a, b, c),(b, c, d)in S_n$ so $ngeq 3$.







group-theory permutations symmetric-groups







share|cite|improve this question











$endgroup$




I was reading a book in Group theory and they said that $(a, b)(c, d)=(a, b, c)(b, c, d),$ but they did not prove it, for some reason (maybe it's too obvious but i can't seem to see it).




How to formally prove that $$(a, b)(c, d)=(a, b, c)(b, c, d),$$ when $(a, b),(c, d),(a, b, c),(b, c, d)in S_n$ so $ngeq 3$.







group-theory permutations symmetric-groups




group-theory permutations symmetric-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 19:41









Shaun

8,818113681




8,818113681










asked Jan 1 at 19:22









vesiivesii

886




886




closed as off-topic by amWhy, Paul Frost, Chris Godsil, KReiser, M. Vinay Jan 2 at 7:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, Chris Godsil, KReiser, M. Vinay

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, Paul Frost, Chris Godsil, KReiser, M. Vinay Jan 2 at 7:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, Chris Godsil, KReiser, M. Vinay

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Please specify which book you are reading, using an edit.
    $endgroup$
    – Shaun
    Jan 1 at 19:24






  • 2




    $begingroup$
    it is obvious if one knows the rules to do it: form right to left. THat's all.
    $endgroup$
    – DonAntonio
    Jan 1 at 19:26


















  • $begingroup$
    Please specify which book you are reading, using an edit.
    $endgroup$
    – Shaun
    Jan 1 at 19:24






  • 2




    $begingroup$
    it is obvious if one knows the rules to do it: form right to left. THat's all.
    $endgroup$
    – DonAntonio
    Jan 1 at 19:26
















$begingroup$
Please specify which book you are reading, using an edit.
$endgroup$
– Shaun
Jan 1 at 19:24




$begingroup$
Please specify which book you are reading, using an edit.
$endgroup$
– Shaun
Jan 1 at 19:24




2




2




$begingroup$
it is obvious if one knows the rules to do it: form right to left. THat's all.
$endgroup$
– DonAntonio
Jan 1 at 19:26




$begingroup$
it is obvious if one knows the rules to do it: form right to left. THat's all.
$endgroup$
– DonAntonio
Jan 1 at 19:26










3 Answers
3






active

oldest

votes


















1












$begingroup$

Hint:



This seems to two products of cycles. The product corresponds to composition, and goes from right to left. So the product of the two $3$-cycles on the right the results in
$$begin{cases}
a mapsto a mapsto b \
bmapsto cmapsto a \
cmapsto dmapsto d \
dmapsto bmapsto c
end{cases}quad text{which is }; (a,b)(c,d).$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint:



    Recall that a permutation is a bijection to & from a (finite) set. What do $a,b,c,$ and $d$ map to under $(a, b)(c, d)$ and under $(a, b, c)(b, c, d)$?






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      They both equate to the permutation: $$begin{pmatrix} a& b & c & d\
      b & a & d&cend{pmatrix}.$$






      share|cite|improve this answer











      $endgroup$




















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        Hint:



        This seems to two products of cycles. The product corresponds to composition, and goes from right to left. So the product of the two $3$-cycles on the right the results in
        $$begin{cases}
        a mapsto a mapsto b \
        bmapsto cmapsto a \
        cmapsto dmapsto d \
        dmapsto bmapsto c
        end{cases}quad text{which is }; (a,b)(c,d).$$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Hint:



          This seems to two products of cycles. The product corresponds to composition, and goes from right to left. So the product of the two $3$-cycles on the right the results in
          $$begin{cases}
          a mapsto a mapsto b \
          bmapsto cmapsto a \
          cmapsto dmapsto d \
          dmapsto bmapsto c
          end{cases}quad text{which is }; (a,b)(c,d).$$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Hint:



            This seems to two products of cycles. The product corresponds to composition, and goes from right to left. So the product of the two $3$-cycles on the right the results in
            $$begin{cases}
            a mapsto a mapsto b \
            bmapsto cmapsto a \
            cmapsto dmapsto d \
            dmapsto bmapsto c
            end{cases}quad text{which is }; (a,b)(c,d).$$






            share|cite|improve this answer









            $endgroup$



            Hint:



            This seems to two products of cycles. The product corresponds to composition, and goes from right to left. So the product of the two $3$-cycles on the right the results in
            $$begin{cases}
            a mapsto a mapsto b \
            bmapsto cmapsto a \
            cmapsto dmapsto d \
            dmapsto bmapsto c
            end{cases}quad text{which is }; (a,b)(c,d).$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 1 at 19:39









            BernardBernard

            119k639112




            119k639112























                0












                $begingroup$

                Hint:



                Recall that a permutation is a bijection to & from a (finite) set. What do $a,b,c,$ and $d$ map to under $(a, b)(c, d)$ and under $(a, b, c)(b, c, d)$?






                share|cite|improve this answer











                $endgroup$


















                  0












                  $begingroup$

                  Hint:



                  Recall that a permutation is a bijection to & from a (finite) set. What do $a,b,c,$ and $d$ map to under $(a, b)(c, d)$ and under $(a, b, c)(b, c, d)$?






                  share|cite|improve this answer











                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Hint:



                    Recall that a permutation is a bijection to & from a (finite) set. What do $a,b,c,$ and $d$ map to under $(a, b)(c, d)$ and under $(a, b, c)(b, c, d)$?






                    share|cite|improve this answer











                    $endgroup$



                    Hint:



                    Recall that a permutation is a bijection to & from a (finite) set. What do $a,b,c,$ and $d$ map to under $(a, b)(c, d)$ and under $(a, b, c)(b, c, d)$?







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 1 at 20:10

























                    answered Jan 1 at 19:28









                    ShaunShaun

                    8,818113681




                    8,818113681























                        0












                        $begingroup$

                        They both equate to the permutation: $$begin{pmatrix} a& b & c & d\
                        b & a & d&cend{pmatrix}.$$






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          They both equate to the permutation: $$begin{pmatrix} a& b & c & d\
                          b & a & d&cend{pmatrix}.$$






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            They both equate to the permutation: $$begin{pmatrix} a& b & c & d\
                            b & a & d&cend{pmatrix}.$$






                            share|cite|improve this answer











                            $endgroup$



                            They both equate to the permutation: $$begin{pmatrix} a& b & c & d\
                            b & a & d&cend{pmatrix}.$$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 1 at 20:11









                            Shaun

                            8,818113681




                            8,818113681










                            answered Jan 1 at 19:47









                            Chris CusterChris Custer

                            11.2k3824




                            11.2k3824















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