Let $sigma,tau in S_n$. Prove that $sigma tau$ and $tau sigma $ have the same cycle type.












5












$begingroup$


Let $sigma,tau in S_n$. Prove that $sigma tau$ and $tau sigma $ have the same cycle type.



I was thinking that you could rewrite $sigma=g_1cdots g_k$ with $g_i$ disjoint cycles and $tau=h_1cdots h_l$ with $h_i$ disjoint cycles. But I don't know what to do next, as $g_i$ and $h_j$ doesn't have to be disjoint.



Edit: Could I prove it like this ?



We can write $σ$ in disjoint cycles: $σ=σ_1...σ_r$ with lenghts $l_1,...l_r$. So you get: $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$.



Which gives:
begin{align*}
τστ^{-1}&=τσ_1...σ_rτ^{-1}\
&=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\
&=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\
&=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\
end{align*}



Therefore $σ$ and $τστ^{-1}$ have the same cycle type. So $τσ$ and $στ$ have the same cycle type.






abstract-algebra group-theory finite-groups permutations







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    Let $sigma,tau in S_n$. Prove that $sigma tau$ and $tau sigma $ have the same cycle type.



    I was thinking that you could rewrite $sigma=g_1cdots g_k$ with $g_i$ disjoint cycles and $tau=h_1cdots h_l$ with $h_i$ disjoint cycles. But I don't know what to do next, as $g_i$ and $h_j$ doesn't have to be disjoint.



    Edit: Could I prove it like this ?



    We can write $σ$ in disjoint cycles: $σ=σ_1...σ_r$ with lenghts $l_1,...l_r$. So you get: $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$.



    Which gives:
    begin{align*}
    τστ^{-1}&=τσ_1...σ_rτ^{-1}\
    &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\
    &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\
    &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\
    end{align*}



    Therefore $σ$ and $τστ^{-1}$ have the same cycle type. So $τσ$ and $στ$ have the same cycle type.






    abstract-algebra group-theory finite-groups permutations







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      4



      $begingroup$


      Let $sigma,tau in S_n$. Prove that $sigma tau$ and $tau sigma $ have the same cycle type.



      I was thinking that you could rewrite $sigma=g_1cdots g_k$ with $g_i$ disjoint cycles and $tau=h_1cdots h_l$ with $h_i$ disjoint cycles. But I don't know what to do next, as $g_i$ and $h_j$ doesn't have to be disjoint.



      Edit: Could I prove it like this ?



      We can write $σ$ in disjoint cycles: $σ=σ_1...σ_r$ with lenghts $l_1,...l_r$. So you get: $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$.



      Which gives:
      begin{align*}
      τστ^{-1}&=τσ_1...σ_rτ^{-1}\
      &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\
      &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\
      &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\
      end{align*}



      Therefore $σ$ and $τστ^{-1}$ have the same cycle type. So $τσ$ and $στ$ have the same cycle type.






      abstract-algebra group-theory finite-groups permutations







      share|cite|improve this question











      $endgroup$




      Let $sigma,tau in S_n$. Prove that $sigma tau$ and $tau sigma $ have the same cycle type.



      I was thinking that you could rewrite $sigma=g_1cdots g_k$ with $g_i$ disjoint cycles and $tau=h_1cdots h_l$ with $h_i$ disjoint cycles. But I don't know what to do next, as $g_i$ and $h_j$ doesn't have to be disjoint.



      Edit: Could I prove it like this ?



      We can write $σ$ in disjoint cycles: $σ=σ_1...σ_r$ with lenghts $l_1,...l_r$. So you get: $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$.



      Which gives:
      begin{align*}
      τστ^{-1}&=τσ_1...σ_rτ^{-1}\
      &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\
      &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\
      &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\
      end{align*}



      Therefore $σ$ and $τστ^{-1}$ have the same cycle type. So $τσ$ and $στ$ have the same cycle type.






      abstract-algebra group-theory finite-groups permutations




      abstract-algebra group-theory finite-groups permutations






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 24 '18 at 21:16









      Shaun

      9,268113684




      9,268113684










      asked Mar 19 '13 at 15:51









      90intuition90intuition

      1,13331230




      1,13331230






















          5 Answers
          5






          active

          oldest

          votes


















          6












          $begingroup$

          Hint: they are conjugate elements of the group.






          share|cite|improve this answer









          $endgroup$









          • 5




            $begingroup$
            My book introduces conjugates in the next chapter.
            $endgroup$
            – 90intuition
            Mar 19 '13 at 15:59



















          4












          $begingroup$

          First consider the case that $tau$ is a single cycle, say $tau = (i_1,i_2,ldots,i_k)$. You should be able to compute $sigmatau$ and $tausigma$ explicitly in this case. Then figure out how to generalize when $tau$ is a product of disjoint cycles: for this try inserting copies of $sigma^{-1}sigma$ in various places of the product $sigmatausigma^{-1}$ when $tau$ is written as a product of disjoint cycles.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This one did it ! Thanks
            $endgroup$
            – 90intuition
            Mar 19 '13 at 20:09










          • $begingroup$
            Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
            $endgroup$
            – 90intuition
            Mar 19 '13 at 20:38










          • $begingroup$
            I hope that this is what you meant.
            $endgroup$
            – 90intuition
            Mar 19 '13 at 20:38










          • $begingroup$
            Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
            $endgroup$
            – Shaun
            Dec 24 '18 at 21:11



















          4












          $begingroup$

          First, it suffices to prove that $tausigmatau^{-1}$ and $sigma$ have the same cycle type.



          Suppse one of the cycles in $sigma$ is $(a_1,a_2,dots,a_n)$. Then, try to prove that there is a corresponding cycle in $tausigmatau^{-1}$ that is $big(tau(a_1),tau(a_2),dots,tau(a_n) big)$.



          If $sigma$ sends $a_1$ to $a_2$, what does $tausigmatau^{-1}$ send $tau(a_1)$ to?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
            $endgroup$
            – whacka
            Aug 2 '15 at 21:06





















          3












          $begingroup$

          In the light of @Robert's answer, if $x$ and $y$ are two permutations of a set $Omega$, and $x=(xi_1,xi_2,...,xi_k)$ then $$y^{-1}xy=(xi_1^y,xi_2^y,...,xi_k^y)$$ In fact, in the product $xy$ we expect to have the result of first applying the mapping $x$ and then the mapping $y$. This is what @Robert's trying to tell you.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nice! $quad ddot smile quad +1 quad$
            $endgroup$
            – amWhy
            Mar 20 '13 at 0:15



















          0












          $begingroup$

          Suppose that $sigma tau=(a_1a_2cdots a_k)(b_1b_2cdots b_{lambda})cdots$.

          Then $tausigma(tau(a_i))=tau(sigmatau(a_i))=tau(a_{i+1})ldots$



          Show that $tausigma=left(tau(a_1)tau(a_2)cdotstau(a_k)right)left(tau(b_1)tau(b_2)cdotstau(b_{lambda})right)cdots$.






          share|cite|improve this answer









          $endgroup$













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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6












            $begingroup$

            Hint: they are conjugate elements of the group.






            share|cite|improve this answer









            $endgroup$









            • 5




              $begingroup$
              My book introduces conjugates in the next chapter.
              $endgroup$
              – 90intuition
              Mar 19 '13 at 15:59
















            6












            $begingroup$

            Hint: they are conjugate elements of the group.






            share|cite|improve this answer









            $endgroup$









            • 5




              $begingroup$
              My book introduces conjugates in the next chapter.
              $endgroup$
              – 90intuition
              Mar 19 '13 at 15:59














            6












            6








            6





            $begingroup$

            Hint: they are conjugate elements of the group.






            share|cite|improve this answer









            $endgroup$



            Hint: they are conjugate elements of the group.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 19 '13 at 15:57









            Robert IsraelRobert Israel

            324k23214468




            324k23214468








            • 5




              $begingroup$
              My book introduces conjugates in the next chapter.
              $endgroup$
              – 90intuition
              Mar 19 '13 at 15:59














            • 5




              $begingroup$
              My book introduces conjugates in the next chapter.
              $endgroup$
              – 90intuition
              Mar 19 '13 at 15:59








            5




            5




            $begingroup$
            My book introduces conjugates in the next chapter.
            $endgroup$
            – 90intuition
            Mar 19 '13 at 15:59




            $begingroup$
            My book introduces conjugates in the next chapter.
            $endgroup$
            – 90intuition
            Mar 19 '13 at 15:59











            4












            $begingroup$

            First consider the case that $tau$ is a single cycle, say $tau = (i_1,i_2,ldots,i_k)$. You should be able to compute $sigmatau$ and $tausigma$ explicitly in this case. Then figure out how to generalize when $tau$ is a product of disjoint cycles: for this try inserting copies of $sigma^{-1}sigma$ in various places of the product $sigmatausigma^{-1}$ when $tau$ is written as a product of disjoint cycles.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This one did it ! Thanks
              $endgroup$
              – 90intuition
              Mar 19 '13 at 20:09










            • $begingroup$
              Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
              $endgroup$
              – 90intuition
              Mar 19 '13 at 20:38










            • $begingroup$
              I hope that this is what you meant.
              $endgroup$
              – 90intuition
              Mar 19 '13 at 20:38










            • $begingroup$
              Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
              $endgroup$
              – Shaun
              Dec 24 '18 at 21:11
















            4












            $begingroup$

            First consider the case that $tau$ is a single cycle, say $tau = (i_1,i_2,ldots,i_k)$. You should be able to compute $sigmatau$ and $tausigma$ explicitly in this case. Then figure out how to generalize when $tau$ is a product of disjoint cycles: for this try inserting copies of $sigma^{-1}sigma$ in various places of the product $sigmatausigma^{-1}$ when $tau$ is written as a product of disjoint cycles.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This one did it ! Thanks
              $endgroup$
              – 90intuition
              Mar 19 '13 at 20:09










            • $begingroup$
              Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
              $endgroup$
              – 90intuition
              Mar 19 '13 at 20:38










            • $begingroup$
              I hope that this is what you meant.
              $endgroup$
              – 90intuition
              Mar 19 '13 at 20:38










            • $begingroup$
              Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
              $endgroup$
              – Shaun
              Dec 24 '18 at 21:11














            4












            4








            4





            $begingroup$

            First consider the case that $tau$ is a single cycle, say $tau = (i_1,i_2,ldots,i_k)$. You should be able to compute $sigmatau$ and $tausigma$ explicitly in this case. Then figure out how to generalize when $tau$ is a product of disjoint cycles: for this try inserting copies of $sigma^{-1}sigma$ in various places of the product $sigmatausigma^{-1}$ when $tau$ is written as a product of disjoint cycles.






            share|cite|improve this answer









            $endgroup$



            First consider the case that $tau$ is a single cycle, say $tau = (i_1,i_2,ldots,i_k)$. You should be able to compute $sigmatau$ and $tausigma$ explicitly in this case. Then figure out how to generalize when $tau$ is a product of disjoint cycles: for this try inserting copies of $sigma^{-1}sigma$ in various places of the product $sigmatausigma^{-1}$ when $tau$ is written as a product of disjoint cycles.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 19 '13 at 16:03









            Santiago CanezSantiago Canez

            2,07111113




            2,07111113












            • $begingroup$
              This one did it ! Thanks
              $endgroup$
              – 90intuition
              Mar 19 '13 at 20:09










            • $begingroup$
              Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
              $endgroup$
              – 90intuition
              Mar 19 '13 at 20:38










            • $begingroup$
              I hope that this is what you meant.
              $endgroup$
              – 90intuition
              Mar 19 '13 at 20:38










            • $begingroup$
              Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
              $endgroup$
              – Shaun
              Dec 24 '18 at 21:11


















            • $begingroup$
              This one did it ! Thanks
              $endgroup$
              – 90intuition
              Mar 19 '13 at 20:09










            • $begingroup$
              Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
              $endgroup$
              – 90intuition
              Mar 19 '13 at 20:38










            • $begingroup$
              I hope that this is what you meant.
              $endgroup$
              – 90intuition
              Mar 19 '13 at 20:38










            • $begingroup$
              Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
              $endgroup$
              – Shaun
              Dec 24 '18 at 21:11
















            $begingroup$
            This one did it ! Thanks
            $endgroup$
            – 90intuition
            Mar 19 '13 at 20:09




            $begingroup$
            This one did it ! Thanks
            $endgroup$
            – 90intuition
            Mar 19 '13 at 20:09












            $begingroup$
            Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
            $endgroup$
            – 90intuition
            Mar 19 '13 at 20:38




            $begingroup$
            Let $σ=σ_1...σ_r$ with $σ_1=(a_1...a_{l_1}), σ_2=(b_1...b_{l_2}), σ_3=...$. Then you get: begin{align*} τστ^{-1}&=τσ_1...σ_rτ^{-1}\ &=τσ_1(τ^{-1}τ)σ_2(τ^{-1}τ)...(τ^{-1}τ)σ_rτ^{-1}\ &=τ(a_1...a_{l_1})τ^{-1}τ(b_1...b_{l_2})τ^{-1}τ...τσ_rτ^{-1}\ &=(τ(a_1)...τ(a_{l_1}))(τ(b_1)...τ(b_{l_2}))...τσ_rτ^{-1}\ end{align*}
            $endgroup$
            – 90intuition
            Mar 19 '13 at 20:38












            $begingroup$
            I hope that this is what you meant.
            $endgroup$
            – 90intuition
            Mar 19 '13 at 20:38




            $begingroup$
            I hope that this is what you meant.
            $endgroup$
            – 90intuition
            Mar 19 '13 at 20:38












            $begingroup$
            Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
            $endgroup$
            – Shaun
            Dec 24 '18 at 21:11




            $begingroup$
            Since this answer worked for you, @90intuition, please accept it by clicking the checkmark :)
            $endgroup$
            – Shaun
            Dec 24 '18 at 21:11











            4












            $begingroup$

            First, it suffices to prove that $tausigmatau^{-1}$ and $sigma$ have the same cycle type.



            Suppse one of the cycles in $sigma$ is $(a_1,a_2,dots,a_n)$. Then, try to prove that there is a corresponding cycle in $tausigmatau^{-1}$ that is $big(tau(a_1),tau(a_2),dots,tau(a_n) big)$.



            If $sigma$ sends $a_1$ to $a_2$, what does $tausigmatau^{-1}$ send $tau(a_1)$ to?






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
              $endgroup$
              – whacka
              Aug 2 '15 at 21:06


















            4












            $begingroup$

            First, it suffices to prove that $tausigmatau^{-1}$ and $sigma$ have the same cycle type.



            Suppse one of the cycles in $sigma$ is $(a_1,a_2,dots,a_n)$. Then, try to prove that there is a corresponding cycle in $tausigmatau^{-1}$ that is $big(tau(a_1),tau(a_2),dots,tau(a_n) big)$.



            If $sigma$ sends $a_1$ to $a_2$, what does $tausigmatau^{-1}$ send $tau(a_1)$ to?






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
              $endgroup$
              – whacka
              Aug 2 '15 at 21:06
















            4












            4








            4





            $begingroup$

            First, it suffices to prove that $tausigmatau^{-1}$ and $sigma$ have the same cycle type.



            Suppse one of the cycles in $sigma$ is $(a_1,a_2,dots,a_n)$. Then, try to prove that there is a corresponding cycle in $tausigmatau^{-1}$ that is $big(tau(a_1),tau(a_2),dots,tau(a_n) big)$.



            If $sigma$ sends $a_1$ to $a_2$, what does $tausigmatau^{-1}$ send $tau(a_1)$ to?






            share|cite|improve this answer











            $endgroup$



            First, it suffices to prove that $tausigmatau^{-1}$ and $sigma$ have the same cycle type.



            Suppse one of the cycles in $sigma$ is $(a_1,a_2,dots,a_n)$. Then, try to prove that there is a corresponding cycle in $tausigmatau^{-1}$ that is $big(tau(a_1),tau(a_2),dots,tau(a_n) big)$.



            If $sigma$ sends $a_1$ to $a_2$, what does $tausigmatau^{-1}$ send $tau(a_1)$ to?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 19 '13 at 18:23

























            answered Mar 19 '13 at 16:51









            NECingNECing

            3,30231628




            3,30231628












            • $begingroup$
              Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
              $endgroup$
              – whacka
              Aug 2 '15 at 21:06




















            • $begingroup$
              Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
              $endgroup$
              – whacka
              Aug 2 '15 at 21:06


















            $begingroup$
            Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
            $endgroup$
            – whacka
            Aug 2 '15 at 21:06






            $begingroup$
            Specifically, it suffices to show $alphabetaalpha^{-1}$ and $beta$ have the same cycle type, because then one can set $beta=tausigma$ and $alpha=sigma^{-1}$ to conclude $sigmatau$ and $tausigma$ have the same cycle type. Which is also the hint of Robert's answer. (Sorry for the anachronistic comment.)
            $endgroup$
            – whacka
            Aug 2 '15 at 21:06













            3












            $begingroup$

            In the light of @Robert's answer, if $x$ and $y$ are two permutations of a set $Omega$, and $x=(xi_1,xi_2,...,xi_k)$ then $$y^{-1}xy=(xi_1^y,xi_2^y,...,xi_k^y)$$ In fact, in the product $xy$ we expect to have the result of first applying the mapping $x$ and then the mapping $y$. This is what @Robert's trying to tell you.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Nice! $quad ddot smile quad +1 quad$
              $endgroup$
              – amWhy
              Mar 20 '13 at 0:15
















            3












            $begingroup$

            In the light of @Robert's answer, if $x$ and $y$ are two permutations of a set $Omega$, and $x=(xi_1,xi_2,...,xi_k)$ then $$y^{-1}xy=(xi_1^y,xi_2^y,...,xi_k^y)$$ In fact, in the product $xy$ we expect to have the result of first applying the mapping $x$ and then the mapping $y$. This is what @Robert's trying to tell you.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Nice! $quad ddot smile quad +1 quad$
              $endgroup$
              – amWhy
              Mar 20 '13 at 0:15














            3












            3








            3





            $begingroup$

            In the light of @Robert's answer, if $x$ and $y$ are two permutations of a set $Omega$, and $x=(xi_1,xi_2,...,xi_k)$ then $$y^{-1}xy=(xi_1^y,xi_2^y,...,xi_k^y)$$ In fact, in the product $xy$ we expect to have the result of first applying the mapping $x$ and then the mapping $y$. This is what @Robert's trying to tell you.






            share|cite|improve this answer









            $endgroup$



            In the light of @Robert's answer, if $x$ and $y$ are two permutations of a set $Omega$, and $x=(xi_1,xi_2,...,xi_k)$ then $$y^{-1}xy=(xi_1^y,xi_2^y,...,xi_k^y)$$ In fact, in the product $xy$ we expect to have the result of first applying the mapping $x$ and then the mapping $y$. This is what @Robert's trying to tell you.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 19 '13 at 16:51









            mrsmrs

            1




            1












            • $begingroup$
              Nice! $quad ddot smile quad +1 quad$
              $endgroup$
              – amWhy
              Mar 20 '13 at 0:15


















            • $begingroup$
              Nice! $quad ddot smile quad +1 quad$
              $endgroup$
              – amWhy
              Mar 20 '13 at 0:15
















            $begingroup$
            Nice! $quad ddot smile quad +1 quad$
            $endgroup$
            – amWhy
            Mar 20 '13 at 0:15




            $begingroup$
            Nice! $quad ddot smile quad +1 quad$
            $endgroup$
            – amWhy
            Mar 20 '13 at 0:15











            0












            $begingroup$

            Suppose that $sigma tau=(a_1a_2cdots a_k)(b_1b_2cdots b_{lambda})cdots$.

            Then $tausigma(tau(a_i))=tau(sigmatau(a_i))=tau(a_{i+1})ldots$



            Show that $tausigma=left(tau(a_1)tau(a_2)cdotstau(a_k)right)left(tau(b_1)tau(b_2)cdotstau(b_{lambda})right)cdots$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Suppose that $sigma tau=(a_1a_2cdots a_k)(b_1b_2cdots b_{lambda})cdots$.

              Then $tausigma(tau(a_i))=tau(sigmatau(a_i))=tau(a_{i+1})ldots$



              Show that $tausigma=left(tau(a_1)tau(a_2)cdotstau(a_k)right)left(tau(b_1)tau(b_2)cdotstau(b_{lambda})right)cdots$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Suppose that $sigma tau=(a_1a_2cdots a_k)(b_1b_2cdots b_{lambda})cdots$.

                Then $tausigma(tau(a_i))=tau(sigmatau(a_i))=tau(a_{i+1})ldots$



                Show that $tausigma=left(tau(a_1)tau(a_2)cdotstau(a_k)right)left(tau(b_1)tau(b_2)cdotstau(b_{lambda})right)cdots$.






                share|cite|improve this answer









                $endgroup$



                Suppose that $sigma tau=(a_1a_2cdots a_k)(b_1b_2cdots b_{lambda})cdots$.

                Then $tausigma(tau(a_i))=tau(sigmatau(a_i))=tau(a_{i+1})ldots$



                Show that $tausigma=left(tau(a_1)tau(a_2)cdotstau(a_k)right)left(tau(b_1)tau(b_2)cdotstau(b_{lambda})right)cdots$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 19 '13 at 20:04









                P..P..

                13.4k22348




                13.4k22348






























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