On the product of two cycles (and its conjugates)












1














So this is from Charles C. Pinter's "A Book of Abstract Algebra"- specifically, it's from the second chapter on permutations.
The question is:



Let $alpha_1$ and $alpha_2$ be cycles of the same length. Let $beta_1$ and $beta_2$ be cycles of the same length. Let $alpha_1$ and $beta_1$ be disjoint, and let $alpha_2$ and $beta_2$ be disjoint. [Prove that] There is a permutation, $piin S_n$, such that $alpha_1beta_1=pialpha_2beta_2pi^{-1}$.



That's the question.



But this is actually the last of a five part question, sooooo here are the previous parts - the parts build on each other and it looks like part 4 (which can be found near the bottom) is of greatest relevance but I can't quite see how to use it here.



Part 1 was:



Let $alpha=(a_1,...,a_s)$ be a cycle and let $pi$ be a permutation in $S_n$. Then $pialphapi^{-1}$ is the cycle $(pi(alpha_1),...,pi(alpha_s))$.



(a solution can be found here Proof for conjugate cycles)



Part 2 was:



Conclude from part 1: Any two cycles of the same length are conjugates of each other.



The idea here is not too hard to formulate and some formalization is given in a hint which suggests defining a permutation, $pi$, as sending any element not in either cycle to itself, sending all of the elements of one of the cycles, denoted by A, to the other, denoted by B. Finally, define $pi$ over B-A as some bijection between B-A and A-B. Now, it is easy to show that $pi$ is a permutation and by part 1, that's enough.



Parts 3 and 4 are straightforward.



Part 3:



If $alpha$ and $beta$ are disjoint cycles, then $pialphapi^{-1}$ and $pibetapi^{-1}$ are disjoint cycles.

(this follows directly from part 1)



And Part 4:



Let $sigma$ be a product $alpha_1...alpha_t$ of t disjoint cycles of lengths $l_1,...,l_t$, respectively. Then $pisigmapi^{-1}$ is also a product of t disjoint cycles of lengths $l_1,...,l_t$.

(an easy generalization of part 3)



In the fifth part, I see how one can find a $pi$ such that $alpha_1=pialpha_2pi^{-1}$. I also see how one can find a $pi$ such that $beta_1=pibeta_2pi^{-1}$ (we can get this via an immediate application of part 2). What I can't see is how we can ensure that these two $pi$'s will be the same (which is what the question seems to be asking). I also can't see how to use Part 4 here as, on its own, it doesn't seem enough- it looks to me like we would need, in addition, to show that the act of taking a conjugate could be made 'onto' in a sense (and I don't get how to do that).



Any help appreciated.










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    1














    So this is from Charles C. Pinter's "A Book of Abstract Algebra"- specifically, it's from the second chapter on permutations.
    The question is:



    Let $alpha_1$ and $alpha_2$ be cycles of the same length. Let $beta_1$ and $beta_2$ be cycles of the same length. Let $alpha_1$ and $beta_1$ be disjoint, and let $alpha_2$ and $beta_2$ be disjoint. [Prove that] There is a permutation, $piin S_n$, such that $alpha_1beta_1=pialpha_2beta_2pi^{-1}$.



    That's the question.



    But this is actually the last of a five part question, sooooo here are the previous parts - the parts build on each other and it looks like part 4 (which can be found near the bottom) is of greatest relevance but I can't quite see how to use it here.



    Part 1 was:



    Let $alpha=(a_1,...,a_s)$ be a cycle and let $pi$ be a permutation in $S_n$. Then $pialphapi^{-1}$ is the cycle $(pi(alpha_1),...,pi(alpha_s))$.



    (a solution can be found here Proof for conjugate cycles)



    Part 2 was:



    Conclude from part 1: Any two cycles of the same length are conjugates of each other.



    The idea here is not too hard to formulate and some formalization is given in a hint which suggests defining a permutation, $pi$, as sending any element not in either cycle to itself, sending all of the elements of one of the cycles, denoted by A, to the other, denoted by B. Finally, define $pi$ over B-A as some bijection between B-A and A-B. Now, it is easy to show that $pi$ is a permutation and by part 1, that's enough.



    Parts 3 and 4 are straightforward.



    Part 3:



    If $alpha$ and $beta$ are disjoint cycles, then $pialphapi^{-1}$ and $pibetapi^{-1}$ are disjoint cycles.

    (this follows directly from part 1)



    And Part 4:



    Let $sigma$ be a product $alpha_1...alpha_t$ of t disjoint cycles of lengths $l_1,...,l_t$, respectively. Then $pisigmapi^{-1}$ is also a product of t disjoint cycles of lengths $l_1,...,l_t$.

    (an easy generalization of part 3)



    In the fifth part, I see how one can find a $pi$ such that $alpha_1=pialpha_2pi^{-1}$. I also see how one can find a $pi$ such that $beta_1=pibeta_2pi^{-1}$ (we can get this via an immediate application of part 2). What I can't see is how we can ensure that these two $pi$'s will be the same (which is what the question seems to be asking). I also can't see how to use Part 4 here as, on its own, it doesn't seem enough- it looks to me like we would need, in addition, to show that the act of taking a conjugate could be made 'onto' in a sense (and I don't get how to do that).



    Any help appreciated.










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      1












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      1


      1





      So this is from Charles C. Pinter's "A Book of Abstract Algebra"- specifically, it's from the second chapter on permutations.
      The question is:



      Let $alpha_1$ and $alpha_2$ be cycles of the same length. Let $beta_1$ and $beta_2$ be cycles of the same length. Let $alpha_1$ and $beta_1$ be disjoint, and let $alpha_2$ and $beta_2$ be disjoint. [Prove that] There is a permutation, $piin S_n$, such that $alpha_1beta_1=pialpha_2beta_2pi^{-1}$.



      That's the question.



      But this is actually the last of a five part question, sooooo here are the previous parts - the parts build on each other and it looks like part 4 (which can be found near the bottom) is of greatest relevance but I can't quite see how to use it here.



      Part 1 was:



      Let $alpha=(a_1,...,a_s)$ be a cycle and let $pi$ be a permutation in $S_n$. Then $pialphapi^{-1}$ is the cycle $(pi(alpha_1),...,pi(alpha_s))$.



      (a solution can be found here Proof for conjugate cycles)



      Part 2 was:



      Conclude from part 1: Any two cycles of the same length are conjugates of each other.



      The idea here is not too hard to formulate and some formalization is given in a hint which suggests defining a permutation, $pi$, as sending any element not in either cycle to itself, sending all of the elements of one of the cycles, denoted by A, to the other, denoted by B. Finally, define $pi$ over B-A as some bijection between B-A and A-B. Now, it is easy to show that $pi$ is a permutation and by part 1, that's enough.



      Parts 3 and 4 are straightforward.



      Part 3:



      If $alpha$ and $beta$ are disjoint cycles, then $pialphapi^{-1}$ and $pibetapi^{-1}$ are disjoint cycles.

      (this follows directly from part 1)



      And Part 4:



      Let $sigma$ be a product $alpha_1...alpha_t$ of t disjoint cycles of lengths $l_1,...,l_t$, respectively. Then $pisigmapi^{-1}$ is also a product of t disjoint cycles of lengths $l_1,...,l_t$.

      (an easy generalization of part 3)



      In the fifth part, I see how one can find a $pi$ such that $alpha_1=pialpha_2pi^{-1}$. I also see how one can find a $pi$ such that $beta_1=pibeta_2pi^{-1}$ (we can get this via an immediate application of part 2). What I can't see is how we can ensure that these two $pi$'s will be the same (which is what the question seems to be asking). I also can't see how to use Part 4 here as, on its own, it doesn't seem enough- it looks to me like we would need, in addition, to show that the act of taking a conjugate could be made 'onto' in a sense (and I don't get how to do that).



      Any help appreciated.










      share|cite|improve this question









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      Cardioid_Ass_22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      So this is from Charles C. Pinter's "A Book of Abstract Algebra"- specifically, it's from the second chapter on permutations.
      The question is:



      Let $alpha_1$ and $alpha_2$ be cycles of the same length. Let $beta_1$ and $beta_2$ be cycles of the same length. Let $alpha_1$ and $beta_1$ be disjoint, and let $alpha_2$ and $beta_2$ be disjoint. [Prove that] There is a permutation, $piin S_n$, such that $alpha_1beta_1=pialpha_2beta_2pi^{-1}$.



      That's the question.



      But this is actually the last of a five part question, sooooo here are the previous parts - the parts build on each other and it looks like part 4 (which can be found near the bottom) is of greatest relevance but I can't quite see how to use it here.



      Part 1 was:



      Let $alpha=(a_1,...,a_s)$ be a cycle and let $pi$ be a permutation in $S_n$. Then $pialphapi^{-1}$ is the cycle $(pi(alpha_1),...,pi(alpha_s))$.



      (a solution can be found here Proof for conjugate cycles)



      Part 2 was:



      Conclude from part 1: Any two cycles of the same length are conjugates of each other.



      The idea here is not too hard to formulate and some formalization is given in a hint which suggests defining a permutation, $pi$, as sending any element not in either cycle to itself, sending all of the elements of one of the cycles, denoted by A, to the other, denoted by B. Finally, define $pi$ over B-A as some bijection between B-A and A-B. Now, it is easy to show that $pi$ is a permutation and by part 1, that's enough.



      Parts 3 and 4 are straightforward.



      Part 3:



      If $alpha$ and $beta$ are disjoint cycles, then $pialphapi^{-1}$ and $pibetapi^{-1}$ are disjoint cycles.

      (this follows directly from part 1)



      And Part 4:



      Let $sigma$ be a product $alpha_1...alpha_t$ of t disjoint cycles of lengths $l_1,...,l_t$, respectively. Then $pisigmapi^{-1}$ is also a product of t disjoint cycles of lengths $l_1,...,l_t$.

      (an easy generalization of part 3)



      In the fifth part, I see how one can find a $pi$ such that $alpha_1=pialpha_2pi^{-1}$. I also see how one can find a $pi$ such that $beta_1=pibeta_2pi^{-1}$ (we can get this via an immediate application of part 2). What I can't see is how we can ensure that these two $pi$'s will be the same (which is what the question seems to be asking). I also can't see how to use Part 4 here as, on its own, it doesn't seem enough- it looks to me like we would need, in addition, to show that the act of taking a conjugate could be made 'onto' in a sense (and I don't get how to do that).



      Any help appreciated.







      abstract-algebra permutation-cycles






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      edited Jan 4 at 20:39







      Cardioid_Ass_22













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      asked Jan 4 at 19:39









      Cardioid_Ass_22Cardioid_Ass_22

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          Let



          $alpha_1=(alpha^1_1,alpha^1_2,dots,alpha^1_n),alpha_2=(alpha^2_1,alpha^2_2,dots,alpha^2_n),$
          $beta_1=(beta^1_1,beta^1_2,dots,beta^1_m),beta_2=(beta^2_1,beta^2_2,dots,beta^2_m)$.



          Let $pi$ be any permutation which satisfies $pi(alpha^2_i)=alpha^1_i$ and $pi(beta^2_j)=beta^1_j$ for all $1le i le n$ and $1le j le m$. Then $pialpha_2beta_2pi^{-1}=pialpha_2pi^{-1}pibeta_2pi^{-1}=alpha_1beta_1$, using part (1) of your problem.



          Edit: I see now that the crux of your question is why such a permutation $pi$ exists. For any given pair of lists of $k$ distinct numbers $x_1,dots,x_k$ and $y_1,dots,y_k$, there exists a permutation which satisfies $pi(x_i)=y_i$. In fact, there are $(n-k)!$ choices for $pi$. To choose $pi$, let $z_1,dots,z_{n-k}$ be a list of the numbers not present in $x_1,dots,x_k$, then for each $j=1,2,dots,n-k$ in order, set $pi(z_j)$ to be any number unequal to $y_1,dots,y_k,z_1,z_2,dots,z_{j-1}$.






          share|cite|improve this answer























          • I'm sorry but I'm still confused. The process that you have described looks to me to work for single cycles. How can we extend the process for constructing a permutation you describe so that it ensures such a permutation exists for a product of cycles? Or if it already applies to products of cycles (at least for just a product of two cycles), well, how can we show that it works even in that case?
            – Cardioid_Ass_22
            Jan 4 at 20:38










          • The process I described in my edit also works for double cycles. Just let $x_1,dots,x_k$ be the list of $n+m$ numbers $alpha^2_1,dots,alpha^2_n,beta^2_1,dots,beta^2_m$, and let $y_1,dots,y_k$ be the same list but with $alpha^2$ and $beta^2$ replaced with $alpha^1$ and $beta^1$.
            – Mike Earnest
            Jan 4 at 21:24












          • Goodness, that's a lot easier than anything I had in my mind! Yes, that seems an absolutely fair argument. Thank you!
            – Cardioid_Ass_22
            Jan 4 at 21:49











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          1 Answer
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          Let



          $alpha_1=(alpha^1_1,alpha^1_2,dots,alpha^1_n),alpha_2=(alpha^2_1,alpha^2_2,dots,alpha^2_n),$
          $beta_1=(beta^1_1,beta^1_2,dots,beta^1_m),beta_2=(beta^2_1,beta^2_2,dots,beta^2_m)$.



          Let $pi$ be any permutation which satisfies $pi(alpha^2_i)=alpha^1_i$ and $pi(beta^2_j)=beta^1_j$ for all $1le i le n$ and $1le j le m$. Then $pialpha_2beta_2pi^{-1}=pialpha_2pi^{-1}pibeta_2pi^{-1}=alpha_1beta_1$, using part (1) of your problem.



          Edit: I see now that the crux of your question is why such a permutation $pi$ exists. For any given pair of lists of $k$ distinct numbers $x_1,dots,x_k$ and $y_1,dots,y_k$, there exists a permutation which satisfies $pi(x_i)=y_i$. In fact, there are $(n-k)!$ choices for $pi$. To choose $pi$, let $z_1,dots,z_{n-k}$ be a list of the numbers not present in $x_1,dots,x_k$, then for each $j=1,2,dots,n-k$ in order, set $pi(z_j)$ to be any number unequal to $y_1,dots,y_k,z_1,z_2,dots,z_{j-1}$.






          share|cite|improve this answer























          • I'm sorry but I'm still confused. The process that you have described looks to me to work for single cycles. How can we extend the process for constructing a permutation you describe so that it ensures such a permutation exists for a product of cycles? Or if it already applies to products of cycles (at least for just a product of two cycles), well, how can we show that it works even in that case?
            – Cardioid_Ass_22
            Jan 4 at 20:38










          • The process I described in my edit also works for double cycles. Just let $x_1,dots,x_k$ be the list of $n+m$ numbers $alpha^2_1,dots,alpha^2_n,beta^2_1,dots,beta^2_m$, and let $y_1,dots,y_k$ be the same list but with $alpha^2$ and $beta^2$ replaced with $alpha^1$ and $beta^1$.
            – Mike Earnest
            Jan 4 at 21:24












          • Goodness, that's a lot easier than anything I had in my mind! Yes, that seems an absolutely fair argument. Thank you!
            – Cardioid_Ass_22
            Jan 4 at 21:49
















          2














          Let



          $alpha_1=(alpha^1_1,alpha^1_2,dots,alpha^1_n),alpha_2=(alpha^2_1,alpha^2_2,dots,alpha^2_n),$
          $beta_1=(beta^1_1,beta^1_2,dots,beta^1_m),beta_2=(beta^2_1,beta^2_2,dots,beta^2_m)$.



          Let $pi$ be any permutation which satisfies $pi(alpha^2_i)=alpha^1_i$ and $pi(beta^2_j)=beta^1_j$ for all $1le i le n$ and $1le j le m$. Then $pialpha_2beta_2pi^{-1}=pialpha_2pi^{-1}pibeta_2pi^{-1}=alpha_1beta_1$, using part (1) of your problem.



          Edit: I see now that the crux of your question is why such a permutation $pi$ exists. For any given pair of lists of $k$ distinct numbers $x_1,dots,x_k$ and $y_1,dots,y_k$, there exists a permutation which satisfies $pi(x_i)=y_i$. In fact, there are $(n-k)!$ choices for $pi$. To choose $pi$, let $z_1,dots,z_{n-k}$ be a list of the numbers not present in $x_1,dots,x_k$, then for each $j=1,2,dots,n-k$ in order, set $pi(z_j)$ to be any number unequal to $y_1,dots,y_k,z_1,z_2,dots,z_{j-1}$.






          share|cite|improve this answer























          • I'm sorry but I'm still confused. The process that you have described looks to me to work for single cycles. How can we extend the process for constructing a permutation you describe so that it ensures such a permutation exists for a product of cycles? Or if it already applies to products of cycles (at least for just a product of two cycles), well, how can we show that it works even in that case?
            – Cardioid_Ass_22
            Jan 4 at 20:38










          • The process I described in my edit also works for double cycles. Just let $x_1,dots,x_k$ be the list of $n+m$ numbers $alpha^2_1,dots,alpha^2_n,beta^2_1,dots,beta^2_m$, and let $y_1,dots,y_k$ be the same list but with $alpha^2$ and $beta^2$ replaced with $alpha^1$ and $beta^1$.
            – Mike Earnest
            Jan 4 at 21:24












          • Goodness, that's a lot easier than anything I had in my mind! Yes, that seems an absolutely fair argument. Thank you!
            – Cardioid_Ass_22
            Jan 4 at 21:49














          2












          2








          2






          Let



          $alpha_1=(alpha^1_1,alpha^1_2,dots,alpha^1_n),alpha_2=(alpha^2_1,alpha^2_2,dots,alpha^2_n),$
          $beta_1=(beta^1_1,beta^1_2,dots,beta^1_m),beta_2=(beta^2_1,beta^2_2,dots,beta^2_m)$.



          Let $pi$ be any permutation which satisfies $pi(alpha^2_i)=alpha^1_i$ and $pi(beta^2_j)=beta^1_j$ for all $1le i le n$ and $1le j le m$. Then $pialpha_2beta_2pi^{-1}=pialpha_2pi^{-1}pibeta_2pi^{-1}=alpha_1beta_1$, using part (1) of your problem.



          Edit: I see now that the crux of your question is why such a permutation $pi$ exists. For any given pair of lists of $k$ distinct numbers $x_1,dots,x_k$ and $y_1,dots,y_k$, there exists a permutation which satisfies $pi(x_i)=y_i$. In fact, there are $(n-k)!$ choices for $pi$. To choose $pi$, let $z_1,dots,z_{n-k}$ be a list of the numbers not present in $x_1,dots,x_k$, then for each $j=1,2,dots,n-k$ in order, set $pi(z_j)$ to be any number unequal to $y_1,dots,y_k,z_1,z_2,dots,z_{j-1}$.






          share|cite|improve this answer














          Let



          $alpha_1=(alpha^1_1,alpha^1_2,dots,alpha^1_n),alpha_2=(alpha^2_1,alpha^2_2,dots,alpha^2_n),$
          $beta_1=(beta^1_1,beta^1_2,dots,beta^1_m),beta_2=(beta^2_1,beta^2_2,dots,beta^2_m)$.



          Let $pi$ be any permutation which satisfies $pi(alpha^2_i)=alpha^1_i$ and $pi(beta^2_j)=beta^1_j$ for all $1le i le n$ and $1le j le m$. Then $pialpha_2beta_2pi^{-1}=pialpha_2pi^{-1}pibeta_2pi^{-1}=alpha_1beta_1$, using part (1) of your problem.



          Edit: I see now that the crux of your question is why such a permutation $pi$ exists. For any given pair of lists of $k$ distinct numbers $x_1,dots,x_k$ and $y_1,dots,y_k$, there exists a permutation which satisfies $pi(x_i)=y_i$. In fact, there are $(n-k)!$ choices for $pi$. To choose $pi$, let $z_1,dots,z_{n-k}$ be a list of the numbers not present in $x_1,dots,x_k$, then for each $j=1,2,dots,n-k$ in order, set $pi(z_j)$ to be any number unequal to $y_1,dots,y_k,z_1,z_2,dots,z_{j-1}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 4 at 20:17

























          answered Jan 4 at 20:08









          Mike EarnestMike Earnest

          20.7k11950




          20.7k11950












          • I'm sorry but I'm still confused. The process that you have described looks to me to work for single cycles. How can we extend the process for constructing a permutation you describe so that it ensures such a permutation exists for a product of cycles? Or if it already applies to products of cycles (at least for just a product of two cycles), well, how can we show that it works even in that case?
            – Cardioid_Ass_22
            Jan 4 at 20:38










          • The process I described in my edit also works for double cycles. Just let $x_1,dots,x_k$ be the list of $n+m$ numbers $alpha^2_1,dots,alpha^2_n,beta^2_1,dots,beta^2_m$, and let $y_1,dots,y_k$ be the same list but with $alpha^2$ and $beta^2$ replaced with $alpha^1$ and $beta^1$.
            – Mike Earnest
            Jan 4 at 21:24












          • Goodness, that's a lot easier than anything I had in my mind! Yes, that seems an absolutely fair argument. Thank you!
            – Cardioid_Ass_22
            Jan 4 at 21:49


















          • I'm sorry but I'm still confused. The process that you have described looks to me to work for single cycles. How can we extend the process for constructing a permutation you describe so that it ensures such a permutation exists for a product of cycles? Or if it already applies to products of cycles (at least for just a product of two cycles), well, how can we show that it works even in that case?
            – Cardioid_Ass_22
            Jan 4 at 20:38










          • The process I described in my edit also works for double cycles. Just let $x_1,dots,x_k$ be the list of $n+m$ numbers $alpha^2_1,dots,alpha^2_n,beta^2_1,dots,beta^2_m$, and let $y_1,dots,y_k$ be the same list but with $alpha^2$ and $beta^2$ replaced with $alpha^1$ and $beta^1$.
            – Mike Earnest
            Jan 4 at 21:24












          • Goodness, that's a lot easier than anything I had in my mind! Yes, that seems an absolutely fair argument. Thank you!
            – Cardioid_Ass_22
            Jan 4 at 21:49
















          I'm sorry but I'm still confused. The process that you have described looks to me to work for single cycles. How can we extend the process for constructing a permutation you describe so that it ensures such a permutation exists for a product of cycles? Or if it already applies to products of cycles (at least for just a product of two cycles), well, how can we show that it works even in that case?
          – Cardioid_Ass_22
          Jan 4 at 20:38




          I'm sorry but I'm still confused. The process that you have described looks to me to work for single cycles. How can we extend the process for constructing a permutation you describe so that it ensures such a permutation exists for a product of cycles? Or if it already applies to products of cycles (at least for just a product of two cycles), well, how can we show that it works even in that case?
          – Cardioid_Ass_22
          Jan 4 at 20:38












          The process I described in my edit also works for double cycles. Just let $x_1,dots,x_k$ be the list of $n+m$ numbers $alpha^2_1,dots,alpha^2_n,beta^2_1,dots,beta^2_m$, and let $y_1,dots,y_k$ be the same list but with $alpha^2$ and $beta^2$ replaced with $alpha^1$ and $beta^1$.
          – Mike Earnest
          Jan 4 at 21:24






          The process I described in my edit also works for double cycles. Just let $x_1,dots,x_k$ be the list of $n+m$ numbers $alpha^2_1,dots,alpha^2_n,beta^2_1,dots,beta^2_m$, and let $y_1,dots,y_k$ be the same list but with $alpha^2$ and $beta^2$ replaced with $alpha^1$ and $beta^1$.
          – Mike Earnest
          Jan 4 at 21:24














          Goodness, that's a lot easier than anything I had in my mind! Yes, that seems an absolutely fair argument. Thank you!
          – Cardioid_Ass_22
          Jan 4 at 21:49




          Goodness, that's a lot easier than anything I had in my mind! Yes, that seems an absolutely fair argument. Thank you!
          – Cardioid_Ass_22
          Jan 4 at 21:49










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