Decide how many elements who commutate with this symmetric group?











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Let $S_3$ be the symmetric group on $lbrace 1,2,3rbrace.$ Decide how many elements who commutate with $(23)$





Permutation naturally commutes itself, with it's inverse and with the identity permutation. So that's 3.



And then I'm insecure. What shall I do next?










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    up vote
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    down vote

    favorite












    Let $S_3$ be the symmetric group on $lbrace 1,2,3rbrace.$ Decide how many elements who commutate with $(23)$





    Permutation naturally commutes itself, with it's inverse and with the identity permutation. So that's 3.



    And then I'm insecure. What shall I do next?










    share|cite|improve this question


























      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      Let $S_3$ be the symmetric group on $lbrace 1,2,3rbrace.$ Decide how many elements who commutate with $(23)$





      Permutation naturally commutes itself, with it's inverse and with the identity permutation. So that's 3.



      And then I'm insecure. What shall I do next?










      share|cite|improve this question















      Let $S_3$ be the symmetric group on $lbrace 1,2,3rbrace.$ Decide how many elements who commutate with $(23)$





      Permutation naturally commutes itself, with it's inverse and with the identity permutation. So that's 3.



      And then I'm insecure. What shall I do next?







      permutations






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      share|cite|improve this question













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      edited Nov 25 at 20:11









      amWhy

      191k27223439




      191k27223439










      asked Nov 25 at 19:58









      soetirl13

      114




      114






















          1 Answer
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          First of all, the permutation $(23)$ is the inverse of itself so you actually found only $2$ permutations that commute with it so far, not $3$. Next, $S_3$ is a very small group, it has only $6$ elements. So you can just check every element in a direct way.






          share|cite|improve this answer





















          • How is it a inverse of itself? the inverse is (32) ?
            – soetirl13
            Nov 25 at 20:08










          • $(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
            – Mark
            Nov 25 at 20:09










          • Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
            – soetirl13
            Nov 25 at 20:13










          • It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
            – Mark
            Nov 25 at 20:23










          • Ahh 2. :))))) thank you
            – soetirl13
            Nov 25 at 21:06











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          1 Answer
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          1 Answer
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          up vote
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          down vote













          First of all, the permutation $(23)$ is the inverse of itself so you actually found only $2$ permutations that commute with it so far, not $3$. Next, $S_3$ is a very small group, it has only $6$ elements. So you can just check every element in a direct way.






          share|cite|improve this answer





















          • How is it a inverse of itself? the inverse is (32) ?
            – soetirl13
            Nov 25 at 20:08










          • $(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
            – Mark
            Nov 25 at 20:09










          • Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
            – soetirl13
            Nov 25 at 20:13










          • It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
            – Mark
            Nov 25 at 20:23










          • Ahh 2. :))))) thank you
            – soetirl13
            Nov 25 at 21:06















          up vote
          0
          down vote













          First of all, the permutation $(23)$ is the inverse of itself so you actually found only $2$ permutations that commute with it so far, not $3$. Next, $S_3$ is a very small group, it has only $6$ elements. So you can just check every element in a direct way.






          share|cite|improve this answer





















          • How is it a inverse of itself? the inverse is (32) ?
            – soetirl13
            Nov 25 at 20:08










          • $(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
            – Mark
            Nov 25 at 20:09










          • Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
            – soetirl13
            Nov 25 at 20:13










          • It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
            – Mark
            Nov 25 at 20:23










          • Ahh 2. :))))) thank you
            – soetirl13
            Nov 25 at 21:06













          up vote
          0
          down vote










          up vote
          0
          down vote









          First of all, the permutation $(23)$ is the inverse of itself so you actually found only $2$ permutations that commute with it so far, not $3$. Next, $S_3$ is a very small group, it has only $6$ elements. So you can just check every element in a direct way.






          share|cite|improve this answer












          First of all, the permutation $(23)$ is the inverse of itself so you actually found only $2$ permutations that commute with it so far, not $3$. Next, $S_3$ is a very small group, it has only $6$ elements. So you can just check every element in a direct way.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 20:04









          Mark

          5,750415




          5,750415












          • How is it a inverse of itself? the inverse is (32) ?
            – soetirl13
            Nov 25 at 20:08










          • $(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
            – Mark
            Nov 25 at 20:09










          • Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
            – soetirl13
            Nov 25 at 20:13










          • It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
            – Mark
            Nov 25 at 20:23










          • Ahh 2. :))))) thank you
            – soetirl13
            Nov 25 at 21:06


















          • How is it a inverse of itself? the inverse is (32) ?
            – soetirl13
            Nov 25 at 20:08










          • $(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
            – Mark
            Nov 25 at 20:09










          • Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
            – soetirl13
            Nov 25 at 20:13










          • It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
            – Mark
            Nov 25 at 20:23










          • Ahh 2. :))))) thank you
            – soetirl13
            Nov 25 at 21:06
















          How is it a inverse of itself? the inverse is (32) ?
          – soetirl13
          Nov 25 at 20:08




          How is it a inverse of itself? the inverse is (32) ?
          – soetirl13
          Nov 25 at 20:08












          $(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
          – Mark
          Nov 25 at 20:09




          $(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
          – Mark
          Nov 25 at 20:09












          Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
          – soetirl13
          Nov 25 at 20:13




          Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
          – soetirl13
          Nov 25 at 20:13












          It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
          – Mark
          Nov 25 at 20:23




          It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
          – Mark
          Nov 25 at 20:23












          Ahh 2. :))))) thank you
          – soetirl13
          Nov 25 at 21:06




          Ahh 2. :))))) thank you
          – soetirl13
          Nov 25 at 21:06


















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