Order of $C_{S_n}((12)(34)) ; forall ngeq 4$ and its elements?












1












$begingroup$



What is the order of $C_{S_n}((12)(34)) ; forall ngeq 4$. Determine
the elements of this centralizer explicitly




Since $(12)(34)$ is a conjugacy class in $S_n ; forall ngeq 4$, denote this class as $K$



Then we have $$|K| = frac{|S_n|}{|C_{S_n}((12)(34))|}=frac{(C^{n}_{2}cdotfrac{2!}{2})(C^{n-2}_{2}cdotfrac{2!}{2})}{2!} = frac{n(n-1)(n-2)(n-3)}{8}$$



$$Rightarrow |C_{S_n}((12)(34))|=8cdot(n-4)!$$



But how do I determine these elements explicitly?






group-theory group-actions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: $S_n$ has a subgroup isomorphic to $S_4times S_{n-4}$.
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 14:38
















1












$begingroup$



What is the order of $C_{S_n}((12)(34)) ; forall ngeq 4$. Determine
the elements of this centralizer explicitly




Since $(12)(34)$ is a conjugacy class in $S_n ; forall ngeq 4$, denote this class as $K$



Then we have $$|K| = frac{|S_n|}{|C_{S_n}((12)(34))|}=frac{(C^{n}_{2}cdotfrac{2!}{2})(C^{n-2}_{2}cdotfrac{2!}{2})}{2!} = frac{n(n-1)(n-2)(n-3)}{8}$$



$$Rightarrow |C_{S_n}((12)(34))|=8cdot(n-4)!$$



But how do I determine these elements explicitly?






group-theory group-actions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: $S_n$ has a subgroup isomorphic to $S_4times S_{n-4}$.
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 14:38














1












1








1





$begingroup$



What is the order of $C_{S_n}((12)(34)) ; forall ngeq 4$. Determine
the elements of this centralizer explicitly




Since $(12)(34)$ is a conjugacy class in $S_n ; forall ngeq 4$, denote this class as $K$



Then we have $$|K| = frac{|S_n|}{|C_{S_n}((12)(34))|}=frac{(C^{n}_{2}cdotfrac{2!}{2})(C^{n-2}_{2}cdotfrac{2!}{2})}{2!} = frac{n(n-1)(n-2)(n-3)}{8}$$



$$Rightarrow |C_{S_n}((12)(34))|=8cdot(n-4)!$$



But how do I determine these elements explicitly?






group-theory group-actions







share|cite|improve this question









$endgroup$





What is the order of $C_{S_n}((12)(34)) ; forall ngeq 4$. Determine
the elements of this centralizer explicitly




Since $(12)(34)$ is a conjugacy class in $S_n ; forall ngeq 4$, denote this class as $K$



Then we have $$|K| = frac{|S_n|}{|C_{S_n}((12)(34))|}=frac{(C^{n}_{2}cdotfrac{2!}{2})(C^{n-2}_{2}cdotfrac{2!}{2})}{2!} = frac{n(n-1)(n-2)(n-3)}{8}$$



$$Rightarrow |C_{S_n}((12)(34))|=8cdot(n-4)!$$



But how do I determine these elements explicitly?






group-theory group-actions




group-theory group-actions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 8 '18 at 14:25









learning_mathslearning_maths

1228




1228












  • $begingroup$
    Hint: $S_n$ has a subgroup isomorphic to $S_4times S_{n-4}$.
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 14:38


















  • $begingroup$
    Hint: $S_n$ has a subgroup isomorphic to $S_4times S_{n-4}$.
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 14:38
















$begingroup$
Hint: $S_n$ has a subgroup isomorphic to $S_4times S_{n-4}$.
$endgroup$
– Tobias Kildetoft
Dec 8 '18 at 14:38




$begingroup$
Hint: $S_n$ has a subgroup isomorphic to $S_4times S_{n-4}$.
$endgroup$
– Tobias Kildetoft
Dec 8 '18 at 14:38










1 Answer
1






active

oldest

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0












$begingroup$

So we want to have $sigma(12)(34)sigma^{-1}=(12)(34)$. We have the following set $S$ which are in the stabilizer of $(12)(34)$:
$S:={(12)(34),e,(13)(24),(14)(23),(1324),(1423),(12),(34)}$. So let $S'_{n-4} subset S_n$ denote the permutations in $S_n$ that fixes ${1,2,3,4}$. The centralizer should be $Stimes S'_{n-4}:={ssigma|s in S, sigmain S'_{n-4}}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Since your claim does not fit the count given by the OP, you should probably comment on this.
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 14:56






  • 1




    $begingroup$
    The issue is that you are missing some elements that centralize the given permutation (note that it is the square of a $4$-cycle, so it clearly commutes with such a $4$-cycle).
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 15:21










  • $begingroup$
    Right! Thank you!
    $endgroup$
    – nafhgood
    Dec 8 '18 at 15:58











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1 Answer
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$begingroup$

So we want to have $sigma(12)(34)sigma^{-1}=(12)(34)$. We have the following set $S$ which are in the stabilizer of $(12)(34)$:
$S:={(12)(34),e,(13)(24),(14)(23),(1324),(1423),(12),(34)}$. So let $S'_{n-4} subset S_n$ denote the permutations in $S_n$ that fixes ${1,2,3,4}$. The centralizer should be $Stimes S'_{n-4}:={ssigma|s in S, sigmain S'_{n-4}}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Since your claim does not fit the count given by the OP, you should probably comment on this.
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 14:56






  • 1




    $begingroup$
    The issue is that you are missing some elements that centralize the given permutation (note that it is the square of a $4$-cycle, so it clearly commutes with such a $4$-cycle).
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 15:21










  • $begingroup$
    Right! Thank you!
    $endgroup$
    – nafhgood
    Dec 8 '18 at 15:58
















0












$begingroup$

So we want to have $sigma(12)(34)sigma^{-1}=(12)(34)$. We have the following set $S$ which are in the stabilizer of $(12)(34)$:
$S:={(12)(34),e,(13)(24),(14)(23),(1324),(1423),(12),(34)}$. So let $S'_{n-4} subset S_n$ denote the permutations in $S_n$ that fixes ${1,2,3,4}$. The centralizer should be $Stimes S'_{n-4}:={ssigma|s in S, sigmain S'_{n-4}}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Since your claim does not fit the count given by the OP, you should probably comment on this.
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 14:56






  • 1




    $begingroup$
    The issue is that you are missing some elements that centralize the given permutation (note that it is the square of a $4$-cycle, so it clearly commutes with such a $4$-cycle).
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 15:21










  • $begingroup$
    Right! Thank you!
    $endgroup$
    – nafhgood
    Dec 8 '18 at 15:58














0












0








0





$begingroup$

So we want to have $sigma(12)(34)sigma^{-1}=(12)(34)$. We have the following set $S$ which are in the stabilizer of $(12)(34)$:
$S:={(12)(34),e,(13)(24),(14)(23),(1324),(1423),(12),(34)}$. So let $S'_{n-4} subset S_n$ denote the permutations in $S_n$ that fixes ${1,2,3,4}$. The centralizer should be $Stimes S'_{n-4}:={ssigma|s in S, sigmain S'_{n-4}}$.






share|cite|improve this answer











$endgroup$



So we want to have $sigma(12)(34)sigma^{-1}=(12)(34)$. We have the following set $S$ which are in the stabilizer of $(12)(34)$:
$S:={(12)(34),e,(13)(24),(14)(23),(1324),(1423),(12),(34)}$. So let $S'_{n-4} subset S_n$ denote the permutations in $S_n$ that fixes ${1,2,3,4}$. The centralizer should be $Stimes S'_{n-4}:={ssigma|s in S, sigmain S'_{n-4}}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 15:57

























answered Dec 8 '18 at 14:53









nafhgoodnafhgood

1,797422




1,797422












  • $begingroup$
    Since your claim does not fit the count given by the OP, you should probably comment on this.
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 14:56






  • 1




    $begingroup$
    The issue is that you are missing some elements that centralize the given permutation (note that it is the square of a $4$-cycle, so it clearly commutes with such a $4$-cycle).
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 15:21










  • $begingroup$
    Right! Thank you!
    $endgroup$
    – nafhgood
    Dec 8 '18 at 15:58


















  • $begingroup$
    Since your claim does not fit the count given by the OP, you should probably comment on this.
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 14:56






  • 1




    $begingroup$
    The issue is that you are missing some elements that centralize the given permutation (note that it is the square of a $4$-cycle, so it clearly commutes with such a $4$-cycle).
    $endgroup$
    – Tobias Kildetoft
    Dec 8 '18 at 15:21










  • $begingroup$
    Right! Thank you!
    $endgroup$
    – nafhgood
    Dec 8 '18 at 15:58
















$begingroup$
Since your claim does not fit the count given by the OP, you should probably comment on this.
$endgroup$
– Tobias Kildetoft
Dec 8 '18 at 14:56




$begingroup$
Since your claim does not fit the count given by the OP, you should probably comment on this.
$endgroup$
– Tobias Kildetoft
Dec 8 '18 at 14:56




1




1




$begingroup$
The issue is that you are missing some elements that centralize the given permutation (note that it is the square of a $4$-cycle, so it clearly commutes with such a $4$-cycle).
$endgroup$
– Tobias Kildetoft
Dec 8 '18 at 15:21




$begingroup$
The issue is that you are missing some elements that centralize the given permutation (note that it is the square of a $4$-cycle, so it clearly commutes with such a $4$-cycle).
$endgroup$
– Tobias Kildetoft
Dec 8 '18 at 15:21












$begingroup$
Right! Thank you!
$endgroup$
– nafhgood
Dec 8 '18 at 15:58




$begingroup$
Right! Thank you!
$endgroup$
– nafhgood
Dec 8 '18 at 15:58


















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