Finding all permutations which permute with (12)(34)












1












$begingroup$


We are working in the symmetric group of 4, so I called it S4.



For example, how would you go about finding the other permutations of (12)(34) other than itself, identity, (12), (34)?



These are 4 and by conjugacy classes, we observe that there should be 8 permutations that commute with (12)(34). How would you figure them out? S4 is a big set so you cannot try them all out I guess there should be a method.



I have been looking but I cannot find or understand any method so please if you could make it simple for me. Thanks.






abstract-algebra permutations







share|cite|improve this question









$endgroup$












  • $begingroup$
    @Matt Samuel the reason they brought up conjugacy classes, is because there's a theorem that states the size of an orbit of an element equals the index of its stabilizers. Since There are 3 conjugates of $(12)(34)$ we can deduce its centralizer has size 8.
    $endgroup$
    – Melody
    Jan 14 at 11:13










  • $begingroup$
    A systematic way to find all 8 of them to write down the 8 versions of $(12)(34)$ in cycle notation, i.e. $(21)(34)$, $(12)(43)$, ... , $(43)(21)$. For each one, write $(12)(34)$ above it and map "down" to get a permutation that conjugates $(12)(34)$ to itself.
    $endgroup$
    – Ned
    Jan 14 at 13:13










  • $begingroup$
    What do you mean by "write (12)(34) above it and map "down" " @Ned.
    $endgroup$
    – Val
    Jan 14 at 13:55










  • $begingroup$
    @Val, For example, write (12)(34) above (34)(21) vertically lining up the two 2-cycles. Mapping "down" takes 1 to 3, 2 to 4, 3 to 2 and 4 to 1, which is to say the permutation obtained is (1324). Which means that (1324) conjugates (12)(34) to itself [in the form of (34)(12), if you want to think of the relabeling like that]. "Conjugating (12)(34) to itself" is the same as "commuting with (12)(34)." The theme here is Conjugation = Relabeling.
    $endgroup$
    – Ned
    Jan 14 at 22:25
















1












$begingroup$


We are working in the symmetric group of 4, so I called it S4.



For example, how would you go about finding the other permutations of (12)(34) other than itself, identity, (12), (34)?



These are 4 and by conjugacy classes, we observe that there should be 8 permutations that commute with (12)(34). How would you figure them out? S4 is a big set so you cannot try them all out I guess there should be a method.



I have been looking but I cannot find or understand any method so please if you could make it simple for me. Thanks.






abstract-algebra permutations







share|cite|improve this question









$endgroup$












  • $begingroup$
    @Matt Samuel the reason they brought up conjugacy classes, is because there's a theorem that states the size of an orbit of an element equals the index of its stabilizers. Since There are 3 conjugates of $(12)(34)$ we can deduce its centralizer has size 8.
    $endgroup$
    – Melody
    Jan 14 at 11:13










  • $begingroup$
    A systematic way to find all 8 of them to write down the 8 versions of $(12)(34)$ in cycle notation, i.e. $(21)(34)$, $(12)(43)$, ... , $(43)(21)$. For each one, write $(12)(34)$ above it and map "down" to get a permutation that conjugates $(12)(34)$ to itself.
    $endgroup$
    – Ned
    Jan 14 at 13:13










  • $begingroup$
    What do you mean by "write (12)(34) above it and map "down" " @Ned.
    $endgroup$
    – Val
    Jan 14 at 13:55










  • $begingroup$
    @Val, For example, write (12)(34) above (34)(21) vertically lining up the two 2-cycles. Mapping "down" takes 1 to 3, 2 to 4, 3 to 2 and 4 to 1, which is to say the permutation obtained is (1324). Which means that (1324) conjugates (12)(34) to itself [in the form of (34)(12), if you want to think of the relabeling like that]. "Conjugating (12)(34) to itself" is the same as "commuting with (12)(34)." The theme here is Conjugation = Relabeling.
    $endgroup$
    – Ned
    Jan 14 at 22:25














1












1








1





$begingroup$


We are working in the symmetric group of 4, so I called it S4.



For example, how would you go about finding the other permutations of (12)(34) other than itself, identity, (12), (34)?



These are 4 and by conjugacy classes, we observe that there should be 8 permutations that commute with (12)(34). How would you figure them out? S4 is a big set so you cannot try them all out I guess there should be a method.



I have been looking but I cannot find or understand any method so please if you could make it simple for me. Thanks.






abstract-algebra permutations







share|cite|improve this question









$endgroup$




We are working in the symmetric group of 4, so I called it S4.



For example, how would you go about finding the other permutations of (12)(34) other than itself, identity, (12), (34)?



These are 4 and by conjugacy classes, we observe that there should be 8 permutations that commute with (12)(34). How would you figure them out? S4 is a big set so you cannot try them all out I guess there should be a method.



I have been looking but I cannot find or understand any method so please if you could make it simple for me. Thanks.






abstract-algebra permutations




abstract-algebra permutations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 14 at 10:48









ValVal

557




557












  • $begingroup$
    @Matt Samuel the reason they brought up conjugacy classes, is because there's a theorem that states the size of an orbit of an element equals the index of its stabilizers. Since There are 3 conjugates of $(12)(34)$ we can deduce its centralizer has size 8.
    $endgroup$
    – Melody
    Jan 14 at 11:13










  • $begingroup$
    A systematic way to find all 8 of them to write down the 8 versions of $(12)(34)$ in cycle notation, i.e. $(21)(34)$, $(12)(43)$, ... , $(43)(21)$. For each one, write $(12)(34)$ above it and map "down" to get a permutation that conjugates $(12)(34)$ to itself.
    $endgroup$
    – Ned
    Jan 14 at 13:13










  • $begingroup$
    What do you mean by "write (12)(34) above it and map "down" " @Ned.
    $endgroup$
    – Val
    Jan 14 at 13:55










  • $begingroup$
    @Val, For example, write (12)(34) above (34)(21) vertically lining up the two 2-cycles. Mapping "down" takes 1 to 3, 2 to 4, 3 to 2 and 4 to 1, which is to say the permutation obtained is (1324). Which means that (1324) conjugates (12)(34) to itself [in the form of (34)(12), if you want to think of the relabeling like that]. "Conjugating (12)(34) to itself" is the same as "commuting with (12)(34)." The theme here is Conjugation = Relabeling.
    $endgroup$
    – Ned
    Jan 14 at 22:25


















  • $begingroup$
    @Matt Samuel the reason they brought up conjugacy classes, is because there's a theorem that states the size of an orbit of an element equals the index of its stabilizers. Since There are 3 conjugates of $(12)(34)$ we can deduce its centralizer has size 8.
    $endgroup$
    – Melody
    Jan 14 at 11:13










  • $begingroup$
    A systematic way to find all 8 of them to write down the 8 versions of $(12)(34)$ in cycle notation, i.e. $(21)(34)$, $(12)(43)$, ... , $(43)(21)$. For each one, write $(12)(34)$ above it and map "down" to get a permutation that conjugates $(12)(34)$ to itself.
    $endgroup$
    – Ned
    Jan 14 at 13:13










  • $begingroup$
    What do you mean by "write (12)(34) above it and map "down" " @Ned.
    $endgroup$
    – Val
    Jan 14 at 13:55










  • $begingroup$
    @Val, For example, write (12)(34) above (34)(21) vertically lining up the two 2-cycles. Mapping "down" takes 1 to 3, 2 to 4, 3 to 2 and 4 to 1, which is to say the permutation obtained is (1324). Which means that (1324) conjugates (12)(34) to itself [in the form of (34)(12), if you want to think of the relabeling like that]. "Conjugating (12)(34) to itself" is the same as "commuting with (12)(34)." The theme here is Conjugation = Relabeling.
    $endgroup$
    – Ned
    Jan 14 at 22:25
















$begingroup$
@Matt Samuel the reason they brought up conjugacy classes, is because there's a theorem that states the size of an orbit of an element equals the index of its stabilizers. Since There are 3 conjugates of $(12)(34)$ we can deduce its centralizer has size 8.
$endgroup$
– Melody
Jan 14 at 11:13




$begingroup$
@Matt Samuel the reason they brought up conjugacy classes, is because there's a theorem that states the size of an orbit of an element equals the index of its stabilizers. Since There are 3 conjugates of $(12)(34)$ we can deduce its centralizer has size 8.
$endgroup$
– Melody
Jan 14 at 11:13












$begingroup$
A systematic way to find all 8 of them to write down the 8 versions of $(12)(34)$ in cycle notation, i.e. $(21)(34)$, $(12)(43)$, ... , $(43)(21)$. For each one, write $(12)(34)$ above it and map "down" to get a permutation that conjugates $(12)(34)$ to itself.
$endgroup$
– Ned
Jan 14 at 13:13




$begingroup$
A systematic way to find all 8 of them to write down the 8 versions of $(12)(34)$ in cycle notation, i.e. $(21)(34)$, $(12)(43)$, ... , $(43)(21)$. For each one, write $(12)(34)$ above it and map "down" to get a permutation that conjugates $(12)(34)$ to itself.
$endgroup$
– Ned
Jan 14 at 13:13












$begingroup$
What do you mean by "write (12)(34) above it and map "down" " @Ned.
$endgroup$
– Val
Jan 14 at 13:55




$begingroup$
What do you mean by "write (12)(34) above it and map "down" " @Ned.
$endgroup$
– Val
Jan 14 at 13:55












$begingroup$
@Val, For example, write (12)(34) above (34)(21) vertically lining up the two 2-cycles. Mapping "down" takes 1 to 3, 2 to 4, 3 to 2 and 4 to 1, which is to say the permutation obtained is (1324). Which means that (1324) conjugates (12)(34) to itself [in the form of (34)(12), if you want to think of the relabeling like that]. "Conjugating (12)(34) to itself" is the same as "commuting with (12)(34)." The theme here is Conjugation = Relabeling.
$endgroup$
– Ned
Jan 14 at 22:25




$begingroup$
@Val, For example, write (12)(34) above (34)(21) vertically lining up the two 2-cycles. Mapping "down" takes 1 to 3, 2 to 4, 3 to 2 and 4 to 1, which is to say the permutation obtained is (1324). Which means that (1324) conjugates (12)(34) to itself [in the form of (34)(12), if you want to think of the relabeling like that]. "Conjugating (12)(34) to itself" is the same as "commuting with (12)(34)." The theme here is Conjugation = Relabeling.
$endgroup$
– Ned
Jan 14 at 22:25










1 Answer
1






active

oldest

votes


















1












$begingroup$

I'm not sure of an efficient method to find all of them. However, one way that works in this case is to recall that if $sigmain S_4$, then $sigma(12)(34)sigma^{-1}=(sigma(1)sigma(2))(sigma(3)sigma(4)),$ so if $$sigma(1)=3,sigma(2)=4,sigma(3)=1,sigma(4)=2,$$
then $sigma$ works. This tells us another element of the centralizer is $(13)(24).$ To find the others we can just look at the left coset $$(13)(24)cdotleft<(12),(34)right>.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry, I m a bit confused about what σ does and why σ(12)(34)σ−1=(σ(1)σ(2))(σ(3)σ(4)). Also when you want to get new terms for σ(x), where x is between 1 and 4 I guess, how would you pick that image? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:28










  • $begingroup$
    @Val edited my answer. The map $sigma$ is just an arbitrary permutation. The formula I mentioned is just a standard formula for conjugation of a product of disjoint cycles. The reason I mentioned it, is because commuting with an element is equivalent to leaving it fixed under conjugation, so knowing how $(12)(34)$ acts under conjugation can help us determine which elements it commutes with. For example, if $sigma$ commutes and $sigma(1)=3,$ then we must have $sigma(3)=1,2.$
    $endgroup$
    – Melody
    Jan 14 at 11:34












  • $begingroup$
    What do you mean by " knowing how a (12)(34) acts under conjugation". I am sure there are some permutations of order 4 that will commute, how would you find those? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:43












  • $begingroup$
    @Val Conjugation is a group action, so what I mean is basically what $sigma(12)(34)sigma^{-1}$ looks like for an arbitrary permutation. As for how to find the permutations of order 4, they'll be in the coset I mentioned $$(12)(34)cdotleft<(12),(34)right>.$$ Another way you could find them would be to notice that clearly we if $sigma$ commutes with $(12)(34),$ and $sigma$ is order 4, then $sigma(1)=2$ implies $sigma(2)=1$ by the formula I mentioned. So we must have $sigma(1)=3,4.$ Choose $sigma(1)=3,$ then $sigma(3)not=1,$ so $sigma(3)=2$ by my last comment. Then $sigma=(1324).$
    $endgroup$
    – Melody
    Jan 14 at 11:47










  • $begingroup$
    If we were working in the symmetric group of 5, I guess I will have to choose σ(1)=3,4,5 (for an order 4 permutation that commutes) and I choose σ(1)=4, then σ(4)≠1 and then σ(4)=2,3,5 and I choose 5 then we have σ(5)=2,3 and I choose 3 . Then that means σ=(1453) commutes with (12)(34)? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:59












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1 Answer
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1 Answer
1






active

oldest

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oldest

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active

oldest

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1












$begingroup$

I'm not sure of an efficient method to find all of them. However, one way that works in this case is to recall that if $sigmain S_4$, then $sigma(12)(34)sigma^{-1}=(sigma(1)sigma(2))(sigma(3)sigma(4)),$ so if $$sigma(1)=3,sigma(2)=4,sigma(3)=1,sigma(4)=2,$$
then $sigma$ works. This tells us another element of the centralizer is $(13)(24).$ To find the others we can just look at the left coset $$(13)(24)cdotleft<(12),(34)right>.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry, I m a bit confused about what σ does and why σ(12)(34)σ−1=(σ(1)σ(2))(σ(3)σ(4)). Also when you want to get new terms for σ(x), where x is between 1 and 4 I guess, how would you pick that image? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:28










  • $begingroup$
    @Val edited my answer. The map $sigma$ is just an arbitrary permutation. The formula I mentioned is just a standard formula for conjugation of a product of disjoint cycles. The reason I mentioned it, is because commuting with an element is equivalent to leaving it fixed under conjugation, so knowing how $(12)(34)$ acts under conjugation can help us determine which elements it commutes with. For example, if $sigma$ commutes and $sigma(1)=3,$ then we must have $sigma(3)=1,2.$
    $endgroup$
    – Melody
    Jan 14 at 11:34












  • $begingroup$
    What do you mean by " knowing how a (12)(34) acts under conjugation". I am sure there are some permutations of order 4 that will commute, how would you find those? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:43












  • $begingroup$
    @Val Conjugation is a group action, so what I mean is basically what $sigma(12)(34)sigma^{-1}$ looks like for an arbitrary permutation. As for how to find the permutations of order 4, they'll be in the coset I mentioned $$(12)(34)cdotleft<(12),(34)right>.$$ Another way you could find them would be to notice that clearly we if $sigma$ commutes with $(12)(34),$ and $sigma$ is order 4, then $sigma(1)=2$ implies $sigma(2)=1$ by the formula I mentioned. So we must have $sigma(1)=3,4.$ Choose $sigma(1)=3,$ then $sigma(3)not=1,$ so $sigma(3)=2$ by my last comment. Then $sigma=(1324).$
    $endgroup$
    – Melody
    Jan 14 at 11:47










  • $begingroup$
    If we were working in the symmetric group of 5, I guess I will have to choose σ(1)=3,4,5 (for an order 4 permutation that commutes) and I choose σ(1)=4, then σ(4)≠1 and then σ(4)=2,3,5 and I choose 5 then we have σ(5)=2,3 and I choose 3 . Then that means σ=(1453) commutes with (12)(34)? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:59
















1












$begingroup$

I'm not sure of an efficient method to find all of them. However, one way that works in this case is to recall that if $sigmain S_4$, then $sigma(12)(34)sigma^{-1}=(sigma(1)sigma(2))(sigma(3)sigma(4)),$ so if $$sigma(1)=3,sigma(2)=4,sigma(3)=1,sigma(4)=2,$$
then $sigma$ works. This tells us another element of the centralizer is $(13)(24).$ To find the others we can just look at the left coset $$(13)(24)cdotleft<(12),(34)right>.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry, I m a bit confused about what σ does and why σ(12)(34)σ−1=(σ(1)σ(2))(σ(3)σ(4)). Also when you want to get new terms for σ(x), where x is between 1 and 4 I guess, how would you pick that image? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:28










  • $begingroup$
    @Val edited my answer. The map $sigma$ is just an arbitrary permutation. The formula I mentioned is just a standard formula for conjugation of a product of disjoint cycles. The reason I mentioned it, is because commuting with an element is equivalent to leaving it fixed under conjugation, so knowing how $(12)(34)$ acts under conjugation can help us determine which elements it commutes with. For example, if $sigma$ commutes and $sigma(1)=3,$ then we must have $sigma(3)=1,2.$
    $endgroup$
    – Melody
    Jan 14 at 11:34












  • $begingroup$
    What do you mean by " knowing how a (12)(34) acts under conjugation". I am sure there are some permutations of order 4 that will commute, how would you find those? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:43












  • $begingroup$
    @Val Conjugation is a group action, so what I mean is basically what $sigma(12)(34)sigma^{-1}$ looks like for an arbitrary permutation. As for how to find the permutations of order 4, they'll be in the coset I mentioned $$(12)(34)cdotleft<(12),(34)right>.$$ Another way you could find them would be to notice that clearly we if $sigma$ commutes with $(12)(34),$ and $sigma$ is order 4, then $sigma(1)=2$ implies $sigma(2)=1$ by the formula I mentioned. So we must have $sigma(1)=3,4.$ Choose $sigma(1)=3,$ then $sigma(3)not=1,$ so $sigma(3)=2$ by my last comment. Then $sigma=(1324).$
    $endgroup$
    – Melody
    Jan 14 at 11:47










  • $begingroup$
    If we were working in the symmetric group of 5, I guess I will have to choose σ(1)=3,4,5 (for an order 4 permutation that commutes) and I choose σ(1)=4, then σ(4)≠1 and then σ(4)=2,3,5 and I choose 5 then we have σ(5)=2,3 and I choose 3 . Then that means σ=(1453) commutes with (12)(34)? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:59














1












1








1





$begingroup$

I'm not sure of an efficient method to find all of them. However, one way that works in this case is to recall that if $sigmain S_4$, then $sigma(12)(34)sigma^{-1}=(sigma(1)sigma(2))(sigma(3)sigma(4)),$ so if $$sigma(1)=3,sigma(2)=4,sigma(3)=1,sigma(4)=2,$$
then $sigma$ works. This tells us another element of the centralizer is $(13)(24).$ To find the others we can just look at the left coset $$(13)(24)cdotleft<(12),(34)right>.$$






share|cite|improve this answer











$endgroup$



I'm not sure of an efficient method to find all of them. However, one way that works in this case is to recall that if $sigmain S_4$, then $sigma(12)(34)sigma^{-1}=(sigma(1)sigma(2))(sigma(3)sigma(4)),$ so if $$sigma(1)=3,sigma(2)=4,sigma(3)=1,sigma(4)=2,$$
then $sigma$ works. This tells us another element of the centralizer is $(13)(24).$ To find the others we can just look at the left coset $$(13)(24)cdotleft<(12),(34)right>.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 14 at 11:29

























answered Jan 14 at 11:19









MelodyMelody

1,31212




1,31212












  • $begingroup$
    Sorry, I m a bit confused about what σ does and why σ(12)(34)σ−1=(σ(1)σ(2))(σ(3)σ(4)). Also when you want to get new terms for σ(x), where x is between 1 and 4 I guess, how would you pick that image? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:28










  • $begingroup$
    @Val edited my answer. The map $sigma$ is just an arbitrary permutation. The formula I mentioned is just a standard formula for conjugation of a product of disjoint cycles. The reason I mentioned it, is because commuting with an element is equivalent to leaving it fixed under conjugation, so knowing how $(12)(34)$ acts under conjugation can help us determine which elements it commutes with. For example, if $sigma$ commutes and $sigma(1)=3,$ then we must have $sigma(3)=1,2.$
    $endgroup$
    – Melody
    Jan 14 at 11:34












  • $begingroup$
    What do you mean by " knowing how a (12)(34) acts under conjugation". I am sure there are some permutations of order 4 that will commute, how would you find those? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:43












  • $begingroup$
    @Val Conjugation is a group action, so what I mean is basically what $sigma(12)(34)sigma^{-1}$ looks like for an arbitrary permutation. As for how to find the permutations of order 4, they'll be in the coset I mentioned $$(12)(34)cdotleft<(12),(34)right>.$$ Another way you could find them would be to notice that clearly we if $sigma$ commutes with $(12)(34),$ and $sigma$ is order 4, then $sigma(1)=2$ implies $sigma(2)=1$ by the formula I mentioned. So we must have $sigma(1)=3,4.$ Choose $sigma(1)=3,$ then $sigma(3)not=1,$ so $sigma(3)=2$ by my last comment. Then $sigma=(1324).$
    $endgroup$
    – Melody
    Jan 14 at 11:47










  • $begingroup$
    If we were working in the symmetric group of 5, I guess I will have to choose σ(1)=3,4,5 (for an order 4 permutation that commutes) and I choose σ(1)=4, then σ(4)≠1 and then σ(4)=2,3,5 and I choose 5 then we have σ(5)=2,3 and I choose 3 . Then that means σ=(1453) commutes with (12)(34)? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:59


















  • $begingroup$
    Sorry, I m a bit confused about what σ does and why σ(12)(34)σ−1=(σ(1)σ(2))(σ(3)σ(4)). Also when you want to get new terms for σ(x), where x is between 1 and 4 I guess, how would you pick that image? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:28










  • $begingroup$
    @Val edited my answer. The map $sigma$ is just an arbitrary permutation. The formula I mentioned is just a standard formula for conjugation of a product of disjoint cycles. The reason I mentioned it, is because commuting with an element is equivalent to leaving it fixed under conjugation, so knowing how $(12)(34)$ acts under conjugation can help us determine which elements it commutes with. For example, if $sigma$ commutes and $sigma(1)=3,$ then we must have $sigma(3)=1,2.$
    $endgroup$
    – Melody
    Jan 14 at 11:34












  • $begingroup$
    What do you mean by " knowing how a (12)(34) acts under conjugation". I am sure there are some permutations of order 4 that will commute, how would you find those? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:43












  • $begingroup$
    @Val Conjugation is a group action, so what I mean is basically what $sigma(12)(34)sigma^{-1}$ looks like for an arbitrary permutation. As for how to find the permutations of order 4, they'll be in the coset I mentioned $$(12)(34)cdotleft<(12),(34)right>.$$ Another way you could find them would be to notice that clearly we if $sigma$ commutes with $(12)(34),$ and $sigma$ is order 4, then $sigma(1)=2$ implies $sigma(2)=1$ by the formula I mentioned. So we must have $sigma(1)=3,4.$ Choose $sigma(1)=3,$ then $sigma(3)not=1,$ so $sigma(3)=2$ by my last comment. Then $sigma=(1324).$
    $endgroup$
    – Melody
    Jan 14 at 11:47










  • $begingroup$
    If we were working in the symmetric group of 5, I guess I will have to choose σ(1)=3,4,5 (for an order 4 permutation that commutes) and I choose σ(1)=4, then σ(4)≠1 and then σ(4)=2,3,5 and I choose 5 then we have σ(5)=2,3 and I choose 3 . Then that means σ=(1453) commutes with (12)(34)? @Melody
    $endgroup$
    – Val
    Jan 14 at 11:59
















$begingroup$
Sorry, I m a bit confused about what σ does and why σ(12)(34)σ−1=(σ(1)σ(2))(σ(3)σ(4)). Also when you want to get new terms for σ(x), where x is between 1 and 4 I guess, how would you pick that image? @Melody
$endgroup$
– Val
Jan 14 at 11:28




$begingroup$
Sorry, I m a bit confused about what σ does and why σ(12)(34)σ−1=(σ(1)σ(2))(σ(3)σ(4)). Also when you want to get new terms for σ(x), where x is between 1 and 4 I guess, how would you pick that image? @Melody
$endgroup$
– Val
Jan 14 at 11:28












$begingroup$
@Val edited my answer. The map $sigma$ is just an arbitrary permutation. The formula I mentioned is just a standard formula for conjugation of a product of disjoint cycles. The reason I mentioned it, is because commuting with an element is equivalent to leaving it fixed under conjugation, so knowing how $(12)(34)$ acts under conjugation can help us determine which elements it commutes with. For example, if $sigma$ commutes and $sigma(1)=3,$ then we must have $sigma(3)=1,2.$
$endgroup$
– Melody
Jan 14 at 11:34






$begingroup$
@Val edited my answer. The map $sigma$ is just an arbitrary permutation. The formula I mentioned is just a standard formula for conjugation of a product of disjoint cycles. The reason I mentioned it, is because commuting with an element is equivalent to leaving it fixed under conjugation, so knowing how $(12)(34)$ acts under conjugation can help us determine which elements it commutes with. For example, if $sigma$ commutes and $sigma(1)=3,$ then we must have $sigma(3)=1,2.$
$endgroup$
– Melody
Jan 14 at 11:34














$begingroup$
What do you mean by " knowing how a (12)(34) acts under conjugation". I am sure there are some permutations of order 4 that will commute, how would you find those? @Melody
$endgroup$
– Val
Jan 14 at 11:43






$begingroup$
What do you mean by " knowing how a (12)(34) acts under conjugation". I am sure there are some permutations of order 4 that will commute, how would you find those? @Melody
$endgroup$
– Val
Jan 14 at 11:43














$begingroup$
@Val Conjugation is a group action, so what I mean is basically what $sigma(12)(34)sigma^{-1}$ looks like for an arbitrary permutation. As for how to find the permutations of order 4, they'll be in the coset I mentioned $$(12)(34)cdotleft<(12),(34)right>.$$ Another way you could find them would be to notice that clearly we if $sigma$ commutes with $(12)(34),$ and $sigma$ is order 4, then $sigma(1)=2$ implies $sigma(2)=1$ by the formula I mentioned. So we must have $sigma(1)=3,4.$ Choose $sigma(1)=3,$ then $sigma(3)not=1,$ so $sigma(3)=2$ by my last comment. Then $sigma=(1324).$
$endgroup$
– Melody
Jan 14 at 11:47




$begingroup$
@Val Conjugation is a group action, so what I mean is basically what $sigma(12)(34)sigma^{-1}$ looks like for an arbitrary permutation. As for how to find the permutations of order 4, they'll be in the coset I mentioned $$(12)(34)cdotleft<(12),(34)right>.$$ Another way you could find them would be to notice that clearly we if $sigma$ commutes with $(12)(34),$ and $sigma$ is order 4, then $sigma(1)=2$ implies $sigma(2)=1$ by the formula I mentioned. So we must have $sigma(1)=3,4.$ Choose $sigma(1)=3,$ then $sigma(3)not=1,$ so $sigma(3)=2$ by my last comment. Then $sigma=(1324).$
$endgroup$
– Melody
Jan 14 at 11:47












$begingroup$
If we were working in the symmetric group of 5, I guess I will have to choose σ(1)=3,4,5 (for an order 4 permutation that commutes) and I choose σ(1)=4, then σ(4)≠1 and then σ(4)=2,3,5 and I choose 5 then we have σ(5)=2,3 and I choose 3 . Then that means σ=(1453) commutes with (12)(34)? @Melody
$endgroup$
– Val
Jan 14 at 11:59




$begingroup$
If we were working in the symmetric group of 5, I guess I will have to choose σ(1)=3,4,5 (for an order 4 permutation that commutes) and I choose σ(1)=4, then σ(4)≠1 and then σ(4)=2,3,5 and I choose 5 then we have σ(5)=2,3 and I choose 3 . Then that means σ=(1453) commutes with (12)(34)? @Melody
$endgroup$
– Val
Jan 14 at 11:59


















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