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
asked Jan 14 at 10:48
ValVal
557
$endgroup$
add a comment |
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
asked Jan 14 at 10:48
ValVal
557
$endgroup$
$begingroup$
@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
add a comment |
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
asked Jan 14 at 10:48
ValVal
557
$endgroup$
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
asked Jan 14 at 10:48
ValVal
557
asked Jan 14 at 10:48
ValVal
557
asked Jan 14 at 10:48
ValVal
557
asked Jan 14 at 10:48
ValVal
557
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
add a comment |
$begingroup$
@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
add a comment |
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 $$(13)(24)cdotleft<(12),(34)right>.$$
answered Jan 14 at 11:19
MelodyMelody
1,31212
$endgroup$
$begingroup$
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
|
show 1 more comment
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3073097%2ffinding-all-permutations-which-permute-with-1234%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 $$(13)(24)cdotleft<(12),(34)right>.$$
answered Jan 14 at 11:19
MelodyMelody
1,31212
$endgroup$
$begingroup$
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
|
show 1 more comment
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 $$(13)(24)cdotleft<(12),(34)right>.$$
answered Jan 14 at 11:19
MelodyMelody
1,31212
$endgroup$
$begingroup$
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
|
show 1 more comment
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 $$(13)(24)cdotleft<(12),(34)right>.$$
answered Jan 14 at 11:19
MelodyMelody
1,31212
$endgroup$
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 $$(13)(24)cdotleft<(12),(34)right>.$$
answered Jan 14 at 11:19
MelodyMelody
1,31212
edited Jan 14 at 11:29
answered Jan 14 at 11:19
MelodyMelody
1,31212
answered Jan 14 at 11:19
MelodyMelody
1,31212
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3073097%2ffinding-all-permutations-which-permute-with-1234%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
@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