Find the conjugacy class $O_{(234)}$ if $G=A_{4}$ is acting on itself by conjugation.












2












$begingroup$



Find the conjugacy class $O_{(234)}$ if $G=A_{4}$ is acting on itself by conjugation.




I calculated $(x)(234)$ for all $xin A_{4}$ and got the set $O_{(234)}=left {(234), (143), (142), (123), (132)right }$.



This is because $(1)(234) = (234)$, $(123)(234) = (143)$, $(132)(234) = (241) = (142)$, $(142)(234) = (321) = (123)$, and $(143)(234) = (132)$.



Am I missing anything or doing anything wrong?






abstract-algebra group-actions permutation-cycles







share|cite|improve this question









$endgroup$












  • $begingroup$
    Orbit-Stabilizer theorem tells you the size of conjugacy class needs to divide $lvert A_4rvert=12$.
    $endgroup$
    – user10354138
    Dec 5 '18 at 19:30






  • 2




    $begingroup$
    Don't you need to calculate all $x(234)x^{-1}$ (not just $x(234)$).
    $endgroup$
    – Chris Custer
    Dec 5 '18 at 19:47










  • $begingroup$
    @ChrisCuster Oh I see. Maybe this will really help me!
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:52










  • $begingroup$
    @ChrisCuster How would I compute $(12)(34)(234)((12)(34))^{-1}$?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 20:01






  • 1




    $begingroup$
    $((12)(34))^{-1}=(12)(34)$. Check what happens to each element, moving from right to left. You get $(143)$.
    $endgroup$
    – Chris Custer
    Dec 5 '18 at 20:06
















2












$begingroup$



Find the conjugacy class $O_{(234)}$ if $G=A_{4}$ is acting on itself by conjugation.




I calculated $(x)(234)$ for all $xin A_{4}$ and got the set $O_{(234)}=left {(234), (143), (142), (123), (132)right }$.



This is because $(1)(234) = (234)$, $(123)(234) = (143)$, $(132)(234) = (241) = (142)$, $(142)(234) = (321) = (123)$, and $(143)(234) = (132)$.



Am I missing anything or doing anything wrong?






abstract-algebra group-actions permutation-cycles







share|cite|improve this question









$endgroup$












  • $begingroup$
    Orbit-Stabilizer theorem tells you the size of conjugacy class needs to divide $lvert A_4rvert=12$.
    $endgroup$
    – user10354138
    Dec 5 '18 at 19:30






  • 2




    $begingroup$
    Don't you need to calculate all $x(234)x^{-1}$ (not just $x(234)$).
    $endgroup$
    – Chris Custer
    Dec 5 '18 at 19:47










  • $begingroup$
    @ChrisCuster Oh I see. Maybe this will really help me!
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:52










  • $begingroup$
    @ChrisCuster How would I compute $(12)(34)(234)((12)(34))^{-1}$?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 20:01






  • 1




    $begingroup$
    $((12)(34))^{-1}=(12)(34)$. Check what happens to each element, moving from right to left. You get $(143)$.
    $endgroup$
    – Chris Custer
    Dec 5 '18 at 20:06














2












2








2





$begingroup$



Find the conjugacy class $O_{(234)}$ if $G=A_{4}$ is acting on itself by conjugation.




I calculated $(x)(234)$ for all $xin A_{4}$ and got the set $O_{(234)}=left {(234), (143), (142), (123), (132)right }$.



This is because $(1)(234) = (234)$, $(123)(234) = (143)$, $(132)(234) = (241) = (142)$, $(142)(234) = (321) = (123)$, and $(143)(234) = (132)$.



Am I missing anything or doing anything wrong?






abstract-algebra group-actions permutation-cycles







share|cite|improve this question









$endgroup$





Find the conjugacy class $O_{(234)}$ if $G=A_{4}$ is acting on itself by conjugation.




I calculated $(x)(234)$ for all $xin A_{4}$ and got the set $O_{(234)}=left {(234), (143), (142), (123), (132)right }$.



This is because $(1)(234) = (234)$, $(123)(234) = (143)$, $(132)(234) = (241) = (142)$, $(142)(234) = (321) = (123)$, and $(143)(234) = (132)$.



Am I missing anything or doing anything wrong?






abstract-algebra group-actions permutation-cycles




abstract-algebra group-actions permutation-cycles






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 19:21









numericalorangenumericalorange

1,775311




1,775311












  • $begingroup$
    Orbit-Stabilizer theorem tells you the size of conjugacy class needs to divide $lvert A_4rvert=12$.
    $endgroup$
    – user10354138
    Dec 5 '18 at 19:30






  • 2




    $begingroup$
    Don't you need to calculate all $x(234)x^{-1}$ (not just $x(234)$).
    $endgroup$
    – Chris Custer
    Dec 5 '18 at 19:47










  • $begingroup$
    @ChrisCuster Oh I see. Maybe this will really help me!
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:52










  • $begingroup$
    @ChrisCuster How would I compute $(12)(34)(234)((12)(34))^{-1}$?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 20:01






  • 1




    $begingroup$
    $((12)(34))^{-1}=(12)(34)$. Check what happens to each element, moving from right to left. You get $(143)$.
    $endgroup$
    – Chris Custer
    Dec 5 '18 at 20:06


















  • $begingroup$
    Orbit-Stabilizer theorem tells you the size of conjugacy class needs to divide $lvert A_4rvert=12$.
    $endgroup$
    – user10354138
    Dec 5 '18 at 19:30






  • 2




    $begingroup$
    Don't you need to calculate all $x(234)x^{-1}$ (not just $x(234)$).
    $endgroup$
    – Chris Custer
    Dec 5 '18 at 19:47










  • $begingroup$
    @ChrisCuster Oh I see. Maybe this will really help me!
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:52










  • $begingroup$
    @ChrisCuster How would I compute $(12)(34)(234)((12)(34))^{-1}$?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 20:01






  • 1




    $begingroup$
    $((12)(34))^{-1}=(12)(34)$. Check what happens to each element, moving from right to left. You get $(143)$.
    $endgroup$
    – Chris Custer
    Dec 5 '18 at 20:06
















$begingroup$
Orbit-Stabilizer theorem tells you the size of conjugacy class needs to divide $lvert A_4rvert=12$.
$endgroup$
– user10354138
Dec 5 '18 at 19:30




$begingroup$
Orbit-Stabilizer theorem tells you the size of conjugacy class needs to divide $lvert A_4rvert=12$.
$endgroup$
– user10354138
Dec 5 '18 at 19:30




2




2




$begingroup$
Don't you need to calculate all $x(234)x^{-1}$ (not just $x(234)$).
$endgroup$
– Chris Custer
Dec 5 '18 at 19:47




$begingroup$
Don't you need to calculate all $x(234)x^{-1}$ (not just $x(234)$).
$endgroup$
– Chris Custer
Dec 5 '18 at 19:47












$begingroup$
@ChrisCuster Oh I see. Maybe this will really help me!
$endgroup$
– numericalorange
Dec 5 '18 at 19:52




$begingroup$
@ChrisCuster Oh I see. Maybe this will really help me!
$endgroup$
– numericalorange
Dec 5 '18 at 19:52












$begingroup$
@ChrisCuster How would I compute $(12)(34)(234)((12)(34))^{-1}$?
$endgroup$
– numericalorange
Dec 5 '18 at 20:01




$begingroup$
@ChrisCuster How would I compute $(12)(34)(234)((12)(34))^{-1}$?
$endgroup$
– numericalorange
Dec 5 '18 at 20:01




1




1




$begingroup$
$((12)(34))^{-1}=(12)(34)$. Check what happens to each element, moving from right to left. You get $(143)$.
$endgroup$
– Chris Custer
Dec 5 '18 at 20:06




$begingroup$
$((12)(34))^{-1}=(12)(34)$. Check what happens to each element, moving from right to left. You get $(143)$.
$endgroup$
– Chris Custer
Dec 5 '18 at 20:06










1 Answer
1






active

oldest

votes


















1












$begingroup$

Given that $|A_4|=12$ it isn't too much work to explicitly write out all conjugations. A few checks for your work:




  • As noted in the comments, the size of the conjugacy class needs to divide $|A_4|=12$.

  • Cycle type is preserved by conjugation, so all elements of the conjugacy class are $3$-cycles.


A bit more thought gets you even further; as there are precisely eight $3$-cycles in $A_4$, there are at most eight conjugates. As the conjugacy classes partition the $3$-cycles and $S_4$ acts transitively on them (because it acts transitively on $3$-cycles), the size of the conjugacy classes must also divide $8$. So the number of conjugates divides $gcd(8,12)=4$.



Alternatively, by the orbit-stabilizer theorem, as the subgroup generated by $(2 3 4)$ clearly stabilizes $(2 3 4)$, and this subgroup contains $3$ elements, there are at most $frac{|A_4|}{|langle(2 3 4)rangle|}=frac{12}{3}=4$ conjugates.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! I realized I have to remove an element from my set in order for it to divide $|A_{4}|$. But I am not sure which element to remove?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:43










  • $begingroup$
    Do the computations and you will see for yourself. You should be able to do such a computation with certainty yourself. If you are unsure, share your computations and I can try to point out a mistake if there is one.
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:46












  • $begingroup$
    I feel like the element $(123)$ should be removed, but does $(124)=(142)$?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:51






  • 1




    $begingroup$
    You can write out how $(1 2 4)$ and $(1 4 2)$ act on ${1,2,3,4}$ and see that they are different. Or note that $(1 2 4)(1 2 4)=(1 4 2)$ and so they cannot be equal. By the looks of it either $(1 2 3)$ or $(1 3 2)$ should not be in the conjugacy class, but I can't be bothered to check.
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:53













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

Given that $|A_4|=12$ it isn't too much work to explicitly write out all conjugations. A few checks for your work:




  • As noted in the comments, the size of the conjugacy class needs to divide $|A_4|=12$.

  • Cycle type is preserved by conjugation, so all elements of the conjugacy class are $3$-cycles.


A bit more thought gets you even further; as there are precisely eight $3$-cycles in $A_4$, there are at most eight conjugates. As the conjugacy classes partition the $3$-cycles and $S_4$ acts transitively on them (because it acts transitively on $3$-cycles), the size of the conjugacy classes must also divide $8$. So the number of conjugates divides $gcd(8,12)=4$.



Alternatively, by the orbit-stabilizer theorem, as the subgroup generated by $(2 3 4)$ clearly stabilizes $(2 3 4)$, and this subgroup contains $3$ elements, there are at most $frac{|A_4|}{|langle(2 3 4)rangle|}=frac{12}{3}=4$ conjugates.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! I realized I have to remove an element from my set in order for it to divide $|A_{4}|$. But I am not sure which element to remove?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:43










  • $begingroup$
    Do the computations and you will see for yourself. You should be able to do such a computation with certainty yourself. If you are unsure, share your computations and I can try to point out a mistake if there is one.
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:46












  • $begingroup$
    I feel like the element $(123)$ should be removed, but does $(124)=(142)$?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:51






  • 1




    $begingroup$
    You can write out how $(1 2 4)$ and $(1 4 2)$ act on ${1,2,3,4}$ and see that they are different. Or note that $(1 2 4)(1 2 4)=(1 4 2)$ and so they cannot be equal. By the looks of it either $(1 2 3)$ or $(1 3 2)$ should not be in the conjugacy class, but I can't be bothered to check.
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:53


















1












$begingroup$

Given that $|A_4|=12$ it isn't too much work to explicitly write out all conjugations. A few checks for your work:




  • As noted in the comments, the size of the conjugacy class needs to divide $|A_4|=12$.

  • Cycle type is preserved by conjugation, so all elements of the conjugacy class are $3$-cycles.


A bit more thought gets you even further; as there are precisely eight $3$-cycles in $A_4$, there are at most eight conjugates. As the conjugacy classes partition the $3$-cycles and $S_4$ acts transitively on them (because it acts transitively on $3$-cycles), the size of the conjugacy classes must also divide $8$. So the number of conjugates divides $gcd(8,12)=4$.



Alternatively, by the orbit-stabilizer theorem, as the subgroup generated by $(2 3 4)$ clearly stabilizes $(2 3 4)$, and this subgroup contains $3$ elements, there are at most $frac{|A_4|}{|langle(2 3 4)rangle|}=frac{12}{3}=4$ conjugates.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! I realized I have to remove an element from my set in order for it to divide $|A_{4}|$. But I am not sure which element to remove?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:43










  • $begingroup$
    Do the computations and you will see for yourself. You should be able to do such a computation with certainty yourself. If you are unsure, share your computations and I can try to point out a mistake if there is one.
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:46












  • $begingroup$
    I feel like the element $(123)$ should be removed, but does $(124)=(142)$?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:51






  • 1




    $begingroup$
    You can write out how $(1 2 4)$ and $(1 4 2)$ act on ${1,2,3,4}$ and see that they are different. Or note that $(1 2 4)(1 2 4)=(1 4 2)$ and so they cannot be equal. By the looks of it either $(1 2 3)$ or $(1 3 2)$ should not be in the conjugacy class, but I can't be bothered to check.
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:53
















1












1








1





$begingroup$

Given that $|A_4|=12$ it isn't too much work to explicitly write out all conjugations. A few checks for your work:




  • As noted in the comments, the size of the conjugacy class needs to divide $|A_4|=12$.

  • Cycle type is preserved by conjugation, so all elements of the conjugacy class are $3$-cycles.


A bit more thought gets you even further; as there are precisely eight $3$-cycles in $A_4$, there are at most eight conjugates. As the conjugacy classes partition the $3$-cycles and $S_4$ acts transitively on them (because it acts transitively on $3$-cycles), the size of the conjugacy classes must also divide $8$. So the number of conjugates divides $gcd(8,12)=4$.



Alternatively, by the orbit-stabilizer theorem, as the subgroup generated by $(2 3 4)$ clearly stabilizes $(2 3 4)$, and this subgroup contains $3$ elements, there are at most $frac{|A_4|}{|langle(2 3 4)rangle|}=frac{12}{3}=4$ conjugates.






share|cite|improve this answer











$endgroup$



Given that $|A_4|=12$ it isn't too much work to explicitly write out all conjugations. A few checks for your work:




  • As noted in the comments, the size of the conjugacy class needs to divide $|A_4|=12$.

  • Cycle type is preserved by conjugation, so all elements of the conjugacy class are $3$-cycles.


A bit more thought gets you even further; as there are precisely eight $3$-cycles in $A_4$, there are at most eight conjugates. As the conjugacy classes partition the $3$-cycles and $S_4$ acts transitively on them (because it acts transitively on $3$-cycles), the size of the conjugacy classes must also divide $8$. So the number of conjugates divides $gcd(8,12)=4$.



Alternatively, by the orbit-stabilizer theorem, as the subgroup generated by $(2 3 4)$ clearly stabilizes $(2 3 4)$, and this subgroup contains $3$ elements, there are at most $frac{|A_4|}{|langle(2 3 4)rangle|}=frac{12}{3}=4$ conjugates.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 19:45

























answered Dec 5 '18 at 19:42









ServaesServaes

26.8k34098




26.8k34098












  • $begingroup$
    Thanks! I realized I have to remove an element from my set in order for it to divide $|A_{4}|$. But I am not sure which element to remove?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:43










  • $begingroup$
    Do the computations and you will see for yourself. You should be able to do such a computation with certainty yourself. If you are unsure, share your computations and I can try to point out a mistake if there is one.
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:46












  • $begingroup$
    I feel like the element $(123)$ should be removed, but does $(124)=(142)$?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:51






  • 1




    $begingroup$
    You can write out how $(1 2 4)$ and $(1 4 2)$ act on ${1,2,3,4}$ and see that they are different. Or note that $(1 2 4)(1 2 4)=(1 4 2)$ and so they cannot be equal. By the looks of it either $(1 2 3)$ or $(1 3 2)$ should not be in the conjugacy class, but I can't be bothered to check.
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:53




















  • $begingroup$
    Thanks! I realized I have to remove an element from my set in order for it to divide $|A_{4}|$. But I am not sure which element to remove?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:43










  • $begingroup$
    Do the computations and you will see for yourself. You should be able to do such a computation with certainty yourself. If you are unsure, share your computations and I can try to point out a mistake if there is one.
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:46












  • $begingroup$
    I feel like the element $(123)$ should be removed, but does $(124)=(142)$?
    $endgroup$
    – numericalorange
    Dec 5 '18 at 19:51






  • 1




    $begingroup$
    You can write out how $(1 2 4)$ and $(1 4 2)$ act on ${1,2,3,4}$ and see that they are different. Or note that $(1 2 4)(1 2 4)=(1 4 2)$ and so they cannot be equal. By the looks of it either $(1 2 3)$ or $(1 3 2)$ should not be in the conjugacy class, but I can't be bothered to check.
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:53


















$begingroup$
Thanks! I realized I have to remove an element from my set in order for it to divide $|A_{4}|$. But I am not sure which element to remove?
$endgroup$
– numericalorange
Dec 5 '18 at 19:43




$begingroup$
Thanks! I realized I have to remove an element from my set in order for it to divide $|A_{4}|$. But I am not sure which element to remove?
$endgroup$
– numericalorange
Dec 5 '18 at 19:43












$begingroup$
Do the computations and you will see for yourself. You should be able to do such a computation with certainty yourself. If you are unsure, share your computations and I can try to point out a mistake if there is one.
$endgroup$
– Servaes
Dec 5 '18 at 19:46






$begingroup$
Do the computations and you will see for yourself. You should be able to do such a computation with certainty yourself. If you are unsure, share your computations and I can try to point out a mistake if there is one.
$endgroup$
– Servaes
Dec 5 '18 at 19:46














$begingroup$
I feel like the element $(123)$ should be removed, but does $(124)=(142)$?
$endgroup$
– numericalorange
Dec 5 '18 at 19:51




$begingroup$
I feel like the element $(123)$ should be removed, but does $(124)=(142)$?
$endgroup$
– numericalorange
Dec 5 '18 at 19:51




1




1




$begingroup$
You can write out how $(1 2 4)$ and $(1 4 2)$ act on ${1,2,3,4}$ and see that they are different. Or note that $(1 2 4)(1 2 4)=(1 4 2)$ and so they cannot be equal. By the looks of it either $(1 2 3)$ or $(1 3 2)$ should not be in the conjugacy class, but I can't be bothered to check.
$endgroup$
– Servaes
Dec 5 '18 at 19:53






$begingroup$
You can write out how $(1 2 4)$ and $(1 4 2)$ act on ${1,2,3,4}$ and see that they are different. Or note that $(1 2 4)(1 2 4)=(1 4 2)$ and so they cannot be equal. By the looks of it either $(1 2 3)$ or $(1 3 2)$ should not be in the conjugacy class, but I can't be bothered to check.
$endgroup$
– Servaes
Dec 5 '18 at 19:53




















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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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