Find the conjugacy class $O_{(234)}$ if $G=A_{4}$ is acting on itself by conjugation.
$begingroup$
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
asked Dec 5 '18 at 19:21
numericalorangenumericalorange
1,775311
$endgroup$
|
show 4 more comments
$begingroup$
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
asked Dec 5 '18 at 19:21
numericalorangenumericalorange
1,775311
$endgroup$
$begingroup$
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
|
show 4 more comments
$begingroup$
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
asked Dec 5 '18 at 19:21
numericalorangenumericalorange
1,775311
$endgroup$
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
asked Dec 5 '18 at 19:21
numericalorangenumericalorange
1,775311
asked Dec 5 '18 at 19:21
numericalorangenumericalorange
1,775311
asked Dec 5 '18 at 19:21
numericalorangenumericalorange
1,775311
asked Dec 5 '18 at 19:21
numericalorangenumericalorange
1,775311
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
|
show 4 more comments
$begingroup$
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
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 5 '18 at 19:42
ServaesServaes
26.8k34098
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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$begingroup$
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.
answered Dec 5 '18 at 19:42
ServaesServaes
26.8k34098
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 5 '18 at 19:42
ServaesServaes
26.8k34098
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 5 '18 at 19:42
ServaesServaes
26.8k34098
$endgroup$
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.
answered Dec 5 '18 at 19:42
ServaesServaes
26.8k34098
edited Dec 5 '18 at 19:45
answered Dec 5 '18 at 19:42
ServaesServaes
26.8k34098
answered Dec 5 '18 at 19:42
ServaesServaes
26.8k34098
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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