Mixing
Finding all permutations which permute with a certain one
$begingroup$
I have found this topic here: How to find all the permutations which commute with a certain one, but I still could not understand it. Could anyone please explain how to think about doing it?
If I have (12)(34), working in the symmetric group 4 for instance, how would I find every possible permutation that permutes with it?
Thanks.
permutations
asked Jan 12 at 22:00
ValVal
557
$endgroup$
add a comment |
$begingroup$
I have found this topic here: How to find all the permutations which commute with a certain one, but I still could not understand it. Could anyone please explain how to think about doing it?
If I have (12)(34), working in the symmetric group 4 for instance, how would I find every possible permutation that permutes with it?
Thanks.
permutations
asked Jan 12 at 22:00
ValVal
557
$endgroup$
$begingroup$
Look for conjugates of a permutation. A link that can help is math.stackexchange.com/questions/1532745/… For your problem, you want $k(12)(34)k^{-1}=(12)(34)$.
$endgroup$
– Anurag A
Jan 12 at 22:08
$begingroup$
I am not really sure about what "(k(1)k(3)k(5))(k(2)k(4))" is meaning and how would I use that. It's from the post you send.
$endgroup$
– Val
Jan 12 at 22:26
$begingroup$
You should work with the so useful decomposition of a parmutation into a product of disjoint cycles.
$endgroup$
– Jean Marie
Jan 12 at 22:49
$begingroup$
@BurLeXyOOnuTz if you read the answer carefully you will see $k(1)$ is the image of $1$ under the (permutation) function $k$.
$endgroup$
– Anurag A
Jan 13 at 1:44
add a comment |
$begingroup$
I have found this topic here: How to find all the permutations which commute with a certain one, but I still could not understand it. Could anyone please explain how to think about doing it?
If I have (12)(34), working in the symmetric group 4 for instance, how would I find every possible permutation that permutes with it?
Thanks.
permutations
asked Jan 12 at 22:00
ValVal
557
$endgroup$
I have found this topic here: How to find all the permutations which commute with a certain one, but I still could not understand it. Could anyone please explain how to think about doing it?
If I have (12)(34), working in the symmetric group 4 for instance, how would I find every possible permutation that permutes with it?
Thanks.
permutations
permutations
asked Jan 12 at 22:00
ValVal
557
asked Jan 12 at 22:00
ValVal
557
asked Jan 12 at 22:00
ValVal
557
asked Jan 12 at 22:00
ValVal
557
asked Jan 12 at 22:00
ValVal
557
557
$begingroup$
Look for conjugates of a permutation. A link that can help is math.stackexchange.com/questions/1532745/… For your problem, you want $k(12)(34)k^{-1}=(12)(34)$.
$endgroup$
– Anurag A
Jan 12 at 22:08
$begingroup$
I am not really sure about what "(k(1)k(3)k(5))(k(2)k(4))" is meaning and how would I use that. It's from the post you send.
$endgroup$
– Val
Jan 12 at 22:26
$begingroup$
You should work with the so useful decomposition of a parmutation into a product of disjoint cycles.
$endgroup$
– Jean Marie
Jan 12 at 22:49
$begingroup$
@BurLeXyOOnuTz if you read the answer carefully you will see $k(1)$ is the image of $1$ under the (permutation) function $k$.
$endgroup$
– Anurag A
Jan 13 at 1:44
add a comment |
$begingroup$
Look for conjugates of a permutation. A link that can help is math.stackexchange.com/questions/1532745/… For your problem, you want $k(12)(34)k^{-1}=(12)(34)$.
$endgroup$
– Anurag A
Jan 12 at 22:08
$begingroup$
I am not really sure about what "(k(1)k(3)k(5))(k(2)k(4))" is meaning and how would I use that. It's from the post you send.
$endgroup$
– Val
Jan 12 at 22:26
$begingroup$
You should work with the so useful decomposition of a parmutation into a product of disjoint cycles.
$endgroup$
– Jean Marie
Jan 12 at 22:49
$begingroup$
@BurLeXyOOnuTz if you read the answer carefully you will see $k(1)$ is the image of $1$ under the (permutation) function $k$.
$endgroup$
– Anurag A
Jan 13 at 1:44
$begingroup$
Look for conjugates of a permutation. A link that can help is math.stackexchange.com/questions/1532745/… For your problem, you want $k(12)(34)k^{-1}=(12)(34)$.
$endgroup$
– Anurag A
Jan 12 at 22:08
$begingroup$
Look for conjugates of a permutation. A link that can help is math.stackexchange.com/questions/1532745/… For your problem, you want $k(12)(34)k^{-1}=(12)(34)$.
$endgroup$
– Anurag A
Jan 12 at 22:08
$begingroup$
I am not really sure about what "(k(1)k(3)k(5))(k(2)k(4))" is meaning and how would I use that. It's from the post you send.
$endgroup$
– Val
Jan 12 at 22:26
$begingroup$
I am not really sure about what "(k(1)k(3)k(5))(k(2)k(4))" is meaning and how would I use that. It's from the post you send.
$endgroup$
– Val
Jan 12 at 22:26
$begingroup$
You should work with the so useful decomposition of a parmutation into a product of disjoint cycles.
$endgroup$
– Jean Marie
Jan 12 at 22:49
$begingroup$
You should work with the so useful decomposition of a parmutation into a product of disjoint cycles.
$endgroup$
– Jean Marie
Jan 12 at 22:49
$begingroup$
@BurLeXyOOnuTz if you read the answer carefully you will see $k(1)$ is the image of $1$ under the (permutation) function $k$.
$endgroup$
– Anurag A
Jan 13 at 1:44
$begingroup$
@BurLeXyOOnuTz if you read the answer carefully you will see $k(1)$ is the image of $1$ under the (permutation) function $k$.
$endgroup$
– Anurag A
Jan 13 at 1:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You want all the permutations $k in S_4$ such that
$$k(12)(34)k^{-1}=(12)(34).$$
Since
$$
k(12)(34)k^{-1} =k(12)color{red}k^{-1} , color{red}k(34)k^{-1}
= (k(1) ,, k(2)) ,,,, (k(3) ,, k(4))
$$
Where $k(i)$ means the image of $i$ under permutation $k$.
You want this to be
$$(k(1) ,, k(2)) ,,,, (k(3) ,, k(4)) = (1 ,, 2) ,,, (3,, 4)$$
So our permutation $k$ can be such that
$$(k(1) ,, k(2))=(1 ,, 2) quad text{ and } quad (k(3) ,, k(4))=(3,, 4)$$
OR
$$(k(1) ,, k(2))=(3 ,, 4) quad text{ and } quad (k(3) ,, k(4))=(1,, 2)$$
From the first scenario, we get either $k(1)=1,k(2)=2,k(3)=3,k(4)=4$ (identity permutation) or $k(1)=1,k(2)=2,k(3)=4,k(4)=3$ (this refers to the permutation $(34)$) or.....
Hope you can take it from here.
Added response:
I am adding this to answer the question you have raised in your comment.
Observe from the second scenario I have mentioned in my answer above, we can have $k(1)=3, k(2)=4$ and $k(3)=2, k(4)=1$. So ask yourself, what will be the cycle notation for this permutation?
we have ($1 xrightarrow{k} 3 xrightarrow{k} 2 xrightarrow{k} 4 xrightarrow{k} 1$) and this corresponds to $(1324)$.
answered Jan 13 at 1:53
Anurag AAnurag A
26.1k12250
$endgroup$
$begingroup$
I saw in the answers that (1324) is a permutation that satisfies commutativity too, but I don't understand how would you get that using this technique. Could you please explain how would you get (1324) please? I would really appreciate it.
$endgroup$
– Val
Jan 13 at 2:18
$begingroup$
@BurLeXyOOnuTz I have added some remarks to my answer, so please take a look.
$endgroup$
– Anurag A
Jan 13 at 2:48
$begingroup$
excuse me I do not feel like I understand k. As a question, can k(1) be 2,3,4 then k(2)=1,3,4 then k(3)=1,2,4 and k(4)=1,2,3 and we are just rearranging these maps in all possible ways?
$endgroup$
– Val
Jan 13 at 18:45
add a comment |
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$begingroup$
You want all the permutations $k in S_4$ such that
$$k(12)(34)k^{-1}=(12)(34).$$
Since
$$
k(12)(34)k^{-1} =k(12)color{red}k^{-1} , color{red}k(34)k^{-1}
= (k(1) ,, k(2)) ,,,, (k(3) ,, k(4))
$$
Where $k(i)$ means the image of $i$ under permutation $k$.
You want this to be
$$(k(1) ,, k(2)) ,,,, (k(3) ,, k(4)) = (1 ,, 2) ,,, (3,, 4)$$
So our permutation $k$ can be such that
$$(k(1) ,, k(2))=(1 ,, 2) quad text{ and } quad (k(3) ,, k(4))=(3,, 4)$$
OR
$$(k(1) ,, k(2))=(3 ,, 4) quad text{ and } quad (k(3) ,, k(4))=(1,, 2)$$
From the first scenario, we get either $k(1)=1,k(2)=2,k(3)=3,k(4)=4$ (identity permutation) or $k(1)=1,k(2)=2,k(3)=4,k(4)=3$ (this refers to the permutation $(34)$) or.....
Hope you can take it from here.
Added response:
I am adding this to answer the question you have raised in your comment.
Observe from the second scenario I have mentioned in my answer above, we can have $k(1)=3, k(2)=4$ and $k(3)=2, k(4)=1$. So ask yourself, what will be the cycle notation for this permutation?
we have ($1 xrightarrow{k} 3 xrightarrow{k} 2 xrightarrow{k} 4 xrightarrow{k} 1$) and this corresponds to $(1324)$.
answered Jan 13 at 1:53
Anurag AAnurag A
26.1k12250
$endgroup$
$begingroup$
I saw in the answers that (1324) is a permutation that satisfies commutativity too, but I don't understand how would you get that using this technique. Could you please explain how would you get (1324) please? I would really appreciate it.
$endgroup$
– Val
Jan 13 at 2:18
$begingroup$
@BurLeXyOOnuTz I have added some remarks to my answer, so please take a look.
$endgroup$
– Anurag A
Jan 13 at 2:48
$begingroup$
excuse me I do not feel like I understand k. As a question, can k(1) be 2,3,4 then k(2)=1,3,4 then k(3)=1,2,4 and k(4)=1,2,3 and we are just rearranging these maps in all possible ways?
$endgroup$
– Val
Jan 13 at 18:45
add a comment |
$begingroup$
You want all the permutations $k in S_4$ such that
$$k(12)(34)k^{-1}=(12)(34).$$
Since
$$
k(12)(34)k^{-1} =k(12)color{red}k^{-1} , color{red}k(34)k^{-1}
= (k(1) ,, k(2)) ,,,, (k(3) ,, k(4))
$$
Where $k(i)$ means the image of $i$ under permutation $k$.
You want this to be
$$(k(1) ,, k(2)) ,,,, (k(3) ,, k(4)) = (1 ,, 2) ,,, (3,, 4)$$
So our permutation $k$ can be such that
$$(k(1) ,, k(2))=(1 ,, 2) quad text{ and } quad (k(3) ,, k(4))=(3,, 4)$$
OR
$$(k(1) ,, k(2))=(3 ,, 4) quad text{ and } quad (k(3) ,, k(4))=(1,, 2)$$
From the first scenario, we get either $k(1)=1,k(2)=2,k(3)=3,k(4)=4$ (identity permutation) or $k(1)=1,k(2)=2,k(3)=4,k(4)=3$ (this refers to the permutation $(34)$) or.....
Hope you can take it from here.
Added response:
I am adding this to answer the question you have raised in your comment.
Observe from the second scenario I have mentioned in my answer above, we can have $k(1)=3, k(2)=4$ and $k(3)=2, k(4)=1$. So ask yourself, what will be the cycle notation for this permutation?
we have ($1 xrightarrow{k} 3 xrightarrow{k} 2 xrightarrow{k} 4 xrightarrow{k} 1$) and this corresponds to $(1324)$.
answered Jan 13 at 1:53
Anurag AAnurag A
26.1k12250
$endgroup$
$begingroup$
I saw in the answers that (1324) is a permutation that satisfies commutativity too, but I don't understand how would you get that using this technique. Could you please explain how would you get (1324) please? I would really appreciate it.
$endgroup$
– Val
Jan 13 at 2:18
$begingroup$
@BurLeXyOOnuTz I have added some remarks to my answer, so please take a look.
$endgroup$
– Anurag A
Jan 13 at 2:48
$begingroup$
excuse me I do not feel like I understand k. As a question, can k(1) be 2,3,4 then k(2)=1,3,4 then k(3)=1,2,4 and k(4)=1,2,3 and we are just rearranging these maps in all possible ways?
$endgroup$
– Val
Jan 13 at 18:45
add a comment |
$begingroup$
You want all the permutations $k in S_4$ such that
$$k(12)(34)k^{-1}=(12)(34).$$
Since
$$
k(12)(34)k^{-1} =k(12)color{red}k^{-1} , color{red}k(34)k^{-1}
= (k(1) ,, k(2)) ,,,, (k(3) ,, k(4))
$$
Where $k(i)$ means the image of $i$ under permutation $k$.
You want this to be
$$(k(1) ,, k(2)) ,,,, (k(3) ,, k(4)) = (1 ,, 2) ,,, (3,, 4)$$
So our permutation $k$ can be such that
$$(k(1) ,, k(2))=(1 ,, 2) quad text{ and } quad (k(3) ,, k(4))=(3,, 4)$$
OR
$$(k(1) ,, k(2))=(3 ,, 4) quad text{ and } quad (k(3) ,, k(4))=(1,, 2)$$
From the first scenario, we get either $k(1)=1,k(2)=2,k(3)=3,k(4)=4$ (identity permutation) or $k(1)=1,k(2)=2,k(3)=4,k(4)=3$ (this refers to the permutation $(34)$) or.....
Hope you can take it from here.
Added response:
I am adding this to answer the question you have raised in your comment.
Observe from the second scenario I have mentioned in my answer above, we can have $k(1)=3, k(2)=4$ and $k(3)=2, k(4)=1$. So ask yourself, what will be the cycle notation for this permutation?
we have ($1 xrightarrow{k} 3 xrightarrow{k} 2 xrightarrow{k} 4 xrightarrow{k} 1$) and this corresponds to $(1324)$.
answered Jan 13 at 1:53
Anurag AAnurag A
26.1k12250
$endgroup$
You want all the permutations $k in S_4$ such that
$$k(12)(34)k^{-1}=(12)(34).$$
Since
$$
k(12)(34)k^{-1} =k(12)color{red}k^{-1} , color{red}k(34)k^{-1}
= (k(1) ,, k(2)) ,,,, (k(3) ,, k(4))
$$
Where $k(i)$ means the image of $i$ under permutation $k$.
You want this to be
$$(k(1) ,, k(2)) ,,,, (k(3) ,, k(4)) = (1 ,, 2) ,,, (3,, 4)$$
So our permutation $k$ can be such that
$$(k(1) ,, k(2))=(1 ,, 2) quad text{ and } quad (k(3) ,, k(4))=(3,, 4)$$
OR
$$(k(1) ,, k(2))=(3 ,, 4) quad text{ and } quad (k(3) ,, k(4))=(1,, 2)$$
From the first scenario, we get either $k(1)=1,k(2)=2,k(3)=3,k(4)=4$ (identity permutation) or $k(1)=1,k(2)=2,k(3)=4,k(4)=3$ (this refers to the permutation $(34)$) or.....
Hope you can take it from here.
Added response:
I am adding this to answer the question you have raised in your comment.
Observe from the second scenario I have mentioned in my answer above, we can have $k(1)=3, k(2)=4$ and $k(3)=2, k(4)=1$. So ask yourself, what will be the cycle notation for this permutation?
we have ($1 xrightarrow{k} 3 xrightarrow{k} 2 xrightarrow{k} 4 xrightarrow{k} 1$) and this corresponds to $(1324)$.
answered Jan 13 at 1:53
Anurag AAnurag A
26.1k12250
edited Jan 13 at 2:47
answered Jan 13 at 1:53
Anurag AAnurag A
26.1k12250
answered Jan 13 at 1:53
Anurag AAnurag A
26.1k12250
answered Jan 13 at 1:53
Anurag AAnurag A
26.1k12250
26.1k12250
$begingroup$
I saw in the answers that (1324) is a permutation that satisfies commutativity too, but I don't understand how would you get that using this technique. Could you please explain how would you get (1324) please? I would really appreciate it.
$endgroup$
– Val
Jan 13 at 2:18
$begingroup$
@BurLeXyOOnuTz I have added some remarks to my answer, so please take a look.
$endgroup$
– Anurag A
Jan 13 at 2:48
$begingroup$
excuse me I do not feel like I understand k. As a question, can k(1) be 2,3,4 then k(2)=1,3,4 then k(3)=1,2,4 and k(4)=1,2,3 and we are just rearranging these maps in all possible ways?
$endgroup$
– Val
Jan 13 at 18:45
add a comment |
$begingroup$
I saw in the answers that (1324) is a permutation that satisfies commutativity too, but I don't understand how would you get that using this technique. Could you please explain how would you get (1324) please? I would really appreciate it.
$endgroup$
– Val
Jan 13 at 2:18
$begingroup$
@BurLeXyOOnuTz I have added some remarks to my answer, so please take a look.
$endgroup$
– Anurag A
Jan 13 at 2:48
$begingroup$
excuse me I do not feel like I understand k. As a question, can k(1) be 2,3,4 then k(2)=1,3,4 then k(3)=1,2,4 and k(4)=1,2,3 and we are just rearranging these maps in all possible ways?
$endgroup$
– Val
Jan 13 at 18:45
$begingroup$
I saw in the answers that (1324) is a permutation that satisfies commutativity too, but I don't understand how would you get that using this technique. Could you please explain how would you get (1324) please? I would really appreciate it.
$endgroup$
– Val
Jan 13 at 2:18
$begingroup$
I saw in the answers that (1324) is a permutation that satisfies commutativity too, but I don't understand how would you get that using this technique. Could you please explain how would you get (1324) please? I would really appreciate it.
$endgroup$
– Val
Jan 13 at 2:18
$begingroup$
@BurLeXyOOnuTz I have added some remarks to my answer, so please take a look.
$endgroup$
– Anurag A
Jan 13 at 2:48
$begingroup$
@BurLeXyOOnuTz I have added some remarks to my answer, so please take a look.
$endgroup$
– Anurag A
Jan 13 at 2:48
$begingroup$
excuse me I do not feel like I understand k. As a question, can k(1) be 2,3,4 then k(2)=1,3,4 then k(3)=1,2,4 and k(4)=1,2,3 and we are just rearranging these maps in all possible ways?
$endgroup$
– Val
Jan 13 at 18:45
$begingroup$
excuse me I do not feel like I understand k. As a question, can k(1) be 2,3,4 then k(2)=1,3,4 then k(3)=1,2,4 and k(4)=1,2,3 and we are just rearranging these maps in all possible ways?
$endgroup$
– Val
Jan 13 at 18:45
add a comment |
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$begingroup$
Look for conjugates of a permutation. A link that can help is math.stackexchange.com/questions/1532745/… For your problem, you want $k(12)(34)k^{-1}=(12)(34)$.
$endgroup$
– Anurag A
Jan 12 at 22:08
$begingroup$
I am not really sure about what "(k(1)k(3)k(5))(k(2)k(4))" is meaning and how would I use that. It's from the post you send.
$endgroup$
– Val
Jan 12 at 22:26
$begingroup$
You should work with the so useful decomposition of a parmutation into a product of disjoint cycles.
$endgroup$
– Jean Marie
Jan 12 at 22:49
$begingroup$
@BurLeXyOOnuTz if you read the answer carefully you will see $k(1)$ is the image of $1$ under the (permutation) function $k$.
$endgroup$
– Anurag A
Jan 13 at 1:44