Compute all $X_{g}$ and all $G_{x}$ for $X = left {1, 2, 3right }$, $G = S_{3} = left {(1), (12), (13), (23),...
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Compute all $X_{g}$ and all $G_{x}$ for $X = left {1, 2, 3right }$, $G = S_{3} = left {(1), (12), (13), (23), (123), (132)right }$.
Can someone give me a head start to this problem?
$X_{g}=left {xin X: gx=xright }$. So how do I find $X_{(12)}$? Do I calculate $(12)(1),(12)(2),(12)(3)$?
abstract-algebra group-theory permutations group-actions permutation-cycles
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Compute all $X_{g}$ and all $G_{x}$ for $X = left {1, 2, 3right }$, $G = S_{3} = left {(1), (12), (13), (23), (123), (132)right }$.
Can someone give me a head start to this problem?
$X_{g}=left {xin X: gx=xright }$. So how do I find $X_{(12)}$? Do I calculate $(12)(1),(12)(2),(12)(3)$?
abstract-algebra group-theory permutations group-actions permutation-cycles
1
Yes. But then, you see, $(12)$ is the permutation that affects only $1$ and $2$, so it doesn't affect any other number. So $X_{(12)}$ should be immediate.
– Scientifica
Dec 5 at 19:01
And so $X_{(12)}=left {3right }$?
– numericalorange
Dec 5 at 19:17
1
That's absolutely right!
– Scientifica
Dec 5 at 19:30
@Scientifica Oh, wow, I feel so happy that you helped me understand this so easily!!
– numericalorange
Dec 5 at 19:31
1
I'm gonna post this as an answer so that your question is answered.
– Scientifica
Dec 5 at 19:32
|
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up vote
1
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up vote
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down vote
favorite
Compute all $X_{g}$ and all $G_{x}$ for $X = left {1, 2, 3right }$, $G = S_{3} = left {(1), (12), (13), (23), (123), (132)right }$.
Can someone give me a head start to this problem?
$X_{g}=left {xin X: gx=xright }$. So how do I find $X_{(12)}$? Do I calculate $(12)(1),(12)(2),(12)(3)$?
abstract-algebra group-theory permutations group-actions permutation-cycles
Compute all $X_{g}$ and all $G_{x}$ for $X = left {1, 2, 3right }$, $G = S_{3} = left {(1), (12), (13), (23), (123), (132)right }$.
Can someone give me a head start to this problem?
$X_{g}=left {xin X: gx=xright }$. So how do I find $X_{(12)}$? Do I calculate $(12)(1),(12)(2),(12)(3)$?
abstract-algebra group-theory permutations group-actions permutation-cycles
abstract-algebra group-theory permutations group-actions permutation-cycles
edited Dec 5 at 19:34
Scientifica
6,34141333
6,34141333
asked Dec 5 at 18:56
numericalorange
1,719311
1,719311
1
Yes. But then, you see, $(12)$ is the permutation that affects only $1$ and $2$, so it doesn't affect any other number. So $X_{(12)}$ should be immediate.
– Scientifica
Dec 5 at 19:01
And so $X_{(12)}=left {3right }$?
– numericalorange
Dec 5 at 19:17
1
That's absolutely right!
– Scientifica
Dec 5 at 19:30
@Scientifica Oh, wow, I feel so happy that you helped me understand this so easily!!
– numericalorange
Dec 5 at 19:31
1
I'm gonna post this as an answer so that your question is answered.
– Scientifica
Dec 5 at 19:32
|
show 1 more comment
1
Yes. But then, you see, $(12)$ is the permutation that affects only $1$ and $2$, so it doesn't affect any other number. So $X_{(12)}$ should be immediate.
– Scientifica
Dec 5 at 19:01
And so $X_{(12)}=left {3right }$?
– numericalorange
Dec 5 at 19:17
1
That's absolutely right!
– Scientifica
Dec 5 at 19:30
@Scientifica Oh, wow, I feel so happy that you helped me understand this so easily!!
– numericalorange
Dec 5 at 19:31
1
I'm gonna post this as an answer so that your question is answered.
– Scientifica
Dec 5 at 19:32
1
1
Yes. But then, you see, $(12)$ is the permutation that affects only $1$ and $2$, so it doesn't affect any other number. So $X_{(12)}$ should be immediate.
– Scientifica
Dec 5 at 19:01
Yes. But then, you see, $(12)$ is the permutation that affects only $1$ and $2$, so it doesn't affect any other number. So $X_{(12)}$ should be immediate.
– Scientifica
Dec 5 at 19:01
And so $X_{(12)}=left {3right }$?
– numericalorange
Dec 5 at 19:17
And so $X_{(12)}=left {3right }$?
– numericalorange
Dec 5 at 19:17
1
1
That's absolutely right!
– Scientifica
Dec 5 at 19:30
That's absolutely right!
– Scientifica
Dec 5 at 19:30
@Scientifica Oh, wow, I feel so happy that you helped me understand this so easily!!
– numericalorange
Dec 5 at 19:31
@Scientifica Oh, wow, I feel so happy that you helped me understand this so easily!!
– numericalorange
Dec 5 at 19:31
1
1
I'm gonna post this as an answer so that your question is answered.
– Scientifica
Dec 5 at 19:32
I'm gonna post this as an answer so that your question is answered.
– Scientifica
Dec 5 at 19:32
|
show 1 more comment
1 Answer
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1
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Yes that's how you do it.
You can also easily see it as follows: the permutation $(12)$ affects $1$ and $2$, but no other element. So $X_{(12)}$ is immediate. The same remark allows you to quickly determine the $G_x$.
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1 Answer
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Yes that's how you do it.
You can also easily see it as follows: the permutation $(12)$ affects $1$ and $2$, but no other element. So $X_{(12)}$ is immediate. The same remark allows you to quickly determine the $G_x$.
add a comment |
up vote
1
down vote
accepted
Yes that's how you do it.
You can also easily see it as follows: the permutation $(12)$ affects $1$ and $2$, but no other element. So $X_{(12)}$ is immediate. The same remark allows you to quickly determine the $G_x$.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes that's how you do it.
You can also easily see it as follows: the permutation $(12)$ affects $1$ and $2$, but no other element. So $X_{(12)}$ is immediate. The same remark allows you to quickly determine the $G_x$.
Yes that's how you do it.
You can also easily see it as follows: the permutation $(12)$ affects $1$ and $2$, but no other element. So $X_{(12)}$ is immediate. The same remark allows you to quickly determine the $G_x$.
answered Dec 5 at 19:34
Scientifica
6,34141333
6,34141333
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Yes. But then, you see, $(12)$ is the permutation that affects only $1$ and $2$, so it doesn't affect any other number. So $X_{(12)}$ should be immediate.
– Scientifica
Dec 5 at 19:01
And so $X_{(12)}=left {3right }$?
– numericalorange
Dec 5 at 19:17
1
That's absolutely right!
– Scientifica
Dec 5 at 19:30
@Scientifica Oh, wow, I feel so happy that you helped me understand this so easily!!
– numericalorange
Dec 5 at 19:31
1
I'm gonna post this as an answer so that your question is answered.
– Scientifica
Dec 5 at 19:32