Representatives for all conjugacy classes of elements of order 15 in A11
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I am trying to find representatives for all conjugacy classes of elements of order 15 in $A_{11}$.
It is not hard to see that $(12345)(678)$ and $(12345)(678)(9 10 11)$ are the representatives for the conjugacy classes of order $15$ in $S_{11}$.
In the second case, the same element is a representative for $A_{11}$ as this conjugacy classes does not split in $A_{11}$ because we have two cycles of the same length. However, in the first case, this conjugacy class splits in $A_{11}$, and I am having a hard time finding representatives of each of these conjugacy classes. Any thoughts?
abstract-algebra finite-groups permutations symmetric-groups group-actions
asked Dec 24 '18 at 5:48
PeterPeter
1106
$endgroup$
add a comment |
$begingroup$
I am trying to find representatives for all conjugacy classes of elements of order 15 in $A_{11}$.
It is not hard to see that $(12345)(678)$ and $(12345)(678)(9 10 11)$ are the representatives for the conjugacy classes of order $15$ in $S_{11}$.
In the second case, the same element is a representative for $A_{11}$ as this conjugacy classes does not split in $A_{11}$ because we have two cycles of the same length. However, in the first case, this conjugacy class splits in $A_{11}$, and I am having a hard time finding representatives of each of these conjugacy classes. Any thoughts?
abstract-algebra finite-groups permutations symmetric-groups group-actions
asked Dec 24 '18 at 5:48
PeterPeter
1106
$endgroup$
1
$begingroup$
The first permutation has four cycles of length one, no?
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:04
$begingroup$
Only 3, but I do not understand how cycles on length 1 are important when considering the sign of a permutation. I know I am missing something.
$endgroup$
– Peter
Dec 24 '18 at 6:14
1
$begingroup$
Yeah, it's only three. But, that is at least two cycles of the same length, right?
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:19
$begingroup$
You are totally right; it's the same argument!
$endgroup$
– Peter
Dec 24 '18 at 6:21
add a comment |
$begingroup$
I am trying to find representatives for all conjugacy classes of elements of order 15 in $A_{11}$.
It is not hard to see that $(12345)(678)$ and $(12345)(678)(9 10 11)$ are the representatives for the conjugacy classes of order $15$ in $S_{11}$.
In the second case, the same element is a representative for $A_{11}$ as this conjugacy classes does not split in $A_{11}$ because we have two cycles of the same length. However, in the first case, this conjugacy class splits in $A_{11}$, and I am having a hard time finding representatives of each of these conjugacy classes. Any thoughts?
abstract-algebra finite-groups permutations symmetric-groups group-actions
asked Dec 24 '18 at 5:48
PeterPeter
1106
$endgroup$
I am trying to find representatives for all conjugacy classes of elements of order 15 in $A_{11}$.
It is not hard to see that $(12345)(678)$ and $(12345)(678)(9 10 11)$ are the representatives for the conjugacy classes of order $15$ in $S_{11}$.
In the second case, the same element is a representative for $A_{11}$ as this conjugacy classes does not split in $A_{11}$ because we have two cycles of the same length. However, in the first case, this conjugacy class splits in $A_{11}$, and I am having a hard time finding representatives of each of these conjugacy classes. Any thoughts?
abstract-algebra finite-groups permutations symmetric-groups group-actions
abstract-algebra finite-groups permutations symmetric-groups group-actions
asked Dec 24 '18 at 5:48
PeterPeter
1106
asked Dec 24 '18 at 5:48
PeterPeter
1106
asked Dec 24 '18 at 5:48
PeterPeter
1106
asked Dec 24 '18 at 5:48
PeterPeter
1106
asked Dec 24 '18 at 5:48
PeterPeter
1106
1106
1
$begingroup$
The first permutation has four cycles of length one, no?
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:04
$begingroup$
Only 3, but I do not understand how cycles on length 1 are important when considering the sign of a permutation. I know I am missing something.
$endgroup$
– Peter
Dec 24 '18 at 6:14
1
$begingroup$
Yeah, it's only three. But, that is at least two cycles of the same length, right?
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:19
$begingroup$
You are totally right; it's the same argument!
$endgroup$
– Peter
Dec 24 '18 at 6:21
add a comment |
1
$begingroup$
The first permutation has four cycles of length one, no?
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:04
$begingroup$
Only 3, but I do not understand how cycles on length 1 are important when considering the sign of a permutation. I know I am missing something.
$endgroup$
– Peter
Dec 24 '18 at 6:14
1
$begingroup$
Yeah, it's only three. But, that is at least two cycles of the same length, right?
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:19
$begingroup$
You are totally right; it's the same argument!
$endgroup$
– Peter
Dec 24 '18 at 6:21
1
1
$begingroup$
The first permutation has four cycles of length one, no?
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:04
$begingroup$
The first permutation has four cycles of length one, no?
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:04
$begingroup$
Only 3, but I do not understand how cycles on length 1 are important when considering the sign of a permutation. I know I am missing something.
$endgroup$
– Peter
Dec 24 '18 at 6:14
$begingroup$
Only 3, but I do not understand how cycles on length 1 are important when considering the sign of a permutation. I know I am missing something.
$endgroup$
– Peter
Dec 24 '18 at 6:14
1
1
$begingroup$
Yeah, it's only three. But, that is at least two cycles of the same length, right?
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:19
$begingroup$
Yeah, it's only three. But, that is at least two cycles of the same length, right?
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:19
$begingroup$
You are totally right; it's the same argument!
$endgroup$
– Peter
Dec 24 '18 at 6:21
$begingroup$
You are totally right; it's the same argument!
$endgroup$
– Peter
Dec 24 '18 at 6:21
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Let $sigmain S_n$ be an even permutation. The size of its conjugcay class in $S_n$ is, by the orbit-stabilizer theorem, equal to $$|[sigma]_{S_n}|=|S_n|/|C_{S_n}(sigma)|.qquad(*)$$
Similarly, in $A_n$ we get
$$
|[sigma]_{A_n}|=|A_n|/|C_{A_n}(sigma)|.qquad(**)
$$
Here $|A_n|=|S_n|/2$ and $C_{A_n}(sigma)=C_{S_n}(sigma)cap A_n$. In other words, $C_{A_n}(sigma)$ consists of the even permutations in $C_{S_n}(sigma)$. It follows that either $C_{A_n}(sigma)=C_{S_n}(sigma)$, or $C_{A_n}(sigma)$ is an index two subgroup of $C_{S_n}(sigma)$. The conjugacy class splits in two in the former case (see equations $(*)$ and $(**)$). We also see that the latter case (=no splitting) occurs, when there are odd permutations in $C_{S_n}(sigma)$.
The conjugacy classes of an even permutation in $A_n$ is equal to its conjugacy class in $S_n$ if and only if it is centralized by at least one odd permutation. OTOH, if the centralizer only contains even permutations, then the conjugacy in $S_n$ splits into two equal size conjugacy classes of $A_n$.
The 2-cycle $(9;10)$ is in the centralizer $C_{S_{11}}((12345)(678))$. This means that the conjugacy class of $sigma=(12345)(678)$ does not split in $A_{11}$.
$endgroup$
$begingroup$
See here. Close to being a duplicate :-(
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:29
$begingroup$
Switching to CW now that I can no longer delete. Should have found that answer earlier.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:30
add a comment |
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$begingroup$
Let $sigmain S_n$ be an even permutation. The size of its conjugcay class in $S_n$ is, by the orbit-stabilizer theorem, equal to $$|[sigma]_{S_n}|=|S_n|/|C_{S_n}(sigma)|.qquad(*)$$
Similarly, in $A_n$ we get
$$
|[sigma]_{A_n}|=|A_n|/|C_{A_n}(sigma)|.qquad(**)
$$
Here $|A_n|=|S_n|/2$ and $C_{A_n}(sigma)=C_{S_n}(sigma)cap A_n$. In other words, $C_{A_n}(sigma)$ consists of the even permutations in $C_{S_n}(sigma)$. It follows that either $C_{A_n}(sigma)=C_{S_n}(sigma)$, or $C_{A_n}(sigma)$ is an index two subgroup of $C_{S_n}(sigma)$. The conjugacy class splits in two in the former case (see equations $(*)$ and $(**)$). We also see that the latter case (=no splitting) occurs, when there are odd permutations in $C_{S_n}(sigma)$.
The conjugacy classes of an even permutation in $A_n$ is equal to its conjugacy class in $S_n$ if and only if it is centralized by at least one odd permutation. OTOH, if the centralizer only contains even permutations, then the conjugacy in $S_n$ splits into two equal size conjugacy classes of $A_n$.
The 2-cycle $(9;10)$ is in the centralizer $C_{S_{11}}((12345)(678))$. This means that the conjugacy class of $sigma=(12345)(678)$ does not split in $A_{11}$.
$endgroup$
$begingroup$
See here. Close to being a duplicate :-(
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:29
$begingroup$
Switching to CW now that I can no longer delete. Should have found that answer earlier.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:30
add a comment |
$begingroup$
Let $sigmain S_n$ be an even permutation. The size of its conjugcay class in $S_n$ is, by the orbit-stabilizer theorem, equal to $$|[sigma]_{S_n}|=|S_n|/|C_{S_n}(sigma)|.qquad(*)$$
Similarly, in $A_n$ we get
$$
|[sigma]_{A_n}|=|A_n|/|C_{A_n}(sigma)|.qquad(**)
$$
Here $|A_n|=|S_n|/2$ and $C_{A_n}(sigma)=C_{S_n}(sigma)cap A_n$. In other words, $C_{A_n}(sigma)$ consists of the even permutations in $C_{S_n}(sigma)$. It follows that either $C_{A_n}(sigma)=C_{S_n}(sigma)$, or $C_{A_n}(sigma)$ is an index two subgroup of $C_{S_n}(sigma)$. The conjugacy class splits in two in the former case (see equations $(*)$ and $(**)$). We also see that the latter case (=no splitting) occurs, when there are odd permutations in $C_{S_n}(sigma)$.
The conjugacy classes of an even permutation in $A_n$ is equal to its conjugacy class in $S_n$ if and only if it is centralized by at least one odd permutation. OTOH, if the centralizer only contains even permutations, then the conjugacy in $S_n$ splits into two equal size conjugacy classes of $A_n$.
The 2-cycle $(9;10)$ is in the centralizer $C_{S_{11}}((12345)(678))$. This means that the conjugacy class of $sigma=(12345)(678)$ does not split in $A_{11}$.
$endgroup$
$begingroup$
See here. Close to being a duplicate :-(
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:29
$begingroup$
Switching to CW now that I can no longer delete. Should have found that answer earlier.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:30
add a comment |
$begingroup$
Let $sigmain S_n$ be an even permutation. The size of its conjugcay class in $S_n$ is, by the orbit-stabilizer theorem, equal to $$|[sigma]_{S_n}|=|S_n|/|C_{S_n}(sigma)|.qquad(*)$$
Similarly, in $A_n$ we get
$$
|[sigma]_{A_n}|=|A_n|/|C_{A_n}(sigma)|.qquad(**)
$$
Here $|A_n|=|S_n|/2$ and $C_{A_n}(sigma)=C_{S_n}(sigma)cap A_n$. In other words, $C_{A_n}(sigma)$ consists of the even permutations in $C_{S_n}(sigma)$. It follows that either $C_{A_n}(sigma)=C_{S_n}(sigma)$, or $C_{A_n}(sigma)$ is an index two subgroup of $C_{S_n}(sigma)$. The conjugacy class splits in two in the former case (see equations $(*)$ and $(**)$). We also see that the latter case (=no splitting) occurs, when there are odd permutations in $C_{S_n}(sigma)$.
The conjugacy classes of an even permutation in $A_n$ is equal to its conjugacy class in $S_n$ if and only if it is centralized by at least one odd permutation. OTOH, if the centralizer only contains even permutations, then the conjugacy in $S_n$ splits into two equal size conjugacy classes of $A_n$.
The 2-cycle $(9;10)$ is in the centralizer $C_{S_{11}}((12345)(678))$. This means that the conjugacy class of $sigma=(12345)(678)$ does not split in $A_{11}$.
$endgroup$
Let $sigmain S_n$ be an even permutation. The size of its conjugcay class in $S_n$ is, by the orbit-stabilizer theorem, equal to $$|[sigma]_{S_n}|=|S_n|/|C_{S_n}(sigma)|.qquad(*)$$
Similarly, in $A_n$ we get
$$
|[sigma]_{A_n}|=|A_n|/|C_{A_n}(sigma)|.qquad(**)
$$
Here $|A_n|=|S_n|/2$ and $C_{A_n}(sigma)=C_{S_n}(sigma)cap A_n$. In other words, $C_{A_n}(sigma)$ consists of the even permutations in $C_{S_n}(sigma)$. It follows that either $C_{A_n}(sigma)=C_{S_n}(sigma)$, or $C_{A_n}(sigma)$ is an index two subgroup of $C_{S_n}(sigma)$. The conjugacy class splits in two in the former case (see equations $(*)$ and $(**)$). We also see that the latter case (=no splitting) occurs, when there are odd permutations in $C_{S_n}(sigma)$.
The conjugacy classes of an even permutation in $A_n$ is equal to its conjugacy class in $S_n$ if and only if it is centralized by at least one odd permutation. OTOH, if the centralizer only contains even permutations, then the conjugacy in $S_n$ splits into two equal size conjugacy classes of $A_n$.
The 2-cycle $(9;10)$ is in the centralizer $C_{S_{11}}((12345)(678))$. This means that the conjugacy class of $sigma=(12345)(678)$ does not split in $A_{11}$.
edited Dec 24 '18 at 6:27
community wiki
2 revs
Jyrki Lahtonen
$begingroup$
See here. Close to being a duplicate :-(
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:29
$begingroup$
Switching to CW now that I can no longer delete. Should have found that answer earlier.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:30
add a comment |
$begingroup$
See here. Close to being a duplicate :-(
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:29
$begingroup$
Switching to CW now that I can no longer delete. Should have found that answer earlier.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:30
$begingroup$
See here. Close to being a duplicate :-(
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:29
$begingroup$
See here. Close to being a duplicate :-(
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:29
$begingroup$
Switching to CW now that I can no longer delete. Should have found that answer earlier.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:30
$begingroup$
Switching to CW now that I can no longer delete. Should have found that answer earlier.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:30
add a comment |
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1
$begingroup$
The first permutation has four cycles of length one, no?
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:04
$begingroup$
Only 3, but I do not understand how cycles on length 1 are important when considering the sign of a permutation. I know I am missing something.
$endgroup$
– Peter
Dec 24 '18 at 6:14
1
$begingroup$
Yeah, it's only three. But, that is at least two cycles of the same length, right?
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:19
$begingroup$
You are totally right; it's the same argument!
$endgroup$
– Peter
Dec 24 '18 at 6:21