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Permutations that conjugate in $S_5$ but not in $A_5$
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Find two permutations that conjugate in $S_5$, but not in $A_5$.
I can't understand why is it possible - in order for two permutations to conjugate, they must have the same cycle structure.
If two permutations are conjugate in $S_5$, this means they have the same cycle structure, and therefore will have the same structure in $A_5$, and will be still conjugate in $A_5$...
What am I missing?
group-theory symmetric-groups
asked Aug 6 at 5:14
ChikChak
650216
 |Â
show 9 more comments
up vote
2
down vote
favorite
Find two permutations that conjugate in $S_5$, but not in $A_5$.
I can't understand why is it possible - in order for two permutations to conjugate, they must have the same cycle structure.
If two permutations are conjugate in $S_5$, this means they have the same cycle structure, and therefore will have the same structure in $A_5$, and will be still conjugate in $A_5$...
What am I missing?
group-theory symmetric-groups
asked Aug 6 at 5:14
ChikChak
650216
See for example here. I think we have even better threads, but that came up first in my search.
– Jyrki Lahtonen
Aug 6 at 5:21
What's happening is that the conjugacy class of $S_n$ may or may not split in two conjugacy classes in $A_n$. See here.
– Jyrki Lahtonen
Aug 6 at 5:23
Here the corresponding exercise in Dummit & Foote is discussed. I won't pick a dupe target because my search is biased by my own history. But I suspect others explained this earlier and better and I simply can't find a link to those threads quickly.
– Jyrki Lahtonen
Aug 6 at 5:26
2
Just for you to think concretely: Are the $3$-cycles $(1,2,3)$ and $(1,3,2)$ conjugate in $A_3$?
– Ted Shifrin
Aug 6 at 5:36
@Ted Shifrin In general, I'm not sure how can I prove that something does not conjugate, rather than looking at the cycle structure or trying all possiblities, which obviously isn't a good idea.
– ChikChak
Aug 6 at 5:48
 |Â
show 9 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Find two permutations that conjugate in $S_5$, but not in $A_5$.
I can't understand why is it possible - in order for two permutations to conjugate, they must have the same cycle structure.
If two permutations are conjugate in $S_5$, this means they have the same cycle structure, and therefore will have the same structure in $A_5$, and will be still conjugate in $A_5$...
What am I missing?
group-theory symmetric-groups
asked Aug 6 at 5:14
ChikChak
650216
Find two permutations that conjugate in $S_5$, but not in $A_5$.
I can't understand why is it possible - in order for two permutations to conjugate, they must have the same cycle structure.
If two permutations are conjugate in $S_5$, this means they have the same cycle structure, and therefore will have the same structure in $A_5$, and will be still conjugate in $A_5$...
What am I missing?
group-theory symmetric-groups
asked Aug 6 at 5:14
ChikChak
650216
asked Aug 6 at 5:14
ChikChak
650216
asked Aug 6 at 5:14
ChikChak
650216
asked Aug 6 at 5:14
ChikChak
650216
650216
See for example here. I think we have even better threads, but that came up first in my search.
– Jyrki Lahtonen
Aug 6 at 5:21
What's happening is that the conjugacy class of $S_n$ may or may not split in two conjugacy classes in $A_n$. See here.
– Jyrki Lahtonen
Aug 6 at 5:23
Here the corresponding exercise in Dummit & Foote is discussed. I won't pick a dupe target because my search is biased by my own history. But I suspect others explained this earlier and better and I simply can't find a link to those threads quickly.
– Jyrki Lahtonen
Aug 6 at 5:26
2
Just for you to think concretely: Are the $3$-cycles $(1,2,3)$ and $(1,3,2)$ conjugate in $A_3$?
– Ted Shifrin
Aug 6 at 5:36
@Ted Shifrin In general, I'm not sure how can I prove that something does not conjugate, rather than looking at the cycle structure or trying all possiblities, which obviously isn't a good idea.
– ChikChak
Aug 6 at 5:48
 |Â
show 9 more comments
See for example here. I think we have even better threads, but that came up first in my search.
– Jyrki Lahtonen
Aug 6 at 5:21
What's happening is that the conjugacy class of $S_n$ may or may not split in two conjugacy classes in $A_n$. See here.
– Jyrki Lahtonen
Aug 6 at 5:23
Here the corresponding exercise in Dummit & Foote is discussed. I won't pick a dupe target because my search is biased by my own history. But I suspect others explained this earlier and better and I simply can't find a link to those threads quickly.
– Jyrki Lahtonen
Aug 6 at 5:26
2
Just for you to think concretely: Are the $3$-cycles $(1,2,3)$ and $(1,3,2)$ conjugate in $A_3$?
– Ted Shifrin
Aug 6 at 5:36
@Ted Shifrin In general, I'm not sure how can I prove that something does not conjugate, rather than looking at the cycle structure or trying all possiblities, which obviously isn't a good idea.
– ChikChak
Aug 6 at 5:48
See for example here. I think we have even better threads, but that came up first in my search.
– Jyrki Lahtonen
Aug 6 at 5:21
See for example here. I think we have even better threads, but that came up first in my search.
– Jyrki Lahtonen
Aug 6 at 5:21
What's happening is that the conjugacy class of $S_n$ may or may not split in two conjugacy classes in $A_n$. See here.
– Jyrki Lahtonen
Aug 6 at 5:23
What's happening is that the conjugacy class of $S_n$ may or may not split in two conjugacy classes in $A_n$. See here.
– Jyrki Lahtonen
Aug 6 at 5:23
Here the corresponding exercise in Dummit & Foote is discussed. I won't pick a dupe target because my search is biased by my own history. But I suspect others explained this earlier and better and I simply can't find a link to those threads quickly.
– Jyrki Lahtonen
Aug 6 at 5:26
Here the corresponding exercise in Dummit & Foote is discussed. I won't pick a dupe target because my search is biased by my own history. But I suspect others explained this earlier and better and I simply can't find a link to those threads quickly.
– Jyrki Lahtonen
Aug 6 at 5:26
2
2
Just for you to think concretely: Are the $3$-cycles $(1,2,3)$ and $(1,3,2)$ conjugate in $A_3$?
– Ted Shifrin
Aug 6 at 5:36
Just for you to think concretely: Are the $3$-cycles $(1,2,3)$ and $(1,3,2)$ conjugate in $A_3$?
– Ted Shifrin
Aug 6 at 5:36
@Ted Shifrin In general, I'm not sure how can I prove that something does not conjugate, rather than looking at the cycle structure or trying all possiblities, which obviously isn't a good idea.
– ChikChak
Aug 6 at 5:48
@Ted Shifrin In general, I'm not sure how can I prove that something does not conjugate, rather than looking at the cycle structure or trying all possiblities, which obviously isn't a good idea.
– ChikChak
Aug 6 at 5:48
 |Â
show 9 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
As you said, being conjugate in $S_5$ is equivalent to having the same cycle structure. But this is not true in $A_5$. Two permutations that are conjugate in $A_5$ will have the same cycle structure, but the converse is not necessarily true. $x,y in A_5$ and $y=gxg^-1$ where $g in S_5 backslash A_5$ then it implies $x$ and $y$ are conjugate in $S^5$ but they might not be conjugate in $A^5$.
answered Aug 6 at 5:18
Carl
2,0281926
add a comment |Â
up vote
2
down vote
If $alpha$ is a $5$-cycle in $S_5$ then the permutations in $S_5$
which commute with it are the powers of $alpha$ which are all even.
If $beta$ is an odd permutation, and $alpha'=betaalphabeta^-1$
then $alpha'$ is a conjugate of $alpha$ in $S_5$ but not in $A_5$.
For if $alpha'=beta'alphabeta'^-1$ for $beta'in A_5$ then
$beta^-1beta'$ commutes with $alpha$, so lies in $A_5$.
That forces $betain A_5$, a contradiction.
I like this geometric picture. $A_5$ is isomorphic to the
rotation group of the regular icosahedron. That has a rotation
of order $5$ with angle $2pi/5$. Its square is a rotation of
order $5$ with angle $4pi/5$. These two rotations cannot be conjugate;
these correspond to non-conjugate $5$-cycles in $A_5$.
answered Aug 6 at 6:19
Lord Shark the Unknown
86k951112
Therefore, $5$-cycles can never conjugate on $A_5$?
– ChikChak
Aug 6 at 15:41
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
As you said, being conjugate in $S_5$ is equivalent to having the same cycle structure. But this is not true in $A_5$. Two permutations that are conjugate in $A_5$ will have the same cycle structure, but the converse is not necessarily true. $x,y in A_5$ and $y=gxg^-1$ where $g in S_5 backslash A_5$ then it implies $x$ and $y$ are conjugate in $S^5$ but they might not be conjugate in $A^5$.
answered Aug 6 at 5:18
Carl
2,0281926
add a comment |Â
up vote
2
down vote
As you said, being conjugate in $S_5$ is equivalent to having the same cycle structure. But this is not true in $A_5$. Two permutations that are conjugate in $A_5$ will have the same cycle structure, but the converse is not necessarily true. $x,y in A_5$ and $y=gxg^-1$ where $g in S_5 backslash A_5$ then it implies $x$ and $y$ are conjugate in $S^5$ but they might not be conjugate in $A^5$.
answered Aug 6 at 5:18
Carl
2,0281926
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As you said, being conjugate in $S_5$ is equivalent to having the same cycle structure. But this is not true in $A_5$. Two permutations that are conjugate in $A_5$ will have the same cycle structure, but the converse is not necessarily true. $x,y in A_5$ and $y=gxg^-1$ where $g in S_5 backslash A_5$ then it implies $x$ and $y$ are conjugate in $S^5$ but they might not be conjugate in $A^5$.
answered Aug 6 at 5:18
Carl
2,0281926
As you said, being conjugate in $S_5$ is equivalent to having the same cycle structure. But this is not true in $A_5$. Two permutations that are conjugate in $A_5$ will have the same cycle structure, but the converse is not necessarily true. $x,y in A_5$ and $y=gxg^-1$ where $g in S_5 backslash A_5$ then it implies $x$ and $y$ are conjugate in $S^5$ but they might not be conjugate in $A^5$.
answered Aug 6 at 5:18
Carl
2,0281926
answered Aug 6 at 5:18
Carl
2,0281926
answered Aug 6 at 5:18
Carl
2,0281926
answered Aug 6 at 5:18
Carl
2,0281926
2,0281926
add a comment |Â
add a comment |Â
up vote
2
down vote
If $alpha$ is a $5$-cycle in $S_5$ then the permutations in $S_5$
which commute with it are the powers of $alpha$ which are all even.
If $beta$ is an odd permutation, and $alpha'=betaalphabeta^-1$
then $alpha'$ is a conjugate of $alpha$ in $S_5$ but not in $A_5$.
For if $alpha'=beta'alphabeta'^-1$ for $beta'in A_5$ then
$beta^-1beta'$ commutes with $alpha$, so lies in $A_5$.
That forces $betain A_5$, a contradiction.
I like this geometric picture. $A_5$ is isomorphic to the
rotation group of the regular icosahedron. That has a rotation
of order $5$ with angle $2pi/5$. Its square is a rotation of
order $5$ with angle $4pi/5$. These two rotations cannot be conjugate;
these correspond to non-conjugate $5$-cycles in $A_5$.
answered Aug 6 at 6:19
Lord Shark the Unknown
86k951112
Therefore, $5$-cycles can never conjugate on $A_5$?
– ChikChak
Aug 6 at 15:41
add a comment |Â
up vote
2
down vote
If $alpha$ is a $5$-cycle in $S_5$ then the permutations in $S_5$
which commute with it are the powers of $alpha$ which are all even.
If $beta$ is an odd permutation, and $alpha'=betaalphabeta^-1$
then $alpha'$ is a conjugate of $alpha$ in $S_5$ but not in $A_5$.
For if $alpha'=beta'alphabeta'^-1$ for $beta'in A_5$ then
$beta^-1beta'$ commutes with $alpha$, so lies in $A_5$.
That forces $betain A_5$, a contradiction.
I like this geometric picture. $A_5$ is isomorphic to the
rotation group of the regular icosahedron. That has a rotation
of order $5$ with angle $2pi/5$. Its square is a rotation of
order $5$ with angle $4pi/5$. These two rotations cannot be conjugate;
these correspond to non-conjugate $5$-cycles in $A_5$.
answered Aug 6 at 6:19
Lord Shark the Unknown
86k951112
Therefore, $5$-cycles can never conjugate on $A_5$?
– ChikChak
Aug 6 at 15:41
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If $alpha$ is a $5$-cycle in $S_5$ then the permutations in $S_5$
which commute with it are the powers of $alpha$ which are all even.
If $beta$ is an odd permutation, and $alpha'=betaalphabeta^-1$
then $alpha'$ is a conjugate of $alpha$ in $S_5$ but not in $A_5$.
For if $alpha'=beta'alphabeta'^-1$ for $beta'in A_5$ then
$beta^-1beta'$ commutes with $alpha$, so lies in $A_5$.
That forces $betain A_5$, a contradiction.
I like this geometric picture. $A_5$ is isomorphic to the
rotation group of the regular icosahedron. That has a rotation
of order $5$ with angle $2pi/5$. Its square is a rotation of
order $5$ with angle $4pi/5$. These two rotations cannot be conjugate;
these correspond to non-conjugate $5$-cycles in $A_5$.
answered Aug 6 at 6:19
Lord Shark the Unknown
86k951112
If $alpha$ is a $5$-cycle in $S_5$ then the permutations in $S_5$
which commute with it are the powers of $alpha$ which are all even.
If $beta$ is an odd permutation, and $alpha'=betaalphabeta^-1$
then $alpha'$ is a conjugate of $alpha$ in $S_5$ but not in $A_5$.
For if $alpha'=beta'alphabeta'^-1$ for $beta'in A_5$ then
$beta^-1beta'$ commutes with $alpha$, so lies in $A_5$.
That forces $betain A_5$, a contradiction.
I like this geometric picture. $A_5$ is isomorphic to the
rotation group of the regular icosahedron. That has a rotation
of order $5$ with angle $2pi/5$. Its square is a rotation of
order $5$ with angle $4pi/5$. These two rotations cannot be conjugate;
these correspond to non-conjugate $5$-cycles in $A_5$.
answered Aug 6 at 6:19
Lord Shark the Unknown
86k951112
answered Aug 6 at 6:19
Lord Shark the Unknown
86k951112
answered Aug 6 at 6:19
Lord Shark the Unknown
86k951112
answered Aug 6 at 6:19
Lord Shark the Unknown
86k951112
86k951112
Therefore, $5$-cycles can never conjugate on $A_5$?
– ChikChak
Aug 6 at 15:41
add a comment |Â
Therefore, $5$-cycles can never conjugate on $A_5$?
– ChikChak
Aug 6 at 15:41
Therefore, $5$-cycles can never conjugate on $A_5$?
– ChikChak
Aug 6 at 15:41
Therefore, $5$-cycles can never conjugate on $A_5$?
– ChikChak
Aug 6 at 15:41
add a comment |Â
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See for example here. I think we have even better threads, but that came up first in my search.
– Jyrki Lahtonen
Aug 6 at 5:21
What's happening is that the conjugacy class of $S_n$ may or may not split in two conjugacy classes in $A_n$. See here.
– Jyrki Lahtonen
Aug 6 at 5:23
Here the corresponding exercise in Dummit & Foote is discussed. I won't pick a dupe target because my search is biased by my own history. But I suspect others explained this earlier and better and I simply can't find a link to those threads quickly.
– Jyrki Lahtonen
Aug 6 at 5:26
2
Just for you to think concretely: Are the $3$-cycles $(1,2,3)$ and $(1,3,2)$ conjugate in $A_3$?
– Ted Shifrin
Aug 6 at 5:36
@Ted Shifrin In general, I'm not sure how can I prove that something does not conjugate, rather than looking at the cycle structure or trying all possiblities, which obviously isn't a good idea.
– ChikChak
Aug 6 at 5:48