Commuting permutations are multiples of each other
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Assumptions: Let $alpha$ be a permutation represented as disjoint cycles $alpha=c_1circ dots circ c_r$ with respective length $lambda_1 ,...,lambda_r$ such that $lambda_ineq lambda_j$ if $ineq j$.
Statement: If now $betaalpha=alphabeta$ then $beta=c_1^{z_1}circ...circ c_r^{z_r}$ for integers $z_1,dots,z_r$.
My attempt: The commututing relation $betaalpha=alphabeta$ says that the permutations are conjugate and hence share the same signature/cycle type. My hope was, that one can then compare cycles of same lenght but I go stuck there.
Help appreciated.
group-theory
asked Nov 26 at 20:24
Hectorx
161
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show 2 more comments
up vote
2
down vote
favorite
Assumptions: Let $alpha$ be a permutation represented as disjoint cycles $alpha=c_1circ dots circ c_r$ with respective length $lambda_1 ,...,lambda_r$ such that $lambda_ineq lambda_j$ if $ineq j$.
Statement: If now $betaalpha=alphabeta$ then $beta=c_1^{z_1}circ...circ c_r^{z_r}$ for integers $z_1,dots,z_r$.
My attempt: The commututing relation $betaalpha=alphabeta$ says that the permutations are conjugate and hence share the same signature/cycle type. My hope was, that one can then compare cycles of same lenght but I go stuck there.
Help appreciated.
group-theory
asked Nov 26 at 20:24
Hectorx
161
Does the commuting relation really say that the permutations are conjugate? Can you say what you’re using there?
– Lubin
Nov 26 at 20:58
Oh I see, $alphabetaalpha^{-1}=beta$ only says that $beta$ is conjugate to itself via $alpha$. This seems to be of no help. However, I know that if $alpha=(a_{11}...a_{1l_1})...(a_{s1}...a_{1l_s})$ then $(beta a_{11}...beta a_{1l_1})...(beta a_{s1}...beta a_{1l_s})=betaalphabeta^{-1}$. Can I continue from here?
– Hectorx
Nov 26 at 21:32
I don’t think it’s at all true that commuting permutations are powers of each other. I hope you don’t need this for other aims.
– Lubin
Nov 26 at 22:37
I thought the assumption $lambda_ineqlamba_j$ would force them to be multiples as I can't think of a counter example. However I appreciate your opinion.
– Hectorx
Nov 26 at 22:48
You’re not looking for simple-enough examples: consider $(1,2)$ and $(3,4)$.
– Lubin
Nov 27 at 0:42
|
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Assumptions: Let $alpha$ be a permutation represented as disjoint cycles $alpha=c_1circ dots circ c_r$ with respective length $lambda_1 ,...,lambda_r$ such that $lambda_ineq lambda_j$ if $ineq j$.
Statement: If now $betaalpha=alphabeta$ then $beta=c_1^{z_1}circ...circ c_r^{z_r}$ for integers $z_1,dots,z_r$.
My attempt: The commututing relation $betaalpha=alphabeta$ says that the permutations are conjugate and hence share the same signature/cycle type. My hope was, that one can then compare cycles of same lenght but I go stuck there.
Help appreciated.
group-theory
asked Nov 26 at 20:24
Hectorx
161
Assumptions: Let $alpha$ be a permutation represented as disjoint cycles $alpha=c_1circ dots circ c_r$ with respective length $lambda_1 ,...,lambda_r$ such that $lambda_ineq lambda_j$ if $ineq j$.
Statement: If now $betaalpha=alphabeta$ then $beta=c_1^{z_1}circ...circ c_r^{z_r}$ for integers $z_1,dots,z_r$.
My attempt: The commututing relation $betaalpha=alphabeta$ says that the permutations are conjugate and hence share the same signature/cycle type. My hope was, that one can then compare cycles of same lenght but I go stuck there.
Help appreciated.
group-theory
group-theory
asked Nov 26 at 20:24
Hectorx
161
asked Nov 26 at 20:24
Hectorx
161
asked Nov 26 at 20:24
Hectorx
161
asked Nov 26 at 20:24
Hectorx
161
asked Nov 26 at 20:24
Hectorx
161
161
Does the commuting relation really say that the permutations are conjugate? Can you say what you’re using there?
– Lubin
Nov 26 at 20:58
Oh I see, $alphabetaalpha^{-1}=beta$ only says that $beta$ is conjugate to itself via $alpha$. This seems to be of no help. However, I know that if $alpha=(a_{11}...a_{1l_1})...(a_{s1}...a_{1l_s})$ then $(beta a_{11}...beta a_{1l_1})...(beta a_{s1}...beta a_{1l_s})=betaalphabeta^{-1}$. Can I continue from here?
– Hectorx
Nov 26 at 21:32
I don’t think it’s at all true that commuting permutations are powers of each other. I hope you don’t need this for other aims.
– Lubin
Nov 26 at 22:37
I thought the assumption $lambda_ineqlamba_j$ would force them to be multiples as I can't think of a counter example. However I appreciate your opinion.
– Hectorx
Nov 26 at 22:48
You’re not looking for simple-enough examples: consider $(1,2)$ and $(3,4)$.
– Lubin
Nov 27 at 0:42
|
show 2 more comments
Does the commuting relation really say that the permutations are conjugate? Can you say what you’re using there?
– Lubin
Nov 26 at 20:58
Oh I see, $alphabetaalpha^{-1}=beta$ only says that $beta$ is conjugate to itself via $alpha$. This seems to be of no help. However, I know that if $alpha=(a_{11}...a_{1l_1})...(a_{s1}...a_{1l_s})$ then $(beta a_{11}...beta a_{1l_1})...(beta a_{s1}...beta a_{1l_s})=betaalphabeta^{-1}$. Can I continue from here?
– Hectorx
Nov 26 at 21:32
I don’t think it’s at all true that commuting permutations are powers of each other. I hope you don’t need this for other aims.
– Lubin
Nov 26 at 22:37
I thought the assumption $lambda_ineqlamba_j$ would force them to be multiples as I can't think of a counter example. However I appreciate your opinion.
– Hectorx
Nov 26 at 22:48
You’re not looking for simple-enough examples: consider $(1,2)$ and $(3,4)$.
– Lubin
Nov 27 at 0:42
Does the commuting relation really say that the permutations are conjugate? Can you say what you’re using there?
– Lubin
Nov 26 at 20:58
Does the commuting relation really say that the permutations are conjugate? Can you say what you’re using there?
– Lubin
Nov 26 at 20:58
Oh I see, $alphabetaalpha^{-1}=beta$ only says that $beta$ is conjugate to itself via $alpha$. This seems to be of no help. However, I know that if $alpha=(a_{11}...a_{1l_1})...(a_{s1}...a_{1l_s})$ then $(beta a_{11}...beta a_{1l_1})...(beta a_{s1}...beta a_{1l_s})=betaalphabeta^{-1}$. Can I continue from here?
– Hectorx
Nov 26 at 21:32
Oh I see, $alphabetaalpha^{-1}=beta$ only says that $beta$ is conjugate to itself via $alpha$. This seems to be of no help. However, I know that if $alpha=(a_{11}...a_{1l_1})...(a_{s1}...a_{1l_s})$ then $(beta a_{11}...beta a_{1l_1})...(beta a_{s1}...beta a_{1l_s})=betaalphabeta^{-1}$. Can I continue from here?
– Hectorx
Nov 26 at 21:32
I don’t think it’s at all true that commuting permutations are powers of each other. I hope you don’t need this for other aims.
– Lubin
Nov 26 at 22:37
I don’t think it’s at all true that commuting permutations are powers of each other. I hope you don’t need this for other aims.
– Lubin
Nov 26 at 22:37
I thought the assumption $lambda_ineqlamba_j$ would force them to be multiples as I can't think of a counter example. However I appreciate your opinion.
– Hectorx
Nov 26 at 22:48
I thought the assumption $lambda_ineqlamba_j$ would force them to be multiples as I can't think of a counter example. However I appreciate your opinion.
– Hectorx
Nov 26 at 22:48
You’re not looking for simple-enough examples: consider $(1,2)$ and $(3,4)$.
– Lubin
Nov 27 at 0:42
You’re not looking for simple-enough examples: consider $(1,2)$ and $(3,4)$.
– Lubin
Nov 27 at 0:42
|
show 2 more comments
1 Answer
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This is true. Because if $O$ is an orbit of $alpha$, then so is $beta(O)$. By the condition on cycle lengths, $beta(O)=O$. So $O$ is left invariant by $beta$.
On $O$, $alpha$ acts as an $n$-cycle, where $|O|=n$. The only permutations commuting with this cycle are its powers. So that is how $beta$ must act.
Since $O$ was an arbitrary orbit, this proves your claim.
answered Nov 28 at 8:28
Hempelicious
7510
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
This is true. Because if $O$ is an orbit of $alpha$, then so is $beta(O)$. By the condition on cycle lengths, $beta(O)=O$. So $O$ is left invariant by $beta$.
On $O$, $alpha$ acts as an $n$-cycle, where $|O|=n$. The only permutations commuting with this cycle are its powers. So that is how $beta$ must act.
Since $O$ was an arbitrary orbit, this proves your claim.
answered Nov 28 at 8:28
Hempelicious
7510
add a comment |
up vote
0
down vote
This is true. Because if $O$ is an orbit of $alpha$, then so is $beta(O)$. By the condition on cycle lengths, $beta(O)=O$. So $O$ is left invariant by $beta$.
On $O$, $alpha$ acts as an $n$-cycle, where $|O|=n$. The only permutations commuting with this cycle are its powers. So that is how $beta$ must act.
Since $O$ was an arbitrary orbit, this proves your claim.
answered Nov 28 at 8:28
Hempelicious
7510
add a comment |
up vote
0
down vote
up vote
0
down vote
This is true. Because if $O$ is an orbit of $alpha$, then so is $beta(O)$. By the condition on cycle lengths, $beta(O)=O$. So $O$ is left invariant by $beta$.
On $O$, $alpha$ acts as an $n$-cycle, where $|O|=n$. The only permutations commuting with this cycle are its powers. So that is how $beta$ must act.
Since $O$ was an arbitrary orbit, this proves your claim.
answered Nov 28 at 8:28
Hempelicious
7510
This is true. Because if $O$ is an orbit of $alpha$, then so is $beta(O)$. By the condition on cycle lengths, $beta(O)=O$. So $O$ is left invariant by $beta$.
On $O$, $alpha$ acts as an $n$-cycle, where $|O|=n$. The only permutations commuting with this cycle are its powers. So that is how $beta$ must act.
Since $O$ was an arbitrary orbit, this proves your claim.
answered Nov 28 at 8:28
Hempelicious
7510
answered Nov 28 at 8:28
Hempelicious
7510
answered Nov 28 at 8:28
Hempelicious
7510
answered Nov 28 at 8:28
Hempelicious
7510
7510
add a comment |
add a comment |
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Does the commuting relation really say that the permutations are conjugate? Can you say what you’re using there?
– Lubin
Nov 26 at 20:58
Oh I see, $alphabetaalpha^{-1}=beta$ only says that $beta$ is conjugate to itself via $alpha$. This seems to be of no help. However, I know that if $alpha=(a_{11}...a_{1l_1})...(a_{s1}...a_{1l_s})$ then $(beta a_{11}...beta a_{1l_1})...(beta a_{s1}...beta a_{1l_s})=betaalphabeta^{-1}$. Can I continue from here?
– Hectorx
Nov 26 at 21:32
I don’t think it’s at all true that commuting permutations are powers of each other. I hope you don’t need this for other aims.
– Lubin
Nov 26 at 22:37
I thought the assumption $lambda_ineqlamba_j$ would force them to be multiples as I can't think of a counter example. However I appreciate your opinion.
– Hectorx
Nov 26 at 22:48
You’re not looking for simple-enough examples: consider $(1,2)$ and $(3,4)$.
– Lubin
Nov 27 at 0:42