Proof of center of symmetric group
Can you help me finish this proof? This is what I have so far, I don't know if it's correct. Any tips on my writing or others are welcome.
"Prove that, for $n geq 3 $, the center of $S_n$ only contains $I_n$ (the identity)".
Suppose we have $sigma in S_n$ such that $sigma = (a_1,...,a_k)$.
Let $varphi in Z(S_n)$ be such that $varphi neq sigma$. We can do that because $(a_1,...,a_k)(a_1,a_2) neq (a_1,a_2)(a_1,...,a_k)$, and $(a_1,...,a_k) notin Z(S_n)$.
Then, we can write $varphi sigma = sigma varphi$. Multiplying by $varphi^{-1}$ the both terms we obtain:
$varphi sigma varphi^{-1} = sigma varphi varphi^{-1}$.
We also know that:
$varphi (a_1,...,a_k) varphi^{-1} = (varphi (a_1),...,varphi (a_k))$.
Then,
$(varphi (a_1),...,varphi (a_k)) = sigma = (a_1,...,a_k)$.
Which means that $varphi = I_n$.
New proof: (I hope this works now)
Suppose $varphi in Z(S_n)$ such that $varphi neq I_n$. We can assume that $exists i: varphi (i) neq i$ and $exists j neq i: varphi (j) = i$.
Now consider $sigma in S_n$ such that $sigma = (j,k), k in { 1,...,n } setminus { i,j }$.
Then, we have the equality $varphi sigma = sigma varphi$. Computing at $j$:
$sigma(varphi(j)) = sigma(i) = i$.
$varphi(sigma(j)) = varphi(k) neq i$, as $varphi$ is injective and $kneq j$.
Then, $varphi$ must be the identity.
abstract-algebra group-theory
asked Nov 28 '18 at 12:13
Albelaski
737
add a comment |
Can you help me finish this proof? This is what I have so far, I don't know if it's correct. Any tips on my writing or others are welcome.
"Prove that, for $n geq 3 $, the center of $S_n$ only contains $I_n$ (the identity)".
Suppose we have $sigma in S_n$ such that $sigma = (a_1,...,a_k)$.
Let $varphi in Z(S_n)$ be such that $varphi neq sigma$. We can do that because $(a_1,...,a_k)(a_1,a_2) neq (a_1,a_2)(a_1,...,a_k)$, and $(a_1,...,a_k) notin Z(S_n)$.
Then, we can write $varphi sigma = sigma varphi$. Multiplying by $varphi^{-1}$ the both terms we obtain:
$varphi sigma varphi^{-1} = sigma varphi varphi^{-1}$.
We also know that:
$varphi (a_1,...,a_k) varphi^{-1} = (varphi (a_1),...,varphi (a_k))$.
Then,
$(varphi (a_1),...,varphi (a_k)) = sigma = (a_1,...,a_k)$.
Which means that $varphi = I_n$.
New proof: (I hope this works now)
Suppose $varphi in Z(S_n)$ such that $varphi neq I_n$. We can assume that $exists i: varphi (i) neq i$ and $exists j neq i: varphi (j) = i$.
Now consider $sigma in S_n$ such that $sigma = (j,k), k in { 1,...,n } setminus { i,j }$.
Then, we have the equality $varphi sigma = sigma varphi$. Computing at $j$:
$sigma(varphi(j)) = sigma(i) = i$.
$varphi(sigma(j)) = varphi(k) neq i$, as $varphi$ is injective and $kneq j$.
Then, $varphi$ must be the identity.
abstract-algebra group-theory
asked Nov 28 '18 at 12:13
Albelaski
737
add a comment |
Can you help me finish this proof? This is what I have so far, I don't know if it's correct. Any tips on my writing or others are welcome.
"Prove that, for $n geq 3 $, the center of $S_n$ only contains $I_n$ (the identity)".
Suppose we have $sigma in S_n$ such that $sigma = (a_1,...,a_k)$.
Let $varphi in Z(S_n)$ be such that $varphi neq sigma$. We can do that because $(a_1,...,a_k)(a_1,a_2) neq (a_1,a_2)(a_1,...,a_k)$, and $(a_1,...,a_k) notin Z(S_n)$.
Then, we can write $varphi sigma = sigma varphi$. Multiplying by $varphi^{-1}$ the both terms we obtain:
$varphi sigma varphi^{-1} = sigma varphi varphi^{-1}$.
We also know that:
$varphi (a_1,...,a_k) varphi^{-1} = (varphi (a_1),...,varphi (a_k))$.
Then,
$(varphi (a_1),...,varphi (a_k)) = sigma = (a_1,...,a_k)$.
Which means that $varphi = I_n$.
New proof: (I hope this works now)
Suppose $varphi in Z(S_n)$ such that $varphi neq I_n$. We can assume that $exists i: varphi (i) neq i$ and $exists j neq i: varphi (j) = i$.
Now consider $sigma in S_n$ such that $sigma = (j,k), k in { 1,...,n } setminus { i,j }$.
Then, we have the equality $varphi sigma = sigma varphi$. Computing at $j$:
$sigma(varphi(j)) = sigma(i) = i$.
$varphi(sigma(j)) = varphi(k) neq i$, as $varphi$ is injective and $kneq j$.
Then, $varphi$ must be the identity.
abstract-algebra group-theory
asked Nov 28 '18 at 12:13
Albelaski
737
Can you help me finish this proof? This is what I have so far, I don't know if it's correct. Any tips on my writing or others are welcome.
"Prove that, for $n geq 3 $, the center of $S_n$ only contains $I_n$ (the identity)".
Suppose we have $sigma in S_n$ such that $sigma = (a_1,...,a_k)$.
Let $varphi in Z(S_n)$ be such that $varphi neq sigma$. We can do that because $(a_1,...,a_k)(a_1,a_2) neq (a_1,a_2)(a_1,...,a_k)$, and $(a_1,...,a_k) notin Z(S_n)$.
Then, we can write $varphi sigma = sigma varphi$. Multiplying by $varphi^{-1}$ the both terms we obtain:
$varphi sigma varphi^{-1} = sigma varphi varphi^{-1}$.
We also know that:
$varphi (a_1,...,a_k) varphi^{-1} = (varphi (a_1),...,varphi (a_k))$.
Then,
$(varphi (a_1),...,varphi (a_k)) = sigma = (a_1,...,a_k)$.
Which means that $varphi = I_n$.
New proof: (I hope this works now)
Suppose $varphi in Z(S_n)$ such that $varphi neq I_n$. We can assume that $exists i: varphi (i) neq i$ and $exists j neq i: varphi (j) = i$.
Now consider $sigma in S_n$ such that $sigma = (j,k), k in { 1,...,n } setminus { i,j }$.
Then, we have the equality $varphi sigma = sigma varphi$. Computing at $j$:
$sigma(varphi(j)) = sigma(i) = i$.
$varphi(sigma(j)) = varphi(k) neq i$, as $varphi$ is injective and $kneq j$.
Then, $varphi$ must be the identity.
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 28 '18 at 12:13
Albelaski
737
asked Nov 28 '18 at 12:13
Albelaski
737
edited Nov 28 '18 at 17:48
asked Nov 28 '18 at 12:13
Albelaski
737
asked Nov 28 '18 at 12:13
Albelaski
737
asked Nov 28 '18 at 12:13
Albelaski
737
737
add a comment |
add a comment |
1 Answer
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Your proof is flawed. Just because $(varphi(a_1),ldots,varphi(a_k)) = (a_1, ldots, a_k)$, that does not mean that $varphi$ didn't touch the $a_i$'s; it may have cyclically permuted them.
In addition to cyclically permute the $a_i$ (or actually not doing anything to them), $varphi$ may permute all the other elements of $S_n$ however it likes, and it will still commute with $sigma$. As an example illustrating both of these flaws, take $n = 5$ with $sigma = (123)$. Then $varphi = (123)(45)$ commutes with $sigma$, and we do have $varphisigmavarphi^{-1} = sigma$ as you (correctly) claim, but $varphi$ is not the identity permutation.
Instead of starting with a $sigmain S_n$, I would start with a $varphiin Z(S_n)$, and if it's not the identity, construct (or find) a $sigmain S_n$ which doesn't commute with it.
answered Nov 28 '18 at 12:19
Arthur
111k7105186
add a comment |
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Your proof is flawed. Just because $(varphi(a_1),ldots,varphi(a_k)) = (a_1, ldots, a_k)$, that does not mean that $varphi$ didn't touch the $a_i$'s; it may have cyclically permuted them.
In addition to cyclically permute the $a_i$ (or actually not doing anything to them), $varphi$ may permute all the other elements of $S_n$ however it likes, and it will still commute with $sigma$. As an example illustrating both of these flaws, take $n = 5$ with $sigma = (123)$. Then $varphi = (123)(45)$ commutes with $sigma$, and we do have $varphisigmavarphi^{-1} = sigma$ as you (correctly) claim, but $varphi$ is not the identity permutation.
Instead of starting with a $sigmain S_n$, I would start with a $varphiin Z(S_n)$, and if it's not the identity, construct (or find) a $sigmain S_n$ which doesn't commute with it.
answered Nov 28 '18 at 12:19
Arthur
111k7105186
add a comment |
Your proof is flawed. Just because $(varphi(a_1),ldots,varphi(a_k)) = (a_1, ldots, a_k)$, that does not mean that $varphi$ didn't touch the $a_i$'s; it may have cyclically permuted them.
In addition to cyclically permute the $a_i$ (or actually not doing anything to them), $varphi$ may permute all the other elements of $S_n$ however it likes, and it will still commute with $sigma$. As an example illustrating both of these flaws, take $n = 5$ with $sigma = (123)$. Then $varphi = (123)(45)$ commutes with $sigma$, and we do have $varphisigmavarphi^{-1} = sigma$ as you (correctly) claim, but $varphi$ is not the identity permutation.
Instead of starting with a $sigmain S_n$, I would start with a $varphiin Z(S_n)$, and if it's not the identity, construct (or find) a $sigmain S_n$ which doesn't commute with it.
answered Nov 28 '18 at 12:19
Arthur
111k7105186
add a comment |
Your proof is flawed. Just because $(varphi(a_1),ldots,varphi(a_k)) = (a_1, ldots, a_k)$, that does not mean that $varphi$ didn't touch the $a_i$'s; it may have cyclically permuted them.
In addition to cyclically permute the $a_i$ (or actually not doing anything to them), $varphi$ may permute all the other elements of $S_n$ however it likes, and it will still commute with $sigma$. As an example illustrating both of these flaws, take $n = 5$ with $sigma = (123)$. Then $varphi = (123)(45)$ commutes with $sigma$, and we do have $varphisigmavarphi^{-1} = sigma$ as you (correctly) claim, but $varphi$ is not the identity permutation.
Instead of starting with a $sigmain S_n$, I would start with a $varphiin Z(S_n)$, and if it's not the identity, construct (or find) a $sigmain S_n$ which doesn't commute with it.
answered Nov 28 '18 at 12:19
Arthur
111k7105186
Your proof is flawed. Just because $(varphi(a_1),ldots,varphi(a_k)) = (a_1, ldots, a_k)$, that does not mean that $varphi$ didn't touch the $a_i$'s; it may have cyclically permuted them.
In addition to cyclically permute the $a_i$ (or actually not doing anything to them), $varphi$ may permute all the other elements of $S_n$ however it likes, and it will still commute with $sigma$. As an example illustrating both of these flaws, take $n = 5$ with $sigma = (123)$. Then $varphi = (123)(45)$ commutes with $sigma$, and we do have $varphisigmavarphi^{-1} = sigma$ as you (correctly) claim, but $varphi$ is not the identity permutation.
Instead of starting with a $sigmain S_n$, I would start with a $varphiin Z(S_n)$, and if it's not the identity, construct (or find) a $sigmain S_n$ which doesn't commute with it.
answered Nov 28 '18 at 12:19
Arthur
111k7105186
answered Nov 28 '18 at 12:19
Arthur
111k7105186
answered Nov 28 '18 at 12:19
Arthur
111k7105186
answered Nov 28 '18 at 12:19
Arthur
111k7105186
111k7105186
add a comment |
add a comment |
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