Mixing
The normalizer of permutation
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Let $sigma = (1 2 dots 9) in S_10$.
a) Calculate the size of the normalizer $N_S_10(<sigma >)$.
b) Describe exactly the elements in $N_S_10(<sigma >)$.
I am not sure how to approach this. I understand that we look for permutations that fixes $10$, and yet I can't see what to do further...
group-theory permutations normal-subgroups
asked Jul 25 at 5:04
ChikChak
659214
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up vote
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favorite
Let $sigma = (1 2 dots 9) in S_10$.
a) Calculate the size of the normalizer $N_S_10(<sigma >)$.
b) Describe exactly the elements in $N_S_10(<sigma >)$.
I am not sure how to approach this. I understand that we look for permutations that fixes $10$, and yet I can't see what to do further...
group-theory permutations normal-subgroups
asked Jul 25 at 5:04
ChikChak
659214
Let $g in N = N_S_10(langle sigma rangle)$. Since $sigma in N$, by multiplying $g$ by a power of $sigma$ you can assume $g(1)=1$. Now $g^i(1)=i+1$, so $gsigma g^-1 = g^i Leftrightarrow g(2)=i+1$, so $g(2)=2,3,5,6,8$ or $9$.
– Derek Holt
Jul 25 at 7:48
@Derek Holt It is really difficult for me to follow all of the assumptions you made... Could you please elaborate?
– ChikChak
Jul 28 at 21:05
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $sigma = (1 2 dots 9) in S_10$.
a) Calculate the size of the normalizer $N_S_10(<sigma >)$.
b) Describe exactly the elements in $N_S_10(<sigma >)$.
I am not sure how to approach this. I understand that we look for permutations that fixes $10$, and yet I can't see what to do further...
group-theory permutations normal-subgroups
asked Jul 25 at 5:04
ChikChak
659214
Let $sigma = (1 2 dots 9) in S_10$.
a) Calculate the size of the normalizer $N_S_10(<sigma >)$.
b) Describe exactly the elements in $N_S_10(<sigma >)$.
I am not sure how to approach this. I understand that we look for permutations that fixes $10$, and yet I can't see what to do further...
group-theory permutations normal-subgroups
asked Jul 25 at 5:04
ChikChak
659214
asked Jul 25 at 5:04
ChikChak
659214
asked Jul 25 at 5:04
ChikChak
659214
asked Jul 25 at 5:04
ChikChak
659214
659214
Let $g in N = N_S_10(langle sigma rangle)$. Since $sigma in N$, by multiplying $g$ by a power of $sigma$ you can assume $g(1)=1$. Now $g^i(1)=i+1$, so $gsigma g^-1 = g^i Leftrightarrow g(2)=i+1$, so $g(2)=2,3,5,6,8$ or $9$.
– Derek Holt
Jul 25 at 7:48
@Derek Holt It is really difficult for me to follow all of the assumptions you made... Could you please elaborate?
– ChikChak
Jul 28 at 21:05
add a comment |Â
Let $g in N = N_S_10(langle sigma rangle)$. Since $sigma in N$, by multiplying $g$ by a power of $sigma$ you can assume $g(1)=1$. Now $g^i(1)=i+1$, so $gsigma g^-1 = g^i Leftrightarrow g(2)=i+1$, so $g(2)=2,3,5,6,8$ or $9$.
– Derek Holt
Jul 25 at 7:48
@Derek Holt It is really difficult for me to follow all of the assumptions you made... Could you please elaborate?
– ChikChak
Jul 28 at 21:05
Let $g in N = N_S_10(langle sigma rangle)$. Since $sigma in N$, by multiplying $g$ by a power of $sigma$ you can assume $g(1)=1$. Now $g^i(1)=i+1$, so $gsigma g^-1 = g^i Leftrightarrow g(2)=i+1$, so $g(2)=2,3,5,6,8$ or $9$.
– Derek Holt
Jul 25 at 7:48
Let $g in N = N_S_10(langle sigma rangle)$. Since $sigma in N$, by multiplying $g$ by a power of $sigma$ you can assume $g(1)=1$. Now $g^i(1)=i+1$, so $gsigma g^-1 = g^i Leftrightarrow g(2)=i+1$, so $g(2)=2,3,5,6,8$ or $9$.
– Derek Holt
Jul 25 at 7:48
@Derek Holt It is really difficult for me to follow all of the assumptions you made... Could you please elaborate?
– ChikChak
Jul 28 at 21:05
@Derek Holt It is really difficult for me to follow all of the assumptions you made... Could you please elaborate?
– ChikChak
Jul 28 at 21:05
add a comment |Â
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Let $g in N = N_S_10(langle sigma rangle)$. Since $sigma in N$, by multiplying $g$ by a power of $sigma$ you can assume $g(1)=1$. Now $g^i(1)=i+1$, so $gsigma g^-1 = g^i Leftrightarrow g(2)=i+1$, so $g(2)=2,3,5,6,8$ or $9$.
– Derek Holt
Jul 25 at 7:48
@Derek Holt It is really difficult for me to follow all of the assumptions you made... Could you please elaborate?
– ChikChak
Jul 28 at 21:05