Mixing
stab(S) is isomorphic to $ S_k times S_{n-k} $
$begingroup$
Show that the group $S_n$ (right) acts on the set of subsets of ${1,2,ldots n}$ by $rho(S,sigma)=Ssigma$.
Show that there are $n+1$ orbits, one for each possible value of $|S|$. Show that if $|S|=k$ then $mathrm{Stab}(S)cong S_ktimes S_{n-k}$.
I have shown that there are n+1 orbits but now I am stuck with the last bit.
my initial thought is that if I let $S ={1,2,3,...,k}$ then $mathrm{Stab}(S)$ would involve $(1)(2)(3)...(k)$ i.e. mapping to itself and then $k+1$ to $n$ mapped in any ways ($S_{n-k}$)
I feel like $mathrm{Stab}$ is isomorphic to $S_{n-k}$ what am I misunderstanding here? How would I prove the last part?
Thanks a lot
abstract-algebra group-theory
edited Dec 18 '14 at 13:12
rschwieb
107k12102251
asked May 13 '14 at 22:29
user120227user120227
510617
$endgroup$
add a comment |
$begingroup$
Show that the group $S_n$ (right) acts on the set of subsets of ${1,2,ldots n}$ by $rho(S,sigma)=Ssigma$.
Show that there are $n+1$ orbits, one for each possible value of $|S|$. Show that if $|S|=k$ then $mathrm{Stab}(S)cong S_ktimes S_{n-k}$.
I have shown that there are n+1 orbits but now I am stuck with the last bit.
my initial thought is that if I let $S ={1,2,3,...,k}$ then $mathrm{Stab}(S)$ would involve $(1)(2)(3)...(k)$ i.e. mapping to itself and then $k+1$ to $n$ mapped in any ways ($S_{n-k}$)
I feel like $mathrm{Stab}$ is isomorphic to $S_{n-k}$ what am I misunderstanding here? How would I prove the last part?
Thanks a lot
abstract-algebra group-theory
edited Dec 18 '14 at 13:12
rschwieb
107k12102251
asked May 13 '14 at 22:29
user120227user120227
510617
$endgroup$
add a comment |
$begingroup$
Show that the group $S_n$ (right) acts on the set of subsets of ${1,2,ldots n}$ by $rho(S,sigma)=Ssigma$.
Show that there are $n+1$ orbits, one for each possible value of $|S|$. Show that if $|S|=k$ then $mathrm{Stab}(S)cong S_ktimes S_{n-k}$.
I have shown that there are n+1 orbits but now I am stuck with the last bit.
my initial thought is that if I let $S ={1,2,3,...,k}$ then $mathrm{Stab}(S)$ would involve $(1)(2)(3)...(k)$ i.e. mapping to itself and then $k+1$ to $n$ mapped in any ways ($S_{n-k}$)
I feel like $mathrm{Stab}$ is isomorphic to $S_{n-k}$ what am I misunderstanding here? How would I prove the last part?
Thanks a lot
abstract-algebra group-theory
edited Dec 18 '14 at 13:12
rschwieb
107k12102251
asked May 13 '14 at 22:29
user120227user120227
510617
$endgroup$
Show that the group $S_n$ (right) acts on the set of subsets of ${1,2,ldots n}$ by $rho(S,sigma)=Ssigma$.
Show that there are $n+1$ orbits, one for each possible value of $|S|$. Show that if $|S|=k$ then $mathrm{Stab}(S)cong S_ktimes S_{n-k}$.
I have shown that there are n+1 orbits but now I am stuck with the last bit.
my initial thought is that if I let $S ={1,2,3,...,k}$ then $mathrm{Stab}(S)$ would involve $(1)(2)(3)...(k)$ i.e. mapping to itself and then $k+1$ to $n$ mapped in any ways ($S_{n-k}$)
I feel like $mathrm{Stab}$ is isomorphic to $S_{n-k}$ what am I misunderstanding here? How would I prove the last part?
Thanks a lot
abstract-algebra group-theory
abstract-algebra group-theory
edited Dec 18 '14 at 13:12
rschwieb
107k12102251
asked May 13 '14 at 22:29
user120227user120227
510617
edited Dec 18 '14 at 13:12
rschwieb
107k12102251
asked May 13 '14 at 22:29
user120227user120227
510617
edited Dec 18 '14 at 13:12
rschwieb
107k12102251
edited Dec 18 '14 at 13:12
rschwieb
107k12102251
edited Dec 18 '14 at 13:12
rschwieb
107k12102251
107k12102251
asked May 13 '14 at 22:29
user120227user120227
510617
asked May 13 '14 at 22:29
user120227user120227
510617
asked May 13 '14 at 22:29
user120227user120227
510617
510617
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're missing that $sigma$ need not fix $S$ pointwise, but rather only setwise. That's where you get the initial $S_k$ factor.
answered May 13 '14 at 22:53
jdcjdc
2,7861330
$endgroup$
$begingroup$
Aha! Thanks for your reply
$endgroup$
– user120227
May 13 '14 at 23:02
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
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votes
$begingroup$
You're missing that $sigma$ need not fix $S$ pointwise, but rather only setwise. That's where you get the initial $S_k$ factor.
answered May 13 '14 at 22:53
jdcjdc
2,7861330
$endgroup$
$begingroup$
Aha! Thanks for your reply
$endgroup$
– user120227
May 13 '14 at 23:02
add a comment |
$begingroup$
You're missing that $sigma$ need not fix $S$ pointwise, but rather only setwise. That's where you get the initial $S_k$ factor.
answered May 13 '14 at 22:53
jdcjdc
2,7861330
$endgroup$
$begingroup$
Aha! Thanks for your reply
$endgroup$
– user120227
May 13 '14 at 23:02
add a comment |
$begingroup$
You're missing that $sigma$ need not fix $S$ pointwise, but rather only setwise. That's where you get the initial $S_k$ factor.
answered May 13 '14 at 22:53
jdcjdc
2,7861330
$endgroup$
You're missing that $sigma$ need not fix $S$ pointwise, but rather only setwise. That's where you get the initial $S_k$ factor.
answered May 13 '14 at 22:53
jdcjdc
2,7861330
answered May 13 '14 at 22:53
jdcjdc
2,7861330
answered May 13 '14 at 22:53
jdcjdc
2,7861330
answered May 13 '14 at 22:53
jdcjdc
2,7861330
2,7861330
$begingroup$
Aha! Thanks for your reply
$endgroup$
– user120227
May 13 '14 at 23:02
add a comment |
$begingroup$
Aha! Thanks for your reply
$endgroup$
– user120227
May 13 '14 at 23:02
$begingroup$
Aha! Thanks for your reply
$endgroup$
– user120227
May 13 '14 at 23:02
$begingroup$
Aha! Thanks for your reply
$endgroup$
– user120227
May 13 '14 at 23:02
add a comment |
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