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Prove that $operatornameStab_G(y) congoperatornameStab_G(x).$ [closed]

Let $G$ be a group and $X$ set. $a,b in X$ are in the same orbit, so show that $stab(a) cong stab (b)$$newcommandstaboperatornamestab$If $a, b$ at the same orbit, then $stab(a) cong stab(b)$stab(S) is isomorphic to $ S_k times S_n-k $Let $G$ be a group and $X$ set. $a,b in X$ are in the same orbit, so show that $stab(a) cong stab (b)$Prove $operatornamestab_G(g cdot x) = goperatornamestab_G(x)g^-1$Prove that $operatornameInn(operatornameAut(G)) cong operatornameAut(G)$For G acting on $Omega$, let $alpha in Omega$, show that $operatornameStab_G(alpha) vartriangleleft G$Show that all the orbits have the cardinality [N:N ∩ Stab a] for some a∈ O1Let $G$ be transitive on $S$. Show that the action is primitive if and only if every $operatornameStab_G(a), ain S$, is a maximal subgroup of $G$.Computing $| operatornameStab_S_n(x)$|Equivariance theorem on group actions

0

1

$begingroup$

Suppose that $G$ acts on a set $X$. Let $x,y in X$ and suppose that $y in Gcdot x$.

Prove that
$$operatornameStab_G(y) congoperatornameStab_G(x).$$

Can anyone help me with this proof?

group-theory group-actions

share|cite|improve this question

edited Mar 13 at 20:53

Shaun

9,585113684

asked Mar 13 at 18:43

Sammy.dSammy.d

325

$endgroup$

closed as off-topic by verret, Leucippus, eyeballfrog, Lord Shark the Unknown, Lee David Chung Lin Mar 14 at 4:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – verret, Leucippus, eyeballfrog, Lord Shark the Unknown

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should know by now that asking questions without context is generally frowned upon. Please edit the question to include your thoughts on the problem.
  $endgroup$
  – Shaun
  Mar 13 at 18:50
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  $begingroup$
  What do you denote $=sim$?
  $endgroup$
  – Bernard
  Mar 13 at 18:52
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Let $G$ be a group and $X$ set. $a,b in X$ are in the same orbit, so show that $stab(a) cong stab (b)$
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  – Yanior Weg
  Mar 13 at 21:52
  

  
  
  
  

add a comment | 

0

1

$begingroup$

Suppose that $G$ acts on a set $X$. Let $x,y in X$ and suppose that $y in Gcdot x$.

Prove that
$$operatornameStab_G(y) congoperatornameStab_G(x).$$

Can anyone help me with this proof?

group-theory group-actions

share|cite|improve this question

edited Mar 13 at 20:53

Shaun

9,585113684

asked Mar 13 at 18:43

Sammy.dSammy.d

325

$endgroup$

closed as off-topic by verret, Leucippus, eyeballfrog, Lord Shark the Unknown, Lee David Chung Lin Mar 14 at 4:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – verret, Leucippus, eyeballfrog, Lord Shark the Unknown

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should know by now that asking questions without context is generally frowned upon. Please edit the question to include your thoughts on the problem.
  $endgroup$
  – Shaun
  Mar 13 at 18:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What do you denote $=sim$?
  $endgroup$
  – Bernard
  Mar 13 at 18:52
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Let $G$ be a group and $X$ set. $a,b in X$ are in the same orbit, so show that $stab(a) cong stab (b)$
  $endgroup$
  – Yanior Weg
  Mar 13 at 21:52
  

  
  
  
  

add a comment | 

$begingroup$

Suppose that $G$ acts on a set $X$. Let $x,y in X$ and suppose that $y in Gcdot x$.

Prove that
$$operatornameStab_G(y) congoperatornameStab_G(x).$$

Can anyone help me with this proof?

group-theory group-actions

share|cite|improve this question

edited Mar 13 at 20:53

Shaun

9,585113684

asked Mar 13 at 18:43

Sammy.dSammy.d

325

$endgroup$

share|cite|improve this question

edited Mar 13 at 20:53

Shaun

9,585113684

asked Mar 13 at 18:43

Sammy.dSammy.d

325

share|cite|improve this question

edited Mar 13 at 20:53

Shaun

9,585113684

asked Mar 13 at 18:43

Sammy.dSammy.d

325

edited Mar 13 at 20:53

Shaun

9,585113684

edited Mar 13 at 20:53

Shaun

9,585113684

asked Mar 13 at 18:43

Sammy.dSammy.d

325

asked Mar 13 at 18:43

Sammy.dSammy.d

325

1 Answer
1

active

oldest

votes

3

$begingroup$

It is given that $yin Gx$ which means that there exists $hin G$ such that $y=hx$.

Thus, if $gin textStab_G(y)$ then $gy=y$ hence $ghx=hx$ and so $h^-1ghx=x$.

In particular $gin textStab_G(y)$ implies that $h^-1g h in textStab_G(x)$.

Let $varphi:textStab_G(y)rightarrow textStab_G(x)$ be such that $varphi(g)=h^-1gh$ we just saw above that $varphi$ is well defined. It is also easy to check that $varphi$ a homomorphism.

Furthermore, the inverse of $varphi$ is given by $varphi^-1(g) = hgh^-1$. The reason that $varphi^-1$ is well defined is similar to that of $varphi$ reversing the roles of $x,y$ .

share|cite|improve this answer

answered Mar 13 at 18:52

YankoYanko

7,8801830

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for such a clear informative explanation!
  $endgroup$
  – Sammy.d
  Mar 13 at 19:03
  

  
  
  
  

add a comment | 

3

$begingroup$

It is given that $yin Gx$ which means that there exists $hin G$ such that $y=hx$.

Thus, if $gin textStab_G(y)$ then $gy=y$ hence $ghx=hx$ and so $h^-1ghx=x$.

In particular $gin textStab_G(y)$ implies that $h^-1g h in textStab_G(x)$.

Let $varphi:textStab_G(y)rightarrow textStab_G(x)$ be such that $varphi(g)=h^-1gh$ we just saw above that $varphi$ is well defined. It is also easy to check that $varphi$ a homomorphism.

Furthermore, the inverse of $varphi$ is given by $varphi^-1(g) = hgh^-1$. The reason that $varphi^-1$ is well defined is similar to that of $varphi$ reversing the roles of $x,y$ .

share|cite|improve this answer

answered Mar 13 at 18:52

YankoYanko

7,8801830

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for such a clear informative explanation!
  $endgroup$
  – Sammy.d
  Mar 13 at 19:03
  

  
  
  
  

add a comment | 

3

$begingroup$

It is given that $yin Gx$ which means that there exists $hin G$ such that $y=hx$.

Thus, if $gin textStab_G(y)$ then $gy=y$ hence $ghx=hx$ and so $h^-1ghx=x$.

In particular $gin textStab_G(y)$ implies that $h^-1g h in textStab_G(x)$.

Let $varphi:textStab_G(y)rightarrow textStab_G(x)$ be such that $varphi(g)=h^-1gh$ we just saw above that $varphi$ is well defined. It is also easy to check that $varphi$ a homomorphism.

Furthermore, the inverse of $varphi$ is given by $varphi^-1(g) = hgh^-1$. The reason that $varphi^-1$ is well defined is similar to that of $varphi$ reversing the roles of $x,y$ .

share|cite|improve this answer

answered Mar 13 at 18:52

YankoYanko

7,8801830

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for such a clear informative explanation!
  $endgroup$
  – Sammy.d
  Mar 13 at 19:03
  

  
  
  
  

add a comment | 

$begingroup$

It is given that $yin Gx$ which means that there exists $hin G$ such that $y=hx$.

Thus, if $gin textStab_G(y)$ then $gy=y$ hence $ghx=hx$ and so $h^-1ghx=x$.

In particular $gin textStab_G(y)$ implies that $h^-1g h in textStab_G(x)$.

Let $varphi:textStab_G(y)rightarrow textStab_G(x)$ be such that $varphi(g)=h^-1gh$ we just saw above that $varphi$ is well defined. It is also easy to check that $varphi$ a homomorphism.

Furthermore, the inverse of $varphi$ is given by $varphi^-1(g) = hgh^-1$. The reason that $varphi^-1$ is well defined is similar to that of $varphi$ reversing the roles of $x,y$ .

share|cite|improve this answer

answered Mar 13 at 18:52

YankoYanko

7,8801830

$endgroup$

share|cite|improve this answer

answered Mar 13 at 18:52

YankoYanko

7,8801830

answered Mar 13 at 18:52

YankoYanko

7,8801830

answered Mar 13 at 18:52

YankoYanko

7,8801830