Group action inverse?
$begingroup$
Suppose $G$ acts on $Omega.$ It's easy to see that if $g in G$ is arbitrary, then the function $sigma_g: Omega to Omega$ defined by $(alpha)sigma_g = alpha cdot g$ has an inverse: the function $sigma_{g^{-1}}$. Therefore, $sigma_g$ is a permutation of the set $Omega$, which means $sigma_g$ is both injective and surjective, and thus $sigma_g$ lies on the symmetric group $text{Sym}(Omega)$ consisting of all permutations of of $Omega$. [...]
My question is what's the function $sigma_{g^{-1}}$? Is it the function $sigma_{g^{-1}}: Omega to Omega$ defined by $$(alpha)sigma_{g^{-1}} = alpha cdot g^{-1}$$
In which case we would have $alpha sigma_g alpha sigma_{g^{-1}} = alpha g alpha g^{-1}$ which doesn't seem to be the identity. I think I'm being confused by the notation.
The only way this would make sense to me is that if I define $sigma_{g^{-1}}: Omega to Omega$ by $$sigma_{g^{-1}} = (alpha cdot g)^{-1} = g^{-1}alpha^{-1}.$$
But that does not appear to be consistent with the author's notation.
abstract-algebra group-theory finite-groups group-actions
edited Dec 21 '18 at 16:30
Shaun
9,739113684
asked Dec 20 '18 at 2:27
Art VandelayArt Vandelay
133
$endgroup$
add a comment |
$begingroup$
Suppose $G$ acts on $Omega.$ It's easy to see that if $g in G$ is arbitrary, then the function $sigma_g: Omega to Omega$ defined by $(alpha)sigma_g = alpha cdot g$ has an inverse: the function $sigma_{g^{-1}}$. Therefore, $sigma_g$ is a permutation of the set $Omega$, which means $sigma_g$ is both injective and surjective, and thus $sigma_g$ lies on the symmetric group $text{Sym}(Omega)$ consisting of all permutations of of $Omega$. [...]
My question is what's the function $sigma_{g^{-1}}$? Is it the function $sigma_{g^{-1}}: Omega to Omega$ defined by $$(alpha)sigma_{g^{-1}} = alpha cdot g^{-1}$$
In which case we would have $alpha sigma_g alpha sigma_{g^{-1}} = alpha g alpha g^{-1}$ which doesn't seem to be the identity. I think I'm being confused by the notation.
The only way this would make sense to me is that if I define $sigma_{g^{-1}}: Omega to Omega$ by $$sigma_{g^{-1}} = (alpha cdot g)^{-1} = g^{-1}alpha^{-1}.$$
But that does not appear to be consistent with the author's notation.
abstract-algebra group-theory finite-groups group-actions
edited Dec 21 '18 at 16:30
Shaun
9,739113684
asked Dec 20 '18 at 2:27
Art VandelayArt Vandelay
133
$endgroup$
$begingroup$
Where is the quote from? Please edit the question to include it.
$endgroup$
– Shaun
Dec 21 '18 at 16:31
add a comment |
$begingroup$
Suppose $G$ acts on $Omega.$ It's easy to see that if $g in G$ is arbitrary, then the function $sigma_g: Omega to Omega$ defined by $(alpha)sigma_g = alpha cdot g$ has an inverse: the function $sigma_{g^{-1}}$. Therefore, $sigma_g$ is a permutation of the set $Omega$, which means $sigma_g$ is both injective and surjective, and thus $sigma_g$ lies on the symmetric group $text{Sym}(Omega)$ consisting of all permutations of of $Omega$. [...]
My question is what's the function $sigma_{g^{-1}}$? Is it the function $sigma_{g^{-1}}: Omega to Omega$ defined by $$(alpha)sigma_{g^{-1}} = alpha cdot g^{-1}$$
In which case we would have $alpha sigma_g alpha sigma_{g^{-1}} = alpha g alpha g^{-1}$ which doesn't seem to be the identity. I think I'm being confused by the notation.
The only way this would make sense to me is that if I define $sigma_{g^{-1}}: Omega to Omega$ by $$sigma_{g^{-1}} = (alpha cdot g)^{-1} = g^{-1}alpha^{-1}.$$
But that does not appear to be consistent with the author's notation.
abstract-algebra group-theory finite-groups group-actions
edited Dec 21 '18 at 16:30
Shaun
9,739113684
asked Dec 20 '18 at 2:27
Art VandelayArt Vandelay
133
$endgroup$
Suppose $G$ acts on $Omega.$ It's easy to see that if $g in G$ is arbitrary, then the function $sigma_g: Omega to Omega$ defined by $(alpha)sigma_g = alpha cdot g$ has an inverse: the function $sigma_{g^{-1}}$. Therefore, $sigma_g$ is a permutation of the set $Omega$, which means $sigma_g$ is both injective and surjective, and thus $sigma_g$ lies on the symmetric group $text{Sym}(Omega)$ consisting of all permutations of of $Omega$. [...]
My question is what's the function $sigma_{g^{-1}}$? Is it the function $sigma_{g^{-1}}: Omega to Omega$ defined by $$(alpha)sigma_{g^{-1}} = alpha cdot g^{-1}$$
In which case we would have $alpha sigma_g alpha sigma_{g^{-1}} = alpha g alpha g^{-1}$ which doesn't seem to be the identity. I think I'm being confused by the notation.
The only way this would make sense to me is that if I define $sigma_{g^{-1}}: Omega to Omega$ by $$sigma_{g^{-1}} = (alpha cdot g)^{-1} = g^{-1}alpha^{-1}.$$
But that does not appear to be consistent with the author's notation.
abstract-algebra group-theory finite-groups group-actions
abstract-algebra group-theory finite-groups group-actions
edited Dec 21 '18 at 16:30
Shaun
9,739113684
asked Dec 20 '18 at 2:27
Art VandelayArt Vandelay
133
edited Dec 21 '18 at 16:30
Shaun
9,739113684
asked Dec 20 '18 at 2:27
Art VandelayArt Vandelay
133
edited Dec 21 '18 at 16:30
Shaun
9,739113684
edited Dec 21 '18 at 16:30
Shaun
9,739113684
edited Dec 21 '18 at 16:30
Shaun
9,739113684
9,739113684
asked Dec 20 '18 at 2:27
Art VandelayArt Vandelay
133
asked Dec 20 '18 at 2:27
Art VandelayArt Vandelay
133
asked Dec 20 '18 at 2:27
Art VandelayArt Vandelay
133
133
$begingroup$
Where is the quote from? Please edit the question to include it.
$endgroup$
– Shaun
Dec 21 '18 at 16:31
add a comment |
$begingroup$
Where is the quote from? Please edit the question to include it.
$endgroup$
– Shaun
Dec 21 '18 at 16:31
$begingroup$
Where is the quote from? Please edit the question to include it.
$endgroup$
– Shaun
Dec 21 '18 at 16:31
$begingroup$
Where is the quote from? Please edit the question to include it.
$endgroup$
– Shaun
Dec 21 '18 at 16:31
add a comment |
1 Answer
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$begingroup$
Note that $sigma$ is a function. So it only makes sense for function composition. Then $(alpha)(sigma_g circ sigma_{g^{-1}}) = (alpha cdot g)cdot g^{-1} = alpha$.
answered Dec 20 '18 at 2:39
kminikmini
913
$endgroup$
add a comment |
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$begingroup$
Note that $sigma$ is a function. So it only makes sense for function composition. Then $(alpha)(sigma_g circ sigma_{g^{-1}}) = (alpha cdot g)cdot g^{-1} = alpha$.
answered Dec 20 '18 at 2:39
kminikmini
913
$endgroup$
add a comment |
$begingroup$
Note that $sigma$ is a function. So it only makes sense for function composition. Then $(alpha)(sigma_g circ sigma_{g^{-1}}) = (alpha cdot g)cdot g^{-1} = alpha$.
answered Dec 20 '18 at 2:39
kminikmini
913
$endgroup$
add a comment |
$begingroup$
Note that $sigma$ is a function. So it only makes sense for function composition. Then $(alpha)(sigma_g circ sigma_{g^{-1}}) = (alpha cdot g)cdot g^{-1} = alpha$.
answered Dec 20 '18 at 2:39
kminikmini
913
$endgroup$
Note that $sigma$ is a function. So it only makes sense for function composition. Then $(alpha)(sigma_g circ sigma_{g^{-1}}) = (alpha cdot g)cdot g^{-1} = alpha$.
answered Dec 20 '18 at 2:39
kminikmini
913
answered Dec 20 '18 at 2:39
kminikmini
913
answered Dec 20 '18 at 2:39
kminikmini
913
answered Dec 20 '18 at 2:39
kminikmini
913
913
add a comment |
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$begingroup$
Where is the quote from? Please edit the question to include it.
$endgroup$
– Shaun
Dec 21 '18 at 16:31