Left action of a group on permutation representation
$begingroup$
I am studying my course on permutation representation, and I am stuck at understanding the left action of a finite group on the permutation representation $F(X,Bbb C)$. In my course it is given for $(g,phi)in Gtimes F(X,Bbb C) $ by:
$g*phi(xin X)to phi(g^{-1}x)$.
Is this a left group action?
I have: $h*(g*phi(x))= h*phi(g^{-1}x)= phi(h^{-1}g^{-1}x) = phi((gh)^{-1}x)= (gh)*phi(x) neq (hg)*phi(x)$.
I must be missing something.
abstract-algebra group-theory representation-theory group-actions
edited yesterday
Peter Mortensen
563310
asked yesterday
PerelManPerelMan
803414
$endgroup$
add a comment |
$begingroup$
I am studying my course on permutation representation, and I am stuck at understanding the left action of a finite group on the permutation representation $F(X,Bbb C)$. In my course it is given for $(g,phi)in Gtimes F(X,Bbb C) $ by:
$g*phi(xin X)to phi(g^{-1}x)$.
Is this a left group action?
I have: $h*(g*phi(x))= h*phi(g^{-1}x)= phi(h^{-1}g^{-1}x) = phi((gh)^{-1}x)= (gh)*phi(x) neq (hg)*phi(x)$.
I must be missing something.
abstract-algebra group-theory representation-theory group-actions
edited yesterday
Peter Mortensen
563310
asked yesterday
PerelManPerelMan
803414
$endgroup$
add a comment |
$begingroup$
I am studying my course on permutation representation, and I am stuck at understanding the left action of a finite group on the permutation representation $F(X,Bbb C)$. In my course it is given for $(g,phi)in Gtimes F(X,Bbb C) $ by:
$g*phi(xin X)to phi(g^{-1}x)$.
Is this a left group action?
I have: $h*(g*phi(x))= h*phi(g^{-1}x)= phi(h^{-1}g^{-1}x) = phi((gh)^{-1}x)= (gh)*phi(x) neq (hg)*phi(x)$.
I must be missing something.
abstract-algebra group-theory representation-theory group-actions
edited yesterday
Peter Mortensen
563310
asked yesterday
PerelManPerelMan
803414
$endgroup$
I am studying my course on permutation representation, and I am stuck at understanding the left action of a finite group on the permutation representation $F(X,Bbb C)$. In my course it is given for $(g,phi)in Gtimes F(X,Bbb C) $ by:
$g*phi(xin X)to phi(g^{-1}x)$.
Is this a left group action?
I have: $h*(g*phi(x))= h*phi(g^{-1}x)= phi(h^{-1}g^{-1}x) = phi((gh)^{-1}x)= (gh)*phi(x) neq (hg)*phi(x)$.
I must be missing something.
abstract-algebra group-theory representation-theory group-actions
abstract-algebra group-theory representation-theory group-actions
edited yesterday
Peter Mortensen
563310
asked yesterday
PerelManPerelMan
803414
edited yesterday
Peter Mortensen
563310
asked yesterday
PerelManPerelMan
803414
edited yesterday
Peter Mortensen
563310
edited yesterday
Peter Mortensen
563310
edited yesterday
Peter Mortensen
563310
563310
asked yesterday
PerelManPerelMan
803414
asked yesterday
PerelManPerelMan
803414
asked yesterday
PerelManPerelMan
803414
803414
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^{-1}x)$, so the calculation is the following:
$$begin{align}
(h*(g*phi))(x)&= (g*phi)(h^{-1}x)\
&= phi(g^{-1}h^{-1}x)\
&= phi((hg)^{-1}x)\
&= (hg*phi)(x).
end{align}$$
Thus $h*(g*phi)= hg*phi$.
edited yesterday
Shaun
10.6k113687
answered yesterday
SMMSMM
3,268512
$endgroup$
add a comment |
$begingroup$
In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*phi)(x) := phi(g^{-1}x)$, we have $g*phi = x to phi(g^{-1}x)$, therefore
$$begin{align}
(h*(g*phi))(x)
&= (h*(ttophi(g^{-1}t)))(x)\
&= (ttophi(g^{-1}t))(h^{-1}x)\
&= phi(g^{-1}(h^{-1}x))\
&= (hg*phi)(x).
end{align}$$
answered yesterday
Micha WiedenmannMicha Wiedenmann
1396
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^{-1}x)$, so the calculation is the following:
$$begin{align}
(h*(g*phi))(x)&= (g*phi)(h^{-1}x)\
&= phi(g^{-1}h^{-1}x)\
&= phi((hg)^{-1}x)\
&= (hg*phi)(x).
end{align}$$
Thus $h*(g*phi)= hg*phi$.
edited yesterday
Shaun
10.6k113687
answered yesterday
SMMSMM
3,268512
$endgroup$
add a comment |
$begingroup$
Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^{-1}x)$, so the calculation is the following:
$$begin{align}
(h*(g*phi))(x)&= (g*phi)(h^{-1}x)\
&= phi(g^{-1}h^{-1}x)\
&= phi((hg)^{-1}x)\
&= (hg*phi)(x).
end{align}$$
Thus $h*(g*phi)= hg*phi$.
edited yesterday
Shaun
10.6k113687
answered yesterday
SMMSMM
3,268512
$endgroup$
add a comment |
$begingroup$
Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^{-1}x)$, so the calculation is the following:
$$begin{align}
(h*(g*phi))(x)&= (g*phi)(h^{-1}x)\
&= phi(g^{-1}h^{-1}x)\
&= phi((hg)^{-1}x)\
&= (hg*phi)(x).
end{align}$$
Thus $h*(g*phi)= hg*phi$.
edited yesterday
Shaun
10.6k113687
answered yesterday
SMMSMM
3,268512
$endgroup$
Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^{-1}x)$, so the calculation is the following:
$$begin{align}
(h*(g*phi))(x)&= (g*phi)(h^{-1}x)\
&= phi(g^{-1}h^{-1}x)\
&= phi((hg)^{-1}x)\
&= (hg*phi)(x).
end{align}$$
Thus $h*(g*phi)= hg*phi$.
edited yesterday
Shaun
10.6k113687
answered yesterday
SMMSMM
3,268512
edited yesterday
Shaun
10.6k113687
edited yesterday
Shaun
10.6k113687
edited yesterday
Shaun
10.6k113687
10.6k113687
answered yesterday
SMMSMM
3,268512
answered yesterday
SMMSMM
3,268512
answered yesterday
SMMSMM
3,268512
3,268512
add a comment |
add a comment |
$begingroup$
In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*phi)(x) := phi(g^{-1}x)$, we have $g*phi = x to phi(g^{-1}x)$, therefore
$$begin{align}
(h*(g*phi))(x)
&= (h*(ttophi(g^{-1}t)))(x)\
&= (ttophi(g^{-1}t))(h^{-1}x)\
&= phi(g^{-1}(h^{-1}x))\
&= (hg*phi)(x).
end{align}$$
answered yesterday
Micha WiedenmannMicha Wiedenmann
1396
$endgroup$
add a comment |
$begingroup$
In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*phi)(x) := phi(g^{-1}x)$, we have $g*phi = x to phi(g^{-1}x)$, therefore
$$begin{align}
(h*(g*phi))(x)
&= (h*(ttophi(g^{-1}t)))(x)\
&= (ttophi(g^{-1}t))(h^{-1}x)\
&= phi(g^{-1}(h^{-1}x))\
&= (hg*phi)(x).
end{align}$$
answered yesterday
Micha WiedenmannMicha Wiedenmann
1396
$endgroup$
add a comment |
$begingroup$
In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*phi)(x) := phi(g^{-1}x)$, we have $g*phi = x to phi(g^{-1}x)$, therefore
$$begin{align}
(h*(g*phi))(x)
&= (h*(ttophi(g^{-1}t)))(x)\
&= (ttophi(g^{-1}t))(h^{-1}x)\
&= phi(g^{-1}(h^{-1}x))\
&= (hg*phi)(x).
end{align}$$
answered yesterday
Micha WiedenmannMicha Wiedenmann
1396
$endgroup$
In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*phi)(x) := phi(g^{-1}x)$, we have $g*phi = x to phi(g^{-1}x)$, therefore
$$begin{align}
(h*(g*phi))(x)
&= (h*(ttophi(g^{-1}t)))(x)\
&= (ttophi(g^{-1}t))(h^{-1}x)\
&= phi(g^{-1}(h^{-1}x))\
&= (hg*phi)(x).
end{align}$$
answered yesterday
Micha WiedenmannMicha Wiedenmann
1396
answered yesterday
Micha WiedenmannMicha Wiedenmann
1396
answered yesterday
Micha WiedenmannMicha Wiedenmann
1396
answered yesterday
Micha WiedenmannMicha Wiedenmann
1396
1396
add a comment |
add a comment |
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