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Left action of a group on permutation representation

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3

$begingroup$

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:

$g*phi(xin X)to phi(g^-1x)$.

Is this a left group action?

I have: $h*(g*phi(x))= h*phi(g^-1x)= phi(h^-1g^-1x) = phi((gh)^-1x)= (gh)*phi(x) neq (hg)*phi(x)$.

I must be missing something.

abstract-algebra group-theory representation-theory group-actions

share|cite|improve this question

edited Apr 15 at 13:54

Peter Mortensen

58233 silver badges1010 bronze badges

asked Apr 15 at 11:05

PerelManPerelMan

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$endgroup$

add a comment
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3

$begingroup$

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:

$g*phi(xin X)to phi(g^-1x)$.

Is this a left group action?

I have: $h*(g*phi(x))= h*phi(g^-1x)= phi(h^-1g^-1x) = phi((gh)^-1x)= (gh)*phi(x) neq (hg)*phi(x)$.

I must be missing something.

abstract-algebra group-theory representation-theory group-actions

share|cite|improve this question

edited Apr 15 at 13:54

Peter Mortensen

58233 silver badges1010 bronze badges

asked Apr 15 at 11:05

PerelManPerelMan

1,04444 silver badges1414 bronze badges

$endgroup$

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:

$g*phi(xin X)to phi(g^-1x)$.

Is this a left group action?

I have: $h*(g*phi(x))= h*phi(g^-1x)= phi(h^-1g^-1x) = phi((gh)^-1x)= (gh)*phi(x) neq (hg)*phi(x)$.

I must be missing something.

abstract-algebra group-theory representation-theory group-actions

share|cite|improve this question

edited Apr 15 at 13:54

Peter Mortensen

58233 silver badges1010 bronze badges

asked Apr 15 at 11:05

PerelManPerelMan

1,04444 silver badges1414 bronze badges

$endgroup$

share|cite|improve this question

edited Apr 15 at 13:54

Peter Mortensen

58233 silver badges1010 bronze badges

asked Apr 15 at 11:05

PerelManPerelMan

1,04444 silver badges1414 bronze badges

share|cite|improve this question

edited Apr 15 at 13:54

Peter Mortensen

58233 silver badges1010 bronze badges

asked Apr 15 at 11:05

PerelManPerelMan

1,04444 silver badges1414 bronze badges

edited Apr 15 at 13:54

Peter Mortensen

58233 silver badges1010 bronze badges

edited Apr 15 at 13:54

Peter Mortensen

58233 silver badges1010 bronze badges

asked Apr 15 at 11:05

PerelManPerelMan

1,04444 silver badges1414 bronze badges

asked Apr 15 at 11:05

PerelManPerelMan

1,04444 silver badges1414 bronze badges

2 Answers
2

active

oldest

votes

7

$begingroup$

Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^-1x)$, so the calculation is the following:
$$beginalign
(h*(g*phi))(x)&= (g*phi)(h^-1x)\
&= phi(g^-1h^-1x)\
&= phi((hg)^-1x)\
&= (hg*phi)(x).
endalign$$
Thus $h*(g*phi)= hg*phi$.

share|cite|improve this answer

edited Apr 15 at 12:56

Shaun

13.1k1212 gold badges3838 silver badges125125 bronze badges

answered Apr 15 at 11:29

SMMSMM

3,68855 silver badges1313 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*phi)(x) := phi(g^-1x)$, we have $g*phi = x to phi(g^-1x)$, therefore

$$beginalign
(h*(g*phi))(x)
&= (h*(ttophi(g^-1t)))(x)\
&= (ttophi(g^-1t))(h^-1x)\
&= phi(g^-1(h^-1x))\
&= (hg*phi)(x).
endalign$$

share|cite|improve this answer

answered Apr 15 at 14:44

Micha WiedenmannMicha Wiedenmann

14777 bronze badges

$endgroup$

add a comment
 | 

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7

$begingroup$

Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^-1x)$, so the calculation is the following:
$$beginalign
(h*(g*phi))(x)&= (g*phi)(h^-1x)\
&= phi(g^-1h^-1x)\
&= phi((hg)^-1x)\
&= (hg*phi)(x).
endalign$$
Thus $h*(g*phi)= hg*phi$.

share|cite|improve this answer

edited Apr 15 at 12:56

Shaun

13.1k1212 gold badges3838 silver badges125125 bronze badges

answered Apr 15 at 11:29

SMMSMM

3,68855 silver badges1313 bronze badges

$endgroup$

add a comment
 | 

7

$begingroup$

Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^-1x)$, so the calculation is the following:
$$beginalign
(h*(g*phi))(x)&= (g*phi)(h^-1x)\
&= phi(g^-1h^-1x)\
&= phi((hg)^-1x)\
&= (hg*phi)(x).
endalign$$
Thus $h*(g*phi)= hg*phi$.

share|cite|improve this answer

edited Apr 15 at 12:56

Shaun

13.1k1212 gold badges3838 silver badges125125 bronze badges

answered Apr 15 at 11:29

SMMSMM

3,68855 silver badges1313 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^-1x)$, so the calculation is the following:
$$beginalign
(h*(g*phi))(x)&= (g*phi)(h^-1x)\
&= phi(g^-1h^-1x)\
&= phi((hg)^-1x)\
&= (hg*phi)(x).
endalign$$
Thus $h*(g*phi)= hg*phi$.

share|cite|improve this answer

edited Apr 15 at 12:56

Shaun

13.1k1212 gold badges3838 silver badges125125 bronze badges

answered Apr 15 at 11:29

SMMSMM

3,68855 silver badges1313 bronze badges

$endgroup$

share|cite|improve this answer

edited Apr 15 at 12:56

Shaun

13.1k1212 gold badges3838 silver badges125125 bronze badges

answered Apr 15 at 11:29

SMMSMM

3,68855 silver badges1313 bronze badges

edited Apr 15 at 12:56

Shaun

13.1k1212 gold badges3838 silver badges125125 bronze badges

edited Apr 15 at 12:56

Shaun

13.1k1212 gold badges3838 silver badges125125 bronze badges

answered Apr 15 at 11:29

SMMSMM

3,68855 silver badges1313 bronze badges

answered Apr 15 at 11:29

SMMSMM

3,68855 silver badges1313 bronze badges

1

$begingroup$

In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*phi)(x) := phi(g^-1x)$, we have $g*phi = x to phi(g^-1x)$, therefore

$$beginalign
(h*(g*phi))(x)
&= (h*(ttophi(g^-1t)))(x)\
&= (ttophi(g^-1t))(h^-1x)\
&= phi(g^-1(h^-1x))\
&= (hg*phi)(x).
endalign$$

share|cite|improve this answer

answered Apr 15 at 14:44

Micha WiedenmannMicha Wiedenmann

14777 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*phi)(x) := phi(g^-1x)$, we have $g*phi = x to phi(g^-1x)$, therefore

$$beginalign
(h*(g*phi))(x)
&= (h*(ttophi(g^-1t)))(x)\
&= (ttophi(g^-1t))(h^-1x)\
&= phi(g^-1(h^-1x))\
&= (hg*phi)(x).
endalign$$

share|cite|improve this answer

answered Apr 15 at 14:44

Micha WiedenmannMicha Wiedenmann

14777 bronze badges

$endgroup$

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 $(g*phi)(x) := phi(g^-1x)$, we have $g*phi = x to phi(g^-1x)$, therefore

$$beginalign
(h*(g*phi))(x)
&= (h*(ttophi(g^-1t)))(x)\
&= (ttophi(g^-1t))(h^-1x)\
&= phi(g^-1(h^-1x))\
&= (hg*phi)(x).
endalign$$

share|cite|improve this answer

answered Apr 15 at 14:44

Micha WiedenmannMicha Wiedenmann

14777 bronze badges

$endgroup$

share|cite|improve this answer

answered Apr 15 at 14:44

Micha WiedenmannMicha Wiedenmann

14777 bronze badges

answered Apr 15 at 14:44

Micha WiedenmannMicha Wiedenmann

14777 bronze badges

answered Apr 15 at 14:44

Micha WiedenmannMicha Wiedenmann

14777 bronze badges