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Calculate the characters of the left and right regular representationsof an arbitrary finite group.

Computing the inner product $(chi_R,chi_R)$.Does regular representation of a finite group contain all irreducible representations?Characters on $Cleft( mathbbR^nright)$Computing Brauer characters of a finite groupSimple components and the irreducible characters of the group ring $K[G]$Characters of a finite groupCan each of the following be a character of a finite group $G$Why do the characters of an abelian group form a group?Characters and finite group representationsCharacters of representation for finite and arbitrary groupDifference between 2 questions on the alternating group $A_4$.

1

$begingroup$

Calculate the characters of the left and right regular representations of an arbitrary finite group.

The answer of the question is given below:

But I do not know why the character of the left and right regular representations takes this form ..... could anyone explain for me why?

Thanks!

Edit:

representation-theory characters

share|cite|improve this question

edited Mar 13 at 14:03

Intuition

asked Mar 13 at 7:52

IntuitionIntuition

1,108926

$endgroup$

add a comment | 

1

$begingroup$

Calculate the characters of the left and right regular representations of an arbitrary finite group.

The answer of the question is given below:

But I do not know why the character of the left and right regular representations takes this form ..... could anyone explain for me why?

Thanks!

Edit:

representation-theory characters

share|cite|improve this question

edited Mar 13 at 14:03

Intuition

asked Mar 13 at 7:52

IntuitionIntuition

1,108926

$endgroup$

add a comment | 

$begingroup$

Calculate the characters of the left and right regular representations of an arbitrary finite group.

The answer of the question is given below:

But I do not know why the character of the left and right regular representations takes this form ..... could anyone explain for me why?

Thanks!

Edit:

representation-theory characters

share|cite|improve this question

edited Mar 13 at 14:03

Intuition

asked Mar 13 at 7:52

IntuitionIntuition

1,108926

$endgroup$

share|cite|improve this question

edited Mar 13 at 14:03

Intuition

asked Mar 13 at 7:52

IntuitionIntuition

1,108926

share|cite|improve this question

edited Mar 13 at 14:03

Intuition

asked Mar 13 at 7:52

IntuitionIntuition

1,108926

asked Mar 13 at 7:52

IntuitionIntuition

1,108926

asked Mar 13 at 7:52

IntuitionIntuition

1,108926

1 Answer
1

active

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2

$begingroup$

A left regular representation of a finite group $G$ is given by sending each $g in G$ to the permutation matrix of $sigma$, where $sigma$ is the permutation induced on the elements of $G$ by left multiplication by $g$. In other words, we label the elements of $G$ as $g_1,ldots,g_n$, where $n = vert G vert$. Then given $g in G$, we let $sigma_g in S_n$ such that
$$gg_i = g_sigma_g(i)$$
for each $i = 1,ldots,n$. Then a left regular representation for $G$ is given by sending $g$ to the permutation matrix $M_sigma_g$ of $sigma_g$, i.e., the matrix whose $i$th column has $1$ in position $sigma_g(i)$ and $0$ elsewhere. If $g = e$, then of course $M_sigma_g$ is the identity matrix, which has trace equal to $n$. On the other hand, if $g neq e$, then $gg_i neq g_i$ for all $i = 1ldots,n$. Therefore the diagonal entries of $M_sigma_g$ are all $0$, hence $M_sigma_g$ has trace $0$.

The right regular representation is the same except you multiply by $g$ on the right instead of on the left.

share|cite|improve this answer

answered Mar 13 at 8:04

Ethan AlwaiseEthan Alwaise

6,436717

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you please when you have time look at this question math.stackexchange.com/questions/3146156/…
  $endgroup$
  – Intuition
  Mar 13 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not know how to associate the definition you used with the definition of the book I will edit my question to include the book definition and tell me please how they are related.
  $endgroup$
  – Intuition
  Mar 13 at 13:45
  

  
  
  
  

add a comment | 

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2

$begingroup$

A left regular representation of a finite group $G$ is given by sending each $g in G$ to the permutation matrix of $sigma$, where $sigma$ is the permutation induced on the elements of $G$ by left multiplication by $g$. In other words, we label the elements of $G$ as $g_1,ldots,g_n$, where $n = vert G vert$. Then given $g in G$, we let $sigma_g in S_n$ such that
$$gg_i = g_sigma_g(i)$$
for each $i = 1,ldots,n$. Then a left regular representation for $G$ is given by sending $g$ to the permutation matrix $M_sigma_g$ of $sigma_g$, i.e., the matrix whose $i$th column has $1$ in position $sigma_g(i)$ and $0$ elsewhere. If $g = e$, then of course $M_sigma_g$ is the identity matrix, which has trace equal to $n$. On the other hand, if $g neq e$, then $gg_i neq g_i$ for all $i = 1ldots,n$. Therefore the diagonal entries of $M_sigma_g$ are all $0$, hence $M_sigma_g$ has trace $0$.

The right regular representation is the same except you multiply by $g$ on the right instead of on the left.

share|cite|improve this answer

answered Mar 13 at 8:04

Ethan AlwaiseEthan Alwaise

6,436717

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you please when you have time look at this question math.stackexchange.com/questions/3146156/…
  $endgroup$
  – Intuition
  Mar 13 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not know how to associate the definition you used with the definition of the book I will edit my question to include the book definition and tell me please how they are related.
  $endgroup$
  – Intuition
  Mar 13 at 13:45
  

  
  
  
  

add a comment | 

2

$begingroup$

A left regular representation of a finite group $G$ is given by sending each $g in G$ to the permutation matrix of $sigma$, where $sigma$ is the permutation induced on the elements of $G$ by left multiplication by $g$. In other words, we label the elements of $G$ as $g_1,ldots,g_n$, where $n = vert G vert$. Then given $g in G$, we let $sigma_g in S_n$ such that
$$gg_i = g_sigma_g(i)$$
for each $i = 1,ldots,n$. Then a left regular representation for $G$ is given by sending $g$ to the permutation matrix $M_sigma_g$ of $sigma_g$, i.e., the matrix whose $i$th column has $1$ in position $sigma_g(i)$ and $0$ elsewhere. If $g = e$, then of course $M_sigma_g$ is the identity matrix, which has trace equal to $n$. On the other hand, if $g neq e$, then $gg_i neq g_i$ for all $i = 1ldots,n$. Therefore the diagonal entries of $M_sigma_g$ are all $0$, hence $M_sigma_g$ has trace $0$.

The right regular representation is the same except you multiply by $g$ on the right instead of on the left.

share|cite|improve this answer

answered Mar 13 at 8:04

Ethan AlwaiseEthan Alwaise

6,436717

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you please when you have time look at this question math.stackexchange.com/questions/3146156/…
  $endgroup$
  – Intuition
  Mar 13 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not know how to associate the definition you used with the definition of the book I will edit my question to include the book definition and tell me please how they are related.
  $endgroup$
  – Intuition
  Mar 13 at 13:45
  

  
  
  
  

add a comment | 

$begingroup$

A left regular representation of a finite group $G$ is given by sending each $g in G$ to the permutation matrix of $sigma$, where $sigma$ is the permutation induced on the elements of $G$ by left multiplication by $g$. In other words, we label the elements of $G$ as $g_1,ldots,g_n$, where $n = vert G vert$. Then given $g in G$, we let $sigma_g in S_n$ such that
$$gg_i = g_sigma_g(i)$$
for each $i = 1,ldots,n$. Then a left regular representation for $G$ is given by sending $g$ to the permutation matrix $M_sigma_g$ of $sigma_g$, i.e., the matrix whose $i$th column has $1$ in position $sigma_g(i)$ and $0$ elsewhere. If $g = e$, then of course $M_sigma_g$ is the identity matrix, which has trace equal to $n$. On the other hand, if $g neq e$, then $gg_i neq g_i$ for all $i = 1ldots,n$. Therefore the diagonal entries of $M_sigma_g$ are all $0$, hence $M_sigma_g$ has trace $0$.

The right regular representation is the same except you multiply by $g$ on the right instead of on the left.

share|cite|improve this answer

answered Mar 13 at 8:04

Ethan AlwaiseEthan Alwaise

6,436717

$endgroup$

share|cite|improve this answer

answered Mar 13 at 8:04

Ethan AlwaiseEthan Alwaise

6,436717

answered Mar 13 at 8:04

Ethan AlwaiseEthan Alwaise

6,436717

answered Mar 13 at 8:04

Ethan AlwaiseEthan Alwaise

6,436717