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Characters of nonabelian group of order 57

2

1

$begingroup$

I have the following problem:

Let G be the nonabelian group of order 57.

(a.) How many 1-dimensional characters does G have?

(b.) What are the dimensions (aka degrees) of the other irreducible characters of G?

For part (a) I have already found that $G$ has 3 irreducible representations of degree 1 by analyzing the Sylow subgroups of $G$, realizing that the 19-Sylow is normal and the 3-Sylow's are not. Hence $G$ has three 1-dimensional characters.

However for part (b) I am not so sure what to do. I know two things that might be useful:

1. The number of irreducible representations of $G$ must equal the number of conjugacy classes of $G$.


2. The sum of the squares of the degrees of the irreducible representations of $G$ must equal the order of $G$.

So in particular the sum of the squares of the degrees of the remaining representations must equal $57-3 =54$ and none of those may be degree one. However this alone is not enough to give the answer as there are multiple ways of expressing 54 as a sum of squares, for example
$54= 36+9+9 = 25+16+9+4$.

Any help would be appreciated!

abstract-algebra group-theory representation-theory

share|cite|improve this question

asked Mar 24 at 19:34

Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez

2899

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is a well-known theorem that the dimension of an irreducible representation of a finite group must divide the order of the group. Does this help?
  $endgroup$
  – Sameer Kailasa
  Mar 24 at 19:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh shoot that's huge. So all the remaining characters have to be degree 3, which works since 54= (9)(6), right?
  $endgroup$
  – Edgar Jaramillo Rodriguez
  Mar 24 at 19:52
  

  
  
  
  

add a comment | 

2

1

$begingroup$

I have the following problem:

Let G be the nonabelian group of order 57.

(a.) How many 1-dimensional characters does G have?

(b.) What are the dimensions (aka degrees) of the other irreducible characters of G?

For part (a) I have already found that $G$ has 3 irreducible representations of degree 1 by analyzing the Sylow subgroups of $G$, realizing that the 19-Sylow is normal and the 3-Sylow's are not. Hence $G$ has three 1-dimensional characters.

However for part (b) I am not so sure what to do. I know two things that might be useful:

1. The number of irreducible representations of $G$ must equal the number of conjugacy classes of $G$.


2. The sum of the squares of the degrees of the irreducible representations of $G$ must equal the order of $G$.

So in particular the sum of the squares of the degrees of the remaining representations must equal $57-3 =54$ and none of those may be degree one. However this alone is not enough to give the answer as there are multiple ways of expressing 54 as a sum of squares, for example
$54= 36+9+9 = 25+16+9+4$.

Any help would be appreciated!

abstract-algebra group-theory representation-theory

share|cite|improve this question

asked Mar 24 at 19:34

Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez

2899

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is a well-known theorem that the dimension of an irreducible representation of a finite group must divide the order of the group. Does this help?
  $endgroup$
  – Sameer Kailasa
  Mar 24 at 19:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh shoot that's huge. So all the remaining characters have to be degree 3, which works since 54= (9)(6), right?
  $endgroup$
  – Edgar Jaramillo Rodriguez
  Mar 24 at 19:52
  

  
  
  
  

add a comment | 

$begingroup$

I have the following problem:

Let G be the nonabelian group of order 57.

(a.) How many 1-dimensional characters does G have?

(b.) What are the dimensions (aka degrees) of the other irreducible characters of G?

For part (a) I have already found that $G$ has 3 irreducible representations of degree 1 by analyzing the Sylow subgroups of $G$, realizing that the 19-Sylow is normal and the 3-Sylow's are not. Hence $G$ has three 1-dimensional characters.

However for part (b) I am not so sure what to do. I know two things that might be useful:

1. The number of irreducible representations of $G$ must equal the number of conjugacy classes of $G$.


2. The sum of the squares of the degrees of the irreducible representations of $G$ must equal the order of $G$.

So in particular the sum of the squares of the degrees of the remaining representations must equal $57-3 =54$ and none of those may be degree one. However this alone is not enough to give the answer as there are multiple ways of expressing 54 as a sum of squares, for example
$54= 36+9+9 = 25+16+9+4$.

Any help would be appreciated!

abstract-algebra group-theory representation-theory

share|cite|improve this question

asked Mar 24 at 19:34

Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez

2899

$endgroup$

share|cite|improve this question

asked Mar 24 at 19:34

Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez

2899

share|cite|improve this question

asked Mar 24 at 19:34

Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez

2899

asked Mar 24 at 19:34

Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez

2899

asked Mar 24 at 19:34

Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez

2899

1 Answer
1

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4

$begingroup$

A theorem of Ito asserts that if $A$ is an abelian normal subgroup of $G$, then the degree of an irreducible character divides $|G:A|$. Since the Sylow $19$-subgroup is normal, it follows that the non-linear irreducible characters all have degree $3$. Hence there are $6$ of them, since $57$ equals the sum of squares of all degrees.

If you do not want to use Ito, $chi(1)^2 leq |G|$ and $chi(1)$ divides $|G|$, for all $chi in Irr(G)$.

share|cite|improve this answer

edited Mar 24 at 20:05

answered Mar 24 at 19:59

Nicky HeksterNicky Hekster

29.1k63556

$endgroup$

add a comment | 

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4

$begingroup$

A theorem of Ito asserts that if $A$ is an abelian normal subgroup of $G$, then the degree of an irreducible character divides $|G:A|$. Since the Sylow $19$-subgroup is normal, it follows that the non-linear irreducible characters all have degree $3$. Hence there are $6$ of them, since $57$ equals the sum of squares of all degrees.

If you do not want to use Ito, $chi(1)^2 leq |G|$ and $chi(1)$ divides $|G|$, for all $chi in Irr(G)$.

share|cite|improve this answer

edited Mar 24 at 20:05

answered Mar 24 at 19:59

Nicky HeksterNicky Hekster

29.1k63556

$endgroup$

add a comment | 

4

$begingroup$

A theorem of Ito asserts that if $A$ is an abelian normal subgroup of $G$, then the degree of an irreducible character divides $|G:A|$. Since the Sylow $19$-subgroup is normal, it follows that the non-linear irreducible characters all have degree $3$. Hence there are $6$ of them, since $57$ equals the sum of squares of all degrees.

If you do not want to use Ito, $chi(1)^2 leq |G|$ and $chi(1)$ divides $|G|$, for all $chi in Irr(G)$.

share|cite|improve this answer

edited Mar 24 at 20:05

answered Mar 24 at 19:59

Nicky HeksterNicky Hekster

29.1k63556

$endgroup$

add a comment | 

$begingroup$

A theorem of Ito asserts that if $A$ is an abelian normal subgroup of $G$, then the degree of an irreducible character divides $|G:A|$. Since the Sylow $19$-subgroup is normal, it follows that the non-linear irreducible characters all have degree $3$. Hence there are $6$ of them, since $57$ equals the sum of squares of all degrees.

If you do not want to use Ito, $chi(1)^2 leq |G|$ and $chi(1)$ divides $|G|$, for all $chi in Irr(G)$.

share|cite|improve this answer

edited Mar 24 at 20:05

answered Mar 24 at 19:59

Nicky HeksterNicky Hekster

29.1k63556

$endgroup$

share|cite|improve this answer

edited Mar 24 at 20:05

answered Mar 24 at 19:59

Nicky HeksterNicky Hekster

29.1k63556

answered Mar 24 at 19:59

Nicky HeksterNicky Hekster

29.1k63556

answered Mar 24 at 19:59

Nicky HeksterNicky Hekster

29.1k63556