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Prove that every finite group of order larger than 2 has more than two irreducible complex representations. [on hold]

Does regular representation of a finite group contain all irreducible representations?Two dimensional complex group representationsIrreducible representations of group of order $pq$Irreducible representations of finite lamplighter groupEvery Irreducible Representation of a Compact Group is Finite-DimensionalEvery degree more than one irreducible character has a zeroLet $G$ be a finite abelian group. Find all inequivalent irreducible representations of $G$.$S_n, n>1,$ has precisely two $1$-dimensional irreducible representationsCan the fix point set of a nontrivial irreducible complex representation of a finite odd order group be non trivial?Prove that every irreducible real representation of an abelian group is one or two dimensional.

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prove that every finite group of order larger than 2 has more than two irreducible complex representations. could anyone give me a hint on how to solve this please?

representation-theory

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asked 2 days ago

hopefullyhopefully

217114

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put on hold as off-topic by José Carlos Santos, Eevee Trainer, mrtaurho, Thomas Shelby, Paul Frost 2 days ago

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• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Eevee Trainer, mrtaurho, Thomas Shelby, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.

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  $begingroup$
  Are you aware that the number of irreducible complex representations of a finite group is equal to the number of conjugacy classes?
  $endgroup$
  – Ethan Alwaise
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes I know this @EthanAlwaise
  $endgroup$
  – hopefully
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

prove that every finite group of order larger than 2 has more than two irreducible complex representations. could anyone give me a hint on how to solve this please?

representation-theory

share|cite|improve this question

asked 2 days ago

hopefullyhopefully

217114

$endgroup$

put on hold as off-topic by José Carlos Santos, Eevee Trainer, mrtaurho, Thomas Shelby, Paul Frost 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Eevee Trainer, mrtaurho, Thomas Shelby, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Are you aware that the number of irreducible complex representations of a finite group is equal to the number of conjugacy classes?
  $endgroup$
  – Ethan Alwaise
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes I know this @EthanAlwaise
  $endgroup$
  – hopefully
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

prove that every finite group of order larger than 2 has more than two irreducible complex representations. could anyone give me a hint on how to solve this please?

representation-theory

share|cite|improve this question

asked 2 days ago

hopefullyhopefully

217114

$endgroup$

share|cite|improve this question

asked 2 days ago

hopefullyhopefully

217114

share|cite|improve this question

asked 2 days ago

hopefullyhopefully

217114

asked 2 days ago

hopefullyhopefully

217114

asked 2 days ago

hopefullyhopefully

217114

1 Answer
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1

$begingroup$

Let $G$ be a finite group of order $n$ with conjugacy classes $C_1,ldots,C_k$. Then $vert C_i vert$, the order of the conjugacy class $C_i$, divides $n$. The conjugacy classes also partition $G$, so we have
$$vert C_1 vert + cdots + vert C_k vert = n.$$
Since the number of irreducible complex representations of $G$ is equal to $k$, if $G$ has less than three irreducible complex representations, then $k leq 2$. So either $k = 1$ which implies $G$ is trivial, or $k = 2$ and we get the equation
$$vert C_1 vert + vert C_2 vert = n.$$
However, one of the conjugacy classes, WLOG $C_1$, is the class of the identity and thus has size $1$. So we have the equation
$$1 + vert C_2 vert = n,$$
hence $vert C_2 vert = n - 1$. The only $n$ for which $n - 1$ divides $n$ is $n = 2$, so $G$ is the cyclic group of order two.

share|cite|improve this answer

answered 2 days ago

Ethan AlwaiseEthan Alwaise

6,296717

$endgroup$

add a comment | 

1

$begingroup$

Let $G$ be a finite group of order $n$ with conjugacy classes $C_1,ldots,C_k$. Then $vert C_i vert$, the order of the conjugacy class $C_i$, divides $n$. The conjugacy classes also partition $G$, so we have
$$vert C_1 vert + cdots + vert C_k vert = n.$$
Since the number of irreducible complex representations of $G$ is equal to $k$, if $G$ has less than three irreducible complex representations, then $k leq 2$. So either $k = 1$ which implies $G$ is trivial, or $k = 2$ and we get the equation
$$vert C_1 vert + vert C_2 vert = n.$$
However, one of the conjugacy classes, WLOG $C_1$, is the class of the identity and thus has size $1$. So we have the equation
$$1 + vert C_2 vert = n,$$
hence $vert C_2 vert = n - 1$. The only $n$ for which $n - 1$ divides $n$ is $n = 2$, so $G$ is the cyclic group of order two.

share|cite|improve this answer

answered 2 days ago

Ethan AlwaiseEthan Alwaise

6,296717

$endgroup$

add a comment | 

1

$begingroup$

Let $G$ be a finite group of order $n$ with conjugacy classes $C_1,ldots,C_k$. Then $vert C_i vert$, the order of the conjugacy class $C_i$, divides $n$. The conjugacy classes also partition $G$, so we have
$$vert C_1 vert + cdots + vert C_k vert = n.$$
Since the number of irreducible complex representations of $G$ is equal to $k$, if $G$ has less than three irreducible complex representations, then $k leq 2$. So either $k = 1$ which implies $G$ is trivial, or $k = 2$ and we get the equation
$$vert C_1 vert + vert C_2 vert = n.$$
However, one of the conjugacy classes, WLOG $C_1$, is the class of the identity and thus has size $1$. So we have the equation
$$1 + vert C_2 vert = n,$$
hence $vert C_2 vert = n - 1$. The only $n$ for which $n - 1$ divides $n$ is $n = 2$, so $G$ is the cyclic group of order two.

share|cite|improve this answer

answered 2 days ago

Ethan AlwaiseEthan Alwaise

6,296717

$endgroup$

add a comment | 

$begingroup$

Let $G$ be a finite group of order $n$ with conjugacy classes $C_1,ldots,C_k$. Then $vert C_i vert$, the order of the conjugacy class $C_i$, divides $n$. The conjugacy classes also partition $G$, so we have
$$vert C_1 vert + cdots + vert C_k vert = n.$$
Since the number of irreducible complex representations of $G$ is equal to $k$, if $G$ has less than three irreducible complex representations, then $k leq 2$. So either $k = 1$ which implies $G$ is trivial, or $k = 2$ and we get the equation
$$vert C_1 vert + vert C_2 vert = n.$$
However, one of the conjugacy classes, WLOG $C_1$, is the class of the identity and thus has size $1$. So we have the equation
$$1 + vert C_2 vert = n,$$
hence $vert C_2 vert = n - 1$. The only $n$ for which $n - 1$ divides $n$ is $n = 2$, so $G$ is the cyclic group of order two.

share|cite|improve this answer

answered 2 days ago

Ethan AlwaiseEthan Alwaise

6,296717

$endgroup$

share|cite|improve this answer

answered 2 days ago

Ethan AlwaiseEthan Alwaise

6,296717

answered 2 days ago

Ethan AlwaiseEthan Alwaise

6,296717

answered 2 days ago

Ethan AlwaiseEthan Alwaise

6,296717