Prove Symmetry Group of Regular Polygon Has 1 & 2 Dim Irreducible Reps

Click For Summary

Discussion Overview

The discussion revolves around proving that the symmetry group of a regular polygon, specifically the dihedral group, has only 1 and 2-dimensional irreducible representations. The focus is on theoretical aspects of group representation and the properties of the dihedral group.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • Some participants assert that the dihedral group has an abelian subgroup of index 2, which implies that any irreducible representation is at most 2-dimensional.
  • One participant questions whether the rotation subgroup is of index 2, leading to a clarification that the reflection subgroup has order 2.
  • Another participant expresses difficulty in understanding the proof and seeks hints or relevant theorems to aid in their understanding.

Areas of Agreement / Disagreement

Participants generally agree on the properties of the dihedral group and its subgroups, but there is no consensus on the specifics of the proof or theorems applicable to the problem.

Contextual Notes

There are unresolved details regarding the proof and theorems that may be relevant to the discussion, as well as potential missing assumptions about the representations.

wdlang
Messages
306
Reaction score
0
how to prove that the symmetry group of a regular polygon has only 1 and 2 dim irreducible representations?
 
Physics news on Phys.org


The group in question, i.e. the dihedral group, has an abelian subgroup of index 2 (the one generated by the reflection). Thus any irreducible representation is at most 2 dimensional. I'll let you fill in the details. Post back if you need more help!
 
Last edited:


morphism said:
The group in question, i.e. the dihedral group, has an abelian subgroup of index 2 (the one generated by the reflection). Thus any irreducible representation is at most 2 dimensional. I'll let you fill in the details. Post back if you need more help!

Thanks a lot

maybe you mean the rotation subgroup is of index 2?

i will think of it.
 


wdlang said:
maybe you mean the rotation subgroup is of index 2?
Yup - sorry! (The reflection subgroup has order 2!)
 


morphism said:
The group in question, i.e. the dihedral group, has an abelian subgroup of index 2 (the one generated by the reflection). Thus any irreducible representation is at most 2 dimensional. I'll let you fill in the details. Post back if you need more help!

but i still can not figure out the proof

any hint?

is there any theorem?
 

Similar threads

Undergrad How to show ##p(x)=g(x)x\pm 1\in\Bbb{Q}[x]## is irreducible in ##\Bbb{Q}_{\Bbb{Z}}[x]##?
  • · Replies 48 ·
2
Replies
48
Views
6K
Graduate Does an irreducible representation acting on operators imply....
  • · Replies 3 ·
Replies
3
Views
2K
Graduate Representations of finite groups: Irreducible and reducible
  • · Replies 1 ·
Replies
1
Views
2K
Graduate Tensor symmetries and the symmetric groups
  • · Replies 1 ·
Replies
1
Views
2K
Insights Groups, The Path from a Simple Concept to Mysterious Results
  • · Replies 17 ·
Replies
17
Views
10K
Graduate Understanding the Relationship between Orthogonal and Unitary Groups
  • · Replies 13 ·
Replies
13
Views
3K
Graduate Group of symmetries on a regular polygon
  • · Replies 4 ·
Replies
4
Views
13K
Undergrad Understanding 4-Vector Representations in the Lorentz Group
  • · Replies 2 ·
Replies
2
Views
2K
Graduate Hidden symmetries in subgroups of Lie groups?
  • · Replies 1 ·
Replies
1
Views
3K
Graduate Reality conditions on representations of classical groups
  • · Replies 3 ·
Replies
3
Views
2K