# Difference between revisions of "Brauer's permutation lemma"

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Note that since the field has characteristic zero, the irreducible representations are the same as indecomposable representations. | Note that since the field has characteristic zero, the irreducible representations are the same as indecomposable representations. | ||

− | Consider the matrix with rows indexed by indecomposable representations, columns indexed by conjugacy classes, and the entry in row <math>\rho</math> and column <math>c</math> is <math>\chi(c,\rho)</math>. In other words, the matrix is the [[character table]]. The automorphism group acts on the rows and on the columns in a way that the entries are not affected. {{further|[[Conjugacy- | + | Consider the matrix with rows indexed by indecomposable representations, columns indexed by conjugacy classes, and the entry in row <math>\rho</math> and column <math>c</math> is <math>\chi(c,\rho)</math>. In other words, the matrix is the [[character table]]. The automorphism group acts on the rows and on the columns in a way that the entries are not affected. {{further|[[Conjugacy class-representation duality]]}} |

Applying Brauer's permutation lemma to this, we get that for any automorphism, the sizes of orbits of conjugacy classes under the automorphism, are the same as the sizes of orbits of irreducible representations. In particular, the number of invariant conjugacy classes equals the number of invariant irreducible representations. | Applying Brauer's permutation lemma to this, we get that for any automorphism, the sizes of orbits of conjugacy classes under the automorphism, are the same as the sizes of orbits of irreducible representations. In particular, the number of invariant conjugacy classes equals the number of invariant irreducible representations. |

## Revision as of 23:15, 16 September 2007

## Statement

Brauer's permutation lemma has the following equivalent forms:

- If a row permutation and a column permutation have the same effect on a nonsingular matrix, then they must have the same number of cycles of a given length
- The symmetric group is a conjugacy-closed subgroup in the general linear group over any field of characteristic zero
- If two permutation matrices are conjugate in the general linear group over a field of characteristic zero, then they have the same number of cycles of each length, viz, are conjugate in the symmetric group itself
- If two permutation representations of a cyclic group are conjugate in the general linear group over a field of characteristic zero, they are also conjugate in the symmetric group

## Applications

Brauer's permutation lemma helps us exploit the conjugacy class-representation duality in an interesting way. Let denote the set of conjugacy classes of a finite group and denote the set of indecomposable linear representations of . Let denote the trace of where (i.e. the character value).

Note that since the field has characteristic zero, the irreducible representations are the same as indecomposable representations.

Consider the matrix with rows indexed by indecomposable representations, columns indexed by conjugacy classes, and the entry in row and column is . In other words, the matrix is the character table. The automorphism group acts on the rows and on the columns in a way that the entries are not affected. `Further information: Conjugacy class-representation duality`

Applying Brauer's permutation lemma to this, we get that for any automorphism, the sizes of orbits of conjugacy classes under the automorphism, are the same as the sizes of orbits of irreducible representations. In particular, the number of invariant conjugacy classes equals the number of invariant irreducible representations.