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

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Brauer's permutation lemma has the following equivalent forms: | 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 [[cycle type]] i.e., the same number of cycles of each length in their cycle | + | # If a row permutation and a column permutation have the same effect on a nonsingular matrix, then they must have the same [[cycle type]] i.e., the same number of cycles of each length in their [[cycle decomposition]]s. |

# The [[symmetric group]] is a [[conjugacy-closed subgroup]] in the [[general linear group]] over any [[field]] of characteristic zero | # 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 [[cycle type]] (i.e., the same number of cycles of each length in their [[cycle decomposition]]s), i.e., are conjugate in the symmetric group itself | # If two permutation matrices are conjugate in the general linear group over a field of characteristic zero, then they have the same [[cycle type]] (i.e., the same number of cycles of each length in their [[cycle decomposition]]s), i.e., are conjugate in the symmetric group itself | ||

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==Facts== | ==Facts== | ||

+ | |||

+ | ===Opposite facts=== | ||

* [[Analogue of Brauer's permutation lemma fails over rationals for every non-cyclic finite group]]: If <math>G</math> is a non-cyclic finite group, we can find two permutation representations <math>\varphi_1, \varphi_2</math> of <math>G</math> that are equivalent as linear representations over the rational numbers but not as permutation representations. | * [[Analogue of Brauer's permutation lemma fails over rationals for every non-cyclic finite group]]: If <math>G</math> is a non-cyclic finite group, we can find two permutation representations <math>\varphi_1, \varphi_2</math> of <math>G</math> that are equivalent as linear representations over the rational numbers but not as permutation representations. |

## Latest revision as of 03:51, 14 January 2014

This article gives the statement, and proof, of a particular subgroup in a group being conjugacy-closed: in other words, any two elements of the subgroup that are conjugate in the whole group, are also conjugate in the subgroup

View a complete list of such instances/statements

## 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 cycle type i.e., the same number of cycles of each length in their cycle decompositions.
- 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 cycle type (i.e., the same number of cycles of each length in their cycle decompositions), i.e., 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

## Facts

### Opposite facts

- Analogue of Brauer's permutation lemma fails over rationals for every non-cyclic finite group: If is a non-cyclic finite group, we can find two permutation representations of that are equivalent as linear representations over the rational numbers but not as permutation representations.
- Symmetric group of degree six or higher is not weak subset-conjugacy-closed in general linear group over rationals

## Applications

Brauer's permutation lemma helps us exploit the conjugacy class-representation duality in two interesting ways. 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 .

We can now consider two different actions: