Changes

Jump to: navigation, search

Brauer's permutation lemma

1,337 bytes added, 23:14, 16 September 2007
no edit summary
* 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 <math>C(G)</math> denote the set of conjugacy classes of a [[finite group]] <math>G</math> and <math>I(G)</math> denote the set of [[indecomposable linear representation]]s of <math>G</math>. Let <math>\chi(c,\rho)</math> denote the trace of <math>\rho(g)</math> where <math>g \in G</math> (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 <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.
Bureaucrats, emailconfirmed, Administrators
38,867
edits

Navigation menu