proof that dimension of complex irreducible representation divides order of group
Theorem Let be a finite group and an irreducible
complex representation of finite dimension
. Then divides
.
Proof: Given any in the group ring of (denoted ) we may define a sequence of submodules of (regarded as a module over ) by equals the linear span of .
is Noetherian as a module over so we
must have for some . Hence may be
expressed as a linear combination of lower powers of
. In other solves a monic polynomial of
degree with coefficients in .
Given a conjugacy class in , we may set . Then is central in , as given , we have:
Hence applying to induces a linear map
. By Schur’s lemma this must be
multiplication by some
complex number . Then is an algebraic
integer
as it solves the same monic polynomial as .
Also any has finite order so the map it induces on
must have eigenvalues which are roots of unity
and hence algebraic
integers. Hence the sum of the eigenvalues, , must
also be an algebraic integer.
Now is irreducible so,
Title | proof that dimension of complex irreducible representation divides order of group |
---|---|
Canonical name | ProofThatDimensionOfComplexIrreducibleRepresentationDividesOrderOfGroup |
Date of creation | 2013-03-22 17:09:04 |
Last modified on | 2013-03-22 17:09:04 |
Owner | whm22 (2009) |
Last modified by | whm22 (2009) |
Numerical id | 8 |
Author | whm22 (2009) |
Entry type | Proof |
Classification | msc 20C99 |