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 |