5 Characters of Arbitrary Finite Groups 415
is a diagonal matrix. Moreover, since
T−^1 SkT=diag[ω 1 k,...,ωkn],
ω 1 ,...,ωnare allm-th roots of unity. Thus
χ(s)=ω 1 +···+ωn
is a sum ofnm-th roots of unity. Since the inverse of a root of unityωis its complex
conjugateω ̄, it follows that
χ(s−^1 )=ω− 11 +···+ω−n^1 =χ(s).
Now letGbe a group of finite orderg,andletρ:s→A(s)andσ:s→B(s)be
irreducible matrix representations ofGof degreesnandmrespectively. For anyn×m
matrixC, form the matrix
T=
∑
s∈G
A(s)CB(s−^1 ).
Sincetsruns through the elements ofGat the same time ass,
A(t)T=TB(t) for everyt∈G.
Therefore, by Schur’s lemma,T=Oifρis not equivalent toσandT=λIifρ=σ.
In particular, takeCto be any one of themnmatrices which have a single entry 1 and
all other entries 0. Then ifA=(αij),B=(βkl),weget
∑
s∈G
αij(s)βkl(s−^1 )=
{
0ifρ,σare inequivalent,
λjkδil ifρ=σ,
whereδil = 1 or 0 according asi =lori = l(‘Kronecker delta’). Since for
(αij)=(βij)the left side is unchanged wheniis interchanged withkandjwith
l,wemusthaveλjk=λδjk. To determineλseti=l,j=kand sum with respect
tok. Since the matrices representingsands−^1 are inverse, we getg 1 =nλ. Thus
∑
s∈G
αij(s)αkl(s−^1 )=
{
g/n ifj=kandi=l,
0otherwise.
Ifμ,vrun through an index set for the inequivalent irreducible representations of
G, then the relations which have been obtained can be rewritten in the form
∑
s∈G
αij(μ)(s)α(klv)(s−^1 )=
{
g/nμ ifμ=v,j=k,i=l,
0otherwise.
(5)
Theorthogonality relations(5) for the irreducible matrix elements have several corol-
laries:
(i)The functionsαij(μ):G→Care linearly independent.