4 Representations of Arbitrary Finite Groups 413is also a positive definite inner product onVand that it is invariant underG,i.e.
(A(s)u,A(s)v)G=(u,v)G for everys∈G.IfU is a subspace ofV which is invariant underG,andifU⊥is the subspace
consisting of all vectorsv ∈ Vsuch that(u,v)G =0foreveryu∈U,thenU⊥
is also invariant underGandVis the direct sum ofUandU⊥. Thusρis the sum of
its restrictions toUandU⊥.
The basic result for irreducible representations isSchur’s lemma, which comes in
two parts:
Proposition 10(i)Let s→A 1 (s)and s→A 2 (s)be irreducible representations of
a group G by linear transformations of the vector spaces V 1 and V 2. If there exists a
linear transformation T= 0 of V 1 into V 2 such that
TA 1 (s)=A 2 (s)T for every s∈G,then the spaces V 1 and V 2 have the same dimension and T is invertible, so that the
representations are equivalent.
(ii)Let s→A(s)be an irreducible representation of a group G by linear transforma-
tions of a vector space V. A linear transformation T of V has the property
TA(s)=A(s)T for every s∈G (4)if and only if T=λIforsomeλ∈C.
Proof (i) The image ofV 1 underTis a subspace ofV 2 which is invariant under the
second representation. SinceT=0 and the representation is irreducible, it must be
the whole space:TV 1 =V 2. On the other hand, those vectors inV 1 whose image
underTis 0 form a subspace ofV 1 which is invariant under the first representation.
SinceT=0 and the representation is irreducible, it must contain only the zero vector.
Hence distinct vectors ofV 1 have distinct images inV 2 underT. ThusTis a one-to-one
mapping ofV 1 ontoV 2.
(ii) By the fundamental theorem of algebra, there exists a complex numberλsuch
that det(λI−T)=0. HenceT−λIis not invertible. But ifThas the property (4),
so doesT−λI. ThereforeT−λI =0, by (i) withA 1 = A 2. It is obvious that,
conversely, (4) holds ifT=λI.
Corollary 11Every irreducible representation of an abelian group is of degree 1.
Proof By Proposition 10 (ii) all elements of the group must be represented by scalar
multiples of the identity transformation. But such a representation is irreducible only
if its degree is 1.