Number Theory: An Introduction to Mathematics

422 X A Character Study

Consider now the induced representationψ ̃iofG.SinceHis a normal subgroup,
it follows from (12) that

ψ ̃i(s)=ψ ̃iv(s)=0ifs∈G\H,

ψ ̃i(s)=ψ ̃iv(s)=ψi(s)+ψiv(s) ifs∈H.

Henceψ ̃i=ψ ̃ivand

∑

s∈G

ψ ̃i(s)ψ ̃i(s−^1 )=

∑

s∈H

{ψi(s)+ψiv(s)}{ψi(s−^1 )+ψiv(s−^1 )}

=

∑

s∈H

ψi(s)ψi(s−^1 )+

∑

s∈H

ψiv(s)ψiv(s−^1 )

+

∑

s∈H

{ψi(s)ψiv(s−^1 )+ψi(s−^1 )ψvi(s)}.

Consequently, by the orthogonality relations forH,

∑

s∈G

ψ ̃i(s)ψ ̃i(s−^1 )= 2 h+ 2

∑

s∈H

ψi(s)ψiv(s−^1 ).

Ifψiandψivare inequivalent, the second term on the right vanishes and we obtain

∑

s∈G

ψ ̃i(s)ψ ̃i(s−^1 )=g.

Thus the induced representationψ ̃iofGis irreducible, its degree being twice that
ofψi.
On the other hand, ifψiandψivare equivalent, then
∑

s∈G

ψ ̃i(s)ψ ̃i(s−^1 )= 2 g.

Ifψ ̃i=

∑

jmjχjis the decomposition ofψ ̃iinto irreducible charactersχjofG,itfol-
lows from (8) that

∑

jm

2

j=2. This implies that

ψ ̃idecomposes into two inequivalent

irreducible characters ofG,sayψ ̃i=χk+χl. We will show that in factχl=χkλ.

Ifχk(s)=0foralls∈/H,then

∑

s∈H

χk(s)χk(s−^1 )=

∑

s∈G

χk(s)χk(s−^1 )=g= 2 h

and hence, by the same argument as that just used, the restriction ofχktoHdecom-
poses into two inequivalent irreducible characters ofH. Since the restriction ofψ ̃ito
His 2ψi, this is a contradiction. We conclude thatχk(s)=0forsomes∈/H,i.e.
χkλ=χk.Sinceχkoccurs once in the decomposition ofψ ̃i,andψ ̃i(s)=0ifs∈/H,