8Normal subgroups and lifting
II Representation Theory
8 Normal subgroups and lifting
The idea of this section is that there is some correspondence between the
irreducible characters of
G/N
(with
N C G
) and those of
G
itself. In most
cases,
G/N
is a simpler group, and hence we can use this correspondence to find
irreducible characters of G.
The main result is the following lemma:
Lemma.
Let
N C G
. Let
˜ρ
:
G/N → GL
(
V
) be a representation of
G/N
. Then
the composition
ρ : G G/N GL(V )
natural
˜ρ
is a representation of G, where ρ(g) = ˜ρ(gN). Moreover,
(i) ρ is irreducible if and only if ˜ρ is irreducible.
(ii) The corresponding characters satisfy χ(g) = ˜χ(gN).
(iii) deg χ = deg ˜χ.
(iv) The lifting operation ˜χ 7→ χ is a bijection
{irreducibles of G/N} ←→ {irreducibles of G with N in their kernel}.
We say ˜χ lifts to χ.
Proof.
Since a representation of
G
is just a homomorphism
G → GL
(
V
), and
the composition of homomorphisms is a homomorphisms, it follows immediately
that ρ as defined in the lemma is a representation.
(i) We can compute
hχ, χi =
1
|G|
X
g∈G
χ(g)χ(g)
=
1
|G|
X
gN∈G/N
X
k∈N
χ(gk)χ(gk)
=
1
|G|
X
gN∈G/N
X
k∈N
˜χ(gN)˜χ(gN)
=
1
|G|
X
gN∈G/N
|N|
˜χ(gN)˜χ(gN)
=
1
|G/N|
X
gN∈G/N
˜χ(gN)˜χ(gN)
= h˜χ, ˜χi.
So
hχ, χi
= 1 if and only if
h˜χ, ˜χi
= 1. So
ρ
is irreducible if and only if
˜ρ
is irreducible.
(ii) We can directly compute
χ(g) = tr ρ(g) = tr(˜ρ(gN )) = ˜χ(gN)
for all g ∈ G.
(iii) To see that χ and ˜χ have the same degree, we just notice that
deg χ = χ(1) = ˜χ(N) = deg ˜χ.
Alternatively, to show they have the same dimension, just note that
ρ
and
˜ρ map to the general linear group of the same vector space.
(iv)
To show this is a bijection, suppose
˜χ
is a character of
G/N
and
χ
is its
lift to
G
. We need to show the kernel contains
N
. By definition, we know
˜χ
(
N
) =
χ
(1). Also, if
k ∈ N
, then
χ
(
k
) =
˜χ
(
kN
) =
˜χ
(
N
) =
χ
(1). So
N ≤ ker χ.
Now let
χ
be a character of
G
with
N ≤ ker χ
. Suppose
ρ
:
G → GL
(
V
)
affords χ. Define
˜ρ : G/N → GL(V )
gN 7→ ρ(g)
Of course, we need to check this is well-defined. If
gN
=
g
0
N
, then
g
−1
g
0
∈ N
. So
ρ
(
g
) =
ρ
(
g
0
) since
N ≤ ker ρ
. So this is indeed well-defined.
It is also easy to see that
˜ρ
is a homomorphism, hence a representation of
G/N.
Finally, if
˜χ
is a character of
˜ρ
, then
˜χ
(
gN
) =
χ
(
g
) for all
g ∈ G
by
definition. So
˜χ
lifts to
χ
. It is clear that these two operations are inverses
to each other.
There is one particular normal subgroup lying around that proves to be
useful.
Lemma. Given a group G, the derived subgroup or commutator subgroup
G
0
= h[a, b] : a, b ∈ Gi,
where [
a, b
] =
aba
−1
b
−1
, is the unique minimal normal subgroup of
G
such that
G/G
0
is abelian. So if G/N is abelian, then G
0
≤ N .
Moreover,
G
has precisely
`
=
|G
:
G
0
|
representations of dimension 1, all
with kernel containing G
0
, and are obtained by lifting from G/G
0
.
In particular, by Lagrange’s theorem, ` | G.
Proof. Consider [a, b] = aba
−1
b
−1
∈ G
0
. Then for any h ∈ G, we have
h(aba
−1
b
−1
)h
−1
=
(ha)b(ha)
−1
b
−1
bhb
−1
h
−1
= [ha, b][b, h] ∈ G
0
So in general, let [a
1
, b
1
][a
2
, b
2
] ···[a
n
, b
n
] ∈ G
0
. Then
h[a
1
, b
1
][a
2
, b
2
] ···[a
n
, b
n
]h
−1
= (h[a
1
, b
1
]h
−1
)(h[a
2
, b
2
]h
−1
) ···(h[a
n
, b
n
]h
−1
),
which is in G
0
. So G
0
is a normal subgroup.
Let
N C G
. Let
g, h ∈ G
. Then [
g, h
]
∈ N
if and only if
ghg
−1
h
−1
∈ N
if
and only if ghN = hgN, if and only if (gN)(hN) = (hN)(gN) by normality.
Since
G
0
is generated by all [
g, h
], we know
G
0
≤ N
if and only if
G/N
is
abelian.
Since
G/G
0
, is abelian, we know it has exactly
`
irreducible characters,
˜χ
1
, ··· , ˜χ
`
, all of degree 1. The lifts of these to
G
also have degree 1, and by the
previous lemma, these are precisely the irreducible characters
χ
i
of
G
such that
G
0
≤ ker χ
i
.
But any degree 1 character of
G
is a homomorphism
χ
:
G → C
×
, hence
χ
(
ghg
−1
h
−1
) = 1. So for any 1-dimensional character,
χ
, we must have
G
0
≤
ker χ. So the lifts χ
1
, ··· , χ
`
are all 1-dimensional characters of G.
Example.
Let
G
be
S
n
, with
n ≥
3. We claim that
S
0
n
=
A
n
. Since
sgn
(
a
) =
sgn
(
a
−1
), we know that [
a, b
]
∈ A
n
for any
a, b ∈ S
n
. So
S
0
n
≤ A
n
(alternatively,
since S
n
/A
n
∼
=
C
2
is abelian, we must have S
0
n
≤ A
n
). We now notice that
(1 2 3) = (1 3 2)(1 2)(1 3 2)
−1
(1 2)
−1
∈ S
0
n
.
So by symmetry, all 3-cycles are in
S
0
n
. Since the 3-cycles generate
A
n
, we know
we must have S
0
n
= A
n
.
Since
S
0
n
/A
n
∼
=
C
2
, we get
`
= 2. So
S
n
must have exactly two linear
characters, namely the trivial character and the sign.
Lemma. G
is not simple if and only if
χ
(
g
) =
χ
(1) for some irreducible character
χ 6
= 1
G
and some 1
6
=
g ∈ G
. Any normal subgroup of
G
is the intersection of
the kernels of some of the irreducible characters of G, i.e. N =
T
ker χ
i
.
In other words,
G
is simple if and only if all non-trivial irreducible characters
have trivial kernel.
Proof.
Suppose
χ
(
g
) =
χ
(1) for some non-trivial irreducible character
χ
, and
χ
is afforded by
ρ
. Then
g ∈ ker ρ
. So if
g 6
= 1, then 1
6
=
ker ρ C G
, and
ker ρ 6
=
G
.
So G cannot be simple.
If 1
6
=
N C G
is a non-trivial proper subgroup, take an irreducible character
˜χ
of
G/N
, and suppose
˜χ 6
= 1
G/N
. Lift this to get an irreducible character
χ
,
afforded by the representation
ρ
of
G
. Then
N ≤ ker ρ C G
. So
χ
(
g
) =
χ
(1) for
g ∈ N .
Finally, let 1
6
=
N C G
. We claim that
N
is the intersection of the kernels of
the lifts
χ
1
, ··· , χ
`
of all the irreducibles of
G/N
. Clearly, we have
N ≤
T
i
ker χ
i
.
If
g ∈ G \ N
, then
gN 6
=
N
. So
˜χ
(
gN
)
6
=
˜χ
(
N
) for some irreducible
˜χ
of
G/N
.
Lifting
˜χ
to
χ
, we have
χ
(
g
)
6
=
χ
(1). So
g
is not in the intersection of the
kernels.