11Frobenius groups
II Representation Theory
11 Frobenius groups
We will now use character theory to prove some major results in finite group
theory. We first do Frobenius’ theorem here. Later, will prove Burnside’s
p
a
q
b
theorem.
This is a theorem with lots and lots of proofs, but all of these proofs involves
some sort of representation theory — it seems like representation is an unavoidable
ingredient of it.
Theorem
(Frobenius’ theorem (1891))
.
Let
G
be a transitive permutation group
on a finite set
X
, with
|X|
=
n
. Assume that each non-identity element of
G
fixes at most one element of
X
. Then the set of fixed point-free elements
(“derangements”)
K = {1} ∪ {g ∈ G : gα 6= α for all α ∈ X}
is a normal subgroup of G with order n.
On the face of it, it is not clear
K
is even a subgroup at all. It turns out
normality isn’t really hard to prove — the hard part is indeed showing it is a
subgroup.
Note that we did not explicitly say
G
is finite. But these conditions imply
G ≤ S
n
, which forces G to be finite.
Proof.
The idea of the proof is to construct a character Θ whose kernel is
K
.
First note that by definition of K, we have
G = K ∪
[
α∈X
G
α
,
where
G
α
is, as usual, the stabilizer of
α
. Also, we know that
G
α
∩ G
β
=
{
1
}
if
α 6= β by assumption, and by definition of K, we have K ∩ G
α
= {1} as well.
Next note that all the
G
α
are conjugate. Indeed, we know
G
is transitive,
and
gG
α
g
−1
=
G
gα
. We set
H
=
G
α
for some arbitrary choice of
α
. Then the
above tells us that
|G| = |K| − |X|(|H| −1).
On the other hand, by the orbit-stabilizer theorem, we know
|G|
=
|X||H|
. So it
follows that we have
|K| = |X| = n.
We first compute what induced characters look like.
Claim. Let ψ be a character of H. Then
Ind
G
H
ψ(g) =
nψ(1) g = 1
ψ(g) g ∈ H \ {1}
0 g ∈ K \{1}
.
Since every element in
G
is either in
K
or conjugate to an element in
H
, this
uniquely specifies what the induced character is.
This is a matter of computation. Since [
G
:
H
] =
n
, the case
g
= 1
immediately follows. Using the definition of the induced character, since any
non-identity in
K
is not conjugate to any element in
H
, we know the induced
character vanishes on K \ {1}.
Finally, suppose
g ∈ H \{
1
}
. Note that if
x ∈ G
, then
xgx
−1
∈ G
xα
. So this
lies in H if and only if x ∈ H. So we can write the induced character as
Ind
G
H
ψ(g) =
1
|H|
X
g∈G
˚
ψ(xgx
−1
) =
1
|H|
X
h∈H
ψ(hgh
−1
) = ψ(g).
Claim. Let ψ be an irreducible character of H, and define
θ = ψ
G
− ψ(1)(1
H
)
G
+ ψ(1)1
G
.
Then θ is a character, and
θ(g) =
(
ψ(h) h ∈ H
ψ(1) k ∈ K
.
Note that we chose the coefficients exactly so that the final property of
θ
holds. This is a matter of computation:
1 h ∈ H \ {1} K \{1}
ψ
G
nψ(1) ψ(h) 0
ψ(1)(1
H
)
G
nψ(1) ψ(1) 0
ψ(1)1
G
ψ(1) ψ(1) ψ(1)
θ
i
ψ(1) ψ(h) ψ(1)
The less obvious part is that
θ
is a character. From the way we wrote it, we
already know it is a virtual character. We then compute the inner product
hθ, θi
G
=
1
|G|
X
g∈G
|θ(g)|
2
=
1
|G|
X
g∈K
|θ(g)|
2
+
X
g∈G\K
|θ(g)|
2
=
1
|G|
n|ψ(1)|
2
+ n
X
h6=1∈H
|ψ(h)|
2
=
1
|G|
n
X
h∈H
|ψ(h)|
2
!
=
1
|G|
(n|H|hψ, ψi
H
)
= 1.
So either θ or −θ is a character. But θ(1) = ψ(1) > 0. So θ is a character.
Finally, we have
Claim.
Let
ψ
1
, ··· , ψ
t
be the irreducible representations of
H
, and
θ
i
be the
corresponding representations of G constructed above. Set
Θ =
t
X
i=1
θ
i
(1)θ
i
.
Then we have
θ(g) =
(
|H| g ∈ K
0 g 6∈ K
.
From this, it follows that the kernel of the representation affording
θ
is
K
, and
in particular K is a normal subgroup of G.
This is again a computation using column orthogonality. For 1
6
=
h ∈ H
, we
have
Θ(h) =
t
X
i=1
ψ
i
(1)ψ
i
(h) = 0,
and for any y ∈ K, we have
Θ(y) =
t
X
i=1
ψ
i
(1)
2
= |H|.
Definition
(Frobenius group and Frobenius complement)
.
A Frobenius group
is a group
G
having a subgroup
H
such that
H ∩gHg
−1
= 1 for all
g 6∈ H
. We
say H is a Frobenius complement of G.
How does this relate to the previous theorem?
Proposition.
The left action of any finite Frobenius group on the cosets of the
Frobenius complement satisfies the hypothesis of Frobenius’ theorem.
Definition
(Frobenius kernel)
.
The Frobenius kernel of a Frobenius group
G
is
the K obtained from Frobenius’ theorem.
Proof.
Let
G
be a Frobenius group, having a complement
H
. Then the action of
G
on the cosets
G/H
is transitive. Furthermore, if 1
6
=
g ∈ G
fixes
xH
and
yH
,
then we have
g ∈ xHx
−1
∩ yHy
−1
. This implies
H ∩
(
y
−1
x
)
H
(
y
−1
x
)
−1
6
= 1.
Hence xH = yH.
Note that J. Thompson (in his 1959 thesis) proved any finite group having
a fixed-point free automorphism of prime order is nilpotent. This implies
K
is
nilpotent, which means K is a direct product of its Sylow subgroups.