14Burnside's theorem
II Representation Theory
14 Burnside’s theorem
Finally, we get to Burnside’s theorem.
Theorem
(Burside’s
p
a
q
b
theorem)
.
Let
p, q
be primes, and let
|G|
=
p
a
q
b
,
where a, b ∈ Z
≥0
, with a + b ≥ 2. Then G is not simple.
Note that if a + b = 1 or 0, then the group is trivially simple.
In fact even more is true, namely that
G
is soluble, but this follows easily
from above by induction.
This result is the best possible in the sense that
|A
5
|
= 60 = 2
2
·
3
·
5, and
A
5
is simple. So we cannot allow for three different primes factors (in fact, there
are exactly 8 simple groups with three prime factors). Also, if
a
= 0 or
b
= 0,
then
G
is a
p
-group, which have non-trivial center. So these cases trivially work.
Later, in 1963, Feit and Thompson proved that every group of odd order is
soluble. The proof was 255 pages long. We will not prove this.
In 1972, H. Bender found the first proof of Burnside’s theorem without the
use of representation theory, but the proof is much more complicated.
This theorem follows from two lemmas, and one of them involves some Galois
theory, and is hence non-examinable.
Lemma. Suppose
α =
1
m
m
X
j=1
λ
j
,
is an algebraic integer, where
λ
n
j
= 1 for all
j
and some
n
. Then either
α
= 0 or
|α| = 1.
Proof (non-examinable).
Observe
α ∈ F
=
Q
(
ε
), where
ε
=
e
2πi/n
(since
λ
j
∈ F
for all j). We let G = Gal(F/Q). Then
{β ∈ F : σ(β) = β for all σ ∈ G} = Q.
We define the “norm”
N(α) =
Y
σ∈G
σ(α).
Then N(α) is fixed by every element σ ∈ G. So N(α) is rational.
Now
N
(
α
) is an algebraic integer, since Galois conjugates
σ
(
α
) of algebraic
integers are algebraic integers. So in fact
N
(
α
) is an integer. But for
α ∈ G
, we
know
|σ(α)| =
1
m
X
σ(λ
j
)
≤ 1.
So if α 6= 0, then N(α) = ±1. So |α| = 1.
Lemma.
Suppose
χ
is an irreducible character of
G
, and
C
is a conjugacy class
in G such that χ(1) and |C| are coprime. Then for g ∈ C, we have
|χ(g)| = χ(1) or 0.
Proof. Of course, we want to consider the quantity
α =
χ(g)
χ(1)
.
Since
χ
(
g
) is the sum of
deg χ
=
χ
(1) many roots of unity, it suffices to show
that α is an algebraic integer.
By B´ezout’s theorem, there exists a, b ∈ Z such that
aχ(1) + b|C| = 1.
So we can write
α =
χ(g)
χ(1)
= aχ(g) + b
χ(g)
χ(1)
|C|.
Since χ(g) and
χ(g)
χ(1)
|C| are both algebraic integers, we know α is.
Proposition.
If in a finite group, the number of elements in a conjugacy class
C is of (non-trivial) prime power order, then G is not non-abelian simple.
Proof.
Suppose
G
is a non-abelian simple group, and let 1
6
=
g ∈ G
be living in
the conjugacy class
C
of order
p
r
. If
χ 6
= 1
G
is a non-trivial irreducible character
of
G
, then either
χ
(1) and
|C|
=
p
r
are not coprime, in which case
p | χ
(1), or
they are coprime, in which case |χ(g)| = χ(1) or χ(g) = 0.
However, it cannot be that
|χ
(
g
)
|
=
χ
(1). If so, then we must have
ρ
(
g
) =
λI
for some λ. So it commutes with everything, i.e. for all h, we have
ρ(gh) = ρ(g)ρ(h) = ρ(h)ρ(g) = ρ(hg).
Moreover, since
G
is simple,
ρ
must be faithful. So we must have
gh
=
hg
for all
h
. So
Z
(
G
) is non-trivial. This is a contradiction. So either
p | χ
(1) or
χ
(
g
) = 0.
By column orthogonality applied to C and 1, we get
0 = 1 +
X
16=χ irreducible, p|χ(1)
χ(1)χ(g),
where we have deleted the 0 terms. So we get
−
1
p
=
X
χ6=1
χ(1)
p
χ(g).
But this is both an algebraic integer and a rational number, but not integer.
This is a contradiction.
We can now prove Burnside’s theorem.
Theorem
(Burside’s
p
a
q
b
theorem)
.
Let
p, q
be primes, and let
|G|
=
p
a
q
b
,
where a, b ∈ Z
≥0
, with a + b ≥ 2. Then G is not simple.
Proof.
Let
|G|
=
p
a
q
b
. If
a
= 0 or
b
= 0, then the result is trivial. Suppose
a, b >
0. We let
Q ∈ Syl
q
(
G
). Since
Q
is a
p
-group, we know
Z
(
Q
) is non-trivial.
Hence there is some 1
6
=
g ∈ Z
(
Q
). By definition of center, we know
Q ≤ C
G
(
g
).
Also, C
G
(g) is not the whole of G, since the center of G is trivial. So
|C
G
(g)| = |G : C
G
(g)| = p
r
for some 0 < r ≤ a. So done.
This is all we’ve got to say about finite groups.