1Groups

IB Groups, Rings and Modules



1.7 Sylow theorems
We finally get to the big theorem of this part of the course.
Theorem (Sylow theorems). Let
G
be a finite group of order
p
a
· m
, with
p
a
prime and p - m. Then
(i) The set of Sylow p-subgroups of G, given by
Syl
p
(G) = {P G : |P | = p
a
},
is non-empty. In other words, G has a subgroup of order p
a
.
(ii) All elements of Syl
p
(G) are conjugate in G.
(iii)
The number of Sylow
p
-subgroups
n
p
=
|Syl
p
(
G
)
|
satisfies
n
p
1 (
mod p
)
and n
p
| |G| (in fact n
p
| m, since p is not a factor of n
p
).
These are sometimes known as Sylow’s first/second/third theorem respec-
tively.
We will not prove this just yet. We first look at how we can apply this
theorem. We can use it without knowing how to prove it.
Lemma. If n
p
= 1, then the Sylow p-subgroup is normal in G.
Proof.
Let
P
be the unique Sylow
p
-subgroup, and let
g G
, and consider
g
1
P g
. Since this is isomorphic to
P
, we must have
|g
1
P g|
=
p
a
, i.e. it is also
a Sylow
p
-subgroup. Since there is only one, we must have
P
=
g
1
P g
. So
P
is
normal.
Corollary. Let
G
be a non-abelian simple group. Then
|G| |
n
p
!
2
for every prime
p such that p | |G|.
Proof.
The group
G
acts on
Syl
p
(
G
) by conjugation. So it gives a permutation
representation
φ
:
G Sym
(
Syl
p
(
G
))
=
S
n
p
. We know
ker φ C G
. But
G
is
simple. So ker(φ) = {e} or G. We want to show it is not the whole of G.
If we had
G
=
ker
(
φ
), then
g
1
P g
=
P
for all
g G
. Hence
P
is a normal
subgroup. As
G
is simple, either
P
=
{e}
, or
P
=
G
. We know
P
cannot be
trivial since
p | |G|
. But if
G
=
P
, then
G
is a
p
-group, has a non-trivial center,
and hence G is not non-abelian simple. So we must have ker(φ) = {e}.
Then by the first isomorphism theorem, we know
G
=
im φ S
n
p
. We have
proved the theorem without the divide-by-two part. To prove the whole result,
we need to show that in fact
im
(
φ
)
A
n
p
. Consider the following composition
of homomorphisms:
G S
n
p
1}.
φ sgn
If this is surjective, then
ker
(
sgn φ
)
C G
has index 2 (since the index is the size
of the image), and is not the whole of
G
. This means
G
is not simple (the case
where |G| = C
2
is ruled out since it is abelian).
So the kernel must be the whole
G
, and
sgn φ
is the trivial map. In other
words, sgn(φ(g)) = +1. So φ(g) A
n
p
. So in fact we have
G
=
im(φ) A
n
p
.
So we get |G| |
n
p
!
2
.
Example. Suppose
|G|
= 1000. Then
|G|
is not simple. To show this, we need
to factorize 1000. We have
|G|
= 2
3
·
5
3
. We pick our favorite prime to be
p
= 5.
We know
n
5
=
1 (
mod
5), and
n
5
|
2
3
= 8. The only number that satisfies this
is n
5
= 1. So the Sylow 5-subgroup is normal, and hence G is not normal.
Example. Let
|G|
= 132 = 2
2
·
3
·
11. We want to show this is not simple. So
for a contradiction suppose it is.
We start by looking at
p
= 11. We know
n
11
1 (
mod
11). Also
n
11
|
12.
As G is simple, we must have n
11
= 12.
Now look at
p
= 3. We have
n
3
= 1 (
mod
3) and
n
3
|
44. The possible
values of n
3
are 4 and 22.
If
n
3
= 4, then the corollary says
|G| |
4!
2
= 12, which is of course nonsense.
So n
3
= 22.
At this point, we count how many elements of each order there are. This is
particularly useful if
p | |G|
but
p
2
- |G|
, i.e. the Sylow
p
-subgroups have order
p
and hence are cyclic.
As all Sylow 11-subgroups are disjoint, apart from
{e}
, we know there are
12
·
(11
1) = 120 elements of order 11. We do the same thing with the Sylow
3-subgroups. We need 22
·
(3
1) = 44 elements of order 3. But this is more
elements than the group has. This can’t happen. So G must be simple.
We now get to prove our big theorem. This involves some non-trivial amount
of trickery.
Proof of Sylow’s theorem. Let G be a finite group with |G| = p
a
m, and p - m.
(i)
We need to show that
Syl
p
(
G
)
6
=
, i.e. we need to find some subgroup of
order p
a
. As always, we find something clever for G to act on. We let
= {X subset of G : |X| = p
a
}.
We let G act on by
g {g
1
, g
2
, ··· , g
p
a
} = {gg
1
, gg
2
, ··· , gg
p
a
}.
Let Σ be an orbit.
We first note that if
{g
1
, ··· , g
p
a
}
Σ, then by the definition of an orbit,
for every g G,
gg
1
1
{g
1
, ··· , g
p
a
} = {g, gg
1
1
g
2
, ··· , gg
1
1
g
p
a
} Σ.
The important thing is that this set contains
g
. So for each
g
, Σ contains
a set X which contains g. Since each set X has size p
a
, we must have
|Σ|
|G|
p
a
= m.
Suppose
|
Σ
|
=
m
. Then the orbit-stabilizer theorem says the stabilizer
H
of
any
{g
1
, ··· , g
p
a
}
Σ has index
m
, hence
|H|
=
p
a
, and thus
H Syl
p
(
G
).
So we need to show that not every orbit Σ can have size
> m
. Again, by
the orbit-stabilizer, the size of any orbit divides the order of the group,
|G|
=
p
a
m
. So if
|
Σ
| > m
, then
p | |
Σ
|
. Suppose we can show that
p - |
|
.
Then not every orbit Σ can have size
> m
, since is the disjoint union of
all the orbits, and thus we are done.
So we have to show p - ||. This is just some basic counting. We have
|| =
|G|
p
a
=
p
a
m
p
a
=
p
a
1
Y
j=0
=
p
a
m j
p
a
j
.
Now note that the largest power of
p
dividing
p
a
m j
is the largest power
of
p
dividing
j
. Similarly, the largest power of
p
dividing
p
a
j
is also the
largest power of
p
dividing
j
. So we have the same power of
p
on top and
bottom for each item in the product, and they cancel. So the result is not
divisible by p.
This proof is not straightforward. We first needed the clever idea of letting
G
act on Ω. But then if we are given this set, the obvious thing to do
would be to find something in that is also a group. This is not what we
do. Instead, we find an orbit whose stabilizer is a Sylow p-subgroup.
(ii)
We instead prove something stronger: if
Q G
is a
p
-subgroup (i.e.
|Q|
=
p
b
, for
b
not necessarily
a
), and
P G
is a Sylow
p
-subgroup, then
there is a
g G
such that
g
1
Qg P
. Applying this to the case where
Q
is another Sylow
p
-subgroup says there is a
g
such that
g
1
Qg P
, but
since g
1
Qg has the same size as P , they must be equal.
We let Q act on the set of cosets of G/P via
q gP = qgP.
We know the orbits of this action have size dividing
|Q|
, so is either 1 or
divisible by
p
. But they can’t all be divisible by
p
, since
|G/P |
is coprime
to
p
. So at least one of them have size 1, say
{gP }
. In other words, for
every
q Q
, we have
qgP
=
gP
. This means
g
1
qg P
. This holds for
every element q Q. So we have found a g such that g
1
Qg P .
(iii)
Finally, we need to show that
n
p
=
1 (
mod p
) and
n
p
| |G|
, where
n
p
=
|Syl
P
(G)|.
The second part is easier by Sylow’s second theorem, the action of
G
on
Syl
p
(
G
) by conjugation has one orbit. By the orbit-stabilizer theorem,
the size of the orbit, which is
|Syl
p
(
G
)
|
=
n
p
, divides
|G|
. This proves the
second part.
For the first part, let
P Syl
P
(
G
). Consider the action by conjugation
of
P
on
Syl
p
(
G
). Again by the orbit-stabilizer theorem, the orbits each
have size 1 or size divisible by
p
. But we know there is one orbit of size 1,
namely
{P }
itself. To show
n
p
=
|Syl
P
(
G
)
|
=
1 (
mod p
), it is enough to
show there are no other orbits of size 1.
Suppose {Q} is an orbit of size 1. This means for every p P , we get
p
1
Qp = Q.
In other words,
P N
G
(
Q
). Now
N
G
(
Q
) is itself a group, and we
can look at its Sylow
p
-subgroups. We know
Q N
G
(
Q
)
G
. So
p
a
| |N
G
(
Q
)
| | p
a
m
. So
p
a
is the biggest power of
p
that divides
|N
G
(
Q
)
|
.
So Q is a Sylow p-subgroup of N
G
(Q).
Now we know
P N
G
(
Q
) is also a Sylow
p
-subgroup of
N
G
(
Q
). By Sylow’s
second theorem, they must be conjugate in
N
G
(
Q
). But conjugating
anything in
Q
by something in
N
G
(
Q
) does nothing, by definition of
N
G
(
Q
). So we must have
P
=
Q
. So the only orbit of size 1 is
{P }
itself.
So done.
This is all the theories of groups we’ve got. In the remaining time, we will
look at some interesting examples of groups.
Example. Let
G
=
GL
n
(
Z/p
), i.e. the set of invertible
n × n
matrices with
entries in
Z/p
, the integers modulo
p
. Here
p
is obviously a prime. When we do
rings later, we will study this properly.
First of all, we would like to know the size of this group. A matrix
A
GL
n
(
Z/p
) is the same as
n
linearly independent vectors in the vector space
(
Z/p
)
n
. We can just work out how many there are. This is not too difficult,
when you know how.
We can pick the first vector, which can be anything except zero. So there
are
p
n
1 ways of choosing the first vector. Next, we need to pick the second
vector. This can be anything that is not in the span of the first vector, and this
rules out
p
possibilities. So there are
p
n
p
ways of choosing the second vector.
Continuing this chain of thought, we have
|GL
n
(Z/p)| = (p
n
1)(p
n
p)(p
n
p
2
) ···(p
n
p
n1
).
What is a Sylow
p
-subgroup of
GL
n
(
Z/p
)? We first work out what the order of
this is. We can factorize that as
|GL
n
(Z/p)| = (1 · p · p
2
· ··· · p
n1
)((p
n
1)(p
n1
1) ···(p 1)).
So the largest power of
p
that divides
|GL
n
(
Z/p
)
|
is
p
(
n
2
)
. Let’s find a subgroup
of size p
(
n
2
)
. We consider matrices of the form
U =
1 ···
0 1 ···
0 0 1 ···
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 0 ··· 1
GL
n
(Z/p)
.
Then we know
|U|
=
p
(
n
2
)
as each
can be chosen to be anything in
Z/p
, and
there are
n
2
s.
Is the Sylow
p
-subgroup unique? No. We can take the lower triangular
matrices and get another Sylow p-subgroup.
Example. Let’s be less ambitious and consider GL
2
(Z/p). So
|G| = p(p
2
1)(p 1) = p(p 1)
2
(p + 1).
Let
`
be another prime number such that
` | p
1. Suppose the largest power of
` that divides |G| is `
2
. Can we (explicitly) find a Sylow `-subgroup?
First, we want to find an element of order
`
. How is
p
1 related to
p
(apart
from the obvious way)? We know that
(Z/p)
×
= {x Z/p : (y) xy 1 (mod p)}
=
C
p1
.
So as
` | p
1, there is a subgroup
C
`
C
p1
=
(
Z/p
)
×
. Then we immediately
know where to find a subgroup of order `
2
: we have
C
`
× C
`
(Z/p)
× (Z/p)
×
GL
2
(Z/p),
where the final inclusion is the diagonal matrices, identifying
(a, b)
a 0
0 b
.
So this is the Sylow `-subgroup.