1Groups
IB Groups, Rings and Modules
1.5 Finite p-groups
Note that when studying the orders of groups and subgroups, we always talk
about divisibility, since that is what Lagrange’s theorem tells us about. We
never talk about things like the sum of the orders of two subgroups. When it
comes to divisibility, the simplest case would be when the order is a prime, and
we have done that already. The next best thing we can hope for is that the order
is a power of a prime.
Definition (
p
-group). A finite group
G
is a
p
-group if
|G|
=
p
n
for some prime
number p and n ≥ 1.
Theorem. If
G
is a finite
p
-group, then
Z
(
G
) =
{x ∈ G
:
xg
=
gx for all g ∈ G}
is non-trivial.
This immediately tells us that for n ≥ 2, a p group is never simple.
Proof.
Let
G
act on itself by conjugation. The orbits of this action (i.e. the
conjugacy classes) have order dividing
|G|
=
p
n
. So it is either a singleton, or
its size is divisible by p.
Since the conjugacy classes partition
G
, we know the total size of the conjugacy
classes is |G|. In particular,
|G| = number of conjugacy class of size 1
+
X
order of all other conjugacy classes.
We know the second term is divisible by
p
. Also
|G|
=
p
n
is divisible by
p
. Hence
the number of conjugacy classes of size 1 is divisible by
p
. We know
{e}
is a
conjugacy class of size 1. So there must be at least
p
conjugacy classes of size 1.
Since the smallest prime number is 2, there is a conjugacy class {x} 6= {e}.
But if
{x}
is a conjugacy class on its own, then by definition
g
−1
xg
=
x
for
all g ∈ G, i.e. xg = gx for all g ∈ G. So x ∈ Z(G). So Z(G) is non-trivial.
The theorem allows us to prove interesting things about
p
-groups by induction
— we can quotient
G
by
Z
(
G
), and get a smaller
p
-group. One way to do this is
via the below lemma.
Lemma. For any group G, if G/Z(G) is cyclic, then G is abelian.
In other words, if
G/Z
(
G
) is cyclic, then it is in fact trivial, since the center
of an abelian group is the abelian group itself.
Proof.
Let
g ∈ Z
(
G
) be a generator of the cyclic group
G/Z
(
G
). Hence every
coset of
Z
(
G
) is of the form
g
r
Z
(
G
). So every element
x ∈ G
must be of the
form
g
r
z
for
z ∈ Z
(
G
) and
r ∈ Z
. To show
G
is abelian, let
¯x
=
g
¯r
¯z
be another
element, with
¯z ∈ Z
(
G
)
, ¯r ∈ Z
. Note that
z
and
¯z
are in the center, and hence
commute with every element. So we have
x¯x = g
r
zg
¯r
¯z = g
r
g
¯r
z¯z = g
¯r
g
r
¯zz = g
¯r
¯zg
r
z = ¯xx.
So they commute. So G is abelian.
This is a general lemma for groups, but is particularly useful when applied
to p groups.
Corollary. If p is prime and |G| = p
2
, then G is abelian.
Proof.
Since
Z
(
G
)
≤ G
, its order must be 1,
p
or
p
2
. Since it is not trivial, it
can only be
p
or
p
2
. If it has order
p
2
, then it is the whole group and the group
is abelian. Otherwise,
G/Z
(
G
) has order
p
2
/p
=
p
. But then it must be cyclic,
and thus G must be abelian. This is a contradiction. So G is abelian.
Theorem. Let
G
be a group of order
p
a
, where
p
is a prime number. Then it
has a subgroup of order p
b
for any 0 ≤ b ≤ a.
This means there is a subgroup of every conceivable order. This is not true
for general groups. For example,
A
5
has no subgroup of order 30 or else that
would be a normal subgroup.
Proof.
We induct on
a
. If
a
= 1, then
{e}, G
give subgroups of order
p
0
and
p
1
.
So done.
Now suppose
a >
1, and we want to construct a subgroup of order
p
b
. If
b = 0, then this is trivial, namely {e} ≤ G has order 1.
Otherwise, we know
Z
(
G
) is non-trivial. So let
x 6
=
e ∈ Z
(
G
). Since
ord
(
x
)
| |G|
, its order is a power of
p
. If it in fact has order
p
c
, then
x
p
c−1
has
order
p
. So we can suppose, by renaming, that
x
has order
p
. We have thus
generated a subgroup
hxi
of order exactly
p
. Moreover, since
x
is in the center,
hxi
commutes with everything in
G
. So
hxi
is in fact a normal subgroup of
G
.
This is the point of choosing it in the center. Therefore G/hxi has order p
a−1
.
Since this is a strictly smaller group, we can by induction suppose
G/hxi
has
a subgroup of any order. In particular, it has a subgroup
L
of order
p
b−1
. By
the subgroup correspondence, there is some
K ≤ G
such that
L
=
K/hxi
and
H C K. But then K has order p
b
. So done.