5Group actions

IA Groups



5.2 Orbits and Stabilizers
Definition
(Orbit of action)
.
Given an action
G
on
X
, the orbit of an element
x X is
orb(x) = G(x) = {y X : (g G) g(x) = y}.
Intuitively, it is the elements that x can possibly get mapped to.
Definition (Stabilizer of action). The stabilizer of x is
stab(x) = G
x
= {g G : g(x) = x} G.
Intuitively, it is the elements in G that do not change x.
Lemma. stab(x) is a subgroup of G.
Proof.
We know that
e
(
x
) =
x
by definition. So
stab
(
x
) is non-empty. Suppose
g, h stab
(
x
), then
gh
1
(
x
) =
g
(
h
1
(
x
)) =
g
(
x
) =
x
. So
gh
1
stab
(
X
). So
stab(x) is a subgroup.
Example.
(i)
Consider
D
8
acting on the corners of the square
X
=
{
1
,
2
,
3
,
4
}
. Then
orb
(1) =
X
since 1 can go anywhere by rotations.
stab
(1) =
{e,
reflection
in the line through 1}
(ii)
Consider the rotations of a cube acting on the three axes
x, y, z
. Then
orb
(
x
) is everything, and
stab
(
x
) contains
e
, 180
rotations and rotations
about the x axis.
Definition
(Transitive action)
.
An action
G
on
X
is transitive if (
x
)
orb
(
x
) =
X, i.e. you can reach any element from any element.
Lemma. The orbits of an action partition X.
Proof. Firstly, (x)(x orb(x)) as e(x) = x. So every x is in some orbit.
Then suppose
z orb
(
x
) and
z orb
(
y
), we have to show that
orb
(
x
) =
orb
(
y
). We know that
z
=
g
1
(
x
) and
z
=
g
2
(
y
) for some
g
1
, g
2
. Then
g
1
(
x
) =
g
2
(y) and y = g
1
2
g
1
(x).
For any
w
=
g
3
(
y
)
orb
(
y
), we have
w
=
g
3
g
1
2
g
1
(
x
). So
w orb
(
x
). Thus
orb(y) orb(x) and similarly orb(x) orb(y). Therefore orb(x) = orb(y).
Suppose a group
G
acts on
X
. We fix an
x X
. Then by definition of
the orbit, given any
g G
, we have
g
(
x
)
orb
(
x
). So each
g G
gives us
a member of
orb
(
x
). Conversely, every object in
orb
(
x
) arises this way, by
definition of
orb
(
x
). However, different elements in
G
can give us the same orbit.
In particular, if
g stab
(
x
), then
hg
and
h
give us the same object in
orb
(
x
),
since
hg
(
x
) =
h
(
g
(
x
)) =
h
(
x
). So we have a correspondence between things in
orb(x) and members of G, “up to stab(x)”.
Theorem
(Orbit-stabilizer theorem)
.
Let the group
G
act on
X
. Then there
is a bijection between
orb
(
x
) and cosets of
stab
(
x
) in
G
. In particular, if
G
is
finite, then
| orb(x)|| stab(x)| = |G|.
Proof.
We biject the cosets of
stab
(
x
) with elements in the orbit of
x
. Recall
that G : stab(x) is the set of cosets of stab(x). We can define
θ : (G : stab(x)) orb(x)
g stab(x) 7→ g(x).
This is well-defined — if
g stab
(
x
) =
h stab
(
x
), then
h
=
gk
for some
k stab
(
x
).
So h(x) = g(k(x)) = g(x).
This map is surjective since for any
y orb
(
x
), there is some
g G
such
that
g
(
x
) =
y
, by definition. Then
θ
(
g stab
(
x
)) =
y
. It is injective since if
g(x) = h(x), then h
1
g(x) = x. So h
1
g stab(x). So g stab(x) = h stab(x).
Hence the number of cosets is
| orb
(
x
)
|
. Then the result follows from La-
grange’s theorem.
An important application of the orbit-stabilizer theorem is determining group
sizes. To find the order of the symmetry group of, say, a pyramid, we find
something for it to act on, pick a favorite element, and find the orbit and
stabilizer sizes.
Example.
(i)
Suppose we want to know how big
D
2n
is.
D
2n
acts on the vertices
{
1
,
2
,
3
, · · · , n}
transitively. So
| orb
(1)
|
=
n
. Also,
stab
(1) =
{e,
reflection
in the line through 1}. So |D
2n
| = | orb(1)|| stab(1)| = 2n.
Note that if the action is transitive, then all orbits have size
|X|
and thus
all stabilizers have the same size.
(ii)
Let
h
(1 2)
i
act on
{
1
,
2
,
3
}
. Then
orb
(1) =
{
1
,
2
}
and
stab
(1) =
{e}
.
orb(3) = {3} and stab(3) = h(1 2)i.
(iii)
Consider
S
4
acting on
{
1
,
2
,
3
,
4
}
. We know that
orb
(1) =
X
and
|S
4
|
= 24.
So
| stab
(1)
|
=
24
4
= 6. That makes it easier to find
stab
(1). Clearly
S
{2,3,4}
=
S
3
fix 1. So
S
{2,3,4}
stab
(1). However,
|S
3
|
= 6 =
| stab
(1)
|
,
so this is all of the stabilizer.