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 nonempty. 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 (Orbitstabilizer 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 welldefined — 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 orbitstabilizer 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
⟨
(1 2)
⟩
act on
{
1
,
2
,
3
}
. Then
orb
(1) =
{
1
,
2
}
and
stab
(1) =
{e}
.
orb(3) = {3} and stab(3) = ⟨(1 2)⟩.
(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.