5Group actions

IA Groups



5.3 Important actions
Given any group
G
, there are a few important actions we can define. In particular,
we will define the conjugation action, which is a very important concept on
its own. In fact, the whole of the next chapter will be devoted to studying
conjugation in the symmetric groups.
First, we will study some less important examples of actions.
Lemma
(Left regular action)
.
Any group
G
acts on itself by left multiplication.
This action is faithful and transitive.
Proof. We have
1. (g G)(x G) g(x) = g · x G by definition of a group.
2. (x G) e · x = x by definition of a group.
3. g(hx) = (gh)x by associativity.
So it is an action.
To show that it is faithful, we want to know that [(
x X
)
gx
=
x
]
g
=
e
.
This follows directly from the uniqueness of identity.
To show that it is transitive,
x, y G
, then (
yx
1
)(
x
) =
y
. So any
x
can
be sent to any y.
Theorem
(Cayley’s theorem)
.
Every group is isomorphic to some subgroup of
some symmetric group.
Proof.
Take the left regular action of
G
on itself. This gives a group homo-
morphism
ϕ
:
G Sym G
with
ker ϕ
=
{e}
as the action is faithful. By the
isomorphism theorem, G
=
im ϕ Sym G.
Lemma
(Left coset action)
.
Let
H G
. Then
G
acts on the left cosets of
H
by left multiplication transitively.
Proof. First show that it is an action:
0. g(aH) = (ga)H is a coset of H.
1. e(aH) = (ea)H = aH.
2. g
1
(g
2
(aH)) = g
1
((g
2
a)H) = (g
1
g
2
a)H = (g
1
g
2
)(aH).
To show that it is transitive, given
aH, bH
, we know that (
ba
1
)(
aH
) =
bH
.
So any aH can be mapped to bH.
In the boring case where
H
=
{e}
, then this is just the left regular action
since G/{e}
=
G.
Definition
(Conjugation of element)
.
The conjugation of
a G
by
b G
is
given by
bab
1
G
. Given any
a, c
, if there exists some
b
such that
c
=
bab
1
,
then we say a and c are conjugate.
What is conjugation? This
bab
1
form looks familiar from Vectors and
Matrices. It is the formula used for changing basis. If
b
is the change-of-basis
matrix and
a
is a matrix, then the matrix in the new basis is given by
bab
1
. In
this case, bab
1
is the same matrix viewed from a different basis.
In general, two conjugate elements are “the same” in some sense. For example,
we will later show that in
S
n
, two elements are conjugate if and only if they
have the same cycle type. Conjugate elements in general have many properties
in common, such as their order.
Lemma
(Conjugation action)
.
Any group
G
acts on itself by conjugation (i.e.
g(x) = gxg
1
).
Proof. To show that this is an action, we have
0. g(x) = gxg
1
G for all g, x G.
1. e(x) = exe
1
= x
2. g(h(x)) = g(hxh
1
) = ghxh
1
g
1
= (gh)x(gh)
1
= (gh)(x)
Definition
(Conjugacy classes and centralizers)
.
The conjugacy classes are the
orbits of the conjugacy action.
ccl(a) = {b G : (g g) gag
1
= b}.
The centralizers are the stabilizers of this action, i.e. elements that commute
with a.
C
G
(a) = {g G : gag
1
= a} = {g G : ga = ag}.
The centralizer is defined as the elements that commute with a particular
element a. For the whole group G, we can define the center.
Definition
(Center of group)
.
The center of
G
is the elements that commute
with all other elements.
Z(G) = {g G : (a) gag
1
= a} = {g G : (a) ga = ag}.
It is sometimes written as C(G) instead of Z(G).
In many ways, conjugation is related to normal subgroups.
Lemma. Let K C G. Then G acts by conjugation on K.
Proof.
We only have to prove closure as the other properties follow from the
conjugation action. However, by definition of a normal subgroup, for every
g G, k K, we have gkg
1
K. So it is closed.
Proposition.
Normal subgroups are exactly those subgroups which are unions
of conjugacy classes.
Proof.
Let
K C G
. If
k K
, then by definition for every
g G
, we get
gkg
1
K
. So
ccl
(
k
)
K
. So
K
is the union of the conjugacy classes of all its
elements.
Conversely, if
K
is a union of conjugacy classes and a subgroup of
G
, then
for all k K, g G, we have gkg
1
K. So K is normal.
Lemma.
Let
X
be the set of subgroups of
G
. Then
G
acts by conjugation on
X.
Proof. To show that it is an action, we have
0.
If
H G
, then we have to show that
gHg
1
is also a subgroup. We
know that
e H
and thus
geg
1
=
e gHg
1
, so
gHg
1
is non-empty.
For any two elements
gag
1
and
gbg
1
gHg
1
, (
gag
1
)(
gbg
1
)
1
=
g(ab
1
)g
1
gHg
1
. So gHg
1
is a subgroup.
1. eHe
1
= H.
2. g
1
(g
2
Hg
1
2
)g
1
1
= (g
1
g
2
)H(g
1
g
2
)
1
.
Under this action, normal subgroups have singleton orbits.
Definition
(Normalizer of subgroup)
.
The normalizer of a subgroup is the
stabilizer of the (group) conjugation action.
N
G
(H) = {g G : gHg
1
= H}.
We clearly have
H N
G
(
H
). It is easy to show that
N
G
(
H
) is the largest
subgroup of G in which H is a normal subgroup, hence the name.
There is a connection between actions in general and conjugation of subgroups.
Lemma.
Stabilizers of the elements in the same orbit are conjugate, i.e. let
G
act on X and let g G, x X. Then stab(g(x)) = g stab(x)g
1
.