08 - Centralizers, Normalizers, and Stabilizers

stabilizers

If $G$ is a Group acting on a set $S$ and $s$ is some fixed element of $S$ , then the stabilizer of $s$ in $G$ is the set:
$G_{S} = {g \in G | g \cdot s = s}$

Stabilizers are Subgroups

$G_{s} \leq G$ (see Stabilizers).

Proof

$1_{G} \cdot x = x \Rightarrow 1_{G} \in G_{s}$
(inverses) Suppose $g \in G_{x}$ , so $g \cdot x = x$ . Then:
$\begin{aligned} \Rightarrow g^{- 1} \cdot (g \cdot x) & = g^{- 1} \cdot x \\ \Rightarrow (g^{- 1} g) \cdot x & = g^{- 1} \cdot x \\ \Rightarrow x & = g^{- 1} \cdot x \end{aligned}$
So then $g^{- 1} \in G_{x}$ .
(products) Suppose $g_{1}, g_{2} \in G_{x}$ . Then:
$\begin{aligned} (g_{1} g_{2}) \cdot x & = g_{1} \cdot (g_{2} \cdot x) \\ = g_{1} \cdot x \\ = x \end{aligned}$
So $g_{1}, g_{2} \in G_{x}$ .

☐

fixing subgroup

Let $S \subseteq X$ be any subset. Then we can define the fixing subgroup of $S$ :
$Fix (S) = {g \in G | g \cdot x = x \forall x \in S}$
Which is equivalent to:
$= ⋂_{x \in S} {g \in G | g \cdot x = x} = ⋂_{x \in S} G_{x}$

Here we outline a few lemmas of Fixing Subgroups:

Lemma

For any collection ${H_{i}}$ of Subgroups of a Group the intersection:
$⋂_{i} H_{i}$
is a Subgroup of $G$ . (it's the largest subgroup inside all $H_{i}$ )

Lemma

For subsets $S_{1} \subseteq S_{2} \subseteq X$ we have:
$Fix (X) \leq Fix (S_{2}) \leq Fix (S_{1}) \leq Fix (\emptyset) = G$

Consider Group $G$ acting on its own elements by conjugation:

$X = G$
$g \cdot x = g x g^{- 1}$

Let's look at Stabilizers:
$\begin{aligned} G_{x} & = {g \in G | g \cdot x = x} \\ = {g \in G | g x g^{- 1} = x} \\ = {g \in G | g x = x g} \end{aligned}$
which are all the elements in $G$ that commute with $x$ . This is the Centralizer of the element $x$ .

centralizer

For any subset $A \subseteq G$ , define $C_{G} (A) = {g \in G | g a g^{- 1} = a \forall a \in A}$ . This subset of $G$ is called the centralizer of $A$ in $G$ . Since $g a g^{- 1} = a$ iff $g a = a g$ , then $C_{G} (A)$ is the set of elements of $G$ which commute with every element in $A$ .

Note that $Z (G) = C_{G} (G)$ .

Intuition

The idea is that this is all elements such that they commute with all elements in $A$ . In contrast to the Normalizer, this means that:
$g s = s g$
if $g \in C_{G} (S)$ for some set $S \subseteq G$ .

To contrast this with the Normalizer, that will say that if $g \in N_{G} (S)$ then:
$g s = t g$
for some $t \in S$ instead. Namely, the element in $S$ can change as you apply commutativity. This is why we look at the Stabilizers too, since we could define centralizers in terms of a combination of normalizers and stabilizers.

When we have a Centralizer where $S = G$ , then:
$C_{G} (G) = {g \in G | g x = x g \forall x \in G}$
is a subgroup called the center of $G$ . It usually is denoted $Z (G)$ .

In particular, notice that $G$ is an Abelian Group iff $Z (G) = G$ .

Let's look at some simple examples highlighting these properties.

Example 1: $S_{3}$

Let $G = S_{3} = {1, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}$ from our Symmetric Group & Permutations.

To compute the Centralizer $C_{S_{3}} ((1 2))$ then if we do every Conjugation Subgroup:

\begin{aligned} 1 (1 2) 1^{- 1} & = (1 2) & \Rightarrow (1 \in C_{G} (S_{3})) \\ (1 2) \cdot (1 2) (1 2)^{- 1} & = (1 2) & \Rightarrow ((1 2) \in C_{G} (S_{3})) \\ (1 3) \cdot (1 2) (1 3)^{- 1} & = (2 3) & \Rightarrow ((1 3) \notin C_{G} (S_{3})) \\ (2 3) (1 2) (2 3)^{- 1} & = (1 3) & \Rightarrow ((2 3) \notin C_{G} (S_{3})) \\ (1 2 3) (1 2) (1 2 3)^{- 1} & = (2 3) \\ (1 3 2) (1 2) (1 3 2)^{- 1} & = (1 3) \end{aligned}

So then $C_{S_{3}} ((1 2)) = {1, (1 2)}$ .

Here's some secret info for specifically conjugation in $S_{n}$ . Consider wanting to compute:

σ τ σ^{- 1}

for $σ, τ \in S_{n}$ is always the same "cycle type" as $τ$ . To compute it super fast, do what $σ$ says to the entries of $τ$ . For example in $S_{10}$ :

\begin{aligned} (1 3 4) (2 7 5 6) & = τ \\ (1 4 7 2) (8 9) & = σ \\ \Rightarrow σ τ σ^{- 1} & = (\dots) (\dots) \\ = (4 3 7) (1 2 5 6) \end{aligned}

By looking at the entries of $τ$ and using the left $σ$ ony to see where it went (so say the $1$ in our $τ$ must be the entry after it in $σ$ ).

Computing the rest of our entries:

\begin{aligned} 1 (1 2 3) 1^{- 1} & = (1 2 3) \\ (1 2) (1 2 3) (1 2)^{- 1} & = (2 1 3) \\ (1 3) (1 2 3) (1 3)^{- 1} & = (3 2 1) \\ ⋮ \end{aligned}

Doing this gives that $C_{S_{3}} ((1 2 3)) = {1, (1 2 3), (1 3 2)}$ . Notice that since $C_{S_{3}} ((1 2 3)) \cap C_{S_{3}} ((1 2)) = {1}$ then we don't need to check any more elements! Thus our Fixing Subgroup of $S_{3}$ is thus the trivial subgroup.

Example 2

Let $G$ act on its collection $X$ of subsets by conjugation:

X = P (G)

Let $S \in P (G)$ , so $S \subseteq G$ . Then for any $g \in G$ :

g \cdot S = g S g^{- 1} = {g s g^{- 1} | s \in S} \subseteq G

Now if you look at the Stabilizers, then:

G_{S} = {g \in G | g S g^{- 1} = S}

Where $g \in G_{S} ⟺ g s g^{- 1} \in S \forall s \in S$ . This is called the Normalizer of the set $S$ .

normalizer

Define $g A g^{- 1} = {g a g^{- 1} | a \in A}$ . Define the normalizer of $A$ in $G$ to be the set $N_{G} (A) = {g \in G | g A g^{- 1} = A}$ .

Intuition

Example 1: S3

Example 2

Example 1: $S_{3}$