06 - Group Actions

The idea here is that $G$ is a Group while $X$ is a set. The group is moving things around in the set in a group structure.

left action

A (left) action of Group $G$ on set $X$ consists of a set map:
$σ : G \times X \to X$
Where $(g, x) \mapsto σ (g, x) = g \cdot x = g x$ such that:

$g_{1} \cdot (g_{2} \cdot x) = (g_{1} \cdot g_{2}) x$ for all $g_{1}, g_{2} \in G, x \in X$
$1_{G} \cdot x = x$ for all $x \in X$

This definition may seem a bit loose, so mathematicians sometimes use another definition:

left action

A (left) action of $G$ on $X$ is a group Homomorphism (or Group Morphism) (map) $σ : G \to S_{X} = {f X \to X | f is a bijection}$ (see Symmetric Group & Permutations)

We'll later see that these definitions are equivalent. However, let's first look at an example.

Often we'll use $σ_{g}$ to represent how $σ (g) = σ_{g}$ in the alternative definition above. Namely:
$σ_{g} (x) = g \cdot x$

An Example

The trivial action of $G$ on $X$ is $σ : G \to S_{X}$ such that each $g \mapsto 1$ . Namely $g x = x$ for all $g \in G, x \in X$ .
The Dihedral Groups, like $D_{8}$ for the symmetries of a square, acts on the set of vertices of that square ( $X = {v_{1}, v_{2}, v_{3}, v_{4}}$ for simplicity, for example $r v_{1} = v_{2}$ ). Define $σ_{r} = (v_{1} v_{2} v_{3} v_{4}) \equiv (1 2 3 4)$ . Then this is a group action, where $σ : D_{8} \to S_{X} ≅ S_{4}$ where $σ_{r} : r \mapsto (1 2 3 4)$ .

$D_{8}$ can also act on the set $Y = {d_{1}, d_{2}}$ , the diagonals of the square:

Here $σ_{r} = (d_{1} d_{2}) \equiv (1 2)$ . Here $D_{8} \to S_{Y} ≅ S_{2}$ .

One action of the group onto itself is left multiplication by the group operation:
$g ⋆ g^{'} = g g^{'}$

Note

Why left multiplication? Namely, it's because our definition is in terms of terms from the group being on the left.

Is there any group elements that keeps each element fixed? Namely is there some $g \in G$ such that $g ⋆ x = 1_{G}$ for all $x \in X$ ? See the answer on Trivial Group Action.

trivial action

Suppose Group $G$ acts on $X$ . We say an element $g \in G$ acts trivially if $g x = x$ for all $x \in X$ .
The kernel of the Group Action is the set:
$Ker (σ) = {g \in G | g x = x \forall x \in X} = {g \in G | σ_{g} = 1 \in S_{X}}$

More generally, for any group morphism (map) $φ : G \to H$ the kernel is:
$Ker (φ) := {g \in G | φ (g) = 1_{H}}$

Example

The only element $g \in G$ that acts trivially is $1_{G}$ .
Proof

Take any $g \neq 1_{G}$ . If $x \in X$ is fixed by $g$ then:
$g x = x ⟺ g = 1_{G}$
2. A group $G$ can also act on itself by conjugation. Here $x \in G$ (confusing I know):
$g \cdot x := g x g^{- 1}$
Let's check that this is a valid Group Action.
a. $1_{G} \cdot x = 1_{G} x 1_{G}^{- 1} = 1_{G} x 1_{G} = x$
b. $g_{1} \cdot (g_{2} \cdot x) = g_{1} \cdot (g_{2} x g_{2}^{- 1}) = g_{1} (g_{2} x g_{2}^{- 1}) g_{1}^{- 1} = (g_{1} g_{2}) x (g_{2}^{- 1} g_{1}^{- 1}) = (g_{1} g_{2}) x (g_{1} g_{2})^{- 1} = g_{1} g_{2} \cdot x$ .

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Group Action Equivalences

Both definitions in Group Action are equivalent.

Proof

$(1 \to 2)$ : Have $σ : G \times X \to X$ where $(g, x) \mapsto g \cdot x$ with:

$g_{1} \cdot (g_{2} \cdot x) = (g_{1} g_{2}) \cdot x$
$1_{G} \cdot x = x$ .

We first claim that for each fixed $g \in G$ , the map:
$σ_{g} : X \to X, x \mapsto g \cdot x$
is a permutation, namely a bijection.

Injection: Suppose $x_{1}, x_{2} \in X$ are such that $σ_{g} (x_{1}) = σ_{g} (x_{2})$ . Then:
$\begin{aligned} g \cdot x_{1} & = g \cdot x_{2} \\ g^{- 1} \cdot (g \cdot x_{1}) & = g^{- 1} \cdot (g \cdot x_{2}) \\ (g^{- 1} g) \cdot x_{1} & = (g^{- 1} g) \cdot x_{2} \\ 1_{G} \cdot x_{1} & = 1_{G} \cdot x_{2} \\ x_{1} & = x_{2} \end{aligned}$
Surjection: Let $y \in X$ . We want to find some $x \in X$ where $σ_{g} (x) = y \equiv g \cdot x = y$ . Notice that:
$\begin{aligned} g \cdot x & = y \\ g^{- 1} \cdot (g \cdot x) & = g^{- 1} \cdot y \\ (g^{- 1} g) \cdot x & = g^{- 1} \cdot y \\ 1_{G} \cdot x & = g^{- 1} \cdot y \\ x & = g^{- 1} \cdot y \end{aligned}$
So choose $x = g^{- 1} \cdot y$ , we can use the reverse work above to get the desired result.

Say that $σ : G \to S_{X}$ is a map where $g \mapsto σ_{g}$ . We first claim that this is a group Homomorphism (or Group Morphism). Take any $g_{1}, g_{2} \in G$ and we want to show:
$σ (g_{1} g_{2}) = σ (g_{1}) σ (g_{2}) \equiv σ_{g_{1} g_{2}} = σ_{g_{1}} σ_{g_{2}}$
To verify this, take any $x \in X$ and compare:
$\begin{aligned} σ_{g_{1} g_{2}} (x) & = (g_{1} g_{2}) \cdot x \end{aligned}$
with:
$\begin{aligned} (σ_{g_{1}} σ_{g_{2}}) (x) & = σ_{g_{1}} (σ_{g_{2}} (x)) \\ = σ_{g_{1}} (g_{2} \cdot x) \\ = g_{1} \cdot (g_{2} \cdot x) \\ = (g_{1} g_{2}) \cdot x \end{aligned}$
so both are the same.

$(2 \to 1)$ is proven very similarly.

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There is some upcoming language with Group Actions:

"orbits" of elements: $O_{x} = {g \cdot x | g \in G}$
"stabilizers" of elements: $G_{x} = {g \in G | g \cdot x = x}$
"faithful"s: $σ : G \to S_{X}$ is injective (here, it's all elements that get sent to a unique element, no duplicates allowed).