Equivalences of Group Actions

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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