Group Actions of a Group on a Set

# Group Actions of a Group on a Set

Definition: Let $(G, \cdot)$ be a group and let $A$ be a (nonempty) set. A Left Group Action of the group $G$ on the set $A$ is a map $G \times A \to A$ denoted for all $g \in G$ and all $a \in A$ by $(g, a) \to ga$ that satisfies the following properties:1. $g_1(g_2a) = (g_1 \cdot g_2)a$ for all $g_1, g_2 \in G$ and for all $a \in A$.2. $1a = a$ for all $a \in A$ (where $1 \in G$ denotes the identity element).Similarly, a Left Group Action of the group $G$ on the set $A$ is a map $G \times A \to A$ denoted for all $g \in G$ and all $a \in A$ by $(g, a) \to ag$ that satisfies the following properties:1. $(ag_1)g_2 = a(g_1 \cdot g_2)$ for all $g_1, g_2 \in G$ and for all $a \in A$.2. $1a = a$ for all $a \in A$.In either case we say that the group $G$ is (left/right) Acting on the set $A$. |

We begin by stating some basic results regarding (left) group actions of a group on a set.

Proposition 1: Let $(G, \cdot)$ be a group that is left acting on the set $A$. For each $g \in G$ let $\sigma_g : A \to A$ be defined for all $a \in A$ by $\sigma_g(a) = ga$. Then:a) For each $g \in G$, $\sigma_g$ is a permutation of the set $A$, and so $\varphi_g \in S_A$ for all $g \in G$.b) The map $\varphi : G \to S_A$ (called the Associated Permutation Representation) defined for all $g \in G$ by $\varphi(g) = \sigma_g$ is a group homomorphism of $G$ to $S_A$.c) If $\psi : G \to S_A$ is any homomorphism from $G$ to $S_A$ then the map $G \times A \to A$ defined by $(g, a) \to \varphi(g) a$ is a left group action of $G$ on $A$. |

*A similar result can be stated when $(G, \cdot)$ is right acting on a set $A$.*

**Proof of a)**For each $g \in G$ consider $\sigma_g$ and $\sigma_{g^{-1}}$. For all $a \in A$ we have that:

\begin{align} \quad (\sigma_g \circ \sigma_{g^{-1}})(a) &= \sigma_g (\sigma_{g^{-1}}(a)) \\ &= \sigma_g(g^{-1}a) \\ &= g(g^{-1}a) \\ &= (g \cdot g^{-1})a \\ &= 1a \\ &= a \end{align}

(2)
\begin{align} \quad (\sigma_{g^{-1}} \circ \sigma_g)(a) &= \sigma_{g^{-1}} (\sigma_g(a)) \\ &= \sigma_{g^{-1}}(ga) \\ &= g^{-1}(ga) \\ &= (g^{-1} \cdot g)(a) \\ &= 1a \\ &= a \end{align}

- Therefore $\sigma_g \circ \sigma_{g^{-1}} = \mathrm{id}_A = \sigma_{g^{-1}} \circ \sigma_g$, and so each $\sigma_g$ has an inverse $\sigma_{g^{-1}}$. Thus each $\sigma_g$ is a permutation of $A$. $\blacksquare$

**Proof of b)**By part (a) we have that $\varphi$ is indeed a map from $G$ to $S_A$. All that remains to show is that $\varphi$ is a group homomorphism. Let $g_1, g_2 \in G$. We claim that $\sigma_{g_1 \cdot g_2} = \sigma_{g_1} \circ \sigma_{g_2}$. Indeed, for all $a \in A$ we have that:

\begin{align} \quad \sigma_{g_1 \cdot g_2}(a) = (g_1 \cdot g_2)a = g_1(g_2a) = g_1 \sigma_{g_2}(a) = \sigma_{g_1}(\sigma_{g_2}(a)) = (\sigma_{g_1} \circ \sigma_{g_2})(a) \end{align}

- Therefore $\varphi(g_1 \cdot g_2) = \sigma_{g_1 \cdot g_2} = \sigma_{g_1} \circ \sigma_{g_2} = \varphi(g_1) \circ \varphi(g_2)$. So $\varphi : G \to S_A$ is a group homomorphism. $\blacksquare$

**Proof of c)**Let $\psi : G \to S_A$ be a group homomorphism of $G$ to $S_A$. We aim to show that $(g, a) \to [\psi(g)](a)$ is a left group action of $G$ on $A$, i.e., we need to verify the two properties in the definition.

- For all $g_1, g_2 \in G$ and for all $a \in A$ we have that:

\begin{align} \quad g_1(g_2(a)) = g_1([\psi(g_2)](a)] = [\psi(g_1)](\psi(g_2)(a)) = [\psi(g_1) \circ \psi(g_2)](a) = [\psi(g_1 \cdot g_2)](a) = (g_1 \cdot g_2)(a) \end{align}

- And for all $a \in A$ we have that:

\begin{align} \quad 1(a) = [\varphi(1)](a) = [\mathrm{id}_A](a) = a \end{align}

- So indeed, $(g, a) \to [\psi(g)](a)$ is a left group action of $G$ on $A$. $\blacksquare$