Group Actions

Let $(G, \cdot)$ be a group, and X be a set. An action of G on X is a map $G \times X \rightarrow X$ such that (1) for all $g, h \in G$ and $x \in X, g*(h*x) = (g\cdot h)*x$ and (2) for all $x \in X, 1*x = x$.

A note on notation: I will write gx instead of gx if there is no possible confusion but will try to use the gx notation throughout the post.

Examples: $S_n$ acts on $\{0, ..., n-1\}$ by $0*x = \sigma(x), \sigma(\tau(x)) = \sigma \circ \tau(x)$ id(x) = x. $D_{2n}$ acts on the vertices of the ngon ($\{0, ..., n-1\}$) by sending vertices to their image under the transformation. G acts on titself in the following ways: 1) $g*x = g \cdot x$ (action by left translation) 2) $g*x = g \cdot x \cdot g^{-1}$ (action by conjugation). $h*(g*x) = h(gxg^{-1}) = hgxg^{-1}h^{-1} = (hg)x(hg)^{-1} = (hg)*x. 1*x = 1x1^{-1} = x$ G acts on P(x) in the following ways: a) $g*A = gA = \{g \cdot a, a \in A \}$. b) $g*A = gAg^{-1} = \{gag^{-1}, a \in A\}$. If V is a K vector space then scalar multiplication is a group action of K* on V: $\lambda, \mu \in K*, \lambda(\mu x) = (\lambda \mu)x, 1 \cdot x = x$

Let G be a group, X a set. A right action of G on X is a map $X \times G \rightarrow X$ such that (1) for all $g, h \in G$ and $x \in X$ (xg)h = x(gh) and (2) for all $x \in X$, x*1 = x.

Proposition: let G be a group acting on X. 1) for all $g \in G$ the map $fg: x \mapsto g \cdot x$ is a bijection 2) the map $\phi: G \rightarrow S_x g \mapsto fg$ is a group homomorphism. Conversely if $\phi: G \rightarrow S_x$ is a group homomorphism then g*x = fg(x) is a group action.

Proof: 1) $fg(fg^{-1}(x)) = g*(g^{-1}*x) = (g\cdot g^{-1})*x = x. fg^{-1}(fg(x)) = (g^{-1}\cdot g)*x = x$ the two functions are inverse bijections. 2) $\phi(g\cdot h) = fgh \cdot fgh(x) = (gh)*x = g*(h*x) = fg(fh(x)) = (fg \cdot fh)(x)$ so $fgh = fg \cdot fh$ Conversely $g *(h*x) = fg(fh(x)) = (fg\cdot fh)(x) = fgh(x) = (gh)*x. 1*x =$id(x) = x.

Let G be a group acting on X and $A \subset X$. We define the stabilizer of A in G ($S_G(A))$ as $\{g \in G. \forall a \in A, g*a = a\}$

Proposition: $S_G(A) \leq G$

Proof: $1*a = a \forall a \in A$ so $1 \in S_G(A)$. Let $g, h \in S_G(A), (g \cdot h)*a = g*(h*a) = g*a = a$ Let $g \in S_G(A), 1*a = (g^{-1} \cdot g)*a = g^{-1}*(g*A) = g^{-1}*a$ so $g^{-1} \in S_G(A)$.

Remarks: (1) $C_G(A)$ is the stabilizer of A for the action by conjugation. (2) $N_G(A)$ is the stabilizer of $\{A\}$ under the action of G on P(G) by conjugation. (3) $S_G(X)$ is sometimes called the kernel of the action. It is indeed the kernel of $G \rightarrow S_x$ associated with the action.

Let G be a group acting on X and $x \in X$. We call the set $\{g*x: g \in G\} = C_x$ the orbit of x. The orbit by conjucation action is called the conjugation class.