Note - Group Actions on Sets

notegroup-theory

Group Actions on Sets

Examples

1. $G{L}_2(\mathbb{R})$ act on ${\mathbb{R}}^2$ by $A\bullet v=Av$ ← $2\times 2$ matrix and $\left(\begin{array}{c}x\\ y\end{array}\right)$ column vector. $A\cdot v=\left(\begin{array}{c}{a}_{11}x+a_{12}y\\ a_{21}x+a_{22}y\end{array}\right)\leftarrow {\mathbb{R}}^2$

2. $(S_n,\cdot )$ symmetric permutation group acts on $\{1,2,\dots ,n\}$ by $\sigma \cdot i=\sigma (i)$ where $\sigma =\left(\begin{array}{cccc}1& 2& \dots & n\\ \sigma (1)& \sigma (2)& \dots & \sigma (n)\end{array}\right)$

3. $(G,\cdot )$ group. $G$ acts on itself by left multiplication: $g\bullet g^{\prime }=g{g}^{\prime }$ (note that $\bullet$ is the action and $g{g}^{\prime }$ is the product in $G$)

4. $(G,\cdot )$ also acts on $G$ by conjugation: $g\bullet g^{\prime }\stackrel{def}{=}g{g}^{\prime }{g}^{-1}$ ← conjugation in $G$. Verification: $\mu (g,g^{\prime })=g{g}^{\prime }{g}^{-1}\in G$, and $\mu :G\times G\to G$ is well-defined. $e_G\bullet g^{\prime }=e{g}^{\prime }{e}^{-1}=g^{\prime }\forall g^{\prime }\in G$ $g\bullet (g^{\prime }\bullet g^{\prime \prime })=g\bullet (g^{\prime }{g}^{\prime \prime }(g^{\prime }{)}^{-1}=g{g}^{\prime }{g}^{\prime \prime }(g^{\prime }{)}^{-1}(g^{\prime \prime }{)}^{-1}=(g{g}^{\prime })g^{\prime \prime }(g{g}^{\prime }{)}^{-1}=(g{g}^{\prime })\bullet g^{\prime \prime }$.

Remark

Say $X$ is a finite set $X=\{x_1,\dots ,x_n\}$. $\mathrm{Perm}(X)=\{f:X\to X:f\text{ bijection}\}$ $(\mathrm{Perm}(X),\circ )$ is a group.

$\mathrm{Perm}(X)\simeq S_n$ $f:X\to X\mapsto \sigma$ $f(x_i)=x_{\sigma (i)}$ for some permutation of $\{1,\dots ,n\}$

If $(G,\cdot )$ is a group, then an action of $G$ on $X$ is the same as a morphism $M:G\to \mathrm{Perm}(X)$.

If $M:G\to \mathrm{Perm}(X)$ is a morphism, we define $\mu :G\times X\to X$ as $\mu (g,x)=M(g)(x)$ ← This is the $M(g)$ permutation on $X$ applied to $x\in X$.

Conversely, from an action $\mu :G\times X\to X$, we observe that for all $g\in G$, ${\phi }_g:X\to X$ defined by ${\phi }_g(x)=\mu (g,x)=g\bullet x$, ${\phi }_g$ is a bijection, so $({\phi }_g{)}^{-1}={\phi }_{g^{-1}}$

Define $M:G\to \mathrm{Perm}(X)$ as $M(g)={\phi }_g\in \mathrm{Perm}(X)$. $M$ is a morphism.

Proof ($M$ is a morphism)

$M(g{g}^{\prime })\stackrel{?}{=}M(g)\circ M(g^{\prime })$ ${\phi }_{g{g}^{\prime }}\stackrel{?}{=}{\phi }_g\circ {\phi }_{g^{\prime }}$ ${\phi }_{g{g}^{\prime }}(x)\stackrel{?}{=}{\phi }_g({\phi }_{g^{\prime }}(x))={\phi }_g(g^{\prime }\bullet x)=g\bullet (g^{\prime }\bullet x)=(g{g}^{\prime })\bullet x={\phi }_{g{g}^{\prime }}(x).\square$

Remark

If $G$ acts on $X$, $g^{-1}\bullet (g\bullet x)=(g^{-1}g)\bullet x=e_G\bullet x=x$.

Example

$D_4$ acts on square by rigid motions: rotations and reflections.

Take $\tau \in D_4$ a reflection so $\tau ,{\tau }^{-1}$.

$\tau \bullet (\tau \bullet x)=x$ ($\tau \bullet x$ reflects $x$ across axis of symmetry on $\tau$)