group actions and homomorphisms

Notes on group actions and homomorphisms

Let $G$ be a group, $X$ a non-empty set and $S_X$ the symmetric groupof $X$,i.e. the group of all bijective maps on $X$. $\cdot$ may denote a leftgroupaction of $G$ on $X$.

1. For each $g\in G$ and $x\in X$ we define

   $$f_g:X\to X,x\mapsto g\cdot x\text{.}$$

   Since $f_{g^{-}1}(f_g(x))=g^{-1}\cdot (g\cdot x)=x$ for each$x\in X$,$f_{g^{-}1}$ is the inverse of $f_g$. so $f_g$ is bijective and thus element of$S_X$. We define $F:G\to S_X,F(g)=f_g$ for all $g\in G$. Thismapping is a group homomorphism: Let $g,h\in G,x\in X$. Then

   $F(gh)(x)$$=f_{gh}(x)=(gh)\cdot x=g\cdot (h\cdot x)$

   $=(f_g\circ f_h)(x)=(F(g)\circ F(h))(x)$

   for all $x\in X$ implies $F(gh)=F(g)\circ F(h)$. — The same is obviously true for a rightgroup action.

2. Now let $F:G\to S_x$ be a group homomorphism, and let$f:G\times X\to X,(g,x)\mapsto F(g)(x)$ satisfy

   1. $f(1_G,x)=F(1_g)(x)=x$ for all $x\in X$ and

   2. $f(gh,x)=F(gh)(x)=(F(g)\circ F(h)(x)=F(g)(F(h)(x))=f(g,f(h,x))$,

   so $f$ is a group action induced by $F$.

Characterization of group actions

Let $G$ be a group acting on a set $X$.Using the same notation as above, we have for each$g\in \ker (F)$

$$F(g)={\mathrm{id}}_x=f_g\iff g\cdot x=x,\forall x\in X\iff g\in {\cup }_{x\in X}{G}_x$$(1)

and it follows

$$\ker (F)=\bigcap _{x\in X}{G}_x.$$

Let $G$ act transitively on $X$. Then for any $x\in X$, $X$ is theorbit $G(x)$of $x$. As shown in “conjugate stabilizer subgroups’, all stabilizersubgroupsof elements $y\in G(x)$ are conjugate subgroups to $G_x$ in $G$. Fromthe above it follows that

$$\ker (F)=\bigcap _{g\in G}g{G}_x{g}^{-1}.$$

For a faithful operation of $G$ the condition $g\cdot x=x,\forall x\in X\to g=1_G$ is equivalent to

$$\ker (F)=\{1_G\}$$

and therefore $F:G\to S_X$ is a monomorphism.

For the trivial operation of $G$ on $X$ given by $g\cdot x=x,\forall g\in G$the stabilizer subgroup $G_x$ is $G$ for all $x\in X$, and thus

$$\ker (F)=G.$$

If the operation of $G$ on $X$ is free, then$G_x=\{1_G\},\forall x\in X$, thus the kernel of $F$ is$\{1_G\}$–like for a faithful operation. But:

Let $X=\{1,\mathrm{\dots },n\}$ and $G=S_n$. Then the operation of $G$ on $X$given by

$$\pi \cdot i:=\pi (i),\forall i\in X,\pi \in S_n$$

is faithful but not free.