group actions and homomorphisms
Notes on group actions and homomorphisms
Let $G$ be a group, $X$ a nonempty set and ${S}_{X}$ the symmetric group^{} of $X$, i.e. the group of all bijective^{} maps on $X$. $\cdot $ may denote a left group action^{} 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 $g\in G$. This mapping 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 right group 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

(a)
$f({1}_{G},x)=F({1}_{g})(x)=x$ for all $x\in X$ and

(b)
$f(gh,x)=F(gh)(x)=(F(g)\circ F(h)(x)=F(g)(F(h)(x))=f(g,f(h,x))$,
so $f$ is a group action induced by $F$.

(a)
Characterization of group actions
Let $G$ be a group acting on a set $X$. Using the same notation as above, we have for each $g\in \mathrm{ker}(F)$
$$F(g)={\mathrm{id}}_{x}={f}_{g}\iff g\cdot x=x,\forall x\in X\iff g\in {\cup}_{x\in X}{G}_{x}$$  (1) 
and it follows
$$\mathrm{ker}(F)=\bigcap _{x\in X}{G}_{x}.$$ 
Let $G$ act transitively on $X$. Then for any $x\in X$, $X$ is the orbit $G(x)$ of $x$. As shown in “conjugate stabilizer subgroups’, all stabilizer^{} subgroups^{} of elements $y\in G(x)$ are conjugate subgroups^{} to ${G}_{x}$ in $G$. From the above it follows that
$$\mathrm{ker}(F)=\bigcap _{g\in G}g{G}_{x}{g}^{1}.$$ 
For a faithful operation^{} of $G$ the condition $g\cdot x=x,\forall x\in X\to g={1}_{G}$ is equivalent^{} to
$$\mathrm{ker}(F)=\{{1}_{G}\}$$ 
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
$$\mathrm{ker}(F)=G.$$ 
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.
Title  group actions and homomorphisms 

Canonical name  GroupActionsAndHomomorphisms 
Date of creation  20130322 13:18:48 
Last modified on  20130322 13:18:48 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  15 
Author  CWoo (3771) 
Entry type  Derivation^{} 
Classification  msc 20A05 
Related topic  GroupHomomorphism 