# group actions and homomorphisms

Let $G$ be a group, $X$ a non-empty 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. 1.

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

 $f_{g}\colon X\to X,\quad x\mapsto g\cdot x\mbox{.}$

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

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

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

2. 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. (a)

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

2. (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 $F$.

## Characterization of group actions

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

 $F(g)=\operatorname{id}_{x}=f_{g}\Leftrightarrow g\cdot x=x,\quad\forall x\in X% \Leftrightarrow g\in\cup_{x\in X}G_{x}$ (1)

and it follows

 $\operatorname{ker}(F)=\bigcap_{x\in X}G_{x}.$

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

 $\operatorname{ker}(F)=\bigcap_{g\in G}gG_{x}g^{-1}.$

For a faithful operation of $G$ the condition $g\cdot x=x,\;\forall x\in X\rightarrow g=1_{G}$ is equivalent to

 $\operatorname{ker}(F)=\{1_{G}\}$

and therefore $F\colon G\to S_{X}$ 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

 $\operatorname{ker}(F)=G.$

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 $X=\{1,\ldots,n\}$ and $G=S_{n}$. Then the operation of $G$ on $X$ given by

 $\pi\cdot i:=\pi(i),\quad\forall i\in X,\;\pi\in S_{n}$

is faithful but not free.

Title group actions and homomorphisms GroupActionsAndHomomorphisms 2013-03-22 13:18:48 2013-03-22 13:18:48 CWoo (3771) CWoo (3771) 15 CWoo (3771) Derivation msc 20A05 GroupHomomorphism