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 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

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

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.
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
Canonical name	GroupActionsAndHomomorphisms
Date of creation	2013-03-22 13:18:48
Last modified on	2013-03-22 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

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