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. 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 $F$.