Exercise 2.1.3 (Group Action Determines Bijection)

Solution: We show injectivity (i), surjectivity (ii) and homomorphism (iii) :

(i)

Injectivity:
Let $x,y\in X$ and $ḡ$ be our bijection
If we have $ḡ(x)=ḡ(y)$

$$\begin{array}{ccc}\hfill ⇝\mathit{gx}& =\mathit{gy}⇝g^{-1}(\mathit{gx})=g^{-1}(\mathit{gy})⇝(g^{-1}g)x=(g^{-1}g)y⇝\mathit{ex}=\mathit{ey}⇝x=y\square \hfill & \hfill \\ \hfill \end{array}$$

(ii)

Surjectivity:
We know that $f:X\to Y$ is surjective iff $\forall <!--\; FUNCTION\; APPLICATION\; -->y\in Y\exists <!--\; FUNCTION\; APPLICATION\; -->x\in X:f(x)=y$
We let $x\in X$ and $e$ be the identity in $G$ then,
$x=\mathit{ex}=(g{g}^{-1})x=g(g^{-1}x)=\mathit{gy}=ḡ(y)$ where $y=g^{-1}x\in X\square$

Therefore $ḡ$ is both injective and surjective, hence bijective.

(i)

$\mathrm{\Sigma }$ is Homomorphism:
We have the map:

$$\begin{array}{ccc}\hfill \mathrm{\Sigma }:& G\to \mathit{Sym}(X)\hfill & \hfill \\ \hfill & g\mapsto ḡ\hfill \end{array}$$

We know that $ḡ$ is well defined
Then we take $g,h\in G,x\in X$, then by Definition 2.1.1.

$$\begin{array}{ccc}\hfill \mathrm{\Sigma }(\mathit{gh})(x)& =\mathit{gh}(x)=g(\mathit{hx})\hfill & \hfill \\ \hfill & =\mathrm{\Sigma }(g)(\mathrm{\Sigma }(h)(x))=\mathrm{\Sigma }(g)\circ \mathrm{\Sigma }(h)(x)\square \hfill \end{array}$$