$\newcommand{\stab}{\text{Stab}}$
$\newcommand{\orb}{\text{Orb}}$
Situating definitions.
Definition: Let $G$ be a group acting on a set $X$, we say the action is simply transitive if it is transitive and $\stab(x)$ is trivial for all $x \in X$.
Definition: Let $G$ act on two sets $X,Y$. We say that $X$ is $G$-isomorphic to $Y$ if there exists a bijection $f:X\rightarrow Y$ such that $f(g \ast x) = g \ast f(x)$ for all $x \in X$ and $g \in G$.
Problem: Show that if the action of $G$ on $X$ is simply transitive, then it is $G$-isomorphic to $G$ with the left-multiplication action $g \ast h = gh$.
Attempt: Let $x \in X$, then the Orbit-Stabilizer theorem gives a bijection $f:G/\stab(x) \rightarrow \orb(x)$ by $f(g\stab(x)) = g \ast x$. Because the action of $G$ on $X$ is simply transitive we have that $\orb(x) = X$ for all $x \in X$ and that $\stab(x) = \{1_G\}$ for all $x \in X$. This means the bijection is really from $f:G/\{1\} \rightarrow X$. But $\{1\} \unlhd G$ and so $G/\{1\}$ is a quotient group which is evidently isomorphic to $G$. Therefore we have a bijection $f:G \rightarrow X$; to see that it is a $G$-isomorphism, let $g \in G$, $x \in X$ and $a \in X$
\begin{align*}
f(g \ast a) & = f(ga) \\
& = (ga)\ast x \\
&= g \ast (a \ast x)\\
& = g \ast f(a).
\end{align*}
What Im not quite sure about is keeping the same bijection when we use that $G/\{1\} \cong G$. I know that you can certainly compose the isomorphism map $\psi:G/\{1\} \rightarrow G$ so that we have a bijection $f': \psi \circ f$ but will that change how the map $f$ is defined? Because I guess I'm making the assumption that even when $f$ doesn't map from a set of cosets it acts as $g \mapsto g \ast x$ rather than $g\stab(x) \mapsto g \ast x$. Are there any hints/tips for how to clean this up?
Best Answer
(For $x∈X$ fixed) define $f:G→X$ more directly by $f(g)=g∗x$.
$f$ is onto because (since the action is transitive) for any $y\in X$ there exists a $g\in G$ such that $y=g∗x$.
$f$ is one-to-one because for any $g,h\in G$ such that $f(g)=f(h)$, $g^{-1}h$ fixes $x$ hence (since this transitivity is "simple") $g^{-1}h=e_G$, so $g=h.$