Let $G$ be a finite abelian group. Suppose that $G$ acts faithfully and transitively on a set $X$. Show that $|X|=|G|$. Deduce that the action is equivalent to the action of $G$ on itself by left translation.
Because $G$ acts transitively on $X$, the Orbit-Stabilizer Theorem tells us $|G|=|X|\cdot|H_x|$, where $H_x$ is a point stabiliser for $x\in X$. Now let $y\in X$ be fixed and choose any $x\in X$, then $\exists g\in G$ such that $gx=y$ (because of transitivity). Let $h\in H_y$, then $gx=y$ implies $hgx=hy=y=gx$, and because $G$ is abelian: $ghx=gx\Rightarrow hx=x$. So $h\in H_x$ for any $x\in X$. But $\bigcap_{x\in X} H_x =\{e\}$ since $G$ acts faithfully. So $H_x=\{e\}$ for every $x\in X$, and $|G|=|X|\cdot|H_x|$ gives the result.
But how does one deduce that the action is equivalent to the action of $G$ on itself by left translation?
