Let $G$ be a finite abelian group with a transitive faithful action on set $X,|X|=n$. Prove $|G|=n.$
Since the action is transitive $g^{-1}{\rm Stab}_G(x)g={\rm Stab}_G(y)$.
$G$ is abelian $\implies {\rm Stab}_G(x)={\rm Stab}_G(y)$.
Using ${\rm Stab}_G(x)={\rm Stab}_G(y)\implies {\rm Stab}_G(x)=\{{e}\}.$ (If $$\{e\} \neq g\in {\rm Stab}_G(x),\forall x\in X, g\cdot x=x,$$ a contradiction since the action is faithful.)
Using orbit stabilizer theorem, $|O(x)|=\frac{G}{{\rm Stab}_G(x)} \implies n=\frac{G}{1} \implies |G|=n$.
Is my solution correct?
Any ideas how to solve in another ways? (Burnside's lemma and other different ways)
Thanks!
By action's transitivity and group's abelianess, the kernel of the action is given by $\bigcap_{y\in X}\operatorname{Stab}(y)=$ $\bigcap_{g\in G} g\operatorname{Stab}(x)g^{-1}=$ $\operatorname{Stab}(x)$, for any $x\in X$. So, if the action is also faithful, then $\operatorname{Stab}(x)=1$ for every $x\in X$$^\dagger$.
Now, for any given $x\in X$, consider the map $f_x\colon G\to X$, defined by $g\mapsto gx$. By action's transitivity, $f_x$ is surjective. Let $h\in G$ be such that $gx=hx$; then, $g\in h\operatorname{Stab}(x)=\{h\}$, and hence $g=h$. So, $f_x$ is injective as well, and so $G$ and $X$ have the same cardinality (not necessarily finite).
$^\dagger$With the orbit-stabilizer theorem at hands, this suffices to prove the claim. If fact it states $|O(x)|=$ $[G:\operatorname{Stab}(x)]$. Now, plug into this the two information $\operatorname{Stab}(x)=1$ and $O(x)=X$ (transitivity).