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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!

citadel
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Algo
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  • I think you have enough experience to compare your solution to the standard one without our help. This has been asked before, with a solution in the question and comments, see here. – Dietrich Burde Sep 04 '22 at 10:59

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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).

citadel
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