prove that if $\exists a\in A\space:\space stb(a)\not=\{e\}$ then the action is unfaithful let $G$ be abelian group which acts on non empty set $A$. prove that if $\exists a\in A\space:\space stb(a)\not=\\{e\\}$ then the action is unfaithful (the kernel of the action is not trivial). **notation** : \- $e$ is the identity \- $stb(a)=\\{g\in G\space :\space g\cdot a=a\\}$ where $g\cdot a$ means that $g$ acts on $a$. I manged to prove that for any element in some orbit the stabilizer is the same, yet could not proceed from there.

Suppose $G$ acts transitively on $A$, for a moment, and let $g\
eq e$ be in the stabilizer of $a\in A$.

For an arbitrary element $b\in A$, there exists $h\in G$ such that $b=h\cdot a$. Then $g\cdot b=g\cdot(h\cdot a)=h\cdot (g\cdot a)=h\cdot a=b$, so $g$ would also stabilize $b$. Thus $g$ stabilizes everything, and the action isn't faithful.

* * *

Now go back to the general situation and suppose that $G$ acts on $A$ such that $g\
eq e$ stablizes $a\in A$. Form the new set $C$ which is the disjoint union of $A$ and $G$, and let $G$ act on $C$ in the obvious way (by acting on elements according to $A$'s action if they are from $A$, and according to $G$'s action on itself if they are from $G$).

Now, $g$ still stabilizes $a\in A$, but it can't stabilize $e\in G$, so the action is not unfaithful on $C$. This counterexample leads me to believe your question was intended to be for transitive actions only.