Show up to equivalence that there is only one group action of $\mathbb{Z}_{3}$ on $X=\left \{ 1,2,3 \right \}$.
I am not sure where to begin with such a proof. I would assume it is proof by contradiction and I use the definition of equivalent actions? Is there an easier way, because I am not very knowledgable on equivalent actions and how to apply this to the proof! Any help and guidance is appreciated.
Best Answer
This is false although close to truth. There are two possible group actions on $\{1,2,3\}$.
By $G$-set I will understand $(X,\cdot)$ pair where $X$ is a set and $\cdot:G\times X\to X$ is a group action.
First of all note that any orbit is isomorphic to $G/H$ as a $G$-set for some subgroup $H$. Since $G=\mathbb{Z}_3$ has only two possible subgroups, namely $G$ and $0$ then there are only two possible orbits: $G/0$ and $G/G$. Non-isomorphic because of different cardinality.
Now since every $G$-set is a disjoint union of orbits (up to $G$-isomorphism) than $X=\{1,2,3\}$ has two possible $G$-set structures: namely $X\simeq G/0$ or the trivial one $X\simeq G/G\sqcup G/G\sqcup G/G$.