I am learning about group actions. This should be a fairly simple question but I want to check my understanding.
Claim: Fix some integer $n > 2$, let $X=\{1,2,\ldots,n\}$ and let $S_n$ be the Symmetric group of permutation functions on $X$. Consider the group action of $S_n$ on $X$, which will be denoted by the map $* : S_n \times X \to X$. Then, the action of $S_n$ on $X$ is not semiregular.
Proof
Since we know $|X| \ge 3$, we can take arbitrary $x,y,z \in X$. We know there are distinct permutations $\pi_1, \pi_2 \in S_n$ such that $\pi_1(x)= y$, $\pi_1(y)=x$ and $\pi_2(x)=y$, $\pi_2(y)=z$. Thus, we have demonstrated that $\pi_1 * x = y$ and $\pi_2 * x = y$. Thus the action is not semiregular.
The demonstration in terms of $x\in X$ is without loss of generality.
So semiregular must mean no two group elements act the same way on any element of $X$.
This seems ok.
This is also called a free group action.