Transitivity of $A_3$ on $\{1 ,2, 3 ,4\}$

Question

We know that $A_3$ is equal to $$A_3 = \{1,(123), (132) \}.$$ This group is not transitive on $X=\{1,2,3,4\}$ because $\exists 4 \in X$ such that $\forall g \in A_3 $ , $\forall x \in X$ $$g\cdot x \neq 4.$$ Or the action is not defined because $A_3$ does not contain 4. What is an appropriate explanation for the failure of $A_3$ to be transitive on $X=\{1,2,3,4\}$?

Answer 1

The action of $A_3$ on $X$ is a function $\alpha: A_3\times X \to X$ which satisfies two compatibility axioms

$$ \begin{align} &\alpha(1, x) = x \\ &\alpha(g, \alpha(h, x)) = \alpha(gh, x). \end{align} $$

If you don't define the action then it is fair to say the action is not defined and then it is meaningless to ask whether it is transitive. If you do define the action, then the question is meaningful. I suppose that one could argue that the most natural default definition would be

$$ \begin{align} &\alpha(g, x) = g(x) &\quad\text{if}\quad x\in\{1, 2, 3\} \\ &\alpha(g, x) = x &\quad\text{otherwise}, \end{align} $$

but since this is not the only possibility, it is good form to be explicit about the action being considered.

That said, in the particular case of $A_3$ and $X=\{1, 2, 3, 4\}$, it turns out that however you define the action the question of its transitivity always has the negative answer. This follows from the fact that a group action is transitive if and only if it has exactly one orbit. In other words, $\alpha$ is transitive if and only if there exists $x\in X$ such that $\alpha(A_3, x) = X$. However, for every $x\in X$, the orbit $\alpha(A_3, x)$ has at most three elements and $X$ has four. Therefore, they are not equal and so $\alpha$ is not transitive.

Answer 2

If you are asked the question in the form given then the implicit assumption is that you are to regard $A_3$ as a subset of $S_4$ in the natural way.

You can't say "$A_3$ does not contain $4$". What you mean is that no element of $A_3$ moves $4$. As you understand, that shows the action is not transitive.