I'm having a bit of trouble understanding this problem. I am given a set $A = \{a,b,c,d\}$, and a group action $s : \Bbb Z_4 \to S_A$ where the group operation for $\Bbb Z_4$ is addition modulo $4$. I am trying to show that $s(2) = e_A$ (where $e_A$ is the identity permutation on $A$) or that the order of the permutation $s(2)$ is $2$.
I can't really seem to figure out how to do this. The formulas we usually use when it comes to group actions are: $s(e) = e_A$ and $s(f*g) = s(f) \circ s(g)$
The order of $[2]\in\mathbb{Z}/4\mathbb{Z}$ is $2$, i.e. $|[2]| = 2$. I use $[2]$ to denote the equivalence class of $2$ that is an element of $\mathbb{Z}/4\mathbb{Z}$. I also use multiplicative notation to denote the binary operation of $\mathbb{Z}/4\mathbb{Z}$ though it usually is written additively.
So $s([2])^{2} = s([2]^{2}) = s([0]) = e_{A}$, since $s$ is a homomorphism. Thus the element $s([2])$ must have order dividing $2$ in $S_{A}$, so its order is either $1$ (in which case $s([2]) = e_{A}$) or its order is $2$.