I need help with the end of this proof.
If $G \le S_A$ why $\varphi$ turns out to be injective? I was trying to get to a contradiction assuming there exists $x\ne1$ such that $\varphi(x) = 1$. Then for any $a$ we have $\varphi(ax) = \varphi(a)\varphi(x) = \varphi(a)$ but I got stuck.
Thanks in advance.
It's definitional. You start with "$G$ is a permutation group (i.e., $G \le S_A$)". This means, as a matter of definition, that the only element of $G$ that does not move any elements of $A$ is the identity. If the problem simply said "$G$ acts on $A$", this would not be the case.
By the way, there seems to be a typo in the last paragraph of quoted material, where it should say "...by previous exercise the kernel is $\cap_{g\in G}gG_ag^{-1}$".