There is this supposed to be a not-so-difficult proof but somehow I just find it a bit hard to connect the dots.
Suppose that $G$ acts on a set $S$, and let $\phi$ be the associated homomorphism from $G$ to Perm($S$). Show that $G/N$ acts faithfully on $S$ if and only if $N=\ker\phi$.
My effort:
($\Rightarrow$) Suppose $G/N$ acts faithfully on $S$, then from the definition of faithful action, the identity element $e_{G/N}$ of $G/N$ is the only element such that $e_{G/N}\ast s=s$ for all $s\in S$. And am I correct if I say that $\ker\phi=\{e\}$ where $e$ is the identity element of $G$? Then how can we conclude that $N=\ker\phi$?
($\Leftarrow$) Suppose $N=\ker\phi$, then $N=\{e\}$ where $e$ is the identity element of $G$. But if $N=\{e\}$, then $G/N=G$? Then how can we conclude that $G/N$ acts faithfully on $S$?
I understand the definition of each the concepts and terms but I am struggling to fill in the gaps in the proof. I really appreciate any helps. Thanks!
Note that $G$ is not supposed to act faithfully. However, to obtaoin an action of $G/N$ on $S$ in the first place, we need to have that $N\subseteq \ker \phi$. When considering considering the action not elementwise but on the level of homomorphisms to $\operatorname{Perm}(S)$, the result can be seen very straightforward. Noting that faithfulness is by defintition equivalent to the homomorphism to $\operatorname{Perm}(S)$ being injective, the claim translates as:
Let $\phi\colon G\to\operatorname{Perm}(S)$ be a homomorphism. If we assume that $N\subseteq \ker\phi$, this induces a homomorphism $\bar\phi\colon G/N\to \operatorname{Perm}(S)$. Show that the induced homomorphism $\bar \phi$ is injective if and only if $N=\ker\phi$.
In fact, in this form the result holds for arbitrary groups in place of $\operatorname{Perm}(S)$ as a very general result about kernels and quotients.