From page 44 Qn 6 of Abstract Algebra – Dummit&Foote,
Group action is faithful if and only if the kernel of the action is the set consisting of the identity
I will be using the definition of faithful group action below:
An action of $G$ on $A$ is said to be faithful if distinct elements of $G$ induce distinct permutations of $A$, ie the associated permutation representation is injective.
Proof attempt:
$\Rightarrow$: If the kernel is not trivial, then there exists some $g' \in G, g' \neq e$ such that $g' \cdot a = a, \forall a \in A$. The permutation representation associated to the action $\varphi : G \rightarrow S_A, g \mapsto \sigma_g$ is not injective, since $\varphi(g') = \sigma_{g'} = id_{S_A} = \sigma_e = \varphi(e)$.
$\Leftarrow$: Suppose the action is not faithful (so $\varphi$ is not injective) Then there exists $g_1,g_2 \in G, g_1 \neq g_2$ such that $\sigma_{g_1} = \sigma_{g_2}$. Since $\sigma_{g_1}^{-1} \circ \sigma_{g_2} \in S_A$ is the identity permutation, and $\varphi$ is a homomorphism, we have $g_1^{-1}g_2 \in $ Kernel of the group action. Note $g_1^{-1}g_2 \neq e$, otherwise $g_1 = g_2$. Thus, kernel is not trivial.
Any comments on the proof presented?
Edit: Added some definitions
Best Answer
Your proof is fine. Did you have any specific doubts about it?