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

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    $\begingroup$ Please make clear why you think that the claim is not in fact the definition of faithful $\endgroup$ – Hagen von Eitzen Jul 26 '17 at 19:23
  • $\begingroup$ @HagenvonEitzen perhaps i should have written down what the given definition of faithful was $\endgroup$ – Timothy Jul 26 '17 at 19:27
  • $\begingroup$ The "associated premutation representation" is a homomorphism $\phi$ from $G$ to the group of permutations of $A$. By definition, the action is faithful iff $\phi$ is injective. You ought to know already that a group homomorphism $\phi$ is injective if and only if $\ker\phi=1$. (You are essentially repeating the proof of this) $\endgroup$ – Hagen von Eitzen Jul 26 '17 at 22:14

Your proof is fine. Did you have any specific doubts about it?

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  • $\begingroup$ It feels like a pretty long winded proof of something that should be straightforward imo.. But other than that, thanks for your feedback! $\endgroup$ – Timothy Jul 27 '17 at 3:26

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