S3 group action faithful?

I'm struggling with understanding the term "faithful". I read that a group action for example $S_3$ is faithful on {1,2,3}. Does that mean $S_3$ is not faithful on {1,2,3,4} because it never changes 4? I've read that a group action is faithful when the homomorphism is injectiv ergo it has trivial kernel. But I struggle to understand that, what is the kernel of $S_3$ is on {1,2,3,4} or $S_3$ on {A,B} for example. Thanks

A group $G$ acts faithfully on $X$ if the identity of $G$ is the only group element that leaves all elements of $X$ fixed.

The natural action of $S_3$ on $\{1,2,3\}$ is faithful because any non-identity permutation does not leave all elements of $\{1,2,3\}$ fixed. Therefore, the action of $S_3$ on $\{1,2,3,4\}$ given by permuting the first $3$ elements is also faithful.

A group action is faithful if the homomorphism $G \to S_X$ is injective. This is the case for these actions, because different permutations in $S_3$ act "differently" on the elements of $\{1,2,3\}$ and $\{1,2,3,4\}$.

Recall that a group action can be defined as follows. Let $X$ be a set and denote by $Sym(X)$ the group of all permutations of $X.$ Then an action of a group $G$ on the set $X$ is a group homomorphism $\varrho: G \rightarrow Sym(X).$ Then, the action $\varrho$ is called faithful if $ker\ \varrho = \{1_G\},$ as you noted in your question.

Now, the group $S_3$ acts "naturally" on the set $X := \{1,2,3\},$ i.e. that's one way to define $S_3.$ Clearly, an element $\sigma$ of $S_3$ acts trivially (i.e. as the identity) on $X$ iff $\sigma = 1_{S_3}.$ More formally, we have for $\sigma \in S_3$ that $\sigma \in ker\ \varrho = \{\tau \in S_3|\varrho(\tau)=id_X\} \Leftrightarrow \varrho(\sigma) = id_X \Leftrightarrow \sigma$ acts as the identity on $X \Leftrightarrow \sigma = 1_{S_3}.$ So we have shown that $ker\ \varrho = \{1_{S_3}\}$ and thus, this action is faithful.

Moreover, the set $X$ from the preceding paragraph is obviously a subset of the set $Y := \{1,2,3,4\}.$ With this observation, we can extend the action of $S_3$ on $X \subseteq Y$ to an action of $S_3$ on $Y$ by stipulating that any element of $S_3$ fixes any element of $Y\setminus X.$ More formally, we define $\widetilde\varrho:S_3\rightarrow Sym(Y)$ by the following rule. For $\sigma \in S_3$ and $m \in Y,$ we put $(\widetilde\varrho(\sigma))(m) = (\varrho(\sigma))(m)$ if $m \in X \subseteq Y$ and $(\widetilde\varrho(\sigma))(m) = m$ if $m \in Y\setminus X.$ You should convince yourself that $\widetilde\varrho$ is indeed a group homomorphism.

With this extended action of $S_3$ on $Y$, we get for $\sigma \in S_3$ that $\widetilde\varrho(\sigma) = id_Y \Leftrightarrow \sigma$ fixes every element of $Y \Rightarrow \sigma$ fixes every element of $X \Leftrightarrow \sigma = 1_{S_3},$ by our above considerations. Thus we have shown that $\widetilde\varrho$ is faithful.

Finally, you mention an action of $S_3$ on the set $Z := \{A,B\}.$ I have no idea what that would be.

• Re the last remark, perhaps the following was meant: Since $S_{2}$ is a homomorphic image of $S_{3},$ it is the case that there is an action of $S_{3}$ on $\{1,2 \}$ which is not faithful (the kernel is $\langle (123) \rangle$). – Geoff Robinson Dec 15 '14 at 11:40