0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $\{1,2,3,4\}$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\langle (123) \rangle$). $\endgroup$ – Geoff Robinson Dec 15 '14 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.