 Faithful coset action of the symmetric group on three letters

Take the symmetric group $G=S_3$ and the subgroup $H=⟨(12)⟩=\{e,(12)\}$ of group order two. With the rightmost permutation applied first, its three left cosets areThe subscripts record the image of $3$: every member of $C_i$ sends $3$ to $i$. Because $H$ is the stabilizer subgroup of $3$ in the usual group action, $aH\mapsto a(3)$ is a bijection onto $\{1,2,3\}$. It is an equivariant map, since $(ga)(3)=g(a(3))$. The coset action therefore becomes the familiar group action on three letters. In particular,

The three stabilizer subgroups areTheir intersection is $\{e\}$, so the subgroup core and the group kernel are trivial. This proves that the coset action is a faithful group action even though every point has a nontrivial stabilizer subgroup. The orbit-stabilizer theorem reads $6=3\cdot 2$, while the induced group homomorphism gives six distinct permutations of the three cosets. For example, $(12)$ and $e$ agree at $C_3$, but disagree at $C_1$; inspecting one label would miss their difference.

The subgroup need not be a normal subgroup for the coset action to exist: $(123)H(123{)}^{-1}=\{e,(23)\}\ne H$. Here the group action is informative because the three conjugate stabilizer subgroups have no common nonidentity element. For the normal subgroup $A_3$ instead, nonfaithful coset action of the symmetric group on three letters shows that only the sign of a permutation survives.