= Faithful coset action of the symmetric group on three letters
{title2=$\operatorname{Core}_{S_3}(\langle(12)\rangle)=\{e\}$}
Take the <symmetric group> $G=S_3$ and the <subgroup> $H=\langle(12)\rangle=\{e,(12)\}$ of <group order> two. With the rightmost <permutation> applied first, its three left <cosets> are
$$
C_3=H=\{e,(12)\},\qquad C_2=(23)H=\{(23),(132)\},\qquad C_1=(13)H=\{(13),(123)\}.
$$
The 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,
$$
(12):C_1\leftrightarrow C_2,\ C_3\mapsto C_3,\qquad
(123):C_1\mapsto C_2\mapsto C_3\mapsto C_1.
$$
\Image[/s3-three-coset-action.png]
{title=Three left <cosets> turn left multiplication in $S_3$ into its <faithful group action> on three labels}
{id=s3-three-coset-action}
{description=An original computed <coset> partition matches each two-element <coset> to the common image of $3$. The two rows show left multiplication by $(12)$ and $(123)$. The three <stabilizer subgroups> intersect only in the <identity element>.}
The three <stabilizer subgroups> are
$$
G_{C_1}=\{e,(23)\},\qquad G_{C_2}=\{e,(13)\},\qquad G_{C_3}=\{e,(12)\}.
$$
Their <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 $6=3\cdot2$, 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.
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