= Nonfaithful coset action of the symmetric group on three letters
{title2=$\ker\rho=A_3$}
Let $G=S_3$, the <symmetric group> on three letters, and take the <normal subgroup> $N=A_3=\{e,(123),(132)\}$, the <alternating group> of <even permutations>. Its two left <cosets> are
$$
E=N=\{e,(123),(132)\},\qquad O=(12)N=\{(12),(13),(23)\}.
$$
Products of <permutations> apply the rightmost factor first. Left multiplication by an <even permutation> preserves both <cosets>, while left multiplication by an <odd permutation> exchanges them. For example,
$$
(123)\cdot E=E,\quad (123)\cdot O=O,\qquad
(12)\cdot E=O,\quad (12)\cdot O=E.
$$
Consequently all six <group> elements give only two distinct <permutations> of $\{E,O\}$. Both <stabilizer subgroups> equal $A_3$, so their intersection, the <group kernel>, is $A_3$. This is a <transitive group action> because $(12)$ reaches the other <coset>, but it is not a <faithful group action> because $(123)$ fixes both.
\Image[/s3-two-coset-action-kernel.png]
{title=A <transitive group action> of $S_3$ on two <cosets> detects only the <sign of a permutation>}
{id=s3-two-coset-action-kernel}
{description=This computed example groups the six <permutations> by their induced <permutations> of the two <cosets>. All three <even permutations> fix both labels; all three <odd permutations> exchange them. The <group kernel> has three elements.}
The <orbit-stabilizer theorem> still works perfectly: $6=2\cdot3$. It counts reachable labels, without asserting that the induced <group homomorphism> is <injective>. The <quotient group> $S_3/A_3\cong C_2$ captures exactly the two visible operations, identity and exchange. This is the same <group homomorphism> as the <sign of a permutation>, with $+1$ represented by fixing both labels and $-1$ by exchanging them. Contrast the three-label example in <faithful coset action of the symmetric group on three letters>.
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