= Solution
The kernel is a <normal subgroup>, so it contains the identity class and is a union of the classes listed in part (c). Its candidate orders are therefore $1$, $1+3=4$, $1+2=3$ and $1+3+2=6$. <Lagrange's theorem> excludes four, since it does not divide six. The remaining possibilities are
$$
\boxed{K=\{e\},\qquad K=A_3=\{e,(123),(132)\},\qquad K=S_3.}
$$
All three are normal: the first and last trivially, while $A_3$ is the kernel of the map giving the <sign of a permutation>, or the index-two subgroup generated by a 3-cycle.
Because $\Phi$ is onto, the <first isomorphism theorem> gives $G'\cong S_3/K$. Explicitly, the map $gK\mapsto\Phi(g)$ is well defined, multiplicative and surjective; equality of its values means $h^{-1}g\in K$, so it is also injective. Thus
$$
\boxed{G'\cong S_3\quad\text{or}\quad C_2\quad\text{or}\quad\{e\}.}
$$
Each is realized by the identity map, the sign map and the trivial map, respectively. This completes the classification of <normal subgroups and quotients of the symmetric group on three letters>.
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