 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 areAll 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 $G^{\prime }\cong S_3/K$. 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. ThusEach 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.