 Solution

LetThe product of any two distinct nonidentity elements is the third, and each squares to $e$, so $K$ is a Klein four-group. Conjugating a double transposition simply relabels its letters, keeping it in the set. Hence $K$ is a normal subgroup of $A_4$.

Choose $r=(123)$. The cosets $K,rK,r^{2}K$ are distinct and cover the twelve elements. With $\zeta =e^{2\pi i/3}$, defineThe coset decomposition makes this well-defined. Normality givesso $\Phi$ preserves products. It is onto $\{1,\zeta ,{\zeta }^2\}\cong C_3$, andThis is the tetrahedral quotient onto the cyclic group of order three. In terms of (a), it sends $C_0\cup C_2$ to $1$, $C_{+}$ to $\zeta$, and $C_{-}$ to ${\zeta }^2$. The normality check above also follows abstractly because a group homomorphism kernel is normal.