= Solution
The <symmetric group> $S_3$ is generated by transpositions, each of order two. A <group homomorphism> to $\mathbb Z_3$ must map every transposition to zero, because the only target element annihilated by two is zero. Hence \b[the map to $\mathbb Z_3$ is necessarily trivial].
For $S_3\to\mathbb Z_2$, the 3-cycles map to zero because their order divides three. All transpositions are <conjugate group elements>, and a map to an <abelian group> takes conjugate elements to the same value. That common value is either zero or one. Both choices work: they give \b[the trivial map and the sign homomorphism], with an odd permutation sent to $1$ and an even permutation sent to $0$.
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