Prove that there does not exist an epimorphism from $S_3$ to $(\mathbb{Z}_6,+)$

Question

Prove that there does not exist an epimorphism from $S_3$ to $(\mathbb{Z}_6,+)$.

My approach:

Let $\phi: S_3\to \mathbb{Z}_6$ be an epimorphism.

$\bar{1}\in \mathbb{Z}_6$ and $o(\bar{1})=6$. Since $\phi$ is an epimorphism, there exists a pre-image $x$ of $\bar{1}$ in $S_3$ such that $\phi(x)=(\bar1)$. The possible orders of $x$ are $1,2,3$. We know $o(\phi(x))$ divides $o(x)$, which means $6$ divides $1,2$ or $3$, which is not possible. Hence such $\phi$ does not exist.

Is my approach correct?

Answer 1

Alternatively: Suppose $\phi: S_3\to \Bbb{Z_6}$ is an epimorphism. Then $\frac{S_3}{\ker \phi}\cong \Bbb{Z_6}$ by the First Isomorphism Theorem. Then the only possibility is $\ker \phi=\{e\}$. This implies $S_3\cong\Bbb{Z_6}$. This is impossible: $S_3$ has no element of order $6$.

Answer 2

Your proof is fine, although, at this level, I would specify the use of Lagrange's Theorem and the lemma on how to conclude $$o(\phi(x))\mid o(x).$$

---

Alternatively, you can use this lemma:

If $H$ has the same (finite) order as $G$, then any surjective map is necessarily injective.

. . . to conclude that the epimorphism would be an isomorphism, but, since $S_3$ is nonabelian while $\Bbb Z_6$ is cyclic, this is impossible.