homomorphism from $S_3$ to $\mathbb Z/3\mathbb Z$ 
TRUE/FALSE TEST:

*

*There is a non-trivial group homomorphism from $S_3$ to $\mathbb Z/3\mathbb Z.$

My Attempt:

True: Choose $a,b\in S_3$ such that $|a|=3,|b|=2.$
Then $S_3=\{1,a,a^2,b,ba,ba^2\}.$
Define $f:S_3\to\mathbb Z/3\mathbb Z:b^ia^j\mapsto j+3\mathbb Z$
Then $f$ is a nontrivial homomorphism.

Is the attempt correct?

HINTS:


*

*The image of a homomorphism is a subgroup of co-domain. Does $\mathbb{Z}_3$ has any subgroups except $\{\bar{0}\}$ and itself?

*Since $|S_3|=3!=6$ and $|\mathbb{Z}_3|=3$ then any function $\varphi: S_3 \to \mathbb{Z}_3$ will be not one-to-one. So, the kernel must be non-trivial.

*If $\varphi: S_3 \to \mathbb{Z}_3$ is a homomorphism that doesn't send everything to $\bar{0} \in \mathbb{Z}_3$ then it must be surjective (According to what I said in 1). Now, what happens if you use the first isomorphism theorem? Note that $\operatorname{ker{\varphi}}$ is a normal subgroup of $S_3$ but $S_3$ has no normal subgroups of order 2.
I have actually given you more information that you need, but to sum it up, there are no non-trivial homomorphisms from $S_3$ to $\mathbb{Z}_3$. In a fancy way, they write this as $\operatorname{Hom}(S_3,\mathbb{Z}_3)=\{e\}$ where $e: S_3 \to \mathbb{Z}_3$ is defined as $e(\sigma)=\bar{0}$ for any $\sigma \in S_3$.
