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?

Reference: p. 1 https://www.math.okstate.edu/~mantini/4613info/ps04.pdf
Proof: We must use the result of exercise 4.3.19 in this problem. Note that in $\mathrm{Z}_{3}=\{0,1,2\}$, the orders of the elements are $|0|=1$, $o(1)=o(2)=3$.\ In $S_{3}= \{(1),$\ (12)\ ,\ (13)\ ,\ (23)\ $,\ (123), (132)\}$,the orders of the transpositions are each 2. 
Transpositions in $S_3$.
For any transposition $\tau$, $o(f(\tau))$ must divide $o(\tau)=2$, and this says that $o(f(\tau))=1$, since 3 does not divide 2. Therefore $f(\tau)=0$, the identity, for any  transposition $\tau$.
3-cycles in $S_3$.
Note that $(123)=(13)(12)$. This says that $f(123)=f((13)(12))= f(13)+f(12)=0+0=0$.
Similarly, $f(132)=0$, so $f(g)=0$ for any $g\in S_{3}$, if$f$ : $S_{3}\rightarrow \mathrm{Z}_{3}$ is a homomorphism. 
On the whole, there is only the trivial homomorphism from $S_3$ to $\mathbb{Z_3}$: $f(g) = 0$.♥
