Find all the functions \(\displaystyle{f:\left(\mathbb{Z}_{6},+\right)\longrightarrow \left(S_{3},\circ\right)}\) having the property \(\displaystyle{f(x+y)=f(x)\circ f(y)\,,\forall\,x\,,y\in\mathbb{Z}_{6}\,\,\,\,\,\,}\) (homomorphism) .

Every element of symmetric group ${\cal{S}}_{3}=\big\langle{\rho,\,\tau \; \big| \ \;\rho^{3}=\tau^2={\rm{id}}, \ \tau\,\rho=\rho^2\,\tau}\big\rangle$ has order either $1$, or $2$, or $3$. Let $f:{\mathbb{Z}}_{6}\longrightarrow{\cal{S}}_{3}$ be a group homomorphism. Then must hold
\begin{align*}
&\big[\circ\!\big(f([1]_6)\big)=1\; \veebar \;\circ\big(f([1]_6)\big)=2\; \veebar \;\circ\big(f([1]_6)\big)=3\big]\quad\Rightarrow\\\\
&\left[\begin{array}{r}
f([1]_6)={\rm{id}}\; \veebar \;f([1]_6)=\rho\; \veebar \;f([1]_6)=\rho^2\; \veebar\\
f([1]_6)=\tau\; \veebar \;f([1]_6)=\rho\,\tau\; \veebar \;f([1]_6)=\rho^2\,\tau
\end{array}\right]\,.
\end{align*} In each case, demanding for every $[k]_6\in\mathbb{Z}_6\,,\; k\in\{0,1,\ldots,5\}$ that $f\big([k]_6\big)=k\,f([1]_6)$ we have six different homomorphisms, of which none is an epimorphism nor monomorphism. These homomorphisms are all the homomorphisms from ${\mathbb{Z}}_{6}$ to ${\cal{S}}_{3}$.

_________________
Grigorios Kostakos