Reference: http://users.humboldt.edu/pgoetz/Homework%20Solutions/Math%20343/hwsome number 1 to 17 that I forgotsolns.pdf
There are exactly two nontrivial homomorphisms $\mathbb{Z_3}$ to $S_3$.
One is determined by $f([1]_3) = (1,2, 3)$; the other is determined by $g([1]_3) = (1, 3, 2)$.
(1.) How do you magically know there are 2 nontrivial homomorphisms?
(2.) How do you envisage and envision these? Where did they magically loom from?
(3.) What's the blueprint for the fastest way to check these homomorphisms? Do I actually have to pair up all the permutations?
I was thinking to start with $f([2]_3) = f([1]_3 + [1]_3) \\ = f([1]_3) \circ f([1]_3)$ and $f([0]_3) = f([1]_3 + [2]_3) \\ = f([1]_3) \circ f([2]_3)$.
How do you shirk multiplying $f([i]_3) \circ f([j]_3) $ for all $1 \le i, j \le 3$? Same for $g$.
The question for the other direction $S_3$ to $\mathbb{Z_3}$. homomorphism from $S_3$ to $\mathbb Z/3\mathbb Z$.
A homomorphism $\phi:\mathbb Z_3\to G$ is uniquely determined by the image $\phi([1]_3)$ of the standard generator of $\mathbb Z_3$. Also, $g\in G$ can serve as such image $\phi([1]_3)$ if and only if $g^3=1$.
Thus $g$ is either $1$ (which leads to the trivial homomorphism) or an element of order $3$. In other words, the number of nontrivial homomorphisms $\mathbb Z_3\to G$ is the same as the number of elements of $G$ of order $3$. For $G=S_3$ these are precisely the elements $(1\,2\,3)$ and $(1\,3\,2)$.