Prove that $\phi(f)=\hat0$ Assume that $\phi:S_3 \to \mathbb Z_3$ is an homomorphism and $f\in S_3$ is a function which is defined this way:
$f(1)=2$ , $f(2)=1$ , $f(3)=3$  
Prove that $\phi(f)=\hat0$.  
My problem: I don't understand why this should happen! And that's not the problem. I tried to prove what the question wants. But, for example if i write $\phi(f(1))=\phi(2)$, Then i don't know anything about the definition of $\phi$! What is $\phi(2)$? It doesn't help me to prove...
Note 1: $\mathbb Z_3=\{\hat0,\hat1,\hat2\}$ ( The system of remainders of division by $3$ )  and $S_3$ means all of the permutations from $\{1,2,3\}$ to $\{1,2,3\}$.  
Note 2: I've found an answer here at page 62. But it speaks about 2-cycles (that i know nothing about them). Can someone please say it in a more simple way?
 A: Note that $f$ is really the permutation $(1\ 2)$, and so $f$ has order $2$. Thus, $f^{2}=()=e$ (the identity permutation in $S_{3}$), and since $\phi$ is a homomorphism, it sends identities to identities, and we must have
$$\phi(e)=\phi(f^{2})=\phi(f)+\phi(f)=0$$
If $\phi(f)=1$ or $\phi(f)=2$, then $\phi(f^{2})\neq 0$.
You need to be careful, because the elements of $S_{3}$ are not numbers. They are functions which permute the set $\left\{1,2,3\right\}$. Since $S_{3}$ is the domain of $\phi$, then $\phi$ takes functions as an input. The expression $\phi(2)$ doesn't make sense in this context because $2\notin S_{3}$.
These functions of $S_{3}$ make a group under the operation of function composition. So the composition of the permutations $(1\ 2\ 3)$ and $(3\ 2)$ is
$$(1\ 2\ 3)(3\ 2)=(1\ 2).$$
We look at this composition in reverse order. 
In the first cycle, $1$ is fixed and in the second cycle $1\mapsto 2$, so $1\mapsto 2$ in this composition.
In the first cycle, $2\mapsto 3$ and in the second cycle $3\mapsto 1$, so $2\mapsto 1$ in this composition. 
In the first cycle, $3\mapsto 2$ and in the second cycle $2\mapsto 3$, so $3$ is fixed in this composition. 
Also, note that $S_{3}$ is non-abelian, because
$$(3\ 2)(1\ 2\ 3)=(1\ 3)\neq (1\ 2).$$
