Finding number of homomorphisms from $\mathbb{Z} \to S_3$ Here's a question that I'm trying to solve:

Let $a$ be the number of group homomorphisms $\mathbb{Z} \to S_3$ ;
  let $b$ be the number of such one-one group homomorphisms; let $c$ be
  the number of such onto group homomorphisms. Find $(a,b,c)$.

I claim that $c=0$. Proof of my claim: Suppose there were some onto homomorphism $\varphi : \mathbb{Z} \to S_3$. Since $\mathbb{Z}$ is cyclic, then $\varphi (\mathbb{Z})= S_3$ would be cyclic and thus abelian which is a contradiction.
Also that $b=0$ for we cannot find any mapping $\varphi$ which is one-one.
Now, finding $a$ is a big trouble. Here's my idea how to find $a$: Lagrange's Theorem tells us that there can be every subgroup of $S_3$ is either of the orders: $1,2,3,6$. And we know that $\varphi (\mathbb{Z})$ is a subgroup of $S_3$ so I tried manually to find the mappings. The answer I get is $6$. I was wondering if there was some easier way to do instead of explicitly finding the homomorphisms.
Yet another question I want to ask: Is the number of onto homomorphism from $\mathbb{Z} \to \mathbb{Z}_n$ equals the number of generators $\mathbb{Z}_n$ has i.e $\phi (n)$? I don't seek the proof but I was wondering if it was true.
 A: To give a more elaborate answer : note that if $\phi : \Bbb Z \to S_3$ is a homomorphism, then for all $z \in \mathbb Z$, we have that $\phi(z) = (\phi(1))^z$. There is no other restriction : note that $0$ will map to $0$ anyway, and $\phi(1)$ can be any element of $S_3$. This gives us SIX homomorphisms in this direction. 
For $\phi$ to be injective, the kernel of $\phi$ must be trivial. However, note that the kernel cannot be non-trivial. This is because, every element of $S_3$ has order dividing $6$, and therefore, if $\phi$ is any hommomorphism then $\phi(6) = (\phi(1))^6 = e$, so $6 \in \ker \phi$, hence $\phi$ is not injective.
For any homomorphism $\phi$, the image of $\phi$ is generated by $\phi(1)$ i.e. it is $\langle\phi(1)\rangle$. This cannot be equal to $S_3$ since $S_3$ is not cyclic, hence cannot be generated by one element. Consequently, no homomorphism is surjective.

As for the answer to the second question, indeed $\phi : \mathbb Z \to \mathbb Z_n$ is onto, if and only if $\langle\phi(1)\rangle = \mathbb Z_n$, if and only if $\phi(1)$ to be a generator of $\mathbb Z_n$, if and only if (any representative of the equivalence class) $\phi(1)$ is coprime to $n$. There are exactly $\phi(n)$ choices for this $\phi(1)$, and consequently, exactly those many onto homomorphisms.
