Homomorphisms from $\mathbb{Z} \rightarrow S_3$. I recently had my second abstract algebra exam returned to me and I missed this particular question (unfortunately I left the problem blank, just a bad test day). I've been trying to figure out the problem though and could use some help.
a) Does there exist a surjective homomphism $\phi: \mathbb{Z} \rightarrow S_3$? Explain.
What I'm thinking: Let $\phi: \mathbb{Z} \rightarrow S_3$ be a homomorphism.$\mathbb{Z}$ is a cyclic group and thus $\phi(\mathbb{Z})$ is cyclic. However, $S_3$ is not cyclic. Thus $\phi(\mathbb{Z}) \neq S_3$ and $\phi$ cannot be surjective.
b) Assume that there exist homomorphisms $\phi: \mathbb{Z} \rightarrow S_3$ where Ker$(\phi) \neq \mathbb{Z}$. What are the possible kernels? Explain.
What I'm thinking: Since $1$ generates $\mathbb{Z}$, if $1\in$ Ker$(\phi)$, then $\mathbb{Z}=$Ker$(\phi)$. So could Ker$(\phi)$ = $n\mathbb{Z}$ for $n \in \mathbb{Z},  \; n\neq1$?
c)List at least two non-trivial homomorphisms $\phi: \mathbb{Z} \rightarrow S_3$ whose kernals are not $\mathbb{Z}$.
$\phi_1(n) = (1\; 2)^n$
$\phi_2(n) = (1\; 3)^n$
Any input is much appreciated! 
 A: a) is fine. But for b) Note that $n\in\ker \phi$ if $g^n=1$ for $g:=\phi(1)$. As such $g$ can only have orders $2$ or $3$ (or of course  $1$) in $S_3$, only $2\mathbb Z$ and $3\mathbb Z$ (and of course $\mathbb Z$ that was excluded) are possible kernels.
c) is fine - though in the light of b) you were probably supposed to take $n\mapsto (1\,2\,3)^n$ as one of the examples.
A: The nice thing about the group $\mathbb{Z}$ is that it is the free group generated by one element. That is, you can map a generator (1 or -1) of $\mathbb{Z}$ to any element and that defines a (unique) group homomorphism. One way of seeing the existence of such a homomorphism is as follows: 
Let $G$ be a group and $g \in G$. Then the subgroup $\langle g \rangle$ of $G$ is cyclic and hence is isomorphic to a quotient of $\mathbb{Z}$. This gives a surjection $\mathbb{Z} \twoheadrightarrow \langle g \rangle$ which sends $1$ to $g$. Composing this with the inclusion $\langle g \rangle \hookrightarrow G$ yields a group homomorphism from $\mathbb{Z}$ to $G$ that sends $1$ to $g$.
The homomorphism $\mathbb{Z} \rightarrow G$ with $1 \mapsto g$ has zero kernel if and only if $g$ has infinite order. Otherwise the kernel is $n \mathbb{Z}$ where $n$ is the order of $g$. So the nonzero kernels for the group homomorphisms from $\mathbb{Z} \rightarrow G$ are in one-to-one correspondence with the orders of the non-identity elements in $G$.
