Question: 1) Assume that there exist homomorphisms $f:\mathbb{Z} \rightarrow S\_{3}$ where $\ker(f)\neq \mathbb{Z}$? What are the possible kernels? Explain.  

2) Lest at least two non-trivial homomorphisms $f:\mathbb{Z} \rightarrow S\_{3}$ whose kernel are not $\mathbb{Z}$.

Answer: 
Note that if $\ker(f) \neq \mathbb{Z}$, then $f(1) \neq e$ ($e$ referring to the identity of $S\_3$). Why?

With the above established, we have two possibilities: either the order of $f(1)$ is $2$, or the order of $f(1)$ is $3$. In the first case, we have $f(2k) = f(2)^k = e^k = e$, so that $\ker(f) = 2\mathbb{Z}$. Similarly, in the second case, we have $\ker(f) = 3\mathbb{Z}$.

**Examples:**

* $f:\mathbb{Z}\to S\_3, f(k) = (1\quad 2)^k$
* $f:\mathbb{Z}\to S\_3, f(k) = (1\quad 2\quad 3)^k$
Hint : $\mathbb{Z}\_2$ and $\mathbb{Z}\_3$ are subgroups of $S\_3$ (of course upto isomorphism) .