Question: Does there exist nontrivial homomorphism between symmetric group $S\_n$ to set of integer $\mathbb Z$?

If yes, how many?

If not, then why?

Answer: 
An element of finite order is by a homomorphism always mapped to an element of finite order. This is proven quite directly from the defining properties of homomorphisms. So given an $s\in S\_n$ and a homomorhism $h:S\_n\to \Bbb Z$, what can $h(s)$ possibly be? Does this argument change for $D\_n$ instead of $S\_n$?
If $\phi$ is a group homomorphism between two groups $G$ to $H$ , then for any $a \in G$ , $o(\phi(a)) | o(a) \ldots (\*)$ . Since we are considering homomorphisms from $S\_n$ to $\Bbb Z$ and $S\_n$ is a finite group, any element in $S\_n$ is a divisor of $n!$ and hence finite . Thus its image element must also be of finite order (due to $(\*)$), but in $\Bbb Z$ , 0 is the only element of finite order, thus all the elements in $S\_n$ should be mapped to 0 i.e. the trivial 0-homomorphism is the only homomorphism from $S\_n$ to $\Bbb Z$ .

Note that here we are not using any special property of $S\_n$ , in fact the statement is true for any finite group .