Basic question on exact sequences and $\mathbb Z_n$ Suppose we have a short exact sequence:
$$0\rightarrow A\rightarrow \mathbb{Z}_{n}\rightarrow C\rightarrow 0$$
Typically, I've understood short exact sequences as:
$$0\rightarrow A\rightarrow B\rightarrow B/A\rightarrow 0$$
However, my algebra is a bit rusty, and I am confused about what examples for $A$ and $C$ might make the first sequence exact. I am familiar with the the typical $\mathbb Z/n\mathbb Z\cong \mathbb Z_n$ quotient, but am confused how that fits in if we sandwich a cyclic group in the middle as above (what is an example an $A,C$ pair for the first sequence?). On a related note, is it true that $\mathbb Z_{mn}/\mathbb Z_n\cong \mathbb Z_m?$ 
Edit: considering exact sequences among abelian groups.
 A: You have the right ideas. If the cyclic group $C_n$ is in the middle, then by exactness $A$ has to be a subgroup of $C_n$. This implies that $A \cong C_m$ with $m \mid n$, and this in turn implies that $C \cong C_n / C_m \cong C_{n/m}$. (Note that a quotient of a cyclic group is necessarily cyclic, for an epimorphism maps generators to generators.)
A: I suppose we are talking about exact sequences among Abelian groups.
Well, $0\to A\overset f\to B$ being exact means that $f$ is a monomorphism (has trivial kernel), i.e. is injective, so that we can regard $A$ as a subgroup of $B$. 
Then the exactness of the other part $A\overset f\to B\to C\to 0$ indeed implies $C\cong B/A$.
Now, if given $B$, you can choose $A$ to any of its subgroups, and this will also give you the $C=B/A$. 
When $B=\Bbb Z_n$, choosing $A$ is equivalent to choosing a divisor $d$ of $n$ and then set $A=\Bbb Z_d\cong\{0,\frac nd, \frac{2n}d,\dots\}\subseteq\Bbb Z_n$.
About your last question: yes, that's correct: if $x,y,\dots$ generate a group $G$, then they will generate any quotient of $G$. In particular, every quotient of a cyclic group is cyclic. Then, counting orders, it yields to $\Bbb Z_{nm}/\Bbb Z_n\cong\Bbb Z_m$.
Note also that if $n,m$ are coprime (i.e. $\gcd(n,m)=1$) then the short exact sequence splits, that is, $\Bbb Z_{nm}\cong\Bbb Z_n\times\Bbb Z_m$.
