$G$ modulo $N$ is a cyclic group when $G$ is cyclic If G is a cyclic group and N is a subgroup of G, show that G/N (or GmodN) is a cyclic group.
What I have so far:
Since N is a subgroup of the cyclic group G, G/N is a cyclic group.
I think I'm missing details.  What suggestions do you have?
 A: The exact statement goes for this as it did for your last post, you haven't actually stated why the result is true, you just said it (sorry if I sound harsh). Moreover, the same question I stated last time, applies here. More generally if $C$ is cyclic and $f:C\to H$ is a surjective homomorphism then $H$ is cyclic. Why does this help us? Why is it true?
A: Explicitly: $G$ is isomorphic to either $\mathbb{Z}$ or $\mathbb{Z}_n$ for some $n\in\mathbb{Z_{\geq 2}}$. 
In the first case, any subgroup is of the form $m\mathbb{Z}$ for some $m\in\mathbb{Z_{\geq2}}$ and so the quotient group is then isomorphic to $\mathbb{Z_m}$, which is cyclic.
In the second case, any subgroup of $\mathbb{Z}_n$ is isomorphic to $\mathbb{Z}_k$ where $k|n$. The quotient group $\mathbb{Z}_n/\mathbb{Z}_k$ is isomorphic to $\mathbb{Z}_{\frac{n}{k}}$ which again is cyclic.
So basically: all subgroups of cyclic groups are cyclic, and a cyclic group quotiented by a cyclic subgroup is again cyclic.
A: Now $G = \langle g \rangle$ and $N \trianglelefteq G$ since $G$ is abelian. Since $N$ is normal in $G$, the quotient group $G/N$ exists and it makes sense to wonder if it is cyclic. Then you can show that $G/N = \langle gN \rangle$.
