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?
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?
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?
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.
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 $G/N = \langle gN \rangle$.