Given that $G$ is cyclic then $G/N$ is cyclic.

Question

$\newcommand\sg[1]{\langle#1\rangle}$I'm kind of lost in trying to write the proof. My book has an example involving a Cayley Table of how $\mathbb{Z} / 4\mathbb{Z} \approx \mathbb{Z}_4$. So I can see that since one group is isomorphic to the other it must be cyclic as well. I'm having trouble writing the proof however. My idea so far is this:

proof: Let $G$ be a cyclic group, then $\exists \sg a,\sg b\in G$. Since $G$ is cyclic we see that $N$ is cyclic as well because it is a subgroup of G. We see that $$\sg a+H+\sg b+H=\sg{a+b}+H+H=\sg{a+b}+H$$ Since $\sg{a+b}+H$ is cyclic then we see that all the representative of the cosets will also be cyclic.

Excuse my super weak proof. I mainly just desire a hint or a nudge in the right direction.

Answer 1

Sketch:

• Recall that a group is cyclic if and only if it is generated by a single element.

• Let $g$ be a generator for $G$. Show that $g+N$ generates $G/N$.

Answer 2

Recall that a group $G$ is cyclic if and only if there exists a surjective group homomorphism $\gamma:\Bbb Z\to G$. In that case, $\pi\circ\gamma:\Bbb Z\to G/N$, where $\pi:G\to G/N$ is the canonical projection, is a surjective group homomorphism as well, hence $G/N$ is cyclic.