TITLE: If $G$ is cyclic then $G/H$ is cyclic?
QUESTION [12 upvotes]: If $G$ is cyclic, then $G/H$ is cyclic?

The proof I got goes like this: $G$ is cyclic, so $G=<g>$ for some $g\in G$. So any coset in $G/H$ would be of the form $Hg'=Hg^n$ for some $n$. So $Hg$ is an generator of $G/H$. Thus, $G/H$ is cyclic.
I might just be confusing myself, but we have only shown that $Hg'$ is in form of $(Hg)^n$. But what if we are missing some $n\in\mathbb{N}$? That is, there is no quotient group of the form $Hg^2$, for example.
To make myself a little bit clearer, I think what the above proof has done was showing that $\forall Hg'\in G/H, Hg'\in <Hg>$, thus $G/H \subset <Hg>$. I feel that this is not a complete proof.

REPLY [2 votes]: $G$ cyclic
$H \trianglelefteq G$
$G/H = \{aH: a \in G\}$
$\langle x \rangle = G$ where $x \not = e$
$a = x^k \Rightarrow aH = x^kH = (xH)^k$
This was for arbitrary element so done.