If $G$ is cyclic then $G/H$ is cyclic? 
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.
 A: One can make several arguments to the effect that $(Hg)^n$ is always an element of $G/H$. The simplest would be to appeal to the fact that quotient group is, in fact, a group - in particular, meaning that it is closed under products. If $(Hg)^2$ were not in the quotient, but $Hg$ was what would $Hg\cdot Hg$ be?
More elementarily, though, the quotient group $G/H$ is defined by looking at the cosets of $H$. So $Hx$ is in the quotient group for any $x$ in $G$ - so, since $g^n$ is in $G$, it follows that $Hg^n=(Hg)^n$ is in $H$.
A: $G$ 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. 
A: As other answers have noted, the quotient group is a group.  Because $G$ is cyclic, every element of $G$ can be written in the form $g^n$ for some $n \in \mathbb{N}$.  Thus, all the right cosets will be of the form $Hg^n$ for some $n \in \mathbb{N}$, and you can arrive at $Hg^n$ by "multiplying" (as it's been defined) $Hg$ by itself $n$ times.  Also, if $G/H$ is smaller than $G$⁠—which happens whenever ⁠$H$ is a nontrivial normal subgroup—then there will indeed be some $n$ missing, in a sense$^\dagger$: we will have $Hg^n = Hg^m$ for some $n \neq m$.  But this will be of no consequence to the above proof.
Here is a different way of approaching this problem, sans cosets$^\ddagger$:
If $H$ is a subgroup of a cyclic group $G$, then $H$ is necessarily normal since every subgroup of an abelian group is normal (cyclic implies abelian).  As such*, there exists a group $G_1$ and a surjective group homomorphism $\phi:G \rightarrow G_1$ such that $\ker(\phi) = H$.  From the isomorphism theorem, we get $G_1 \cong G/H$.  
So the problem is equivalent to determining whether the homomorphic image of a cyclic group is cyclic.  This is rather immediate: assuming $g$ generates $G$, consider any $a \in G_1$.  We have $a = \phi(g^n)$ for some $n$, and further $\phi(g^n) = \phi(g)^n$.  So we see that $\phi(g)$ generates $G_1$.

$$\underline{\textbf{Footnotes}} \\[0.5em]$$
$^\dagger$This is to say: as we pass to the quotient, some elements are now regarded as "the same" modulo $H$.  For more about this, see my post here for analogous discussion pertaining to rings and their quotients; for groups of course, there is only $1$ operation / operation table.

$^\ddagger$To be perfectly accurate, it's not that we aren't using cosets; rather, we are sweeping them under the rug of the isomorphism theorem.

*Indeed, in general, $G/H$ is a group $\iff H$ is normal.  There are many equivalent notions for normality.  Click here for further discussion.
A: If $g'$ is any element of a group $G,$ and if $H$ is a subgroup of $G$, then $Hg$ is by definition a right coset of $H$ in $G.$ This includes the case that $g'=g^2$ for some $g\in G,$ regardless of whether $G$ is cyclic or $H$ is a normal subgroup of $G.$
You're pretty much done, as long as you can justify your claims.
A: In general homomorphism maps cyclic group to cyclic subgroup. We know that the quotient homomorphism $\pi$ maps $G$ onto $G/H$, i.e. $G/H=\pi(G)$. So if $G$ is cyclic, then $G/H$ must also be cyclic. 
I don't know if this is a more natural way to think about this...
