If $G$ is cyclic then $G/H$ is cyclic?
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 $G_1$.
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$^\dagger$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.
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 $H$.
$G$ 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.