Cyclic quotient group as a subgroup? There are many questions here along the lines of the following: Let $G$ be a finite group, and $H$ a normal subgroup. Does $G$ admit a subgroup which is isomorphic to $G/H$?
This is necessarily true if $G$ is abelian, and there are counterexamples if $G$ is not, the most common counterexample being the quaternion group of order 8 (see here), which has a quotient isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ (but no such subgroup).
My question is: if we additionally assume that the quotient $G/H$ is cyclic, then is it the case that $G$ must admit a subgroup isomorphic to $G/H$? Or are there counterexamples in this situation also?
My gut tells me it's the latter, that it's not necessarily true, but I so far haven't been able to find an example.
 A: This does not hold if $G$ is infinite, but you have the "finite-groups" tag. In this case, this holds.
Suppose $G$ is finite and $G/H$ is cyclic of order $k$. Then there exists an element $g\in G$ such that $k$ is the smallest positive integer such that $g^k\in H$, namely any pre-image of a generator of $G/H$.
I claim that the order of $g$ is a multiple of $k$. Indeed, if the order of $g$ is $n$, then $g^n\in H$, and so by the standard division-with-remainder argument, we conclude that $k\mid n$: write $n=kq+r$, $0\leq r\lt k$. Then $g^n = (g^k)^q g^r\in H$, and since $g^k\in H$ we conclude that $g^r\in H$. Minimality of $k$ now ensures that $r=0$.
Since $G$ has an element or order a multiple of $k$, it also has an element of order $k$: just take $g^{n/k}$. And then $\langle g^{n/k}\rangle\cong G/H$, since they are both cyclic of order $k$.
The infinite case is not true even for abelian groups, since the infinite cyclic group has finite nontrivial cyclic quotients but no finite nontrivial cyclic subgroups.
