$\pi:G\twoheadrightarrow G/X$, $X=\langle x\rangle \subset G$, $A=\langle aX\rangle \subset G/X$, $x,a\in G$, $x\neq e$, $a\neq e.$
$\pi:g\mapsto gX$.
$G$ is finite, abelian.
I want to show that $\pi^{-1}A$ is cyclic.
I assumed naturally that $a$ would generate $\pi^{-1}A$ if it is cyclic. So I tried to show that $\exists n$ such that $a^n=b$ for every $b\in\pi^{-1}A$.
Pick any $b\in \pi^{-1}A.$ then $\pi b\in A\implies bX\in A$. So $bX=a^iX$ for some $i$.
Thus $bx^k=a^i$ for some $k$.
$\implies x^k=b^{-1}a^i\implies e=x^{km}=b^{-m}a^{im}$, where $m=|X|$.
$\implies b^m=a^{im}.$
But I think I can't just take $m$-th root on both sides.
How to proceed from here?
Best Answer
If we replace $G$ by $\pi^{-1}(A)$ then your problem is the following: Suppose that $X$ is a cyclic subgroup of $G$ such that $A=G/X$ is cyclic. Show that $G$ is cyclic. As mentioned in the comments this is not true in general.
However, it is true if $gcd(|X|,|A|)=1$. You can prove it as follows: First use that (by the assumption) there is an element $g\in G$ which maps to a generator of $A$ and has order $|A|$. Then consider the order of $gx^k$, which is $|X|\cdot|A|$ for suitable $k$ (again use here that $gcd(|X|,|A|)=1$).
$G$ can always be of the form $X\times A$ which is not cyclic if $gcd(|X|,|A|)\neq1$, so the statement does not hold in general.
A more structural view: We are interested in the central extensions of the form $$1\longrightarrow X\longrightarrow G\longrightarrow A\longrightarrow1,$$ which are in bijection with the second group cohomology $H^2(A;X)$. In case that $A$ and $X$ are cyclic this group is isomorphic to $\mathbb{Z}_{|X|}/|A|\mathbb{Z}_{|X|}$. In case that $gcd(|X|,|A|)=1$ this group is trivial, so $G=X\times A$, which is cyclic (in this case). In the other cases the group is not trivial and there are several possibilities for $G$.