Explanation why an abelian tower admits a cyclic refinement

Solution 1:

The assumed homomorphism $G\rightarrow G'=G/X$ is the quotient map: $g\mapsto gX$. One of the isomorphism theorems says that this map establishes a bijection between subgroups of $G/X$ and subgroups of $G$ that contain $X$. Moreover, this bijection preserves inclusions, normality, and quotients. In other words if $\bar{U}$ is a subgroup of $G'$ with subgroup $\bar{U}'$, and if the corresponding subgroups of $G$ are $U$ and $U'$, then $U'\leq U$, it is normal in $U$ if and only if $\bar{U}'$ is normal in $\bar{U}$, and if they are, then $\bar{U}/\bar{U}'\cong U/U'$. All this is easy to prove, and Lang is using all of this in his proof.

The subgroup of $G$ corresponding to the trivial subgroup of $G'$ is indeed $X$ - the smallest subgroup of $G$ containing $X$.

As for the last question, it means exactly what it says: all factor groups are cyclic. It would be pointless to introduce this notion if it was simply a reformulation of "$G$ is cyclic". E.g. $1\leq \mathbb{Z}/2\mathbb{Z} \leq \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ is a cyclic tower where the top term is not cyclic.