 Solution

Write $C=⟨c⟩$ with $|c|=n$. For each divisor $k$ the subgroup $⟨c^{n/k}⟩$ has order $k$. Conversely, if $H$ has order $k$, Lagrange gives $k\mid n$, and every $c^j\in H$ satisfies $(c^j{)}^k=1$, so $n\mid jk$. Therefore $n/k\mid j$ and $H\subseteq ⟨c^{n/k}⟩$. Equality follows from the common order. Thus there is exactly one subgroup for each divisor of $n$.

For the Abelian assertion, take a surjection ${\mathbb{Z}}^n\to A$ and let $L$ be the inverse image of $B$. Every subgroup of ${\mathbb{Z}}^n$ has at most $n$ generators, by induction: its projection onto the first coordinate is either zero, reducing to ${\mathbb{Z}}^{n-1}$, or $d\mathbb{Z}$ for some $d>0$. In the latter case choose one element with first coordinate $d$; subtracting its multiples reduces every element to the kernel, generated by at most $n-1$ elements by induction. The base case is the elementary description of subgroups of $\mathbb{Z}$ by their least positive element. Images of these generators generate $B$, proving the claim.

In $S_6$, the commuting disjoint transpositions $(12),(34),(56)$ generate $C_2^3$. This is a three-dimensional vector space over ${\mathbb{F}}_2$, so it requires three generators and cannot be generated by two.

For each $n\ge 3$, take $G=S_{2n+2}$, generated by $(12)$ and the cycle $(12\dots 2n+2)$. Its subgroup generated byis $C_2^{n+1}$. It requires $n+1$ generators by the same dimension argument, giving a two-generated group with a subgroup requiring more than $n$ generators.