= Solution
Write $C=\langle c\rangle$ with $|c|=n$. For each divisor $k$ the <subgroup> $\langle c^{n/k}\rangle$ 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 $H\subseteq\langle c^{n/k}\rangle$. Equality follows from the common order. Thus \b[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 $n\geq3$, take $G=S_{2n+2}$, generated by $(12)$ and the cycle $(1\,2\,\ldots\,2n+2)$. Its <subgroup> generated by
$$
(12),(34),\ldots,(2n+1\,2n+2)
$$
is $C_2^{n+1}$. It requires $n+1$ generators by the same <dimension> argument, giving \b[a two-generated group with a <subgroup> requiring more than $n$ generators].
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