Submodules of dyadic rationals

Question

Let $D = \{j/2^n \mid j \in \mathbb{Z}, n \geq 0\}$ the set of dyadic rationals as a $\mathbb{Z}-$module. If $X = \frac{D}{\mathbb{Z}},$ what about the submodules of $X?$ There is any infinite submodule of $X?$

Answer 1

Assuming $X=D/\Bbb Z$ is the quotient module, we can prove that $X$ has no nontrivial infinite submodule (subgroup - as $\Bbb Z$-modules are just the Abelian groups).

Let $A\le X$ be an infinite subgroup and $n\in\Bbb N$ arbitrary.
Then, as the set $\{\frac a{2^m}\mid 0\le a<2^m,\ m<n\}$ is finite, $A$ must have an element of the form $\frac a{2^k}$ with $a$ odd and $k\ge n$.
Since $a$ is odd, $\gcd(a,2^k)=1$, thus Bezout's identity yield integers $u,v$ such that $au+2^kv=1$, which shows that $\frac 1{2^k}\in A$, and multiplying it by $2^{k-n}$, we also get $\frac1{2^n}\in A$.
Since $n$ was arbitrary, $A$ must contain all elements of $X$.