The cardinal of some set of cosets is countable

Question

Let $G$ be an abelian group and $(G_n)_{n \in \mathbb{Z}}$ be a family of subgroups of $G$ with $G_{n+1} \subset G_n$ for any $n$ and $G=\bigcup_{n \in \mathbb{Z}} G_n$. In addition, we suppose that the index $[G_n:G_{n+1}]$ of $G_{n+1}$ in $G_n$ is finite.

How prove that there are only countably many cosets of $G_j$ in $G$ ?

Answer 1

The coset space $G/G_j$ can be written as $\bigcup_{n\geq j} G_n/G_j$, a countable union of finite sets.

Answer 2

Hint: if you let $H_i$ = $G_{j+i}/G_j$ for $i = 0, 1, \ldots$, then $(H_i)$ is an ascending chain of finite groups whose union is isomorphic to $G/G_j$.