Suppose $\langle G,\oplus\rangle$ is an infinite abelian group. We can define an action of $\Bbb Z$ on $G$ by recursion as $0g:=e,$ $(n+1)g:=(ng)\oplus g$ for $n\ge 0,$ and $(-n)g:=\ominus(ng).$ Moreover, given any $m\in\Bbb Z,$ we have that $mG:=\{mg:g\in G\}$ is a subgroup (so a normal subgroup) of $G.$
If we know that $[G:mG]$ is finite, does it necessarily divide $m$? If not, under what conditions can we draw this conclusion?
When $G=\Bbb Z,$ then we know that $[G:mG]=m,$ but this doesn't always hold. For example, letting $G$ be the countable direct sum of copies of $\Bbb Z_2,$ we have $$[G:mG]=\begin{cases}\infty & 2\mid m\\1 & 2\not\mid m\end{cases},$$ so when $[G:mG]$ is finite, then it divides $m.$
I thought the approach outlined below would work, but Robert Shore pointed out a serious flaw in my first step, as $\langle \Bbb Q,+\rangle$ is an example of an abelian group that can't be written as such a direct sum. On the other hand, I believe that the only subgroup of $\Bbb Q$ of finite index is $\Bbb Q$ itself, so that doesn't necessarily preclude the larger claim from being true.
Since $G$ is abelian, then it is isomorphic to a direct sum of copies of $\Bbb Z$ and copies of $\Bbb Z_{p^k}$ for some primes $p$ (not necessarily distinct) and positive integers $k.$
For any direct sum of abelian groups $G=\sum_{i\in I}G_i,$ we can show that $G/mG\cong\sum_{i\in I}(G_i/mG_i).$ Note that $m\Bbb Z_{p^k}=\Bbb Z_{p^k}$ when $p\not\mid m,$ and that $m\Bbb Z_{p^k}\cong\Bbb Z_{p^{k-l}}$ when $m=p^lj$ for some $j$ coprime to $p$ and some positive integer $l\leq k.$
If there is at least one $g\in G$ of infinite order, then $[G:mG]$ is the least common multiple of $[\Bbb Z:m\Bbb Z]=m$ and the indices of the form $[\Bbb Z_{p^k}:\Bbb Z_{p^{k-l}}]=p^l$ where $m=p^lj,$ and so $[G:mG]=m.$
Otherwise, $[G:mG]$ is the least common multiple of indices of the form $[\Bbb Z_{p^k}:\Bbb Z_{p^{k-l}}]=p^l$ where $m=p^lj,$ so since $m$ is a multiple of each such index, then $[G:mG]$ divides $m.$
If the result is true, is there a slick way to prove it?
If the result is not true, what relationship (if any) can be proved between $[G:mG]$ and $m$ if we know that $[G:mG]$ is finite?
We cannot necessarily show that $[G:mG]$ is finite; take the example of $G=\mathbb Z^{\mathbb N}$ (an infinite direct product of copies of $\mathbb Z$). Then, $$G/mG\cong (\mathbb Z/m\mathbb Z)^{\mathbb N}$$ is infinite.
If $[G:mG]$ is finite, it must divide a power of $m$. To see this, if $p$ is a prime that divides $|G/mG|$ but not $m$, then there exists some $x\in G/mG$ of order exactly $p$. However, $mx=0$ (as is true for all elements of $G/mG$), and $px=0$, so $0=\gcd(m,p)x=x$, a contradiction. So, every prime $p$ dividing $|G/mG|=[G:mG]$ must divide $m$, and $[G:mG]$ must divide a finite power of $m$.
These are all attainable. For any divisor $d=p_1^{e_1}\cdots p_s^{e_s}$ of a power of $m$, $$G=\prod_{i=1}^s(\mathbb Z/p_i\mathbb Z)^{e_i}$$ has $mG=0$ and thus has $[G:mG]=d$.