Call an abelian group $G = (G,+)$ $m$-torsion for some natural number $m$ if one has $m \cdot x = 0$ for all $x \in G$. A subgroup $H$ of $G$ is said to be complemented if one can write $G = H \oplus K$ for some other subgroup $K$ of $G$. I was able to establish the following fact:

Proposition. If $G$ is a finite abelian $m$-torsion group, and $H$ is an index $k$ subgroup of $G$, then $H$ contains a complemented subgroup $H'$ of $G$ of index at most $C(m,k)$, for some quantity $C(m,k)$ that depends only on $m$ and $k$.

For instance, if $G = ({\bf Z}/4{\bf Z})^n$ for some large $n$, then $G$ is $4$-torsion, and the index $2$ subgroup $H := \{ (x_1,\dots,x_n) \in G: 2(x_1+\dots+x_n)=0\}$ is not complemented, but it contains the complemented index $4$ subgroup $H := \{ (x_1,\dots,x_n) \in G: x_1+\dots+x_n=0\}$. The restriction to $G$ being finite is probably not essential, but I include it here for simplicity.

I am not satisfied with my proof because it relies a fair bit on the classification of finite abelian groups. Here is my proof:

- By Pontryagin duality, it suffices to establish the dual claim that every order $k$ subgroup $K$ of $G$ is contained in a complemented subgroup $K'$ of order at most $C(m,k)$.
- By induction on $k$, we may assume without loss of generality that $K$ is simple. Indeed, if $K$ contained a non-trivial normal subgroup $N$, then by induction $N$ is contained in a complemented subgroup $N'$ of bounded size, and after quotienting out by this subgroup $N'$, $K$ will also be contained in a complemented subgroup of bounded size. From this it is easy to place $K$ itself in a complemented subgroup of bounded size.
- By Cauchy's theorem, $N$ is now of some prime order $p$, which must divide $m$. By the Schur-Zassenhaus theorem, we may now pass to the Sylow $p$-group and assume that $G$ is a $p$-group.
- By the classification of finite abelian groups, one can write $G$ as the product of $O_m(1)$ many factors $({\bf Z}/p^i {\bf Z})^{n_i}$, where $i$ is bounded but $n_i$ need not be. By projecting $N$ to each of these factors separately, and then taking Cartesian products of any complemented subgroups located, we may reduce to the case of a single factor $G = ({\bf Z}/p^i {\bf Z})^{n_i}$.
- If we let $e$ be a generator of $N$, one can now perform an explicit computation in coordinates to locate a root $p^{i-1} e' = e$ of $e$. Taking $N'$ to be the group generated by $e'$, one can check by hand that $N'$ is complemented and of order $p^i$, giving the claim.

My (somewhat vague) question is whether there is a more conceptual or high-level proof of this proposition that does not rely so much on the classification of finite abelian groups, or working in explicit coordinates. In a somewhat different direction, I am also curious as to what the correct dependence of the constant $C(m,k)$ on $m,k$ should be; the argument I gave above is quite inefficient in this regard.