Let $G$ be a finitely generated Abelian group. For each subgroup $H$ of $G$, does there exist another subgroup $K$ of $G$ such that $\text{rank}(G)=\text{rank}(H)+\text{rank}(K)$ and $\text{rank}(H\cap K) = 0$?

Edit: For background, given two subgroups $H$ and $K$ of $G$, we say that $K$ is a complement of $H$ in $G$ if $G=H+K$ and $H\cap K = \{ 0 \}$; generally, given $H$, a complement of $H$ in $G$ may not exist, e.g., take the subgroup $H=2\mathbb{Z}$ of the group $G=\mathbb{Z}$. My question concerns a weaker notion, a 'rank-complement'.

  • 2
    $\begingroup$ I suggest to change the title. After all, $K$ is not supposed to be a complement of $H$. It is sort of a "rank-complement". $\endgroup$ Jun 11, 2015 at 8:19
  • $\begingroup$ Have you solved the case that $G$ is free? $\endgroup$ Jun 11, 2015 at 8:22

1 Answer 1



The structure theory of finitely generated Abelian groups tells us that $G$ can be written as a sum of subgroups $G = M + T$, where $M$ is torsion-free and $T$ is finite; it follows that $\text{rank}(M)=\text{rank}(G)$. Now $U = H \cap M$ is also torsion-free with $\text{rank}(U)=\text{rank}(H)$ since the intersection only removes torsion elements. Thus, $U$ admits a basis $B_U$ which can be augmented to obtain a basis $B_V$ of some subgroup $V$ of $M$ with $\text{rank}(V)=\text{rank}(M)$; therefore, the subgroup $K$ spanned by $B_K = B_V \setminus B_U$ is a complement of $U$ in $V$, so $\text{rank}(K)=\text{rank}(V)-\text{rank}(U)$. Putting this all together, $\text{rank}(G) = \text{rank}(V)=\text{rank}(U)+\text{rank}(K)=\text{rank}(H)+\text{rank}(K)$, and $H\cap K = U \cap K = \{0\}$, so $\text{rank}(H \cap K)=0$.

  • $\begingroup$ This is what I have so far. I'd like to reformulate this argument (or a correct version of it) in a basis-independent manner. Additionally, I suspect that torsion could be handled more simply (without passing through the torsion-free subgroup $M$). $\endgroup$
    – nknight
    Jun 12, 2015 at 20:12

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