I switch notation to make it intuitive (at least for me): assume by contradiction that $G$ is a torsion abelian free group of rank 4 (i.e., isomorphic to a subgroup of $\mathbf{Q}^4$ containing $\mathbf{Z}^4$) and
$$A_1\oplus A_2\oplus B=G=A\oplus C,$$
with $A_1,A_2,A$ subgroups of rank 1, $B$ of rank 2, $C$ of rank 3 and $B,C$ indecomposable.

Then $A$ cannot be contained in $A_2\oplus B$ because otherwise we would have $G/A\simeq A_1\oplus (A_2\oplus B)/A\simeq C$, contradicting that $C$ is indecomposable, and similarly $A$ is not contained in any of $A_1\oplus B$, $A_1\oplus A_2$. Again similarly, $A_1\oplus A_2$ is not contained in $C$ since this would imply $G/(A_1\oplus A_2)\simeq B\simeq A\oplus C/(A_1\oplus A_2)$, contradicting that $B$ is indecomposable.

Write $\{1,2\}=\{i,j\}$. Then $A$ being a subgroup of $A_i\oplus (A_j\oplus B)$ not contained in $A_j\oplus B$, it is isomorphic to a subgroup of $A_i$. So $A$ is isomorphic to a subgroup of both $A_1$ and $A_2$.

Also, $A_1\oplus A_2$ being not contained in $C$, there exists $i$ in $\{1,2\}$ such that $A_i$ is not contained in $C$; we assume $i=2$ up to switching notation. Since $A_2\subset G=A\oplus C$, we deduce that $A_2$ is isomorphic to a subgroup of $A$.

If $U,V$ are subgroups of $\mathbf{Q}$, $U$ and $V$ are isomorphic if and only if $U\cap V$ has finite index in both $U$ and $V$ (simple consequence of classification, or easy exercise). In particular, if $U$ is isomorphic to a subgroup of $V$ and $V$ is isomorphic to a subgroup of $V$, then $U,V$ are isomorphic.

So $A$ and $A_2$ are isomorphic, and are isomorphic to a subgroup of $A_1$. We now distinguish two cases.

(a) $A_1$ is contained in $C$. Then $C=A_1\oplus ((A_2\oplus B)\cap C)$: indeed, first it's clear that the sum is direct, and if $c\in C$, write it as $a_1+x$ with $a_1\in A_1$ and $x\in A_2\oplus B$: since $a_1\in C$, we deduce that $x\in C$. So we contradict that $C$ is indecomposable.

(b) $A_1$ is also not contained in $C$. Then, for the same reason as previously (when we checked that $A$ and $A_2$ are isomorphic), $A$ and $A_1$ are isomorphic. Therefore the group $\mathrm{GL}_2(\mathbf{Z})$ acts on $A_1\oplus A_2\simeq A\oplus A$ in the standard way, and this action is transitive on the set of maximal rank 1 subgroups, and also it extends to an action on $G$ (because $A_1\oplus A_2$ is a direct factor in $G$). Therefore, if $A'$ is the projection of $A$ on $A_1\oplus A_2$ modulo $B$, we can use such an automorphism and change $A_1$ and $A_2$ so as to assume that $A'\subset A_1$. This means that $A\subset A_1\oplus B$, and we have shown that this cannot be the case.

infiniterank; I can hardly imagine what can be written about torsion-free abelian groups of finite rank. $\endgroup$ – Alex Degtyarev Oct 28 '16 at 8:17Infinite Abelian Groups, including an exposition of Corner's results, but not the question I asked. Incidentally, this is not simply an isolated problem, but part of a general program to describe sets of partitions of $n$ which can be realised by a rank $n$ torsion-free abelian group. $\endgroup$ – Phill Schultz Oct 28 '16 at 9:04