 Solution

A proper finite-index subgroup of a torsion-free group cannot be a direct summand when the complement is a subgroup of that same free abelian group. Indeed, if $L=M\oplus K$, every $k\in K$ has $18k\in M$ because the quotient has order eighteen. But also $18k\in K$, so $18k=0$ by the trivial intersection. The ambient group ${\mathbb{Z}}^3$ is torsion-free, forcing $k=0$ and hence $K=0$. This would give $M=L$, contrary to index eighteen. Thus $M$ is not a direct summand.