Direct summand

A submodule $N$ of a module $M$ is a direct summand if some submodule $L$ gives $M=N\oplus L$. Equivalently, the inclusion has a left inverse R-module homomorphism, or $N$ is the image of an idempotent endomorphism of $M$.

Integer row vectors forming a square unimodular matrix are a basis of the corresponding free abelian group: the adjugate formula makes the inverse integral. The subgroup spanned by any subset of that basis is therefore a direct summand, complemented by the remaining basis vectors. In contrast, a proper finite-index subgroup of a free abelian group cannot be a direct summand: its nonzero finite quotient would have to embed as a subgroup of a torsion-free group.