Unimodular basis test for an integer direct summand ($\det U=\pm 1$)

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.