= Unimodular basis test for an integer direct summand
{title2=$\det U=\pm1$}
<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.
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