Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 3 / 14F (Groups, Rings and Modules)

Put the generators into the rows of the integer matrixIts determinant is $5-2-21=-18$. A full-rank integer lattice generated by its rows has index equal to the absolute determinant, henceFor a direct verification, $u,v,e_3$ form a unimodular basis because their row determinant is $-1$, and $w=-7u+5v+18e_3$. In these integer coordinates $M=\mathbb{Z}u\oplus \mathbb{Z}v\oplus 18\mathbb{Z}{e}_3$, visibly leaving eighteen cosets.

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.

The matrix with rows $u,v,e_3$ has determinantIts inverse has integer entries by the adjugate formula, so these rows form an integer basis of $L$. The unimodular basis test for an integer direct summand now givesThus $N$ is a direct summand, with the displayed complement. This proves integral splitting, not just real or rational linear independence.

In the integer basis $u,v,e_3$, direct substitution gives $w=-7u+5v+18e_3$. ThereforeThe first two basis coordinates disappear in the quotient and the third is taken modulo eighteen. Equivalently the Smith normal form has diagonal entries $1,1,18$.