 Solution

The group presentation gives the quotient group ${\mathbb{Z}}^3$ by the subgroup generated by the rows ofFor the Smith normal form, the first determinantal divisor is the greatest common divisor of all entries, namely $3$. The nonzero two-by-two matrix minors are $\pm 27$, so the second determinantal divisor is $27$. The third row is the second minus the first, and a nonzero minor shows the rank is exactly two. Consequently the nonzero invariant factors are $3$ and $27/3=9$, with one free generator. The Fundamental theorem of finitely generated abelian groups givesAn element killed by $3$ has zero free coordinate, arbitrary coordinate in $\mathbb{Z}/3\mathbb{Z}$, and coordinate $0,3$ or $6$ in $\mathbb{Z}/9\mathbb{Z}$. There are nine such elements. Removing the identity leaves eight elements of order three.