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Parametrize the matrices byDirect matrix multiplication givesThis proves closure. The identity matrix is $h(0,0,0)$, and the formulagives both a left and a right inverse, as substitution in the product formula verifies. Thus each element is an invertible matrix, and the subgroup test shows that $H$ is a subgroup of ${\mathrm{GL}}_3({\mathbb{Z}}_3)$. It is the unitriangular group ${\mathrm{UT}}_3({\mathbb{F}}_3)$ and has $3^3=27$ elements.

The unitriangular group ${\mathrm{UT}}_n(F)$ consists of upper triangular matrices over a field $F$ with every diagonal entry equal to one. Every such matrix is $I_n+N$ for a strictly upper triangular matrix $N$. Products have the same form, and $N^n=0$ gives the inverseThus these matrices form a subgroup of the general linear group. The case ${\mathrm{UT}}_3({\mathbb{F}}_3)$ is noncommutative but every element has cube equal to the identity matrix: $(I_3+N{)}^3=I_3$ because $N^3=0$ and $3=0$ in the finite field ${\mathbb{F}}_3$.