= Solution
Write $A(a,b,c)$ with $a$ in entry $(1,2)$, $b$ in entry $(1,3)$ and $c$ in entry $(2,3)$, and perform all arithmetic in the <finite field> $\mathbb F_5$. Direct multiplication gives
$$
\boxed{A(a,b,c)A(a',b',c')=A(a+a',\ b+b'+ac',\ c+c').}
$$
Thus the set is closed. <Associativity> comes from <matrix multiplication>. The <identity element> is $A(0,0,0)$, and solving the product formula for it gives
$$
\boxed{A(a,b,c)^{-1}=A(-a,\ ac-b,\ -c).}
$$
Both orders of multiplication with this inverse give the identity. Hence these matrices form a <finite group>, an instance of the <Heisenberg group over a prime field>. Each coordinate has five independent possibilities, with distinct triples giving distinct matrices, so
$$
\boxed{|G|=5^3=125.}
$$
The explicit coordinate convention matters because the central coordinate is $b$ here; other descriptions of the <Heisenberg group over a prime field> may place the central coordinate last.
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