= Solution
\b[An example is the <unitriangular group of degree three over F3>], consisting of
$$
M(a,b,c)=\begin{pmatrix}1&a&c\\0&1&b\\0&0&1\end{pmatrix},\qquad a,b,c\in\mathbb F_3.
$$
This is an <upper unitriangular group> of order $3^3=27$. Direct <matrix multiplication> gives
$$
M(a,b,c)M(a',b',c')=M(a+a',b+b',c+c'+ab').
$$
The formula gives closure, identity $M(0,0,0)$, and inverse $M(-a,-b,ab-c)$; associativity follows from <matrix multiplication>. The two elements $M(1,0,0)$ and $M(0,1,0)$ do not commute: their products have respectively $c=1$ and $c=0$.
Every element has the form $I+N$ with $N^3=0$. Since the <finite field> $\mathbb F_3$ has characteristic three, the binomial expansion gives
$$
(I+N)^3=I+3N+3N^2+N^3=I.
$$
Hence every nonidentity element has order exactly three. \b[This is a nonabelian <finite group> with <exponent of a finite group> equal to three.]
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