= Solution
Label the triangle vertices $1,2,3$ counterclockwise. Let $R$ rotate by $2\pi/3$ counterclockwise and let $m$ reflect across the line through vertex 1 and the midpoint of the opposite side. The <dihedral group> has elements $I,R,R^2,m,Rm,R^2m$, with $R^3=m^2=I$ and $mRm=R^{-1}$.
The action on vertices identifies them with these <permutations>, with the rightmost operation applied first:
$$
\boxed{\begin{array}{c|c|l}
\text{element}&\text{permutation}&\text{geometric action}\\\hline
I&e&\text{identity}\\
R&(123)&\text{rotation through }120^\circ\\
R^2&(132)&\text{rotation through }240^\circ\\
m&(23)&\text{reflection fixing vertex 1}\\
Rm&(12)&\text{reflection fixing vertex 3}\\
R^2m&(13)&\text{reflection fixing vertex 2}
\end{array}.}
$$
The minimal number of generators is two, and $\{R,m\}$ is one minimal generating set. They produce every displayed element. A one-element generating set would make the group cyclic, but every nonidentity element has order two or three, never six.
The vertex-action map preserves composition, hence is a <group homomorphism> from $D_3$ to the <symmetric group> $S_3=\Sigma_3$. The displayed actions are all six distinct permutations of three vertices, so the map is both injective and surjective. Therefore
$$
\boxed{D_3\cong S_3.}
$$
This identification uses the convention $|D_3|=6$, with the subscript counting polygon vertices.
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