= Solution
Label the triangle's vertices counterclockwise as $1,2,3$. Let $r$ rotate it by $2\pi/3$ and let $s$ reflect it in the axis through vertex 1 and the midpoint of the opposite side. Then
$$
r^3=e,\qquad s^2=e,\qquad srs=r^{-1}.
$$
The six <dihedral group> elements are
$$
\boxed{D_3=\{e,r,r^2,s,rs,r^2s\}.}
$$
The first three are the identity and the rotations by $120^\circ,240^\circ$; the last three are reflections fixing vertices $1,3,2$, respectively. These are all possible symmetries: the image of one vertex has three choices, and the order of the other two has two choices.
Use rightmost-first composition. From the generator relation,
$$
(r^as^b)(r^cs^d)=r^{a+(-1)^bc}s^{b+d},
$$
with exponents reduced modulo three and two. The row-times-column <multiplication table> is
$$
\boxed{\begin{array}{c|cccccc}
&e&r&r^2&s&rs&r^2s\\\hline
e&e&r&r^2&s&rs&r^2s\\
r&r&r^2&e&rs&r^2s&s\\
r^2&r^2&e&r&r^2s&s&rs\\
s&s&r^2s&rs&e&r^2&r\\
rs&rs&s&r^2s&r&e&r^2\\
r^2s&r^2s&rs&s&r^2&r&e
\end{array}}
$$
Thus every symmetry is expressed in the two <group generators> $r,s$.
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