= Solution
Map each symmetry to the <permutation> it induces on the three vertices. This is a <group homomorphism>, because composing geometric motions composes their vertex permutations. In the labels chosen in (b), the map is
$$
\boxed{\begin{array}{c|cccccc}
g&e&r&r^2&s&rs&r^2s\\\hline
\Phi(g)&e&(123)&(132)&(23)&(12)&(13)
\end{array}}
$$
All six images are distinct and exhaust the <symmetric group> $S_3$. Equivalently, the <kernel> is trivial because a triangle symmetry fixing all three vertices is the identity. Thus the map is bijective and
$$
\boxed{D_3\cong S_3.}
$$
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