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Take the genus-two surface obtained from a regular hyperbolic octagon fundamental polygon by identifying opposite sides. Rotation of the octagon through $\pi /4$ respects the side pairing and descends to an orientation-preserving isometry of order eight. Thus $G=C_8$ acts on $S_2$ andThe fixed image of the octagon centre explains why this does not contradict part c.

After the $2n$ sides are identified in pairs, the polygonal-schema Euler count gives one face, $n$ edges and some number $V\ge 1$ of vertices. The Euler characteristic of a closed orientable genus-$g$ surface is $2-2g$, so

For genus two, take a regular octagon in the Poincare disc model and identify each side with its opposite side, with the orientation reversed along the boundary. One cyclic labelling isThe quotient has $V=1$, $E=4$, and $F=1$, hence Euler characteristic $-2$ and genus two. All eight vertices become one point, so smoothness requires their angles to sum to $2\pi$. Each interior angle is consequentlyEquivalently, the hyperbolic polygon area formula shows that the regular hyperbolic octagon fundamental polygon has areaas required by the Gauss-Bonnet theorem for a genus-two surface of curvature $-1$.