Solution
= Solution
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 $S_2$ and
$$
|G|=8>1=g-1.
$$
The fixed image of the octagon centre explains why this does not contradict part c.