Question: is there a regular map in the genus 2 surface with three heptagons meeting at each vertex? Answer: no. Proof: if there were, it would (using Euler's formula) need to have 28 vertices, 12 faces, and 42 edges, and therefore a rotational symmetry group of order 84. Now consider the Sylow-7-subgroups of this group. There must be 1 modulo 7 of them, and the only possibility is 1 of them. So the rotation that rotates and fixes one of the heptagons must rotate and fix all the heptagons. This can't work, so no such regular map can exist.
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