Consider an example: the regular map S^{2}:{7,3} with 28 vertices, 12 faces
and 42 edges. Its faces are heptagonal, so the rotational symmetry
group must have some number of Sylow-7-subgroups. A Sylow theorem
shows that this number must be one of (1, 8, 15, ...). Consideration of
the size of the group, which has 84 elements, shows that there can
in fact be only one Sylow-7-subgroup.

So when we rotate this regular map while fixing (and rotating) one of its heptagonal faces, all the other heptagonal faces must also remain fixed. But if we consider what happens at any one vertex, we realise that this is impossible.

If a postulated regular map has Schläfli formula {G,H} and E edges, G is a power of a prime p, and H>(2 for even G, 1 for odd G), then its rotational symmetry group must have more than one Sylow-p-subgroup. If, given that the size of the group is 2E, this is not possible, then there can be no such regular map.

Dually, if a postulated regular map has Schläfli formula {G,H} and E edges, H is a power of a prime p, and G>(2 for even H, 1 for odd H), then its rotational symmetry group must have more than one Sylow-p-subgroup. If, given that the size of the group is 2E, this is not possible, then there can be no such regular map.

Consider an example: the regular map S^{2}:{12,3} with 8 vertices,
2 faces and 12 edges. Choose one of its dodecagonal faces. Its
edges cannot all connect a vertex to itself (if they did the graph
would not be connected), so they must involve some subset of the
vertices in a regular way. They must involve four of the 8 vertices
in the order abcdabcdabcd, or two of them in the order abababab.
But the either of these requires six edges at each vertex, which
is incompatible with the Schläfli formula {12,3}.

If a postulated regular map has Schläfli formula {G,H} and V vertices, then (at least) G/lcm(G,V) edges must connect to each vertex. If this is greater than H, no such regular regular map can exist.

Dually, if a postulated regular map has Schläfli formula {G,H} and F vertices, then (at least) H/lcm(H,F) edges must surround each face. If this is greater than G, no such regular map can exist.

Consider a regular map with one vertex and at least three odd-prime-sided faces. If it exists, each face must border a set of distinct other faces (if they are all the same the map is not connected, and if just some of them are the same it is not edge-transitive). Therefore the graph showing which of the faces border each other must have vertices all of the same odd degree.

Now consider the order in which we visit the faces as we go around the
vertex of the map. This order must traverse each edge of the above-mentioned
graph once. *No, twice, once in each direction. This proof is broken. But
the constraints of this double-Eulerian circuit do help us to construct the
regular map in question. *But no such circuit can be possible for a
uniformly odd-degree graph with more than two vertices.

Dually, a regular map with one face and at least three odd-prime-valent vertices cannot exist.

If the number of faces is less than and prime to the valence of the vertices, then at each vertex there must be at least one face which borders itself, and at least one face which borders another face. Therefore the map is not edge-transitive.

Dually, there can be no regular map having a number of vertices less than and prime to the number of vertices per face..

Consider a regular map with G edges to each face, and F (>2) faces. Each face borders at most F-1 other faces, and must border each of these the same number of times (otherwise the map will not be edge-transitive). So the number of distinct faces that border a given face must be greater than 1 (otherwise the map will not be connected), must be less than F, and must divide G. Therefore a map such that no number strictly between 1 and F divides G cannot be regular.

Dually, a map with H edges to each vertex, and V (>2) vertices, such that no number strictly between 1 and V divides H cannot be regular.

Some regular maps

Some pages on groups

Copyright N.S.Wedd 2009