Consider an infinite planar graph with the following properties.
- Its vertices all have valence $3$.
- The faces all have $5$ edges.
Now put it in cartesian space and require that the faces are all congruent. The whole thing wraps around on itself infinitely many times, but the faces overlap exactly, and none intersect!
This also works if, per vertex, two of the faces have $6$ edges and one has $4$.
I'm surprised how many symmetries work out this way, especially because it's easy to build symmetries that don't.
Change $(6, 6, 4)$ to $(7, 7, 4)$ and you get... I'm not sure exactly what you get, but I'm pretty sure you get intersecting faces.
So how can you tell from the symmetries by which you constrain your graph, whether or not you get something nice in the way I've described above?
If you won't get something nice, how do you figure out the order of the quotient group?
Looking it up in Wikipedia isn't allowed.
Answering my own question here...
You can't have an infinite planar graph with the properties I describe unless they're chosen carefully. You can fill the plane with an infinite number of complexes such as a square face with a vertex at one corner and two edges adjacent to that vertex. Put a bunch side by side and you see the vertices and edges get shared by adjacent faces, and you can build indefinitely. You find
$$F - E + V = 0$$
because each complex has 1 face, 2 edges, and 1 vertex.
This sum gives the Euler characteristic $X$, which depends on the general form of the structure. You'll find $X$ must be $0$ if you try to change it by sticking in edges and vertices. You can't add an edge without splitting a face, and you can't add a vertex without splitting an edge.
Let's say you don't want to use squares. Say your faces have $n$ edges and vertices, and your vertices have valence $v$. Each edge is shared by two faces, so $E = \frac{1}{2} nF$. Each vertex is shared by $v$ faces, so $V = \frac{1}{v} nF$. Do all the substitutions and you find you're restricted to $(n,v) \in \{(3,6), (4,4), (6,3)\}$.
So my specification $(5, 3)$ doesn't have Euler characteristic 0. Plug it in and we find $X=2$.
Just one complex closed off with vertices and edges is a polygon. In this case $X = 1$.
If you have polygons stuck together so they share edges and vertices, this is a planar graph with faces, and still, $X = 1$. You can verify this by starting with any polygon and glomming on new faces.
Now look at the outer boundary of any planar graph. Deform the graph so it's a bowl or snifter shape and the outer boundary is the rim. The rim is a polygon without its own face. Add that face. Now you have a polyhedron and $X = 2$.
In the case where the plane wraps around itself many times, it seems to depend on what happens where it crosses itself. I'm not clear on that part.
My other case with a square and two heptagons at each vertex is impossible because if you walk around a heptagon glomming on squares as needed to meet the constraints, you end up with either two squares or three heptagons at a vertex. If you do the math anyway, you get 48 squares and 42 heptagons. Not sure how to interpret that, but there you go.
Let's try $(4,6,6)$:
$$\begin{align} F_4 + F_6 - F_4 \frac{4}{2} - F_6 \frac{6}{2} + F_4 \frac{4}{3} + F_6 \frac{6}{3} & = 2 \\ \frac{1}{3} F_4 & = 2 \\ F_4 & = 6 \\ \end{align}$$
For every vertex there is one square, shared with 3 other vertices, 4 in all. There also two hexagons, each shared by 6 vertices. So we also have
$$\begin{align} F_4 & = V \frac{1}{4} \\ F_6 & = V \frac{2}{6} \\ & = \frac{4}{3} F_4 \\ & = 8 \end{align}$$
You can find the other properties from there.