Using the notation for regular tessellations $\{p,q\}$ denoting the tessellation consisting of p-gons , q of which meet at each vertex [e.g. $\{3,6\}$ in $\mathbb{R}^{2}$ is the equilateral triangle tessellation.]
My conjecture is that $\{p,q\}$ is a regular tessellation iff $\{q,p\}$ is a regular tessellation (both in the same space). (It is certainly true in $\mathbb{R}^{2}$) But I want to generalise to any two dimensional space i.e Spherical and Hyperbolic.
Any hints on how to prove this or a counter example would be very helpful.
Thank you
This is true, and it is proved using the concept of the "dual tesselation", which works in each of the geometries you mention.
Suppose you have a regular tesselation by regular $p$-gons indexed as $\{T_i\}$ with $q$ meeting at each vertex. For each $p$-gon tile $T_i$ let $C_i$ be its center. If $T_i,T_j$ meet along a common edge, let $E_{ij}$ be the segment connecting $C_i$ to $C_j$, crossing the common edge of $T_i,T_j$ orthogonally at the midpoint. Given a vertex $V_j$ of the tesselation, enumerate the tiles meeting at $V$ in cyclic order as $T_1,…,T_q$. Let $S_j$ be the regular $q$-gon whose sides are $E_{12} E_{23} … E_{q-1,q} E_{q1}$. Then the $S_j$'s form a tesselation by regular $q$-gons of which $p$ meet at each vertex.
Try it out with some pictures, for example: the cube defines a $\{4,3\}$ tesselation of the 2-sphere, and its dual $\{3,4\}$ tesselation is defined by an octahedron. Also, a dodecahedron defines a $\{5,3\}$ tesselation of the 2-sphere, and its dual $\{3,5\}$ tessellation is defined by an icosahedron.