Prove that the lattice graph of $D_{16}$ is not planar How do we prove that the lattice graph of $D_{16}$ is non-planar?
I wanted to prove it using Kuratwoski's Theorem but was unable to do it. 

And to add one more question, are there any interesting implication of planarity in group theory?
 A: Note that the minimum size of a cycle for the graph for $D_{16}$ is four.  This means, if it were to be planar, any "polygons" in the graph would have to have $4$ sides or more, and that $\frac{4}{2}f\le e$.  Since $v+f-e=2$, $v-\frac12e\ge 2\Longrightarrow e\le2v-4$ must be satisfied for the graph to be planar.  There are $19$ vertices and $36$ edges.  $36\not\le 2(19)-4=34$, so it's not planar.
This paper is relevant.
A: Consider the subgraph with the following as vertices and edges.
$\langle sr,r^2 \rangle------D_{16} ------ \langle s,r^2 \rangle $
$ \langle sr,r^2 \rangle------ \langle r^2 \rangle ------  \langle r^4 \rangle$
$ \langle sr,r^2 \rangle------ \langle sr^3,r^4 \rangle------\langle sr^3 \rangle ------  \langle 1 \rangle$
$\langle sr^2,r^4 \rangle ------ \langle s,r^2 \rangle$
$\langle sr^2,r^4 \rangle ------ \langle r^4 \rangle$
$\langle sr^2,r^4 \rangle ------\langle sr^2\rangle------ \langle 1 \rangle$
$\langle s,r^4 \rangle ------ \langle s,r^2 \rangle$
$\langle s,r^4 \rangle ------ \langle r^4 \rangle$
$\langle s,r^4 \rangle ------\langle s\rangle------ \langle 1 \rangle$
Then this subgraph is homeomorphic to the graph $K_{3,3}$ with partite set $H=\{ \langle 1 \rangle,\langle r^4 \rangle,\langle s,r^2 \rangle\}$ and $K=\{\langle s,r^4 \rangle,\langle sr^2,r^4 \rangle,\langle sr,r^2 \rangle\}$. So by kuratowski's Theorem the given graph is not planar.
