TITLE: Showing planarity of graphs
QUESTION [5 upvotes]: I am trying to show $G_3$ is planar. I have constructed $G'_3$ as shown. Is it correct to say that by the Jordan curve theorem, $G_3$ cannot be planar, as any drawing will cause edges to overlap. Therefore it is non-planar?

REPLY [5 votes]: The graph $G_3$ is nonplanar.  It contains as a graph minor $K_{3,3}$.  To see this, see my picture:

The first arrow is a rearrangement of the vertices of $G_3$.  The second arrow is the desired deletion and contraction to obtain the minor $G_3^{\prime}\approx K_{3,3}$.  Hence $G_3$ is not planar by Kuratowski's Theorem.