TITLE: Planar and non-planar graphs, and Kuratowski's Theorem
QUESTION [5 upvotes]: A graph $G$ is planar if and only if every subdivision of $G$ is planar.
A graph $G$ is planar if and only if it contains no subdivision of $K_{3,3}$
  or $K_5$.
A subdivision of an edge $e$ in $G$ is a substitution of a path for $e$.
  We say that a graph $H$ is a subdivision of $G$ if $H$ can be obtained
  from $G$ by a finite sequence of subdivisions.

Using Kuratowski's Theorem, I need to show that the graphs below are non planar

I know how the $K_{3,3}$ and $K_5$ look like but the subdivision part lost me. I'd appreciate as much help as possible. Thank you

REPLY [3 votes]: Note that no subdivision is needed for $G_5$. There is a way to choose three vertices $a,b,c$ from $1,2,3,4$, and three vertices $x,y,z$ from $5,6,7,8,9$, in such a way that each of the vertices $a,b,c$ is adjacent to each of the vertices $x,y,z$. That's a $K_{3,3}$, and you're done. 
$G_6$ doesn't need any subdivisions, either. In fact, there's a way to see it as a $K_{3,4}$ with some extra edges.