# Planar graph whose line graph is non-planar

The line graph $L(G)$ of a simple graph $G$ is defined as follows: There is exactly one vertex $v(e)$ in $L(G)$ for each edge $e$ in G. For any two edges $e$ and $e'$ in $G$, $L(G)$ has an edge between $v(e)$ and $v(e')$, if and only if $e$ and $e'$ are incident with the same vertex in $G$.

Then I came across statement saying:

The line graph of a planar graph is not necessarily planar.

I see this is the case by considering e.g., the star graph $S_5$. However, can we generalize the property of the graph whose line graph results in non-planar graph. Obviously this must be the subgraphs whose line graph translates to either of Kuratowaski's graph $K_5$ or $K_{3,3}$. As pointed out, $L(K_5)=S_6$ which is planar. But $L(K_{3,3})$ is not planar (or can it be drawn as planar too?) as shown in figure below:

So is it like if the planar graph has a subgraph isomorphic to $S_5$, then its line graph is non planar (as it will contain $K_5$). Or is there anything more to this?

• I didn't think $L(K_5)$ was planar? Consider that each vertex in $L(K_5)$ has degree 6, since each vertex in $K_5$ has degree 4, and would mean each edge in $K_5$ is incident to 6 other edges. Thus, $L(K_5)$ has 10 vertices and 30 edges. It can be deduced it's not planar because of it violating $e \leq 3v - 6$ ($30\not\leq 24$) – Joshua Detwiler Mar 23 '18 at 7:57

Sedlacek has shown that a graph $G$ has a planar line graph if and only if $G$ is planar of maximum degree 4 and every vertex of degree 4 is a cut vertex. Using the result, Greenwell and Hemminger [2] proved a graph $G$ has a planar line graph if and only if $G$ has no subgraph homeomorphic to $K_{3,3}$, $K_{1,5}$, $P_4+K_1$, or $K_2+\bar{K}_3$.