# Why line graph is a proper subclass of claw-free graph?

A claw is the graph isomorphic to $K_{1,3}$, that is, a star graph with three edges, three leaves, and one central vertex. (see below)

A graph is claw-free if it does not contain a claw as an induced subgraph.

The line graph of G is the intersection graph of the edges of G, that represents the adjacencies between edges of G.

A graph $F$ is a line graph if there exists a graph $H$ such that $L(H) = F$.

My question is why are line graphs a subclass of claw-free graphs?? Thanks a lot!

• What have you tried? Have you tried to find a graph $H$ such that $L(H) = K_{1,3}$? There are only so many possibilities for $H$, so you should be able to try them all. What happens when you do that? – D.W. Apr 10 '18 at 0:02
• I'm confused. Even the line graph in the picture shown contains claws. – Apiwat Chantawibul Apr 10 '18 at 0:14
• @Billiska "Claw-free" means no induced subgraphs are claws. I've edited to correct the definition in the question. – David Richerby Apr 10 '18 at 0:25
• oh. never mind. induced by vertex. not induced by edge. Yes, I was going insane. thanks. – Apiwat Chantawibul Apr 10 '18 at 1:07
• "why the former is proper subclass ". This is a different question from the one you asked originally. Please create a new question if you want to know something other than your original question. Thanks. – Discrete lizard Apr 10 '18 at 9:20

Wikipedia contains a pictorial description of these nine graphs. One of them is $K_5^-$, obtained from $K_5$ by removing a single edge.