Why is $K(1,3)$ a forbidden subgraph of a line graph $G$? My professor taught about Forbidden subgraphs in class.
"$H$ is forbidden subgraph of $G$ for a property $P$, if $G$ has property $P$ and $G$ cannot contain an induced subgraph isomorphic to $H$"
He gave an example that $K(1,3)$ is a forbidden subgraph of a line graph $G$.
I tried solving by contradiction. $G$ is a line graph of $H$ containing $K(1,3)$ as an induced subgraph.
If $v$ is a vertex if degree $3$ in $K(1,3)$ and $v_1$, $v_2$, $v_3$ are adjacent to $v$ in $K(1,3)$, then edge $e$ corresponding to $v$ in $H$ is adjacent to edges $e_1$, $e_2$, $e_3$ corresponding to vertices $v_1$, $v_2$, $v_3$.
This is where I get stuck. Can anyone please help me. Where did I go wrong? Is this approach correct?
 A: The vertex $v$ has degree $3$ because it came from an edge, $\epsilon$ in $H$ which shared a vertex with $3$ other edges. An edge only has $2$ vertices. That means (pigeonhole principle) that $2$ of those $3$ edges must share one of the endpoints of $\epsilon$, therefore in the line graph $G$, the vertices corresponding to those two edges of $H$ must share an edge due to having shared an endpoint in $G$. So the induced subgraph is not the bipartite $K(1,3)$.
A: Brandon du Preez has actually answered the reason. Let me write it in detail here.
Suppose that a line graph $L(G)$ contains $K(1,3)$ as an induced subgraph. Let $v_1$ be the vertex of degree $3$ in $K(1,3)$.   $v_2,v_3$ and $v_4$ are adjacent to $v_1$ in $K(1,3)$. By the definition of line graphs,  each vertex $v_i$ of $L(G)$ corresponds to an edge of $G$, which may  be noted as $e_i$ for $i\in \{1,2,3,4\}$.
Note that $v_1$ has three neighbours in $L(G)$ that correspond to the three edges adjacent with $e_1$ in the original graph $G$. By the pigeonhole principle, at least two edges (say $e_2,e_3$.) are incident to one of the endpoints of $e_1$.  Then by the definition of  line graphs, $e_2$ is adjacent with $e_3$ in $L(G)$. This contradicts the fact that $K_{1,3}$ is an induced subgraph of $L(G)$.

Edit  when I was editing, toe-pose had not answered. Please ignore my answer now.
