The exercise is given as it follows:
Show that there is no Graph $G$ such that $L(G)$ is isomorphic to $K_{1,3}$ otherwise show a counter example.
Well I'm truly lost about this one, I know that a vertex in a $L(G)$ is an edge in $G$ and also an edge in $L(G)$ is a set given by the form $\{e_1,e_2\}$ such that this edges has a common vertex in G. Given that I find pretty difficult to make the isomorphism, so I'm thinking about a proof by contradiction. Any Hints?
$K_{1,3}$ has vertices $a$, $b_1$, $b_2$, $b_3$ with $a$ being adjacent to each $b_i$. If this is a line graph of $G$ then these correspond to edges $e$, $f_1$, $f_2$, $f_3$ with $e$ sharing a vertex with each $f_i$. So there are (at least) two of the $f_i$ sharing the same vertex with $e$, and so at least two of the $f_i$ share a vertex, etc.