Let a graph $G$ have a cycle that contains a vertex covering of the graph. Prove that $L(G)$ is Hamiltonian

Question

Suppose a graph $G$ have a cycle that contains a vertex covering of the graph. Prove that $L(G)$ is Hamiltonian

Answer 1

Let $c_1,c_2, ... , c_n$ be successive edges of the cycle $C$ of $G$. Then vertices $c_1,c_2, ... , c_n$ form a cycle in $L(G)$

Let $c_1$ and $c_2$ meet at vertex $v$ of $C$. Let $e_1,e_2, ... , e_k$ be the other edges (if any) of $G$ which meet at $v$.

In $L(G)$, the vertices $c_1,c_2,e_1,e_2, ... , e_k$ are then all adjacent and so the cycle $c_1,c_2, ... , c_n$ can be enlarged into the cycle $c_1,e_1,e_2, ... , e_k,c_2, ... , c_n$.

This process can be repeated with each pair of successive edges of $C$ except that we ignore any edges already added into the cycle of $L(G)$. Since $C$ is a vertex covering of $G$, the cycle of $L(G)$ then contains every edge of $G$ and $L(G)$ is therefore Hamiltonian.