# Prove that the line graph of a Hamiltonian simple graph is Hamiltonian.

Prove that the line graph of a Hamiltonian simple graph is Hamiltonian.

My proof

If $G$ is hamiltonian then there is a cycle that traverse all the vertices of $G$ exactly once. Any other edges of $G$ that's not part of this cycle can become chords of $G$ or placed outside the cycle.

How do I proceed from here?How do I use the definition of Hamiltonian to prove that the line graph of $G$ is also Hamiltonian?

• Note that all edges incident to a vertex of $G$ form a clique in the line graph $L(G)$ of graph $G$. – Smylic Feb 16 '17 at 0:05
• @Smylic Every two distinct vertices in clique are adjacent. Do you mean all clique in $L(G)$ together will form a cycle and make $L(G)$ Hamiltonian? – user59036 Feb 16 '17 at 0:25
• I mean that for any vertex $v \in V(G)$ every edge $\{\,u, v\,\}$ not belonging to the selected Hamiltonian cycle $C$ in graph $G$ is adjacent to both edges incident to vertex $v$ and belonging to the cycle $C$. Also all such edges $\{\,u, v\,\}$ (with the same $v$) are adjacent to each other. – Smylic Feb 16 '17 at 9:33
• @Smylic I'm more confused now, how does what you pointed out prove that $L(G)$ is Hamiltonian? – user59036 Feb 16 '17 at 10:02

Let $C_H = v_1v_2\ldots v_nv_1$ be Hamiltonian cycle in $G$. For each edge $\{\,v_i, v_j\,\} \in E(G)$ that doesn't belong to $C_H$ we choose one of two ends and assign this edge to the selected end. All edges from $C_H$ form a simple cycle $C$ in $L(G)$. Now we can insert all edges of $G$ assigned to $v_i$ between edges $\{\,v_{i - 1}, v_i\,\}$ and $\{\,v_i, v_{i + 1}\,\}$ in any order and get longer simple cycle $C'$, because all edges incident to $v_i$ are pairwise adjacent. Proceeding this for each vertex $v_i$ we get simple cycle $C^{(n)}$ in $L(G)$ containing all vertices of $L(G)$ (that are edges of $G$), so $C^{(n)}$ is Hamiltonian cycle in $L(G)$.