Proving that if $G$ has a Hamiltonian circuit, then the line graph $L(G)$ also has a Hamiltonian Circuit I have a graph G = (V, E). I know that the line graph L(G) of G is defined as follows: each vertex in L(G) corresponds to an edge in G, and two vertices are connected by an edge in L(G) if and only if the corresponding edges in G are adjacent.
Now I am trying to (a) prove that if G has a Hamiltonian circuit, then L(G) also has a Hamiltonian circuit, and (b) give an example of a graph G such that G does not have a Hamiltonian circuit but L(G) does.
However, I'm extremely stuck on how to do this, and any help would be greatly appreciated. Thanks so much!

For (a), we're essentially after a sequence of the edges such that consecutive edges share an endpoint (and the first and last edges in the sequence also share an endpoint).
We start off with the sequence of edges defined by the Hamiltonian circuit:
$$(1,2),(2,3),\ldots,(n-1,n),(n,1)$$
after relabeling the vertices such that $(1,2,\ldots,n)$ is the Hamiltonian circuit. Then we add edges outside of the Hamiltonian circuit to the sequence. Here's an illustration as to how we would do it:
The line graph of

has the Hamiltonian circuit:
$$(1,2),\color{orange}{(2,4)},\color{orange}{(2,5)},(2,3),(3,4),\color{blue}{(4,6)},(4,5),(5,6),(6,1).$$
For (b), there's some 4-vertex example. Graphs with cut vertices don't have Hamiltonian circuits.
