Why is the $W_6$ wheel graph not a line graph? I know a line graph of an undirected graph $G$ represents the adjacencies between edges of $G$.
If I assume that $W_6$ is a line graph, where is the contradiction that it cannot represent the adjacencies between edges of any graph (i.e. such a graph $G$ does not exist such that $W_6$ is its corresponding line graph)?
Suppose that there is a graph $G$ such that its line graph $L$ is isomorphic to $W_6$.
Notice $G$ has an edge $e$ that is adjacent to exactly $5$ edges, after we remove this edge from the graph we obtain a graph $G'$ such that its line graph is isomorphic to a cycle. It is easy to show that if the line graph of a graph $H$ is a cycle then $H$ is isomorphic to the same cycle. To see this notice that since the line graph of $H$ contains a cycle then $H$ must also contain a cycle $C$, also $H$ must be connected, now notice that no extra edge can be joined to $C$, as this would force a vertex of the line graph to have degree exceeding $2$, so $H$ is equal to $C$.( this can be generalized, see here)
We conclude that $G$ must be isomorphic to the cycle $C_5$ along with a diagonal, but this clearly does not work.
Let $G$ be a graph with $L(G)$ isomorphic to $W_6$. Since $W_6$ has exactly 6 vertices, $G$ has exactly 6 edges. One vertex in $W_6$ is adjacent to all 5 other vertices, so one edge $e$ of $G$ is adjacent to all 5 other edges of $G$. The remaining vertices of $W_6$ have degree 3, thus the remaining edges of $G$ are adjacent to exactly 3 other edges.
By the pigeonhole principle, one endpoint $v$ of $e$ is incident with at least 3 other edges of $G$. Since those edges are adjacent to exactly 3 edges, $v$ must be incident with exactly 3 other edges of $G$, and those edges are adjacent only to each other and to $e$. But then the other endpoint $w$ of $e$ is incident with exactly 2 other edges of $G$, and those edges are adjacent only to each other and to $e$, a contradiction, since those edges must be adjacent to exactly 3 edges.
