Having an exercise from an old book of graph theory that does not contain the solution. My graph theory skills are at best entry-level. Need the concept of this problem to be explained(if possible) not the solution.
Having a simple non-directed graph that contains cycles of odd length but no $3$-length cycle. If $C$ a minimal cycle of odd (no 3) length of the graph then show:
1)if $x∈V(G) - V(C)$ and $u,w∈V(C)$, with $u,w$ being adjacent to $x$, the vertices $u,w$ are connected by a path length 2 that are made from edges of cycle c.
2)every vertex of $G$ which doen't belong to $V(C)$ have at most 2 neighborhoods in $V(C)$.
Hints:
1) Clearly $u, w$ are connected in the cycle. It's a cycle, after all. Let's say they aren't connected by a length-2 path in $C$. Then the shortest path between them in $C$ is either of length $1$ or of length more than $2$. Show that either of these cases leads to some contradiction.
2) Take an $x\notin V(C)$ with $3$ distinct neighbours $u, v, w\in V(C)$. Use the result of 1) to discuss what $C$ must look like and again reach a contradiction.