Suppose that in a group of 5 people, there are not necessarily three mutual friends or mutual enemies. Question:
Show that in a group of 5 people (where any two people are either friends or enemies), there are not necessarily three mutual friends or three mutual enemies.
After thinking for hours I think it could be related to graph but I have no idea how start this question. May I get some assistance? 
Edited: I changed the word from suppose to show
 A: This picture describes everything!  
A: HINT: Seat $5$ people around a circular table. Suppose that two of these people are friends if and only if they’re sitting next to each other.
A: Hint Show that neither the cycle $C_5$ nor its complement has triangles.
P.S. A more interesting problem is to prove that up to isomorphism this is the only possibility. A hint for this is to show that if you have a Graph $G$ which is a counterexample to you statement, then every vertex in $G$ has degree at most $2$ both in $G$ and $\bar{G}$ (and hence exactly 2).
A: Some elements from graph theory (in case you want to generalise)
A complete (undirected) graph with 5 nodes (noted $K_n$) has 10 edges.
Your edges are of two kinds : ennemies or friends. We will consider that you delete the ennemy edges.
Your question would be : is it possible to find a graph $G=(V,E)$ with 5 nodes ($|V|=5$) such that neither $G$ nor its complement contains $K_3$ as a subgraph?
subgraph : graph obtained by removing nodes
complement : graph obtained by adding all absent edges and removing all present edges.
complete graph : graph $K_n=(V,E)$ with all edges $\{u,v\}, u \in V, v \in V$
