Take the obvious graph interpretation. Assume that for some fixed vertex $A$, for every other vertex $B,C$, there are $3,1$ or $0$ edges among them. However if the graph wasn't connected, then we could partition it joining some connected components and contradicting the problems condition, so graph is connected and $0$ is not an option. Take any neighbor of $A$ and call it $B$. So for every other neighbor of $A$, $B$ has to connected with it, and so neighbors of $A$ make a complete graph(including $A$). Let $C$ be a vertex not neighbor of $A$. By our assumption, $C$ cannot be connected to this complete graph. Also for every vertex $B$ connected to $A$, all of its neighbors must be connected to $A$. So this complete graph is isolated, contradicting the connectivity.

Elementary: interpret the relationships as a graph. Now choose one of the groups to be the vertex $A$ and another to be all the other vertices, then say $A$ has degree $k>0$(this follows from the problem statement). Now let $S$ be the set of all the friends of $A$. If there are $B^{'}$ and $C^{'}$ in $S$, which are not friends, then the problem is solved, so $S\cup A$ is a clique. Now divide all the students into $S\cup A$ and all the rest (this is a group of at least one student, otherwise the graph is itself a clique, a contradiction by the problem statement). Therefore there is $B_{1}\in S$ and a student $C_{1}\not\in S\cup A$, such that $B_{1}$ and $C_{1}$ are friends. We chose $S$ as all the players of $A$, so $A$ and $C_{1}$ are not friends, so $\{A,B_{1},C_{1}\}$ have exactly two friendships.

The corresponding friendship graph $G$ is connected. Let $u_1$ and $u_2$ be non-adjacent vertices. Let $P_i$ be a path from $A$ to $u_i$ (in the case $A=u_j$ for some $j$, ignore about $P_j$). Let $v_i$ be the vertex on $P_i$ adjacent to $A$ and closest to $u_i$. In case of $v_i = u_i$ for $i=1,2$, we are done. So, WLOG, let's say $v_1\ne u_1$. Then, we are done by taking $B=v_1$ and $C$ to be the person on $P_1$ adjacent to $v_1$ that is closer to $u_1$.

Students be vertices in graph $G$. Draw an edge between $2$ vertices when students are friend in respective vertices. If $G$ isn't connected, then divide it into connected components and divide these components into 2 groups and we will get contradiction from satatesment. Then $G$ is connected , but not complete graph, because there exist at least 2 students which aren't friend. Since $G$ is connected take any $2$ vertices $A_{i} ,A_{j}$ which aren't neighbour and draw a path from $A_{i}$ to $A_{j}$ with $min$ degree ( path is in form of $A_{i} - A_{i+1} - ...-A_{j-1} -A_{j}$) . For shake contradiction assume there isn't $3$ vertices holds statesment. Then we will find that $A_{i}$ and $A_{i+2}$ are neighbour which makes contradiction with minimality of degree of path. It means that $\forall$ $2$ vertices are neighbour which means $G$ is complete. Again contradiction $\blacksquare$