NP-complete reduction for a k-dumbbell graph A k-dumbbell is a graph that consists of 2 cliques each of size k with one and only one edge between them. How do I show that finding if a graph is a k-dumbbell is NP-complete?

Proof it is in NP: Given the cliques of size k, I can verify if it they are the cliques of size k.

Proof of NP-hardness: If there is only one edge between them, how can I be sure that this is the edge I need for the reduction?

I'm trying to reduce from the k-clique problem as it seems to be the easiest choice.

I'm new to this so, if I'm wrong somewhere, please help me out !
Hint: Following your idea, you want to reduce the clique problem to the dumbbell problem, to show that the latter is NP-hard. Given an instance $(G,k)$ of the clique problem, it is natural to consider two copies $G',G''$ of your original graph, connected in some way, so that if the original graphs contains a $k$-clique, then the new one contains a $k$-dumbbell.

If the new graph contains a $k$-dumbbell restricted to one of the copies, then the original graph certainly contains a $k$-clique. You want to arrange matters so that if the $k$-dumbbell straddles both copies, then it is composed of a $k$-clique in each one.

This is just a hint, so I'm not giving a full solution. Indeed, there might be several ways of using this idea to get an NP-hardness reduction. Some of them require you to slightly change the value of $k$ from the clique instance to the dumbbell instance.