Given $ k \gt 0 $ , how can i prove the existence of a simple graph $\mathbf G$ that satisfies the following:
$\bullet$ $\delta (G) \ge k.$ ($\delta$ := the the minimum degree of a vertex in G).
$\bullet$ $\mathbf G$ have a unique perfect matching.
i do know that $|G|, \delta(G) \ge 2k-1\Rightarrow \nu(G) \ge k $. ($\nu$ :=size of the maximum matching), and i also know that a graph with a unique matching must has a cut edge.
but i cant figure out how to prove the statement above.
We can build it by induction on $\delta$. For base cases, $G_1$ is $K_2$, a single edge between two vertices. Here is an example $G_2$ with $\delta=2$.
Now assume we have a graph $G_k$ meeting the definition with $\delta=k$ for some positive integer $k$. We shall create a graph $G_{k+1}$. Start with a single edge, which will be our cut edge. Then, for each end of the vertex, create a copy of $G$ and attach the end of the cut edge to every vertex of $G$. That graph has $\delta=k+1$ and, by the induction hypothesis, has a unique matching.
In fact, $G_2$ is an example of using this algorithm from $G_1$. Here is what $G_3$ would look like.