The <Cayley graph> $\operatorname{Cay}_S(G)$ is a connected graph with $|G|$ vertices. A shortest path never repeats a vertex, so it has at most $|G|-1$ edges. Therefore
$$
\boxed{\operatorname{diam}_S(G)\leq|G|-1<|G|.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 149 / 4 / c / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-149/4/c/i/solution