Let G be a group of order 3*2^n. Prove there exists a non-complete non-cyclic Cayley graph of G such that there is a unique shortest path between every pair of vertices, or otherwise prove no such graph exists.
Gemini 2.5 at least replies that it seems unlikely to be false without hallucinating a proof. From its thoughts it gets very close to figuring out that A_4 exists as a subgroup.
Since any group of order 3⋅2n3⋅2n has ∣G∣≥3∣G∣≥3, it cannot admit a Cayley graph which is a tree. Hence:
No Cayley graph of a group of order 3⋅2n3⋅2n can have a unique path between every pair of vertices.My mistake, I said unique path when I should have said unique shortest path.
Also, there are trivial solutions with odd cycles and complete graphs which must be excluded. (So the answer to the prompt as originally stated is wrong too)