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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
by leifmetcalf 1y ago
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.
- bobxmax 1y agoSince 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.
- leifmetcalf 1y agoMy 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)
- leifmetcalf 1y agoGemini 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.