33rd Putnam 1972

Problem B3

A group has elements g, h satisfying: ghg = hg2h, g3 = 1, hn = 1 for some odd n. Prove h = 1.



It is hard to get started on this. Or rather, it took me a long time to find the right way to get started. With hindsight, the correct approach is probably to work systematically through a set of increasingly complex expressions, trying to simplify them more than one way in order to get a new relation. It would not take long to reach hg2hg2h.

gh2 = (ghg)g2h = (hg2h)g2h = (hg2)(hg2h) = hg2(ghg) = h2g. From this point it is easy.

gh2g2 = h2. Hence gh2ng2 = h2n. Choose n so that h2n = h. Then ghg2 = h, or gh = hg. So the given relation implies h = h2. Hence h = 1.



33rd Putnam 1972

© John Scholes
27 Jan 2001