A group has elements g, h satisfying: ghg = hg^{2}h, g^{3} = 1, h^{n} = 1 for some odd n. Prove h = 1.

**Solution**

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 hg^{2}hg^{2}h.

gh^{2} = (ghg)g^{2}h = (hg^{2}h)g^{2}h = (hg^{2})(hg^{2}h) = hg^{2}(ghg) = h^{2}g. From this point it is easy.

gh^{2}g^{2} = h^{2}. Hence gh^{2n}g^{2} = h^{2n}. Choose n so that h^{2n} = h. Then ghg^{2} = h, or gh = hg. So the given relation implies h = h^{2}. Hence h = 1.

© John Scholes

jscholes@kalva.demon.co.uk

27 Jan 2001