G is a group generated by the two elements g, h, which satisfy g^{4} = 1, g^{2} ≠ 1, h^{7} = 1, h ≠ 1, ghg^{-1}h = 1. The only subgroup containing g and h is G itself. Write down all elements of G which are squares.

**Solution**

Answer: 1, g^{2}, h, h^{2}, h^{3}, h^{4}, h^{5}, h^{6}.

Obviously g^{-1} = g^{3}, h^{-1} = h^{6}. So we have gh = h^{6}g (*). This allows us to write any element generated by g and h in the form h^{n}g^{m} (with n = 0, 1, 2, ... , 6; m = 0, 1, 2, 3). Applying (*) we find that (h^{n}g)^{2} = g^{2}, (h^{n}g^{2})^{2} = h^{2n}, (h^{n}g^{3})^{2} = g^{2}. Thus the only possible squares are 1, g^{2}, h^{1} [for example it is the square of h^{4} ], h^{2}, ... , h^{6}. Moreover, these are all distinct. For if two powers of h were equal then we could deduce h = 1 (a contradiction). We are given that g^{2} is not 1. If g^{2} equalled a power of h, then the square of that power would be 1 and hence h would be 1 (a contradiction).

© John Scholes

jscholes@kalva.demon.co.uk

23 Jan 2001