37th Putnam 1976 Problem B2 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
: 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). 37th Putnam 1976 © John Scholes jscholes@kalva.demon.co.uk 23 Jan 2001