G is a group generated by the two elements g, h, which satisfy g4 = 1, g2 ≠ 1, h7 = 1, h ≠ 1, ghg-1h = 1. The only subgroup containing g and h is G itself. Write down all elements of G which are squares.
Answer: 1, g2, h, h2, h3, h4, h5, h6.
Obviously g-1 = g3, h-1 = h6. So we have gh = h6g (*). This allows us to write any element generated by g and h in the form hngm (with n = 0, 1, 2, ... , 6; m = 0, 1, 2, 3). Applying (*) we find that (hng)2 = g2, (hng2)2 = h2n, (hng3)2 = g2. Thus the only possible squares are 1, g2, h1 [for example it is the square of h4 ], h2, ... , h6. 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 g2 is not 1. If g2 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
23 Jan 2001