Since Z_9 is cyclic (with order 9), there is exactly one subgroup for each order dividing 9, each of which are cyclic. Since the divisors of 9 are 1, 3, and 9, we have: (i) Order 9: Z_9 = {0, 1, 2, ..., 8} itself. (ii) Order 3: {0, 3, 6} (iii) Order 1: {0}. I hope this helps!