Show, using Exercise 13, that (a, b: a³ = 1 b² = 1, ba = a²b) gives a group of order 6. Show that it is nonabelian.

Data from Exercise 13

Let S = {aⁱb^j|0 ≤ i < m, 0 ≤ j < n}, that is, S consists of all formal products aⁱb^j starting with a⁰b⁰ and ending with a^m-1b^n-1. Let r be a positive integer, and define multiplication on S by (a^sb^t)(a^ub^v) = a^xb^y, where x is the remainder of s + u(r^t) when divided by m, and y is the remainder of  t + v when divided by n, in the sense of the division algorithm.