Show, using Exercise 13, that (a, b a 3 1 b 2 1, ba a 2 b) gives a group of order 6 Show that it is nonabelian Data from Exercise 13 Let S a i b j 0 i m, 0 j n , that is, S consists of all formal products a i b j starting with a 0 b 0 and ending with a m 1 b n 1 Let r be a positive integer, and def...

To show that a b a3 1 b2 1 ba a2b gives a group of or ...

Show, using Exercise 13, that (a, b: a 3 = 1 b 2 = 1, ba =

Question:

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^jstarting 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.