Another bit to the problem that I attempted, the 2nd part of the question asked for the order of the subgroup of G with the condition that the determinant of the matrix be 1.

This is my train of thought: Consider the matrix

. Our initial matrix can be converted to

by finite row operations. Hence we can go backwards as well. Consider all the possible rows that can result from row operations on

. I think we will get 'p' options for the top row, {1,0}, {1,1}, ... {1,p-1}, if our bottom row is {0,1}. Similarly for our bottom row as {1,1} we get 'p-1' options for the 1st row: {1,0}, {2,1}, ... {p-1,p-2}. (The row {p,p-1} is not allowed because our integers are modulo p). Similarly for all the possible bottom rows we get 'p-2', 'p-3' ... options for the top rows. Hence the order of our group is p + (p-1) + ... + 1 = p*(p-1)/2

Is it right? Am I overcomplicating a simple problem?

PS: Sorry for not using Latex as much as I should. I'm new to this