Introduction:
In this question, we are given a prime number p and we need to find the order of the group G of all 2x2 matrices over Zp (the integers modulo p) with determinant 1 under matrix multiplication.

Solution:
To find the order of G, we need to determine the number of elements in G.

Step 1: Determinant of a 2x2 matrix:
The determinant of a 2x2 matrix [a b; c d] is given by ad - bc. For the determinant to be equal to 1, we must have ad - bc = 1.

Step 2: Counting the elements:
Let's consider the elements of G. Each element in G can be represented as a 2x2 matrix [a b; c d] such that ad - bc = 1.

Case 1: a = 0:
If a = 0, then ad - bc = 1 implies that bc = -1. Since p is a prime number, Zp is a field, and -1 has a multiplicative inverse. Therefore, for any given b and c, there exists a unique d such that bc = -1. So, there are p^2 choices for (b, c) and p choices for d. Hence, in this case, the number of elements is p^2 * p = p^3.

Case 2: a ≠ 0:
If a ≠ 0, then we can solve the equation ad - bc = 1 for d. We get d = (1 + bc) / a.

Subcase 2.1: b = 0:
If b = 0, then d = 1 / a. Since p is a prime number, a has a multiplicative inverse modulo p. Therefore, there are p choices for a and p - 1 choices for d (excluding 0). Thus, in this subcase, the number of elements is p * (p - 1).

Subcase 2.2: b ≠ 0:
If b ≠ 0, then for any given b, there are p choices for c (including 0) such that bc = -1. For each choice of (b, c), there is a unique value of d = (1 + bc) / a. So, there are p choices for b, p choices for c, and p choices for a (excluding 0). Hence, in this subcase, the number of elements is p * p * p = p^3.

Step 3: Total number of elements:
Adding up the number of elements from both cases, we get the total number of elements in G as (p^3) + (p * (p - 1)) = p^3 + p^2 - p.

Therefore, the order of G is p^3 + p^2 - p, which can be written as p * (p^2 + p - 1).

Final Answer:
The order of the group G is (a) p * (p^2 + p - 1)