Consider the set of \(n\)-by-\(n\) matrices whose entries are integers modulo \(p\), for some prime \(p\). The matrices in this set with nonzero determinant are invertible, and thus form a group under matrix multiplication; this group is given the name \(\text{GL}(n,p)\), where the "GL" stands for "general linear."

Given a matrix \(M \in \text{GL}(n,p)\), its *order* is defined to be the smallest integer \(k\) such that \(M^k = I\), where \(I \in \text{GL}(n,p)\) denotes the identity matrix. In this problem, you will compute the maximum possible order of a matrix in \(\text{GL}(n,p)\). That is, you will answer the question "What is the largest integer \(m \in \mathbb{N}\) such that \(m\) is the order of a matrix in \(\text{GL}(n,p)?\)"

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Hint & Proof Sketch:

Let \(A\) be a matrix in \(\text{GL}(n,p)\). Consider the vector space \(V(A,p)\) generated by the powers of \(A\), with coefficients in the field of \(p\) elements. That is, \(V(A,p)\) is the vector space (under addition) of linear combinations \[a_0 + a_1 A + a_2 A^2 + a_3 A^3 + \cdots,\] where the coefficients \(a_i\) are integers modulo \(p\). The Cayley-Hamilton theorem implies \(V(A,p)\) is finite dimensional; what is the largest possible value of its dimension \(\big(\)as \(A\) ranges over the group \(\text{GL}(n,p)\big)?\)

Suppose \(\dim\big(V(A,p)\big) = k\). What does this imply about the order of \(A\) in \(\text{GL}(n,p)?\) To answer this question, note that any power of \(A\) must be in \(V(A,p)\), since the exponents in powers of \(A\) can be reduced using the polynomial relation produced by Cayley-Hamilton.

Let \(K\) denote the largest possible value of \(\dim\big(V(A,p)\big)\) when \(A\) ranges over \(\text{GL}(n,p)\). Can you find a matrix \(B \in \text{GL}(n,p)\) whose order is at least \(K?\) What does this imply in light of the previous two steps?

**As your answer to this problem, submit the maximum possible order of a matrix in \(\text{GL}(4,5)\), i.e.** \[\max_{A \in \text{GL}(4,5)} \text{Order}(A).\]

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