Exhibit the elements of order 5 in GL(2, ZZ/(19))

Determine (up to similarity) the elements of order 5 in \mathsf{GL}_2(\mathbb{Z}/(19)).

It suffices to find the possible rational canonical forms of 2 \times 2 matrices over \mathbb{Z}/(19) whose minimal polynomials divide p(x) = x^5-1.

To accomplish this, we must first factor p(x) over \mathbb{Z}/(19). To this end, note that p(x) = (x-1)(x^4+x^3+x^2+x+1) = (x-1)q(x).

It is straightforward (if tedious) to show that q(x) has no roots in \mathbb{Z}/(19), and so has no linear factors.

Suppose q(x) has a quadratic factor; say q(x) = (x^2+ax+b)(x^2+cx+d). Comparing coefficients, we have (i) a+c = 1, (2) b+d+ac=1, (3) ad+bc = 1, and (4) bd = 1. Without loss of generality, we can assume that b \leq d. (that is, the canonical representatives of b and d in \mathbb{Z}.) There are then ten possibilities for (b,d): (1,1), (2,10), (3,13), (4,5), (6,16), (7,11), (8,12), (9,17), (14,15), and (18,18). Substituting (i) into (ii) and (iii) we have (iv) b+d+a-a^2 = 1 and (v) ad + b(1-a) = 1. If d \neq b, we may solve (v) for a: a = (1-b)/(d-b). Substituting into (iv) and simplifying, we have (vi) (d-b)^2(d+b-1) - (1-b)(1-d) = 0. It is straightforward to show that none of our candidate pairs (b,d) with b < d satisfy this equation. (I suggest using a calculator to verify this; for instance, WolframAlpha can do it with this input. So if q(x) does factor into do quadratics, then (b,d) is either (1,1) or (18,18). Suppose (b,d) = (1,1); then from our coefficient equations (i) – (iv), we have c = 1-a and ac = 18, and so a^2-a+18 = 0. Indeed, we can see that over \mathbb{Z}/(19), x^2-x+18 = (x-5)(x-15), so that either a = 5 or a = 15. Now (a,c) is (5,15) (by symmetry), and indeed we can verify that q(x) = (x^2 + 5x + 1)(x^2 + 15x + 1).

It is straightforward (if tedious) to show that these quadratic factors are irrducible over \mathbb{Z}/(19) since they do not have roots. So p(x) completely factors over \mathbb{Z}/(19) as p(x) = (x-1)(x^2 - 4x - 18)(x^2 - 14x - 18).

If A is a 2 \times 2 matrix over \mathbb{Z}/(19) of order 5, then its minimal polynomial has degree at most 2, divides x^5-1, and does not divide x-1. There are two such divisors, which each comprise the entire list of invariant factors of A. There are thus two corresponding rational canonical form matrices: \begin{bmatrix} 0 & 18 \\ 1 & 4 \end{bmatrix} and \begin{bmatrix} 0 & 18 \\ 1 & 14 \end{bmatrix}. Every element of order 5 in \mathsf{GL}_2(\mathbb{Z}/(19)) is similar to exactly one of these two.

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