## 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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