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## Main Question or Discussion Point

I'm trying to prove that [tex]GL(2,p^n)[/tex] has a cyclic subgroup of order [tex]p^{2n} - 1[/tex]. This should be generated by

[tex]\left( \begin{array}{cc}

0 & 1 \\

-\lambda & -\mu \end{array} \right)[/tex]

where [tex]X^2 + \mu X + \lambda[/tex] is a polynomial over [tex]F_{p^n}[/tex] such that one of its roots has multiplicative order [tex]q^{2n} - 1[/tex] in its splitting field.

Now since the above matrix satisfies its own characteristic polynomial, [tex]X^2 + \mu X + \lambda[/tex], I believe this somehow implies that it has order [tex]q^{2n} - 1[/tex].

The problem is that my Galois Theory isn't up to scratch so I don't have a clue how to justify this.

Here are some of my thoughts: I think I'm right in saying that any splitting field of [tex]F_{p^n}[/tex] is of the form [tex]F_{p^{mn}}[/tex]. For a quadratic is it always [tex]F_{p^{2n}}[/tex]? A primitive element of [tex]F_{p^{2n}}[/tex] would clearly have the desired order.

[tex]\left( \begin{array}{cc}

0 & 1 \\

-\lambda & -\mu \end{array} \right)[/tex]

where [tex]X^2 + \mu X + \lambda[/tex] is a polynomial over [tex]F_{p^n}[/tex] such that one of its roots has multiplicative order [tex]q^{2n} - 1[/tex] in its splitting field.

Now since the above matrix satisfies its own characteristic polynomial, [tex]X^2 + \mu X + \lambda[/tex], I believe this somehow implies that it has order [tex]q^{2n} - 1[/tex].

The problem is that my Galois Theory isn't up to scratch so I don't have a clue how to justify this.

Here are some of my thoughts: I think I'm right in saying that any splitting field of [tex]F_{p^n}[/tex] is of the form [tex]F_{p^{mn}}[/tex]. For a quadratic is it always [tex]F_{p^{2n}}[/tex]? A primitive element of [tex]F_{p^{2n}}[/tex] would clearly have the desired order.