What is the cyclic subgroup order of GL(2,p^n) generated by the given matrix?

[Hello Kitty]

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

\left( \begin{array}{cc}<br /> 0 &amp; 1 \\<br /> -\lambda &amp; -\mu \end{array} \right)

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

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

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 F_{p^n} is of the form F_{p^{mn}}. For a quadratic is it always F_{p^{2n}}? A primitive element of F_{p^{2n}} would clearly have the desired order.

 

[Hello Kitty]

The q's should be p's. Corrected version:

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

\left( \begin{array}{cc}<br /> 0 &amp; 1 \\<br /> -\lambda &amp; -\mu \end{array} \right)

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

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

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 F_{p^n} is of the form F_{p^{mn}}. For a quadratic is it always F_{p^{2n}}? A primitive element of F_{p^{2n}} would clearly have the desired order.

 

[Mystic998]

It seems like you're going about the problem correctly except that obviously the splitting field won't be as described if the quadratic is reducible. Other than that, unless I'm forgetting something (which would hardly be surprising), what you're saying is completely true.

 

[Hurkyl]

Try thinking linear algebra -- if:

. F is a field
. a is algebraic over F
. the minimal polynomial of a is quadratic

Then you know that F(a) is a vector space over F, with basis {1, a}, right? So if you know the number of elements of F, then how many does F(a) have?


Another useful linear algebra viewpoint -- one common way to study an operator is by looking at its eigenvalues. What are the eigenvalues of X?


Finally, a more algebraic viewpoint: you can prove that F[X] (the subalgebra of two by two matrices generated by X) is actually isomorphic to F(a).

 

[Hello Kitty]

OK thanks guys. So F(a) has cardinality the square of F in Hurkyl's question, hence a splitting field of an (irreducible) quadratic over F_{p^n} is always F_{p^{2n}}. So we just need the polynomial X^2 + \mu X + \lambda to have a root which is a primitive element of F_{p^{2n}}.

I guess eigenvalues are the key to showing that \left( \begin{array}{cc}<br /> 0 &amp; 1 \\<br /> -\lambda &amp; -\mu \end{array} \right) then must have order p^{2n}-1.

Indeed diagonalization over F_{p^{2n}} gives \left( \begin{array}{cc}<br /> 0 &amp; 1 \\<br /> -\lambda &amp; -\mu \end{array} \right) = P^{-1}\left( \begin{array}{cc}<br /> r_1 &amp; 0 \\<br /> 0 &amp; r_2 \end{array} \right)P where r_1, r_2 are the roots of X^2 + \mu X + \lambda in F_{p^{2n}}. It immediately follows that our matrix has exact order p^{2n}-1.

Is there a canonical form for such a polynomial? I.e. can we write \mu = \mu(w), \lambda = \lambda(w) where w is a primitive element of the field. I suspect this is probably asking a bit too much though. Is there at least a method of generating examples of these polynomials?

 

[Hurkyl]

Every element w of an algebraic extension of F has a minimal polynomial -- the unique, monic, irreducible polynomial f that has w as a root. If you know w, it's fairly straightforward to compute -- in this case, just compare w^2 to 1 and w. If you know w's conjugates, it's even easier, since the roots of f are the conjugates of w.