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There's a particular property of the elements of $\mathrm{GL}(2,q)$, the general linear group of $n\times n$ matrices over a finite field of order $q$, that I don't understand.

I know that the order of $\mathrm{GL}(2,q)$ is $(q^2-1)(q^2-q)=q(q+1)(q-1)^2$, since there are $q^2-1$ possible vectors for the first column, excluding the $0$ vector, and $q^2-q$ possible vector for the second column, excluding all multiples of the first.

So the order of any element must divide $q(q+1)(q-1)^2$ by Langrange. However, there is a further detail that

any element of $\mathrm{GL}(2,q)$ must have order dividing $q(q-1)$ or $(q-1)(q+1)$.

Is there a reason why one can narrow down the order to divide one of those smaller factors of $q(q+1)(q-1)^2$? Thanks!

user26857
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Norbert Wiener
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1 Answers1

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Say $q=p^n$, with $p$ prime, $n\gt 0$.

If the matrix is diagonalizable, then you can find elements $a$ and $b$ of $\mathbb{F}_q$ such that $$A\approx \left(\begin{array}{cc} a&0\\ 0&b \end{array}\right).$$ Since $ab\neq 0$, then since $a^{q-1}=b^{q-1} = 1$, it follows that $A^{q-1}=I$, so the order divides $q-1$. Selecting $a$ to be a primitive root shows that we cannot get away with anything smaller.

If $A$ is not diagonalizable but has repeated eigenvalue, then it is similar to the matrix $$\left(\begin{array}{cc} a & 1 \\ 0 & a \end{array}\right).$$ The $n$th power of this matrix is $$\left(\begin{array}{cc} a & 1\\ 0 & a\end{array}\right)^n = \left(\begin{array}{cc} a^n & na^{n-1}\\ 0 & a^n \end{array}\right).$$ For this to be the identity, we need $n$ to be a multiple of the order of $a$ in the multiplicative group of units, and for $n$ to be a multiple of the characteristic. The order of $a$ is coprime to the characteristic, so the order is a multiple of $p$ that divides $p(q-1)$; selecting $a$ to be a primitive root shows we can get away with nothing smaller.

If the matrix has no eigenvalues, then it is diagonalizable over the field of $q^2$ elements, and so by the first case above has order dividing $q^2-1$; selecting a primitive polynomial for $\mathbf{F}_{q^2}$ over $\mathbf{F}_q$ will produce a matrix of order exactly $q^2-1$.

So in any case, the order of an element either divides $p(q-1)$ (if it has eigenvalues), or divides $q^2-1$ (if it has no eigenvalues).

user26857
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Arturo Magidin
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  • Ah, so we can actually do better than saying it divides $q(q-1)$ when it has eigenvalues. Thanks again. – Norbert Wiener Nov 04 '11 at 21:58
  • A quick clarification if you don't mind, why does the matrix having no eigenvalues imply it's diagonalizable over the field of $q^2$ elements? – Norbert Wiener Nov 05 '11 at 11:34
  • @NorbertWiener: If the matrix has no eigenvalues, then its characteristic polynomial must be irreducible quadratic; the splitting field of the irreducible quadratic polynomial is a field of degree $2$ over $\mathbf{F}q$; the only such field is the field $\mathbf{F}{q^2}$. That means that the characteristic polynomial of the irreducible quadratic splits over that field, so it certainly has eigenvalues. Moreover, the extension is separable (since finite fields are perfect), so the characteristic polynomial has two distinct roots in $\mathbf{F}_{q^2}$. – Arturo Magidin Nov 05 '11 at 20:42