TITLE: Explicit construction of an element of ${\rm GL}(2, p)$ of order $p+1$
QUESTION [13 upvotes]: It is well-known that the order of $GL(2, p)$ is $(p^2-1)(p^2-p) = (p-1)^2(p+1)p.$
It is easy to construct matrices of orders $(p-1)$ and $p$ (diagonal and parabolic, respectively), but the only way that leaps to mind of constructing an element of order $p+1$ is by constructing the companion matrix of the irreducible polynomial whose root is the multiplicative generator of $GF( p^2)$ and then raising it to $p-1$st power - this is quite non-explicit. I assume there is no explicit construction which works for all $p,$ but I could be wrong: is there?

REPLY [5 votes]: Here is a slightly different way to think about this, though I admit that it is not really "explicit". Let $E$ be the field of order $p^2$, so $E$ is a two-dimensional vector space over the subfield $F$ of order $p$. The multiplicative group of $E$ is cyclic of order $p^2 - 1$, so if $x$ is a generator, then multiplication by $x$ is an $F$-linear transformation of $E$ with multiplicative order $p^2 - 1$. Now by choosing any $F$-basis for $E$, this linear transformation determines a matrix which is an element of $GL(2,p)$ having order $p^2 - 1$. A suitable power of this matrix has order $p+1$.