For fixed prime power $q$ show that the general linear group $GL_n(\mathbb F_q)$ of invertible matrices with entries in the finite field $\mathbb F_q$ has an element of order $q^n-1$.

I tried to show this question with showing diagonal matrix but i can't find element directly competible with order i think i am on wrong way please give me clue ?


2 Answers 2


Hint: Realize $\mathbb{F}_{q^n}^*$ as a subgroup of $\mathrm{GL}_n(\mathbb{F}_q)$.

  • $\begingroup$ +1 Note to OP: This will lead to an existence argument. It will not describe a specific matrix of order $q^n-1$. That amounts to finding a primitive element in the field, and while there are algorithms for finding one, there's no closed form answer. $\endgroup$ Jan 1, 2014 at 20:06
  • $\begingroup$ Well, already the case $n=1$ shows that we have to use that the multiplicative group of a finite group is cyclic. The case of general $n$ then follows from it, using my hint. $\endgroup$ Jan 1, 2014 at 20:25
  • $\begingroup$ How can you show this subgroup is isomorphic to the field? $\endgroup$ Mar 10, 2021 at 14:23

Since my question was marked as duplicate, I will try to add partial answer here. I would like to find explicit matrix $2\times2$ of order $q^2-1$ with elements in field $F_q$.

Let $A=\pmatrix {1&1\\1&0}$. I would like to calculate its order over finite field. This matrix satisfy equation $A^2=A+1$. In case when number $t$ is root of polynomial $f=x^2-x-1$ then we have $A*\pmatrix{t\\1}=\pmatrix{t+1\\t}=\pmatrix{t^2\\t}=t*\pmatrix{t\\1}$. Therefore when $t,s$ are two different roots of $f$ belonging to field $F_q$ then matrix $A$ is diagonalizable and order of $A$ is equal to order of $t$ or $s$. I believe that in this case order of $t$ is equal to order of $s$ or it is two times bigger, because we have $st=-1$.

Now we can analyze when $f$ has roots in $F_q$ and what is the order of the root. Here is some experimental data first.

gap> Filtered(Primes{[1..25]},p->First(AsList(GF(p)), x->x*x=x+One(GF(p)))<>fail );
[ 5, 11, 19, 29, 31, 41, 59, 61, 71, 79, 89 ]
gap> List(last,p->Order(First(AsList(GF(p)), x->x*x=x+One(GF(p)))));
[ 4, 5, 18, 14, 15, 40, 58, 60, 35, 39, 44 ]

From the second line we can see that order of root is either $p-1$ or $\frac{p-1}{2}$. From the first line we conclude that there is root in $F_p$ when $p=5$ or last digit of $p$ is $1$ or $9$.

We should distinguish cases when matrix has two different roots, then it is diagonalizable, or it has one root with multiplicity 2. In second case $(x-t)^2=x^2-x-1$ from which we conclude $p=5$ and $t=-2$.

If polynomial $f$ has no roots in $F_p$ then it has roots in $F_{p^2}$. Again we would like to know the orders. Here is some experimental data:

    gap> List(Filtered(Primes{[1..26]}, p->p mod 10 in [3,7]), p->
       [p,Order(First(AsList(GF(p*p)), x->x*x=x+One(GF(p))))]);
    [ [ 3, 8 ], [ 7, 16 ], [ 13, 28 ], [ 17, 36 ], [ 23, 48 ], [ 37, 76 ], 
[ 43, 88 ], [ 47, 32 ], [ 53, 108 ],  [ 67, 136 ], [ 73, 148 ], [ 83, 168 ], 
[ 97, 196 ] ]

As we can see the order is $2(p+1)$.

Interesting thing is relation of matrix $A$ with Fibonacci sequence.

My next task is to analyze the order matrix $\pmatrix{n&1\\1&0}$ when $n$ is generator of field $F_q$.

To be continued.


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