# $GL_n(\mathbb F_q)$ has an element of order $q^n-1$

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 ?

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

• +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. Jan 1, 2014 at 20:06
• 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. Jan 1, 2014 at 20:25
• How can you show this subgroup is isomorphic to the field? 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.