11
$\begingroup$

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 ?

$\endgroup$

2 Answers 2

11
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ +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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.