# Order of this subgroup:

• November 6th 2008, 03:30 PM
realillusion
Order of this subgroup:
This question was posed in a class recently. We're not very "far" into group theory so I don't think it requires much specialized knowledge...

H = { A | A^2 = I, A in GL(2,7) }
(GL(2,7) is the general linear group of invertible 2x2 matrices with entries from the field Z_7; H is the subgroup of all elements of GL(2,7) with order 2)
What is the order of H?

I took an element A of H,
A=[a, b; c, d]
and squared it, set it to I to get
a^2 + bc = 1
ab + bd = 0
ac + cd = 0
bc + d^2 = 1

This seems a likely spot for me to have made a mistake. I split this system into 4 cases of solutions.

1) b = 0, c = 0, a^2 = 1, d^2 = 1
This case includes 4 solutions, with a and d chosen from {1,6} (since 6 is "negative 1")
2&3) b XOR c is zero. a^2=1 and d=-a
24 solutions (choose b XOR c, choose a value for it, choose a from {1,6}: 2x6x2), distinct from above where b=c=0.
4) b,c both nonzero.
d=-a again, but now
a^2 + bc = 1, rewritten: c=(1-a^2)/b. Since c is nonzero, we get the restriction that 1-a^2 != 0, so a is neither 1 nor 6.
30 more distinct solutions, from the 5 choices of a and 6 choices of b.

So my initial answer is 58 elements. Unfortunately for me, we were given a "hint" from a putnam exam that basically says the order of H is odd unless I misunderstand:
Suppose a finite group (like GL(2,7)) has exactly n elements of order p (I claim 58 elements of order 2). Then either n=0 or p divides n+1.

Somewhere I missed something. Any thoughts?
• November 6th 2008, 07:59 PM
ThePerfectHacker
Quote:

Originally Posted by realillusion
and squared it, set it to I to get
a^2 + bc = 1
ab + bd = 0
ac + cd = 0
bc + d^2 = 1

An inefficient way to solve these equations is to note that you are working over $\mathbb{F}_7$. Therefore, there are finitely many cases to check. You can start with $a=0$ and go to $a=6$ (for those are elements in the finite field) and see if you get any solutions. For example, say $a=0$. The first equation forces $bc=1$ which in particular means $b,c$ are multiplicative inverses. Now look at second equation, $ac+bd=0$ since by assumption $a=0$ it means $bd=0$, but $b$ is invertible and so $d=0$. Therefore, the third equation gives nothing and and the fourth equation $bc+d^2 = 1\implies bc=1$ also does not tells us anything new. Thus, all we have $d=0$ and $bc=1$. Therefore, if $a=0$ then the matrices that have this property have $d=0$ and the entries on $b,c$ such that $bc=1$. These include, $(1,1),(2,4),(3,5),(4,2),(5,3),(6,6)$. This contributes $6$ elements. Now move over to the case $a=1$ and so on. Those cases get more involved but the same idea applies.

As for the hint the theorem goes as follows: let $G$ be a finite group so that $|G|$ is divisible by $p$ then the number of solutions to $x^p = 1$ is a multiple of $p$. However, there is one solution which is trivial, that is $x=1$. That is the only solution that has order $1$. While all other solutions have order $p$. Thus, if $n_p$ is the number of elements of order $p$ then it follows that $n_p \equiv -1 (\bmod p)$.
• November 7th 2008, 12:17 AM
NonCommAlg
Quote:

Originally Posted by realillusion

H = { A | A^2 = I, A in GL(2,7) }

(GL(2,7) is the general linear group of invertible 2x2 matrices with entries from the field Z_7; H is the subgroup of all elements of GL(2,7) with order 2)

What is the order of H?

you didn't miss anything. the point is that the identity matrix belongs to H but its order is 1 not 2. you can extend your solution to solve the general case:

let $p$ be any prime number and define $H_p=\{A \in \text{GL}(2,p): \ A^2=I \}.$ then we have: $|H_p| = \begin{cases} p^2 + p + 2 & \ \text{if} \ \ p \neq 2 \\ 4 & \ \text{if} \ \ p=2 \end{cases}.$

Question: is it possible for $H_p$ to be a subgroup for some prime $p$?
• November 7th 2008, 06:17 AM
ThePerfectHacker
Is it possible to solve this problem using some properties about general linear groups? I am asking because I wish I knew more about the general linear group and was wondering if there are any useful results about them that can produce this solution.
• November 7th 2008, 08:19 AM
realillusion
Great responses; thanks all!

Quote:

Is it possible to solve this problem using some properties about general linear groups?
The only thing that comes to mind is that it is very easy to find the order of GL groups over finite fields, since the matrices are invertible/columns are independent.
General linear group - Wikipedia, the free encyclopedia
For our case, |GL(2,7)|=48*42=2016

That tells us the order of the GL group, but doesn't say anything about the order of its elements.
• November 7th 2008, 08:00 PM
NonCommAlg
it's just a simple counting problem: it's easy to see that $|H_2|=4.$ so i assume p > 2 is a prime. we want to find the number of solutions of the following system of equations in $\mathbb{F}_p$:

$(1) \ \ \ a^2 + bc=d^2+bc=1,$

$(2) \ \ \ b(a+d)=c(a+d)=0.$

from (1) we get: $a=\pm d.$ so we consider two cases:

Case 1: $a = -d.$ in this case (2) is satisfied. so we'll only have: $a^2 + bc=1.$ now if $b=0,$ then $a=\pm 1.$ so the solutions in this case are: $a=\pm 1, \ b = 0, \ d =\mp 1,$ and $c$ can be anything.

this gives us $2p$ solutions. if $b \neq 0,$ then $c=b^{-1}(1-a^2).$ that gives us $p(p-1)$ solutions. so in this case in total we have $p^2+p$ solutions.

Case 2: $a=d.$ if $a=0,$ then $d=0,$ and hence $a=-d,$ which is included in Case 1. if $a \neq 0,$ then (1) and (2) give us $a = \pm 1, \ b=c=0.$ so in this case we have only 2 solutions.

this proves that $|H_p|=p^2+p+2.$ also note that $H_p$ is never a subgroup of $\text{GL}(2,p)$ because it's not closed under multiplication.