How many elements are in the group u={B in B : det(B)=1}?

[maggie56]

Homework Statement

Hi, i need to show how many elements there are in the group u={B in B : det(B)=1}
where B is the group of invertible matrices - B = b(ij) in GLn(Fp)
where Fp is a field with p elements.

Homework Equations

The Attempt at a Solution

I know Fp^n has p^n elements, and the number of elements in GLn(Fp) = \prod (p^n - p^i) for i=1 to n.
But this doesn't include anything about the determinant being 1 and i can't see how to include this.

Thanks any help will be greatly appreciated.

 

[micromass]

Consider the following function

\varphi:GL_n(\mathbb{F}_p)\rightarrow SL_n(\mathbb{F}_p):A\rightarrow \frac{1}{det(A)}A

Use the first isomorphism theorem on this function. But for this, you will have to calculate the kernel of the function...

 

[maggie56]

hi, thanks for your help. i have seen that due to the first isomorphism theorem,

<br /> \varphi:GL_n(\mathbb{F}_p) / SL_n(\mathbb{F}_p) <br />
is the field F excluding 0 so the number of elements in SLn(Fp) is \prod (p^n - p^i) / (p-1)

but this doesn't allow for the fact that my set is the set of upper triangular matrices that are invertible with determinant 1, so how do a allow for the matrix being upper triangular?

Thanks

 

[micromass]

Can't the same method be applied to upper triangular matrices? Or do I misunderstand your question?

 

[maggie56]

i don't understand how to apply it, i guess there are less elements when there is a requirement that they are upper triangular? but i can't see how to include it in my method?

am i any clearer?

 

[micromass]

Yes, but It's not clear to me what the problem is. Is it correct that you have to find number of triangular matrices whose determinant is 1?

In this case, you can use that the determinant is 1 if and only if the product of the diagonal entries is 1. Thus for triangular matrices, you should only care about the diagonal.

 

[maggie56]

so is my answer correct?
i am confused as i have two questions to determine the number of elements in;

1) B = {B=bij \in GLn(Fp), bij = 0 for i < j }
2) U = {B \in B : det(B) = 1}

i haven't managed to do the first part i don't think as all i have proved is that GLn(Fp) has \prod (p^n - p^i) elements, but i don't understand how to find the number of elements for an upper triangular matrix

sorry i should have put this up before.

 

[micromass]

Aaaah, now I understand! I was really confused when you started talking about triangular matrices...

Well, let's first do a simpler question: can you determine the number of upper triangular matrices (which do not necessairily have to be invertible). Please also include your method, since we will try to tweak your method so that it provides an answer to (1).

 

[maggie56]

sorry, all i can think is that you know the amount of zeros is \sum x! for x=1 to n totally irrelevant though.
i really don't know how to do this?

 

[micromass]

Ok, I'll work it out for 2x2-matrices. I'll let you do the general case.

A 2x2-upper triangular matrix has the form

\left(\begin{array}{cc}a &amp; b\\ 0 &amp; c\end{array}\right)

a, b and c are arbitrary elements of \mathbb{F}_p. Thus we have p choices for a, b and c. This gives us that there are p³ upper triangular matrices...

 

[maggie56]

for an nxn matrix there will be n(n+1)/2 choices, so is it the product of i(i+1)/2 for i=1 to n?

 

[micromass]

Yes, so now we have to adjust our argument somehow so that we only count invertible matrices.

Now, what is the general form of an upper triangular matrix that is also invertible?
Of course, it will have the form

\left(\begin{array}{cc} a &amp; b\\ 0 &amp; c\end{array}\right)

but what conditions have to hold for a, b and c for the matrix to be invertible?

 

[maggie56]

they cannot be zero,

 

[maggie56]

so are there p-1 choices for each of them? which gives (p-1)^n choices along the diagonal?

 

[micromass]

Not quite, but you're almost there. Some of the values CAN be zero. But which one can't?

 

[maggie56]

only diagonal values can't be zero

 

[micromass]

Yes, so an upper triangular matrix is invertible if none of it's diagonal values are zero. So, can you use this to provide a formula for the number of invertible upper triangular matrices?

 

[maggie56]

thank you for your patience!
can i find the formula by multiplying the number of choices for each element in the matrix together, in which case i would have p^(n(n-1)/2)p^n choices for each matrix then the product of this over all matrices would be the formula.
i hope I am not too wrong here, but i don't have much confidence in my answer!

 

[micromass]

Yes, that's perfect!

Now, to end the exercise, find all the matrices with determinant 1. This can be done in two ways:
- a combinatorial argument: thus do something like in the above post
- apply the first isomorphism theorem to suitable groups

 

[maggie56]

is it correct that for the determinant to be one in this upper triangular matrix that the diagonal entries must also be one?
in this case will it be \prod (p ^ {i(i-1))/2}) for i = 1 to n

 

[micromass]

No, for example, the matrix

\left(\begin{array}{cc} 2 &amp; 0\\ 0 &amp; 1/2\end{array}\right)

is upper triangular matrix whose determinant is 1. Still it's diagonal entries are not all one...

 

[maggie56]

sorry I am completely stuck again, i think SLn(Fp) = |GLn(Fp)| / |Fp| which would give me
\prod (p^{n(n-1)/2}(p-1)^n) / |Fp| ?

 

[micromass]

Well that is the correct formula, but you shouldn't use the notation SL or GL. These sets are the sets of all invertible matrices (with SL being of determinant 1). However, you are interested in all TRIANGULAR matrices. So for these, you shouldn't be using SL or GL...

However, the formula you wrote down is in fact the formula that you're looking for...

 

[maggie56]

thank you very much for your help