What's the Order of GL_2(F_5)?

[T-O7]

Okay, so I need to find the order of the group $$GL_2(F_5)$$, the group of all invertible 2×2 matrices whose elements are from the field $$F_5=\{0,1,2,3,4\}$$. It's a pretty tedious question, but I'm not too confident about my answer. I get a total of 486. Anyone bored enough to help?

 

[Hurkyl]

I didn't get 486. How did you do the problem?

 

[shmoe]

Hurkyl, I think you made a typo.

T-O7, you've over counted somewhere. If you give some info on how you came to 486, we might be able to find what went wrong.

 

[T-O7]

My method was super-long...
This was how I broke it up:
Case 1: 1 non zero entry - no possible matrices
Case 2: 2 non-zero entries
Possible only if entries are diagonal from each other( i.e.$$\left(\begin{array}{cc}0&a\\b&0\end{array}\right)$$ or $$\left(\begin{array}{cc}a&0\\0&b\end{array}\right)$$), and then $$4^2\times2 = 32$$ of these are possible.
Case 3: 3 non-zero entries
all combinations are possible, and there are $$4^3\times4 = 256$$ in total.
Case 4:4 non-zero entries
There are $$4^4 = 256$$ possible in theory. There are several subcases which must be deleted:
Subcase1: matrices of the form $$\left(\begin{array}{cc}a&b\\a&b\end{array}\right)$$ and $$\left(\begin{array}{cc}a&a\\b&b\end{array}\right)$$. There are $$4^2 +4^2 - 4 =28$$ of these (overcounting 4 matrices, when a=b)
Subcase2: all other matrices that give a determinant congruent to 0 (mod 5). I found 6 "types" of these matrices, each of which could be rearranged into 2,4, or 8 other combinations to give the same determinant. In total I found $$30$$ of these (this is where I'm a little skeptical).

This gives me a total of $$32 + 256 + (256-28-30) = 486$$ elements. Is there something I missed?

 

[Hurkyl]

What about matrices that look like

/ a a \
\ a b /

?


The typical way of approaching this problem is to consider one row at a time...

 

[T-O7]

I think those are taken care of in case 4. A matrix of that form won't have determinant congruent to 0 (mod5).

 

[shmoe]

Hi, your case 4 is a problem. The matrices you find in Subcase1 will have zero determinant mod 5 as well, so why aren't they included in your Subcase2?

Maybe a better way to count your matrices in subcase 4 is to allow anything non-zero for the first row:

$$\left(\begin{array}{cc}a&b\\ *&*\end{array}\right)$$

Then ask how many options do you have for the second row to make the rows linearly dependant? Try thinking of the rows as vectors when you answer this question.

 

[T-O7]

You're right, shmoe, my subcase1 should really be under subcase2, but i separated that case because it was to me the more "trivial" case of 4 non-zero entries having a determinant 0. Subcase2 consists of all other "non-trivial" combinations of the non-zero entries that also give a determinant of zero. Unfortunately, i think i might be missing a couple from subcase2. I'll think about your suggestion!

 

[T-O7]

Success!
I realized i had neglected 6 matrices in Subcase2, so actually there were 36 matrices in that subcase instead of 30. Therefore, i got $$32+256+(256-28-36)=480$$, which I think is now right...cuz I need the order to divide 16. I guess you can't be too careful in these type of calculations. Thanks all!

 

[shmoe]

I'll think about your suggestion!

Do think about it carefully and note it's the same idea as Hurkyl is hinting at, one row at a time, except restricted to the case of all entries non-zero.

With this idea you can actually answer this question in one fell swoop. In fact, you can can come up with a simpe formula for the number of elements in $$GL_n(F_q)$$ for any prime power q and any n by doing the row by row approach.

Good observation realizing the order had to be divisible by 16 though!

 

[T-O7]

So it turns out that I'm required to now calculate the order of the general group $$GL_n(F_p)$$, and after thinking about the linearly independent columns like you guys suggested, it struck me that it actually is surprisingly simple to calculate its order. And yes, I did get a pretty nice formula ($$(p^n-p^0)(p^n-p^1)...(p^n-p^{n-1}))$$, and then just to double check, I plugged in the corresponding numbers for the group I had above, and my original answer of 480 turned out to be the right answer (phew). If only I had thought of this sooner, I wouldn't have had to gone through all of that! But thanks a lot

 

[Kalimaa23]

lol, this is often how mathematics is done. You do a bunch of tedious calculations, and after that you realize that there is a simpler general way of handling the problem

 

[Rizwanur]

Hi T-O7,

I was actually trying to determine the number of elements in both GL_n(F_p) and SL_n(F_p).

I was looking at your investigation into finding the number of elements in GL_2(F_5). Could you please explain why there are 36 elements discarded in subcase 2 of case 4?

 

[Rizwanur]

Can you also explain how you derived the formula (p^n - 1)*(p^n - p)*...*(p^n - p^(n - 1)) please? I get that at first there are p^n-1 options to choose any vector which is non-zero but I'm not sure about the rest. Thanks.

 

[Rizwanur]

Hi Shmoe,

I'm also doing a similar mathematics problem at Exeter University. Firstly, could you please explain why the order of the group GL_2(F_5) is divisible by 16? Secondly, could you also explain how the number of elements in the group GL_n(F_p) is ((p^n - 1)*(p^n - p)*...*(p^n - p^(n - 1)) (as mentioned by T-O7) please? Thanks