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What's the order of THIS group?

  1. Nov 3, 2004 #1
    Okay, so I need to find the order of the group [tex]GL_2(F_5)[/tex], the group of all invertible 2×2 matrices whose elements are from the field [tex]F_5=\{0,1,2,3,4\}[/tex]. 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? :redface:
     
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  3. Nov 3, 2004 #2

    Hurkyl

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    I didn't get 486. How did you do the problem?
     
    Last edited: Nov 3, 2004
  4. Nov 3, 2004 #3

    shmoe

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    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.
     
  5. Nov 3, 2004 #4
    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.[tex]\left(\begin{array}{cc}0&a\\b&0\end{array}\right)[/tex] or [tex]\left(\begin{array}{cc}a&0\\0&b\end{array}\right)[/tex]), and then [tex]4^2\times2 = 32[/tex] of these are possible.
    Case 3: 3 non-zero entries
    all combinations are possible, and there are [tex]4^3\times4 = 256[/tex] in total.
    Case 4:4 non-zero entries
    There are [tex]4^4 = 256[/tex] possible in theory. There are several subcases which must be deleted:
    Subcase1: matrices of the form [tex]\left(\begin{array}{cc}a&b\\a&b\end{array}\right)[/tex] and [tex]\left(\begin{array}{cc}a&a\\b&b\end{array}\right)[/tex]. There are [tex]4^2 +4^2 - 4 =28[/tex] 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 [tex]30[/tex] of these (this is where I'm a little skeptical).

    This gives me a total of [tex]32 + 256 + (256-28-30) = 486 [/tex] elements. Is there something I missed?
     
    Last edited: Nov 3, 2004
  6. Nov 3, 2004 #5

    Hurkyl

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    What about matrices that look like

    Code (Text):

    / a a \
    \ a b /
     
    ?


    The typical way of approaching this problem is to consider one row at a time...
     
  7. Nov 3, 2004 #6
    I think those are taken care of in case 4. A matrix of that form won't have determinant congruent to 0 (mod5).
     
  8. Nov 3, 2004 #7

    shmoe

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    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:

    [tex]\left(\begin{array}{cc}a&b\\ *&*\end{array}\right)[/tex]

    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.
     
  9. Nov 3, 2004 #8
    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!
     
  10. Nov 3, 2004 #9
    Success!! :biggrin:
    I realized i had neglected 6 matrices in Subcase2, so actually there were 36 matrices in that subcase instead of 30. Therefore, i got [tex]32+256+(256-28-36)=480[/tex], which I think is now right....cuz I need the order to divide 16. :biggrin: I guess you can't be too careful in these type of calculations. Thanks all! :smile:
     
  11. Nov 4, 2004 #10

    shmoe

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    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 [tex]GL_n(F_q)[/tex] 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!
     
  12. Nov 17, 2004 #11
    So it turns out that i'm required to now calculate the order of the general group [tex]GL_n(F_p)[/tex], 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 ([tex](p^n-p^0)(p^n-p^1)...(p^n-p^{n-1}))[/tex], 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 :biggrin:
     
    Last edited: Nov 17, 2004
  13. Nov 19, 2004 #12
    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 :smile:
     
  14. Apr 17, 2008 #13
    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?
     
  15. Apr 17, 2008 #14
    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.
     
  16. Apr 17, 2008 #15
    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
     
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