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General Linear Group Order

  1. Sep 9, 2006 #1
    Here is the problem:

    Let [tex] p [/tex] be a prime. Prove that the order of [tex] GL_2 ( \mathbb{Z} / p \mathbb{Z} ) [/tex] is [tex] p^{4} - p^{3} - p^{2} + p [/tex]

    The text suggests subtracting the number of 2 x 2 matrices which are not invertible from the total number of 2 x 2 matrices over [tex] \mathbb{Z} / p \mathbb{Z} [/tex]

    I have been working on this for awhile, but it's not going well.

    First, it seems obvious to me that the total number of 2 x 2 matrices over [tex] \mathbb{Z} / p \mathbb{Z} [/tex] must be [tex] p^{4} [/tex] since each of the 4 entries has [tex] p [/tex] possible values.

    Based on this assumption, I am forced to conclude that there are [tex] p^{3} + p^{2} - p [/tex] of these matrices that are not invertible. However, I'm having a hard time showing that this is true, if indeed it is.

    Any help is greatly appreciated.
     
  2. jcsd
  3. Sep 9, 2006 #2

    StatusX

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    A matrix is not invertible iff its rows are linearly dependent. In two dimensions, this means one row must be a scalar multiple of the other.
     
  4. Sep 9, 2006 #3
    Right, I knew that... So I tried to build all of the non-invertible 2 x 2 matrices as follows:

    The upper-left element has p possible values.

    The upper-right element has only p - 1 possible values, because we wouldn't want them to both be 0 (already this feels wrong).

    And so on...

    But, I know I'll never get [tex] p^{3} + p^{2} - p [/tex] by multiplying linear factors, since this polynomial doesn't factor into nice linear factors with integer roots.
     
  5. Sep 9, 2006 #4

    StatusX

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    It's probably easier to count the invertible matrices. To build one of these, you can put anything but two zeros in the first row, and anything but the multiples of the first row into the second.
     
    Last edited: Sep 9, 2006
  6. Sep 9, 2006 #5
    Well, actually now that I look at it, I was building the invertible matrices up above. So, let's continue.

    The upper-left element has p possible values.

    The upper-right element has only p - 1 possible values, because we wouldn't want them to both be 0.

    The lower-right element has p possible values.

    The lower-left element has p - 2 possible values, since we don't want both lower elements to be 0 and we don't want the second row to be a multiple of the first.

    So, there are [tex] p^2(p - 1)(p - 2) [/tex] possible invertible matrices, which does not match up with the correct answer.

    I'll work more tomorrow and then check back in here again. Thanks.
     
  7. Sep 9, 2006 #6

    StatusX

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    Try this all out for p=2 and see where you're argument breaks down.
     
  8. Sep 9, 2006 #7
    Okay... I think I've gotten somewhere (so much for going to bed). Check out this argument.

    If a 2 x 2 matrix is invertible, it cannot have 4 or 3 zeros. Therefore, it can only have 2, 1 or no zeros.

    I examined each case and came up with the right answer.

    Case 0: No Zeros

    If the 2 x 2 matrix contains no zeros then there are [tex] (p - 1)^{3} (p - 2) [/tex] possible invertible matrices that one can create. The first three elements can be anything besides 0, and the 4th element has the additional stipulation that it cannot make the second row a multiple of the first.

    Case 1: One Zero

    Assume the zero is the upper right-hand element. The other three elements can each take [tex] p - 1 [/tex] possible values, since only 0 is eliminated. There is no possibility of having linear dependence. Thus, since this argument is true for the zero in any of the four positions, there are [tex] 4(p - 1)^{3} [/tex] possible invertible matices with exactly one zero.

    Case 2: Two Zeros

    In this case, the two zeros must be on a diagonal. Assume they are on the main diagonal. Then the other two elements can each take [tex] p - 1 [/tex] possible values for the same reason as Case 1. Since there are two diagonals, there are [tex] 2 (p - 1)^{2} [/tex] possible invertible matrices with exactly two zeros.


    So, the total number is [tex] (p - 1)^{3} (p - 2) + 4(p - 1)^{3} + 2 (p - 1)^{2} [/tex] which, when expanded, gives the correct answer.
     
  9. Sep 9, 2006 #8

    shmoe

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    That looks fine, but if you wanted to generalize this to 3x3 (or nxn) matrices you'd be in for a nightmare.

    Your first row must be non-zero, there are p^2-1 ways of doing this. After picking this first row, how many choices for the second? (With generalizing to nxn matrices in mind, you could approach this by asking how many elements are in the span of the first row?)

    Counting non-invertible matrices as the text suggests isn't so bad either. Consider two cases, the first row [0 0], and the first row non-zero.
     
  10. Sep 10, 2006 #9
    Thanks for the response. The only problem I noticed is that nowhere did I use the fact that p was a prime...

    Any help there?
     
  11. Sep 10, 2006 #10

    matt grime

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    Just count columns. The first is nonzero, the second is not a multiple of the first , etc.
     
  12. Sep 10, 2006 #11

    shmoe

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    You used that implicitly all over the place. For example, in case 0, how did you conclude there was exactly one solution for the fourth entry? Or that the product of any 3 non-zero entries would be nonzero itself?
     
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