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Math Help - Order of Linear Group.

  1. #1
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    Order of Linear Group.

    How many elements are there in SL(2, Fp)?

    Where SL is the special linear group consisting of all matricies with determinant 1.

    Note: Here were talking about 2 by 2 matrices only.

    Answer: (p+1)(p-1)(p)

    Question: Why?

    Could anyone explain this to me?

    Thanks!
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  2. #2
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    Quote Originally Posted by Fulger85 View Post
    How many elements are there in SL(2, Fp)?

    Where SL is the special linear group consisting of all matricies with determinant 1.
    Let \text{GL}(2, \mathbb{F}_p) be the general linear group, that is, the group of all 2\times 2 invertibles matrices over the finite field \mathbb{F}_p.

    For the 2\times 2 matrix A over \mathbb{F}_p to be invertible. For it to be invertible it is necessary and sufficient that the first zero is non-zero and the second row is not a multiple of the first. The number of different possibilities for the first row is p^2 - 1. While the number of multiples for the first row is p therefore the number of rows that are not multiples is p^2 - p. Thus in total there are (p^2 - 1)(p^2 - p) number of invertible matrices.

    Now consider the group homomorphism \det : \text{GL}(2,\mathbb{F}_p) \to \mathbb{F}_p^{\times} defined through the determinant.
    Notice that \ker (\det) = \text{SL}(2, \mathbb{F}_p).
    Therefore, by fundamental homomorphism theorem,
    \text{GL}(2,\mathbb{F}_p)/\text{SL}(2,\mathbb{F}_p) \simeq \mathbb{F}_p^{\times}.

    Thus, |\text{GL}(2,\mathbb{F}_p)| = |\mathbb{F}_p^{\times}| \cdot |\text{SL}(2,\mathbb{F}_p)|.

    Note, that |\mathbb{F}_p^{\times}| = p-1, thus, |\text{SL}(2,\mathbb{F}_p)| = \frac{(p^2 -1)(p^2 - p)}{p-1} = p(p-1)(p+1)
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  3. #3
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    Could you explain to me the process of the first row?

    If the first entry cannot be 0, that leaves p-1 possibilities.
    The second entry on first row can be anything, so that is p possibilities.

    wouldn't that leave the first row at (p-1)*p = P^2 - p possibilities?

    How did you deduce p^2 - 1 ?

    Thanks!
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  4. #4
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    Quote Originally Posted by Fulger85 View Post
    If the first entry cannot be 0, that leaves p-1 possibilities.
    The second entry on first row can be anything, so that is p possibilities.

    wouldn't that leave the first row at (p-1)*p = P^2 - p possibilities?

    How did you deduce p^2 - 1 ?
    The only exception in the first room is the row vector [0 ~ \ ~ 0].
    While there are p \cdot p = p^2 number of all possible row vectors [a ~ \ ~ b].
    Thus, that leaves us with p^2 - 1 possibilities.
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  5. #5
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    Wouldn't [0,1] still be an exception? Considering that no matter what the second row values are, if the first row, first column value is 0, the determinant is zero?

    Or any value [0, x] for that matter?

    Thanks in advance.
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  6. #6
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    Quote Originally Posted by Fulger85 View Post
    Wouldn't [0,1] still be an exception? Considering that no matter what the second row values are, if the first row, first column value is 0, the determinant is zero?

    Or any value [0, x] for that matter?

    Thanks in advance.
    Sorry! I made a mistake. But we not going to let anyone know that.

    In my first and second post replace the word row by vector.
    I think that should fix the error now.
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