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Math Help - Order of a group

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    Senior Member vincisonfire's Avatar
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    Order of a group

    Let  \mathbb F be a field. Prove that  SL_2 (\mathbb F) be the group of 2 x 2 matrices with coefficient in  \mathbb F and determinant equal to 1.
    Question : If F has q elements, how many elements are in the group  SL_2 (\mathbb F) ?
    I know that  |M_2(\mathbb F)| = q^4.
     SL_2 (\mathbb F) is a subgroup of  GL_2 (\mathbb F) the group of units of  M_2(\mathbb F) .
    From this I've tried to solve the problem but I'm not able to find a way to count the elements in any of the groups.
    Last edited by vincisonfire; November 30th 2008 at 03:52 PM.
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    Quote Originally Posted by vincisonfire View Post
    Let  \mathbb F be a field. Prove that  SL_2 (\mathbb F) be the group of 2 x 2 matrices with coefficient in  \mathbb F and determinant equal to 1.
    Question : If F has q elements, how many elements are in the group  SL_2 (\mathbb F) ?
    Hint: \text{GL}_2(\mathbb{F})/\text{SL}_2(\mathbb{F}) \simeq \mathbb{F}^{\times}
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    Senior Member vincisonfire's Avatar
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    |\text{GL}_2(\mathbb{F})/\text{SL}_2(\mathbb{F})| = |\mathbb{F}^{\times}| = q-1 because any element except 0 has a multiplicative inverse in  \mathbb F .
    In other word, there are q-1 equivalent classes (or orbits) of \text{SL}_2(\mathbb{F}) .
    We thus need to find the number of elements in each orbit.
    In an equivalent class we have
     A=\left( \begin{array}{cc}q&a\\0&q^{-1} \end{array} \right) A \in \text{SL}_2(\mathbb{F})
    There are  \frac{q}{2} combinations of q \text{ and } q^{-1} .
    2q combinations of  a because we can swap  a \text{ and } 0 .
    There are therefore q^2 elements in its equivalent classes.
    |\text{SL}_2(\mathbb{F})| = q^2(q-1) = q^3-q
    Am I right?
    Last edited by vincisonfire; November 30th 2008 at 04:54 PM.
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    Quote Originally Posted by vincisonfire View Post
    |\text{GL}_2(\mathbb{F})/\text{SL}_2(\mathbb{F})| = |\mathbb{F}^{\times}| = q-1 because any element except 0 has a multiplicative inverse in  \mathbb F .
    In other word, there are q-1 equivalent classes (or orbits) of \text{SL}_2(\mathbb{F}) .
    We thus need to find the number of elements in each orbit.
    In an equivalent class we have
     A=\left( \begin{array}{cc}q&a\\0&q^{-1} \end{array} \right) A \in \text{SL}_2(\mathbb{F})
    There are  \frac{q}{2} combinations of q \text{ and } q^{-1} .
    2q combinations of  a because we can swap  a \text{ and } 0 .
    There are therefore q^2 elements in its equivalent classes.
    |\text{SL}_2(\mathbb{F})| = q^2(q-1) = q^3-q
    Am I right?
    | \text{GL}_2(\mathbb{F}) | = (q^2 - 1)(q^2 - q)

    Therefore, \text{SL}_2(\mathbb{F}) = \frac{\text{GL}_2(\mathbb{F})}{q-1} = \frac{(q^2-1)(q^2 - q)}{q-1} = (q-1)q(q+1)
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