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Math Help - The Size of GL_2(Z_7)

  1. #1
    Senior Member roninpro's Avatar
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    The Size of GL_2(Z_7)

    Hello everyone. I am asked to find the size of the group G=GL_2(\mathbb{Z}_7) and would like to verify my work.

    This group acts on the vector space \mathbb{Z}_7\times \mathbb{Z}_7 with basis \{(1,0), (0,1)\}. From this, we have the orbit-stabilizer theorem: if x\in \mathbb{Z}_7\times \mathbb{Z}_7, then |G|=|Gx|\cdot |G_x|, that is, the size of G is the size of the orbit of x times the size of the stabilizer of x. Take x=(1,0). If A\in G, Ax can possibly be any nonzero vector. There are 48 of these. Therefore, |Gx|=48. Then, if Ax=x, we must have A(0,1) equal to another vector not a multiple of x, since A is invertible. In other words, A(0,1) cannot be (0,0), (1,0), (2,0),\ldots,(6,0). There are 42 of these, so |G_x|=42.

    We conclude that |G|=48\cdot 42=2016.

    Comments or suggestions?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by roninpro View Post
    Hello everyone. I am asked to find the size of the group G=GL_2(\mathbb{Z}_7) and would like to verify my work.

    This group acts on the vector space \mathbb{Z}_7\times \mathbb{Z}_7 with basis \{(1,0), (0,1)\}. From this, we have the orbit-stabilizer theorem: if x\in \mathbb{Z}_7\times \mathbb{Z}_7, then |G|=|Gx|\cdot |G_x|, that is, the size of G is the size of the orbit of x times the size of the stabilizer of x. Take x=(1,0). If A\in G, Ax can possibly be any nonzero vector. There are 48 of these. Therefore, |Gx|=48. Then, if Ax=x, we must have A(0,1) equal to another vector not a multiple of x, since A is invertible. In other words, A(0,1) cannot be (0,0), (1,0), (2,0),\ldots,(6,0). There are 42 of these, so |G_x|=42.

    We conclude that |G|=48\cdot 42=2016.

    Comments or suggestions?
    Personally I don't see why you can't just count the possibilities based on the columns. In general \displaystyle \text{GL}_n\left(\mathbb{F}_p\right)=\prod_{j=0}^{  n-1}\left(p^n-p^j\right). So, \left|\text{GL}_2\left(\mathbb{Z}_7\right)\right|=  \left(7^2-1\right)\left(7^2-7\right)=2016.

    EDIT: And by my first sentence I really mean I don't know the theorem you're talking about but know that what follows is true haha
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