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$|M_3(\mathbb Z_2)| = 2^{3^2} = 64$

$|GL_3(\mathbb Z_2)| = 64$ minus all singular matrices in $M_3(\mathbb Z_2)$ or all matrices with linearly independent columns.

My question is how do I find all singular matrices or all matrices with linearly independent columns. Please advise.

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I believe you miscalculated the first time. $|M_3(\mathbb Z_2)|=2^{3^2}=2^9=512$. Now, let's calculate the order of $|GL_3(\mathbb Z_2)|$. How many choices do we have for the first column? For the matrix to be invertible, it can be anything except all zeroes, so we have $2^3-1$ choices. For the second column, we need to pick anything that is not a multiple of the first column. Since we are in $\mathbb Z_2$, there are two possible multiples, which would mean we have $2^3-2$ choices for the second column. Similarly, the last column has $2^3-2^2$ choices as we need to avoid picking anything that is a multiple of the first two columns, so the order of the group is

$$(2^3-1)(2^3-2)(2^3-2^2)=168$$ In general, building off of this idea, we have the order of $GL_m(\mathbb Z_q)$ as $$\prod_{k=0}^{m-1}(q^m-q^k).$$

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  • $\begingroup$ So, $|SL_3\mathbb(Z_2)| = 84$? $\endgroup$ – user551155 Mar 31 '20 at 3:28
  • $\begingroup$ Yes, consider the map $GL_3(Z_2)\rightarrow F^{\times}$ via taking the determinant. It has kernel $SL_3(Z_2)$. $\endgroup$ – Vasting Mar 31 '20 at 3:33

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