# Find the orders of $M_3(\mathbb Z_2)$ and $GL_3(\mathbb Z_2)$ [duplicate]

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

$$|GL_3(\mathbb Z_2)| = 64$$ 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.

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 $$\prod_{k=0}^{m-1}(q^m-q^k).$$
• So, $|SL_3\mathbb(Z_2)| = 84$? – user551155 Mar 31 '20 at 3:28
• Yes, consider the map $GL_3(Z_2)\rightarrow F^{\times}$ via taking the determinant. It has kernel $SL_3(Z_2)$. – Vasting Mar 31 '20 at 3:33