Exercise 1.4.1 (Cardinality of $\mathrm{GL}_2(\mathbb{F}_2))$)

Question

Prove that $|\mathrm{GL}_2(\mathbb{F}_2)| = 6$.

Richard Ganaye · Accepted Answer

ewcommand{\F}{\mathbb{F}}

\begin{proof}

To build a matrix $ M \in \mathrm{GL}_2(\F_2)$, we write on the first row a nonzero vector in $\F_2 	imes \F_2$, among $(1, 0),\ (0,1),\ (1,1)$. For each of these vectors $v = (a, b)$, the second row of the matrix is a non collinear vector, taken among two vectors distinct of $(0, 0)$ and $(a, b)$. So

$$|\mathrm{GL}_2(\F_2)|= 3\cdot 2 =6.$$

(This corresponds to the formula
$$|\mathrm{GL}_n(\F_q)| = \prod_{i=0}^{n-1} (q^n - q^i),$$
which gives for $n= 2$ and $q = 2$,
$$|\mathrm{GL}_2(\F_2)|= (2^2 - 1)(2^2-2) =6.)$$
The complete list of the elements is
$$\mathrm{GL}_2(\F_2) =
 \left \{ 
 \begin{pmatrix} 1 & 0\ 0 & 1\end{pmatrix},
 \begin{pmatrix} 1 & 0\ 1 & 1\end{pmatrix},
 \begin{pmatrix} 0 & 1\ 1 & 0\end{pmatrix},
 \begin{pmatrix} 0& 1\ 1 & 1\end{pmatrix},
 \begin{pmatrix} 1&1\ 0 & 1\end{pmatrix},
 \begin{pmatrix} 1 & 1\ 1 & 0\end{pmatrix}
 ight\}
$$
\end{proof}

Exercise 1.4.1 (Cardinality of $\mathrm{GL}_2(\mathbb{F}_2))$)

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Exercise 1.4.1 (Cardinality of $\mathrm{GL}_2(\mathbb{F}_2))$)

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