Exercise 1.4.3 ($\mathrm{GL}_2(\mathbb{F}_2)$ is non-abelian.)

Question

Show that $\mathrm{GL}_2(\mathbb{F}_2)$ is non-abelian.

Richard Ganaye · Accepted Answer

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\begin{proof} 
Put $M = \begin{pmatrix} 0 & 1\ 1 & 0\end{pmatrix} $ and $N =\begin{pmatrix} 1 & 1\ 1 & 0\end{pmatrix}$ in $\mathrm{GL}_2(\F_2)$.

Then
$$ MN = \begin{pmatrix} 0 & 1\ 1 & 0\end{pmatrix}  \begin{pmatrix} 1 & 1\ 1 & 0\end{pmatrix} =  \begin{pmatrix} 1& 0\ 1 & 1\end{pmatrix},$$
and
$$NM = \begin{pmatrix} 1 & 1\ 1 & 0\end{pmatrix}  \begin{pmatrix} 0 & 1\ 1 & 0\end{pmatrix} = \begin{pmatrix} 1 & 1\  0 & 1\end{pmatrix}.$$
Since $MN 
e NM$, the group $\mathrm{GL}_2(\F_2)$ is non-abelian.
\end{proof}
Note: Since $\mathrm{GL}_2(\F_2)$ is a non-abelian group of order $6$, it is isomorphic to $S_3$ (or $D_6$).

An explicit isomorphism is given by $\begin{pmatrix} 0 & 1\ 1 & 0\end{pmatrix} \mapsto (1 \ 2)$\ and $ \begin{pmatrix} 1 & 1\ 1 & 0\end{pmatrix} \mapsto (1\ 2\  3)$.

Exercise 1.4.3 ($\mathrm{GL}_2(\mathbb{F}_2)$ is non-abelian.)

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Exercise 1.4.3 ($\mathrm{GL}_2(\mathbb{F}_2)$ is non-abelian.)

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