Exercise 1.4.8 ($\mathrm{GL}_n(F)$ is non-abelian for any $n\geq 2$ and any $F$)

Question

Show that $\mathrm{GL}_n(F)$ is non-abelian for any $n\geq 2$ and any $F$.

Richard Ganaye · Accepted Answer

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

\begin{proof} 
Let $F$ be any field, where $0$ is the neutral element of $F$, and $1 
e 0$ the neutral element of $F^	imes$.

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
$$NM = \begin{pmatrix} 1 & 1\ 1 & 0\end{pmatrix}  \begin{pmatrix} 0 & 1\ 1 & 0\end{pmatrix} = \begin{pmatrix} 1 & 1\  0 & 1\end{pmatrix},$$
thus $M	imes N 
e N 	imes M$.
Put
$$M' = \begin{pmatrix} M & 0_{2,n-2} \ 0_{n-2,2} & I_{n-2} \end{pmatrix},\qquad N' = \begin{pmatrix}N & 0_{2,n-2} \ 0_{n-2,2} & I_{n-2} \end{pmatrix}.$$
Then $\det(M') = \det(M) 
e 0$ and $\det(N') = \det(N) 
e 0$, so $M', N'$ are in $\mathrm{GL}_n(F)$.
Moreover
$$M'N' = \begin{pmatrix} MN& 0_{2,n-2} \ 0_{n-2,2} & I_{n-2} \end{pmatrix} 
e \begin{pmatrix} NM & 0_{2,n-2} \ 0_{n-2,2} & I_{n-2} \end{pmatrix} = N' M',$$
therefore $\mathrm{GL}_n(F)$ is non-abelian.
\end{proof}

Exercise 1.4.8 ($\mathrm{GL}_n(F)$ is non-abelian for any $n\geq 2$ and any $F$)

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Exercise 1.4.8 ($\mathrm{GL}_n(F)$ is non-abelian for any $n\geq 2$ and any $F$)

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