General Linear Group is not Abelian/Proof 2

Theorem

Let $K$ be a field whose zero is $0_K$ and unity is $1_K$.

Let $\GL {n, K}$ be the general linear group of order $n$ over $K$.

Then $\GL {n, K}$ is not an abelian group.

Let:

$\ds A$

$=$

$\ds \begin {pmatrix} 1 & 1 \\ 0 & 1 \end {pmatrix}$

$\ds B$

$=$

$\ds \begin {pmatrix} 1 & 1 \\ 1 & 1 \end {pmatrix}$

Both $A$ and $B$ are elements of the general linear group.

Then:

$\ds A B$

$=$

$\ds \begin {pmatrix} 1 & 1 \\ 0 & 1 \end {pmatrix} \begin {pmatrix} 1 & 1 \\ 1 & 1 \end {pmatrix}$

$\ds $

$=$

$\ds \begin {pmatrix} 2 & 2 \\ 1 & 1 \end {pmatrix}$

$\ds B A$

$=$

$\ds \begin {pmatrix} 1 & 1 \\ 1 & 1 \end {pmatrix} \begin {pmatrix} 1 & 1 \\ 0 & 1 \end {pmatrix}$

$\ds $

$=$

$\ds \begin {pmatrix} 2 & 1 \\ 2 & 1 \end {pmatrix}$

$\ds \leadsto \ \ $

$\ds AB$

$\ne$

$\ds BA$

and it follows by definition that the general linear group is not abelian.

$\blacksquare$

1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): group
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): group