General Linear Group is not Abelian/Proof 2
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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.
Proof
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.
$\blacksquare$
Sources
- 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