General Linear Group is not Abelian

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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 1

From Special Linear Group is Subgroup of General Linear Group we have that the special linear group $\SL {n, K}$ is a subgroup of $\GL {n, K}$.

From Special Linear Group is not Abelian, $\SL {n, K}$ is not abelian.

From Subgroup of Abelian Group is Abelian it follows by the Rule of Transposition that $\GL {n, K}$ is not abelian.

$\blacksquare$


Proof 2

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

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