Show that $GL_{n}(F)$ is non-abelian for any $n \geq 2$ and any $F$.
(From Dummit's Abstract Algebra)
Now it says $GL_{n}(F)$ is an n by n matrix with entries from F and must be invertible (the determinant is non zero), with matrix multiplication as its binary operation. Non-abelian means that the group elements do not commute under the operation, that is $A \star B \neq B \star A$, which is generally the case for a matrix multiplication. But the question says ANY matrix larger than 2 by 2, with ANY entries as long as the matrix is invertible.
But aren't non-zero diagonal matrices part of the general linear group? Because surely their elements are in F, and surely their determinant is non zero, and surely they commute! What am I misunderstanding?
Best Answer
You just need to find two matrices that don't commute.
\begin{gather} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}= \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \\[6px] \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} \end{gather}
For $n\ge2$ just take these as the upper left block and complete with ones on the diagonal and zero elsewhere. The coefficients at $(1,1)$ are different.