Q:

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

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? 

A:

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.

