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?

• Oh my god, does the word "any" mean like I could choose just any n and F, so need I just show one example of non-commutation?? – VladeKR Mar 14 '17 at 17:49
• Yes, there only needs to be one example of non commutative matrices for each $n \geq 2$ to make that group non abelian. – wgrenard Mar 14 '17 at 17:53
• I now see my misunderstanding, thanks! – VladeKR Mar 14 '17 at 18:00

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.

• so I guess it was just sufficient to give one example of non-commutation to ensure the non-abelian nature of the group. Thanks – VladeKR Mar 14 '17 at 18:03
• @VladeKR Yes, two particular matrices may commute. For the general case I added a hint. – egreg Mar 14 '17 at 18:05

$A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}+I$, $B=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}+I$, $AB \ne BA$.

• Usually $GL$ denotes the invertible matrices. – egreg Mar 14 '17 at 18:03
• @egreg: Thanks for catching that, I had meant to add the identity. – copper.hat Mar 14 '17 at 18:06