1
$\begingroup$

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?

| cite | improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ 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?? $\endgroup$ – VladeKR Mar 14 '17 at 17:49
  • 1
    $\begingroup$ Yes, there only needs to be one example of non commutative matrices for each $n \geq 2$ to make that group non abelian. $\endgroup$ – wgrenard Mar 14 '17 at 17:53
  • $\begingroup$ I now see my misunderstanding, thanks! $\endgroup$ – VladeKR Mar 14 '17 at 18:00
5
$\begingroup$

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.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ so I guess it was just sufficient to give one example of non-commutation to ensure the non-abelian nature of the group. Thanks $\endgroup$ – VladeKR Mar 14 '17 at 18:03
  • $\begingroup$ @VladeKR Yes, two particular matrices may commute. For the general case I added a hint. $\endgroup$ – egreg Mar 14 '17 at 18:05
1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.