# Why are General Linear Groups Non Abelian?

1. Oct 2, 2011

### lola1990

1. The problem statement, all variables and given/known data
Show that if n>1 and F is an arbitrary field, the general linear group defined by n and F is non-abelian

2. Relevant equations
A general linear group is the group of invertible matrices with entries from F

A non abelian group is a group where the binary operation isn't commutative

3. The attempt at a solution
I thought that a good way to go about this problem would be to find two general invertible matrices that don't commute. However, I'm having trouble finding them. Is this the right way to go about it? If not, how can I prove this?

2. Oct 2, 2011

### Bacle

Actually, the probability that two matrices will commute is zero. Take any two
matrices A,B , neither of which is a scalar multiple of the identity, and multiply
them. If they commute, go buy a lottery ticket; you have beaten gigantic odds.

You may even be able to tell that the respective first entries AB1,1 and
BA1,1 are different.

3. Oct 2, 2011

### spamiam

When finding your non-commuting matrices, remember that there are only 2 elements guaranteed to be in your field, namely 0 and 1 (with $0 \neq 1$), since you could even be working over $\mathbb{F}_2$. But I think your approach is good, and as Bacle said, you can show 2 matrices are not equal just by finding a single entry that isn't the same.

4. Oct 2, 2011

### Bacle

Your right, Spamiam, I had not thought of that. I was just thinking you may do a product of 2x2 matrices , and see how difficult it is for them to commute, by checking a single entry, and trying to determine when/how it can be made to commute.

5. Oct 3, 2011

### Deveno

if char(F) is not 2:

[1 1][1 0]....[2 1]
[0 1][1 1] = [1 1],

[1 0][1 1]....[1 1]
[1 1][0 1] = [1 2]

if char(F) = 2:

[1 1][1 0]....[0 1]
[0 1][1 1] = [1 1],

[1 0][1 1]....[1 1]
[1 1][0 1] = [1 0], so in either case we see these two matrices do not commute.

for n > 2, call the first 2x2 matrix above A, the 2nd B (from the top product).

define the following nxn matrices in block form:

[A 0]..[B 0]
[0 I ], [0 I]...it should be clear these are invertible, and do not commute.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}