lus1450
- 40
- 1
Homework Statement
Show that for any field F, for n\ge2, the group GL_{n}(F) is not abelian.
Homework Equations
The Attempt at a Solution
I have found a counter example for all such n. First, for n=2, consider the matrices: A = \left( \begin{array}{ccc}<br /> 1 & 1\\<br /> 1 & 1 \end{array} \right) and B = \left( \begin{array}{ccc}<br /> 0 & 0\\<br /> 1 & 0 \end{array} \right), since all fields contain 0,1. By computing, we find that AB \ne BA, and so the group is not abelian. To generalize for any n, if we have A' = A \oplus I_{n-2} and B' = B \oplus I_{n-2}, then A'B' \ne B'A', since the "upper left" 2 \times 2 block matrix of A'B' is just AB, then there are 1s on the diagonal and 0s everywhere else. It's the same for B'A', except the upper block is BA instead, and so of course the two matrices don't commute. However, I'm having a trouble wording that rigorously. I'm positive my explanation was not rigorous enough, so I was thinking of doing induction, but I'm not sure about how to proceed with that. Would expressing the product in summation notation be the way to go, or is it intuitive enough that my counterexample is enough to show the groups aren't commutative? Thanks for the advice.