# Explaining my counter-example that a group is not abelian inductively

lus1450

## Homework Statement

Show that for any field $F$, for $n\ge2$, the group $GL_{n}(F)$ is not abelian.

## The Attempt at a Solution

I have found a counter example for all such $n$. First, for $n=2$, consider the matrices: $A = \left( \begin{array}{ccc} 1 & 1\\ 1 & 1 \end{array} \right)$ and $B = \left( \begin{array}{ccc} 0 & 0\\ 1 & 0 \end{array} \right)$, 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 $1$s on the diagonal and $0$s 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.

ArcanaNoir
Don't the matrices have to be invertible? I don't think A is valid.

Staff Emeritus