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

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that for any field [itex] F [/itex], for [itex]n\ge2[/itex], the group [itex] GL_{n}(F)[/itex] is not abelian.


    2. Relevant equations

    3. The attempt at a solution
    I have found a counter example for all such [itex]n[/itex]. First, for [itex]n=2[/itex], consider the matrices: [itex]A = \left( \begin{array}{ccc}
    1 & 1\\
    1 & 1 \end{array} \right)[/itex] and [itex]B = \left( \begin{array}{ccc}
    0 & 0\\
    1 & 0 \end{array} \right)[/itex], since all fields contain [itex]0,1[/itex]. By computing, we find that [itex]AB \ne BA[/itex], and so the group is not abelian. To generalize for any [itex]n[/itex], if we have [itex]A' = A \oplus I_{n-2}[/itex] and [itex]B' = B \oplus I_{n-2}[/itex], then [itex]A'B' \ne B'A'[/itex], since the "upper left" [itex]2 \times 2[/itex] block matrix of [itex]A'B'[/itex] is just [itex]AB[/itex], then there are [itex]1[/itex]s on the diagonal and [itex]0[/itex]s everywhere else. It's the same for [itex]B'A'[/itex], except the upper block is [itex]BA[/itex] 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.
     
  2. jcsd
  3. Oct 16, 2013 #2
    Don't the matrices have to be invertible? I don't think A is valid.
     
  4. Oct 16, 2013 #3

    D H

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    Staff Emeritus
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    Yes, they do. B also is not valid.

    The outlined approach is valid given a pair of valid 2x2 matrices as the starting point.
     
  5. Oct 16, 2013 #4
    Indeed, I apologize. Before, I had a working example where the matrices were actually invertible, but it turned out to not work for a certain field (as pointed out somewhere else), so I hastily pieced together different matrices before posting. In my tiredness, I guess I just ignored the very definition of the general linear group when formulating them. And thank you for the verification of the method outlined, and thanks also for reassuring I should double check myself before I appear ignorant of definitions.
     
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