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

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Homework Help Overview

The original poster attempts to demonstrate that the group GL_{n}(F) is not abelian for any field F and for n ≥ 2. They provide a counterexample using specific matrices for n=2 and suggest a generalization for larger n.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster presents a counterexample involving matrices and considers using induction for a more rigorous proof. They question whether their approach is sufficient or if further expression in summation notation is necessary.

Discussion Status

Some participants question the validity of the matrices used in the original poster's example, noting that they must be invertible. The discussion acknowledges the need for valid examples and explores the outlined approach while emphasizing the importance of adhering to definitions.

Contextual Notes

There is a mention of a previous working example that was not valid for a certain field, highlighting the importance of ensuring that examples meet the criteria for the general linear group.

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Homework Statement


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


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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 &amp; 1\\<br /> 1 &amp; 1 \end{array} \right) and B = \left( \begin{array}{ccc}<br /> 0 &amp; 0\\<br /> 1 &amp; 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&#039; = A \oplus I_{n-2} and B&#039; = B \oplus I_{n-2}, then A&#039;B&#039; \ne B&#039;A&#039;, since the "upper left" 2 \times 2 block matrix of A&#039;B&#039; is just AB, then there are 1s on the diagonal and 0s everywhere else. It's the same for B&#039;A&#039;, 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.
 
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Don't the matrices have to be invertible? I don't think A is valid.
 
ArcanaNoir said:
Don't the matrices have to be invertible? I don't think A is valid.
Yes, they do. B also is not valid.

The outlined approach is valid given a pair of valid 2x2 matrices as the starting point.
 
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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