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

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In summary, for any field F and n≥2, the group GL_n(F) is not abelian. This can be shown by finding a counterexample using 2x2 matrices, such as A = (1 1 / 1 1) and B = (0 0 / 1 0), and showing that AB ≠ BA. This can be generalized for any n by constructing A' and B' with similar properties, and using induction to prove that the group is not abelian. It is important to ensure that the matrices used are invertible, as the general linear group consists of invertible matrices.
  • #1
lus1450
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Homework Statement


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


Homework Equations



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.
 
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  • #2
Don't the matrices have to be invertible? I don't think A is valid.
 
  • #3
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.
 
  • #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.
 

1. What is a counter-example in the context of group theory?

A counter-example is a specific case or example that contradicts a general statement or theorem. In the context of group theory, a counter-example is used to show that a certain group does not possess a certain property, such as being abelian.

2. Why is it important to explain a counter-example in group theory?

Explaining a counter-example in group theory is important because it helps to clarify and solidify our understanding of group properties and concepts. It also allows us to identify any potential flaws in a proof or argument, and can lead to further exploration and discovery.

3. How is a group proven to not be abelian inductively?

A group is proven to not be abelian inductively by providing a series of counter-examples that demonstrate the failure of the group to satisfy the commutative property. These counter-examples must be carefully chosen and logically connected in order to form a strong and convincing argument.

4. What is the process for constructing a counter-example for a group to show it is not abelian?

The process for constructing a counter-example to show that a group is not abelian involves analyzing the group's operation and identifying elements that do not commute with each other. These elements can then be used to construct a specific example that violates the commutative property of an abelian group.

5. Can a counter-example be used to disprove a general statement in group theory?

Yes, a counter-example can be used to disprove a general statement in group theory. By providing a specific case that contradicts the general statement, a counter-example can show that the statement is not universally true. However, it is important to note that a single counter-example does not disprove a statement completely, as there may be other cases or conditions where the statement holds true.

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