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Commutators and solvable groups

  1. Aug 6, 2005 #1
    I am not really clear on what is meant by commutators. I know that the commutator of G is ABA^-1B^-1, but I am not sure how to check if a group is solvable by having the commutator eventually equal the trivial group.

    For example, I know that the Heisenberg group of 3x3 upper triangular matrices is two-step solvable, but am not sure how to SHOW that. I know that it means that the first commutator doesn't equal the identity matrix and that the second one does... but how do I show this???

    Also, how do I show that the group GL(2,R) (2x2 invertible matrices) IS NOT solvable???
     
  2. jcsd
  3. Aug 7, 2005 #2

    matt grime

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    you need to describe the commutator, so do it. take two arbitrary matrices in the upper triangular matrices and work out the commutator. see what happens. now take the commutator of this with another generic element of the comutator, see what you get. as it happens it is clearer to see that the lie algebra is solvable.


    example: upper triangular 2x2 matrices with 1 on the diagonals, what is the commutator or two elements?

    [tex]\left(\begin{array}{cc} 1&a\\0&1 \end{array}\right)\left(\begin{array}{cc} 1&b\\0&1 \end{array}\right)
    \left(\begin{array}{cc} 1&-a\\0&1 \end{array}\right)
    \left(\begin{array}{cc} 1&-b\\0&1 \end{array}\right)[/tex]

    well?

    as for the second example, GL contains SL which is simple.
     
  4. Aug 7, 2005 #3
    How can I show that the group G=<a,b,c> with [a,b]=b, [a,c]=1, [b,c]=1 is solvable but not nilpotent?

    A group G is said to be nilpotent if G^i=identity for some i.
    A group G is said to be solvable if it has subnormal series G=GncG3cG2cG1=identity... where all quotient groups are abelian.
     
    Last edited: Aug 7, 2005
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