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Like the title said, I am trying to prove the following claim:

"$B = \{M\in GL_n(\mathbb{R}): M$ is upper triangular $\}$ " is a solvable group.

So my idea was to calculate the derived series of $B$ and show that it terminates at the trivial group $G_k = \{I_n\}$. I managed to compute the first derived group $[B,B]$, or the commutator subgroup of $B$, which is just $[B,B] = B\, \cap SL_n(\mathbb{R})$. However, I am stuck trying to compute the next derived commutator subgroup.

I would mostly appreciate a hint, not a full solution. Thanks!

dezdichado
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    Have you already tried to calculate the commutator of two upper trianglular matrices? Well, do...and observe you get an strictly upper triangular matrix. After that you're done, because when working with the commutator subgroup $;B';$ , you already have a group with all its matrices nilpotent. – DonAntonio Oct 20 '16 at 16:06
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    The commutator of two upper triangular matrices is of the form "I+(strictly upper triangular)". – Keith Kearnes Oct 20 '16 at 17:23
  • donAntonio are you saying that if the all the matrices in $B'$ are nilpotent, then the group itself is solvable? Is there anyway to see this without using the language of Lie algebra? – dezdichado Oct 21 '16 at 05:37

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