0

Why can't we use the same arguments as for solvability to say that If a Lie algebra $L$ has a nilpotent ideal $I$, and $L/I$ is nilpotent, then $L$ is nilpotent. Can someone point out the fallacy in the argument

Deeponjit Bose
  • 347
matzo
  • 9
  • 4
    Look over the solvability argument carefully to see why it doesn't translate. Right now you are asking why we cannot argue that cars have legs, given that horses have legs, and we can say that we ride horses and we ride cars. The argument also doesn't work for nilpotent groups: $S_3$ is an extension of an abelian group by an abelian group, but it is not nilpotent. – Arturo Magidin Mar 25 '23 at 03:52
  • 2
    Agreeing with @ArturoMagidin, and you should take the easiest example of a non-nilpotent Lie algebra which has a nilpotent ideal with nilpotent quotient: The two-dimensional nonabelian one. Go through whatever proof you imagine line by line and check where it breaks in this example. – Torsten Schoeneberg Mar 25 '23 at 04:57
  • The proof fails, because we really need the assumption that ${\rm ad}(x)|_I$ is nilpotent for all $x \in L$ - see this post. – Dietrich Burde Mar 25 '23 at 17:23

0 Answers0