0

Problem: Let $L$ be a Lie algebra, denote $[L L]=L'$. Show that $L/L'$ abelian.

My attempt: $[x,y] = (x+L')(y+L')-(y+L')(x+L') = ((x+y)+L') - (y+x+L') = ((x+y)+L') - ((x+y)+L') = 0$

Is that enough???

Minh
  • 983
  • 1
    Should be $$(x+L')(y+L')-(y+L')(x+L') = xy-yx + L'=L'.$$ – Dzoooks Apr 21 '19 at 15:42
  • 1
    I presume $L$ is some abstract Lie algebra. Then it doesn't make sense to multiply elements of $L/L'$ like you did. – Wojowu Apr 21 '19 at 15:42
  • @Wojowu They're probably matrices in most of his examples, so multiplication is just composition. – Dzoooks Apr 21 '19 at 15:43
  • Dzoooks By definition, $[xy]$ must be equal to zero. – Minh Apr 21 '19 at 15:45
  • @Dzoooks I suppose, but that doesn't mean this proof is valid for an abstract Lie algebras. (before anyone mentions that: I know you can take the universal enveloping algebra, but I would venture a guess OP hasn't had that covered yet.) – Wojowu Apr 21 '19 at 15:46
  • 2
    @Minh $L'=0+L'$ is the zero element in $L/L'$. – Wojowu Apr 21 '19 at 15:46

0 Answers0