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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???

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    $\begingroup$ Should be $$(x+L')(y+L')-(y+L')(x+L') = xy-yx + L'=L'.$$ $\endgroup$ – Dzoooks Apr 21 at 15:42
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    $\begingroup$ I presume $L$ is some abstract Lie algebra. Then it doesn't make sense to multiply elements of $L/L'$ like you did. $\endgroup$ – Wojowu Apr 21 at 15:42
  • $\begingroup$ @Wojowu They're probably matrices in most of his examples, so multiplication is just composition. $\endgroup$ – Dzoooks Apr 21 at 15:43
  • $\begingroup$ Dzoooks By definition, $[xy]$ must be equal to zero. $\endgroup$ – Minh Apr 21 at 15:45
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    $\begingroup$ @Minh $L'=0+L'$ is the zero element in $L/L'$. $\endgroup$ – Wojowu Apr 21 at 15:46

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