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If I am correct, if $K \subseteq L$ is a Lie subalgebra then we define the normaliser of $K$ in $L$ to be $$N_L (K) = \{ x \in L : [x, y] \in K \ \ \forall y \in K \}.$$

Given this, is $ N_L (K) $ a subalgebra of $L$. Clearly this set contains $K $ since $K$ is itself a subalgebra. But is the set itself?

Anonmath101
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  • Do I understand correctly that you are asking if $N_L(K)$ is a subalgebra in this situation? Well, do you know what conditions you have to check? Which of them cause what problems? – Torsten Schoeneberg Jul 11 '22 at 19:00
  • Checking that if $x , y \in N_L(K) $ that $[x,y] \in N_L(K)$. Can’t seem to see if this is true or not. – Anonmath101 Jul 11 '22 at 20:47

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According to comments, you are trying to show that if both $x$ and $y$ are in $N_L(K)$, then $[x,y] \in N_L(K)$.

That is, we have to show that $[[x,y],z] \in K$ for all $z \in K$, using the assumption that $[x,z] \in K$ and $[y,z] \in K$ for all $z \in K$.

Hint: Jacobi identity.

Torsten Schoeneberg
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