Normalizer of a Lie subalgebra

Remarks

$N_{\mathfrak g}(\mathfrak k)$ is a subalgebra of $\mathfrak g$, by the Jacobi identity. It contains $\mathfrak k$, and $\mathfrak k$ is an ideal of it. It is the largest subalgebra of $\mathfrak g$ with this property. $\mathfrak k$ is self-normalizing if $N_{\mathfrak g}(\mathfrak k) = \mathfrak k$.

Example: let $\mathfrak g = \mathfrak{sl}_2(k)$ with $\operatorname{char} k \ne 2$, basis $e, h, f$, and $[h, e] = 2e$, $[h, f] = -2f$, $[e, f] = h$. Let $\mathfrak k = k h$. For $x = ae + bh + cf$, one gets $[x, h] = -2a e + 2c f$, which lies in $k h$ only if $a = c = 0$. So $N_{\mathfrak g}(k h) = k h$, and $k h$ is self-normalizing.

Non-example: the element $e$ is not in $N_{\mathfrak g}(k h)$, because $[e, h] = -2e \notin k h$.

If $\mathfrak k$ is an ideal of $\mathfrak g$, then $N_{\mathfrak g}(\mathfrak k) = \mathfrak g$.