Def. Lie algebra
A lie algebra is a vector space L over a field F together with a multiplication
L × L → L
(x, y) → [x, y] [x, y] = xy - yx
satisfying the following axioms
1) [x, x]=0
2) [[xy], z] + [[yz], x] + [[zx], y] = 0
Def. Sub algebra, Ideal let L is lie-algebra over field F and let H, K are subspace of L. We then set [H, K] := span({[h, k] ∈ L ｜ h ∈ H, k ∈ K } that we have to ensure that this is a subspace of L. Lie sub algebra of L is a subspace H with [H, H] ⊂ H. Lie ideal of L is subspace K with [K, L] ⊂ K.
K⊂L, K is sub algebra The normalizer of a sub algebra K of L is defined by Nl(K) = {x∈L | [x, k]∈K for all k∈K}. The centralizer of subset K of L is Cl(K) = {x∈L | [x, k] = 0 for all k∈K}.
Q : Prove that if K is sub algebra of L, then Cl(K) is ideal of Nl(K).
You need to show that if $a\in C(K)$ and $b\in N(K)$ then $[a,b]\in C(K)$, that is $[[a,b],c]=0$ for all $c\in K$. By the Jacobi identity, $$[[a,b],c]=[a,[b,c]]+[b,[c,a]].$$ Now, $[b,c]\in K$ and so $[a,[b,c]]=0$ and $[c,a]=0$.