= Solution
The space $\operatorname{End}(V)$ becomes a <Lie algebra> under the commutator
$$
[X,Y]=XY-YX.
$$
A Lie subalgebra $L$ is abelian when $[L,L]=0$; nilpotent when its <Lower central series of a Lie algebra> $\gamma_{r+1}=[L,\gamma_r]$ reaches zero; and soluble when its <derived series of a Lie algebra> $L^{(r+1)}=[L^{(r)},L^{(r)}]$ reaches zero.
If $H\subsetneq L$ and $L$ is nilpotent, choose the least $r$ with $\gamma_r(L)\subseteq H$. Then $\gamma_{r-1}(L)\nsubseteq H$, while
$$
[\gamma_{r-1}(L),H]\subseteq\gamma_r(L)\subseteq H.
$$
Thus the <normalizer of a Lie subalgebra> strictly contains $H$. If $H$ is maximal proper, its normalizer must be all of $L$, so $H$ is an ideal. Solubility is insufficient: in the two-dimensional affine Lie algebra $L=\langle x,y\rangle$ with $[x,y]=y$, the maximal subalgebra $\mathbb Cx$ is not an ideal.
A <derivation of a Lie algebra> is a linear map $D$ satisfying
$$
D[x,y]=[Dx,y]+[x,Dy].
$$
It is inner when $D=\operatorname{ad}z$ for some $z\in L$.
Every nonzero finite-dimensional nilpotent Lie algebra has an <outer derivation of a nilpotent Lie algebra>. Choose a codimension-one maximal subalgebra $K$; it is an ideal by the result above, and write $L=K\oplus\mathbb Cz$. The centralizer $C_L(K)$ is nonzero because it contains $Z(L)$. Let $n$ be largest such that
$$
C_L(K)\subseteq\gamma_n(L),
$$
and choose $z_0\in C_L(K)\setminus\gamma_{n+1}(L)$. Define
$$
D(K)=0,\qquad D(z)=z_0.
$$
Because $K$ is an ideal and $z_0$ centralizes $K$, the derivation identity holds on $K\times K$ and on $z\times K$, hence everywhere. If $D=\operatorname{ad}x$, then $D(K)=0$ would put $x$ in $C_L(K)\subseteq\gamma_n(L)$, so
$$
D(z)=[x,z]\in\gamma_{n+1}(L),
$$
contrary to the choice of $z_0$. Thus $D$ is outer.
The analogous assertion fails for soluble algebras. In the affine example, a derivation has
$$
D(x)=by,\qquad D(y)=dy,
$$
and equals $\operatorname{ad}(dx-by)$. Thus every derivation is inner although $L$ is nonzero and soluble.
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