= Solution
Write $D=\operatorname{ad}y$. The <generalized eigenspace> decomposition of the <linear map> $D$ is
$$
L=\bigoplus_\lambda L_{\lambda,y},
\qquad
L_{\lambda,y}=\ker(D-\lambda I)^N
$$
for any sufficiently large $N$. Because $D$ is a <derivation of a Lie algebra>[derivation], the <generalized-eigenspace bracket lemma> gives
$$
[L_{\lambda,y},L_{\mu,y}]\subseteq L_{\lambda+\mu,y}.
$$
Consequently $L_{0,y}$ is a <Lie subalgebra>.
The set $I(K)=\{x\in L:[x,K]\subseteq K\}$ is the <normalizer of a Lie subalgebra> $K$. Certainly $L_{0,y}\subseteq I(L_{0,y})$. Conversely, if $x\in I(L_{0,y})$, then $y\in L_{0,y}$ gives
$$
Dx=[y,x]\in L_{0,y}.
$$
On the direct sum of the nonzero generalized eigenspaces, $D$ is <invertible linear map>[invertible]. Hence the nonzero-eigenvalue component of $x$ vanishes, and
$$
\boxed{I(L_{0,y})=L_{0,y}.}
$$
Now let $K$ be a <Lie subalgebra> containing $L_{0,y}$. Since $y\in K$, the subspace $K$ is $D$-invariant. The generalized zero eigenspace of the induced map on $L/K$ is the image of $L_{0,y}$, hence is zero. If $x\in I(K)$, then $Dx=[y,x]\in K$, so $x+K$ lies in that zero eigenspace. Thus $x\in K$ and
$$
\boxed{I(K)=K.}
$$
A <Nilpotent Lie algebra> is one whose <Lower central series of a Lie algebra>[lower central series]
$$
\gamma_1(L)=L,
\qquad
\gamma_{r+1}(L)=[L,\gamma_r(L)]
$$
eventually reaches zero. Suppose $L$ is nilpotent and $K\subsetneq L$. Choose the least $r\geq2$ for which $\gamma_r(L)\subseteq K$. Then $\gamma_{r-1}(L)\nsubseteq K$, and any
$$
x\in\gamma_{r-1}(L)\setminus K
$$
satisfies $[x,K]\subseteq[L,\gamma_{r-1}(L)]=\gamma_r(L)\subseteq K$. Therefore $x\in I(K)\setminus K$, proving the <normalizer condition for a nilpotent Lie algebra>
$$
\boxed{K\subsetneq I(K).}
$$
It remains to prove the converse needed here. The <Engel lemma> states that if a finite-dimensional <Lie algebra representation>[Lie algebra of linear maps] consists of <nilpotent linear map>[nilpotent maps], then the maps have a common nonzero vector in their kernels. To prove it, induct on the dimension of the algebra. For a maximal proper subalgebra $H$, induction applied to the action of $H$ on $L/H$ produces $x\notin H$ with $[H,x]\subseteq H$. Thus $H$ is an ideal of codimension one. Induction also gives a nonzero common kernel
$$
W=\{v:Hv=0\}.
$$
The ideal property makes $W$ invariant under $L$; a nilpotent representative of a basis of $L/H$ has a nonzero kernel on $W$, yielding a vector killed by all of $L$.
Apply the lemma to the <Adjoint representation>. It produces a nonzero element of the <Center of a Lie algebra>. Induction on $\dim L$, followed by passage to the quotient by this center, proves <Engel theorem>: if every $\operatorname{ad}y$ is nilpotent, then $L$ is nilpotent. The hypothesis $L_{0,y}=L$ says exactly that every $\operatorname{ad}y$ is nilpotent, so
$$
\boxed{L\text{ is nilpotent}.}
$$
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