[proofplan] We first prove the linear Engel lemma: a finite-dimensional Lie algebra of nilpotent endomorphisms of a nonzero [vector space](/page/Vector%20Space) has a common nonzero annihilated vector. The proof of that lemma uses induction on the dimension of the Lie algebra, together with a maximal-subalgebra argument showing that a maximal proper subalgebra has codimension one. Applying the lemma to the adjoint representation gives a nonzero central element of $\mathfrak g$. We then pass to the quotient by the center and use induction on $\dim_F \mathfrak g$ to prove that the lower central series of $\mathfrak g$ terminates. [/proofplan] [step:Prove the linear Engel lemma for nilpotent endomorphism Lie algebras] [claim:Linear Engel lemma] Let $V$ be a nonzero finite-dimensional vector space over $F$, and let $L \subseteq \operatorname{End}_F(V)$ be a finite-dimensional Lie subalgebra under the commutator bracket \begin{align*} [A,B] := A \circ B - B \circ A. \end{align*} Assume that every $X \in L$ is nilpotent as an endomorphism of $V$. Then there exists $0 \neq v \in V$ such that $X(v)=0$ for every $X \in L$. [/claim] [proof] We prove the claim by induction on $d := \dim_F L$. If $d=0$, then every nonzero $v \in V$ is annihilated by all elements of $L$. If $d=1$, write $L=F X_0$ for some $X_0 \in L$. Since $X_0$ is nilpotent and $V \neq 0$, the kernel $\ker X_0$ is nonzero; any $0 \neq v \in \ker X_0$ is annihilated by every scalar multiple of $X_0$. Assume $d \geq 2$ and assume the result for all Lie algebras of dimension strictly smaller than $d$ satisfying the same nilpotence hypothesis. First we record a nilpotence fact for commutators. For $T \in L$, define \begin{align*} \operatorname{ad}^{\operatorname{End}}_T: \operatorname{End}_F(V) &\to \operatorname{End}_F(V) \\ S &\mapsto [T,S]. \end{align*} If $T^a=0$ for some $a \in \mathbb N$, then for every $k \geq 1$ and every $S \in \operatorname{End}_F(V)$, \begin{align*} (\operatorname{ad}^{\operatorname{End}}_T)^k(S) = \sum_{i=0}^{k} (-1)^{k-i}\binom{k}{i} T^i \circ S \circ T^{k-i}. \end{align*} This identity follows by induction on $k$ from the definition of the commutator. Taking $k=2a-1$, every summand has either $i \geq a$ or $k-i \geq a$, hence every summand is zero. Thus $\operatorname{ad}^{\operatorname{End}}_T$ is nilpotent. Since $L$ is closed under commutators, the restriction \begin{align*} \operatorname{ad}^L_T: L &\to L \\ S &\mapsto [T,S] \end{align*} is also nilpotent. Choose a maximal proper Lie subalgebra $M \subsetneq L$, maximal with respect to inclusion. Such an $M$ exists because $L$ is finite-dimensional and contains at least the proper subalgebra $\{0\}$. We show that $M$ is an ideal of $L$ and has codimension one. Let \begin{align*} N_L(M) := \{X \in L : [X,M] \subseteq M\} \end{align*} be the normalizer of $M$ in $L$. Since $M$ is a Lie subalgebra, $M \subseteq N_L(M)$. Suppose, for contradiction, that $N_L(M)=M$. Consider the quotient vector space $L/M$. For each $Y \in M$, define \begin{align*} \rho(Y): L/M &\to L/M \\ X+M &\mapsto [Y,X]+M. \end{align*} This map is well-defined because $[Y,M]\subseteq M$. The assignment \begin{align*} \rho: M &\to \operatorname{End}_F(L/M) \\ Y &\mapsto \rho(Y) \end{align*} is a Lie algebra homomorphism. Each $\rho(Y)$ is nilpotent because it is induced by the nilpotent endomorphism $\operatorname{ad}^L_Y$ of $L$. By the induction hypothesis applied to the Lie algebra $\rho(M) \subseteq \operatorname{End}_F(L/M)$, there exists a nonzero coset $X_0+M \in L/M$ such that \begin{align*} \rho(Y)(X_0+M)=0 \end{align*} for every $Y \in M$. This means $[Y,X_0]\in M$ for every $Y \in M$, so $X_0 \in N_L(M)$. Since $X_0+M \neq M$, we have $X_0 \notin M$, contradicting $N_L(M)=M$. Therefore $N_L(M)$ strictly contains $M$. By maximality of $M$, this gives $N_L(M)=L$, so $M$ is an ideal of $L$. Because $M$ is an ideal, the quotient $L/M$ is a Lie algebra. If $\dim_F(L/M)\geq 2$, choose a one-dimensional subspace $F(X+M)\subset L/M$ with $X\notin M$. Every one-dimensional vector subspace of a Lie algebra is a Lie subalgebra, since $[X+M,X+M]=0$. Its inverse image under the quotient map $L\to L/M$ is a Lie subalgebra strictly between $M$ and $L$, contradicting the maximality of $M$. Hence \begin{align*} \dim_F(L/M)=1. \end{align*} Choose $X_1 \in L \setminus M$. Then \begin{align*} L = M \oplus F X_1 \end{align*} as vector spaces. Now apply the induction hypothesis to the Lie algebra $M \subseteq \operatorname{End}_F(V)$. Every element of $M$ is nilpotent on $V$, and $\dim_F M<d$, so the subspace \begin{align*} W := \{v \in V : Y(v)=0 \text{ for every } Y \in M\} \end{align*} is nonzero. We show that $W$ is invariant under $L$. Let $w \in W$, let $Y \in M$, and let $X \in L$. Since $M$ is an ideal, $[Y,X]\in M$. Therefore \begin{align*} Y(X(w)) = [Y,X](w)+X(Y(w)) = 0+X(0) = 0. \end{align*} Thus $X(w)\in W$, so $W$ is invariant under every $X\in L$. In particular, the restriction \begin{align*} X_1|_W: W &\to W \end{align*} is a nilpotent endomorphism of the nonzero finite-dimensional vector space $W$. Hence $\ker(X_1|_W)$ contains a nonzero vector. Choose $0\neq v\in W$ with $X_1(v)=0$. Since $v\in W$, every element of $M$ annihilates $v$, and since $L=M\oplus F X_1$, every element of $L$ annihilates $v$. This proves the claim. [/proof] [/step] [step:Find a nonzero central element using the adjoint representation] If $\mathfrak g=0$, then $\mathfrak g$ is nilpotent because $\gamma_1(\mathfrak g)=0$. Assume from now on that $\mathfrak g\neq 0$. Define the adjoint image \begin{align*} \operatorname{ad}(\mathfrak g) := \{\operatorname{ad}_x : x \in \mathfrak g\}\subseteq \operatorname{End}_F(\mathfrak g). \end{align*} This is a Lie subalgebra of $\operatorname{End}_F(\mathfrak g)$ because \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]=\operatorname{ad}_{[x,y]} \end{align*} for all $x,y\in \mathfrak g$, by the Jacobi identity. By hypothesis, every element $\operatorname{ad}_x$ of this Lie algebra is nilpotent. Apply the linear Engel lemma to the nonzero finite-dimensional vector space $V:=\mathfrak g$ and the Lie algebra $L:=\operatorname{ad}(\mathfrak g)$. There exists $0\neq z\in \mathfrak g$ such that \begin{align*} \operatorname{ad}_x(z)=0 \end{align*} for every $x\in \mathfrak g$. Equivalently, $[x,z]=0$ for every $x\in \mathfrak g$. By skew-symmetry of the Lie bracket, $[z,x]=0$ for every $x\in \mathfrak g$ as well. Hence $z$ lies in the center \begin{align*} Z(\mathfrak g):=\{u\in \mathfrak g : [u,x]=0 \text{ for every } x\in \mathfrak g\}. \end{align*} Thus $Z(\mathfrak g)\neq 0$. [/step] [step:Pass the nilpotent adjoint hypothesis to the quotient by the center] We prove the theorem by induction on $n:=\dim_F \mathfrak g$. The case $n=0$ was already handled. Assume $n\geq 1$ and assume the theorem holds for all Lie algebras over $F$ of dimension strictly smaller than $n$. Since $Z(\mathfrak g)\neq 0$, the quotient Lie algebra \begin{align*} \overline{\mathfrak g}:=\mathfrak g/Z(\mathfrak g) \end{align*} has dimension strictly smaller than $n$. For each $x\in \mathfrak g$, write \begin{align*} \overline{x}:=x+Z(\mathfrak g)\in \overline{\mathfrak g}. \end{align*} The adjoint endomorphism of the quotient associated to $\overline{x}$ is \begin{align*} \operatorname{ad}^{\overline{\mathfrak g}}_{\overline{x}}: \overline{\mathfrak g} &\to \overline{\mathfrak g} \\ \overline{y} &\mapsto \overline{[x,y]}. \end{align*} This map is induced by $\operatorname{ad}_x$ because $Z(\mathfrak g)$ is an ideal. If $(\operatorname{ad}_x)^N=0$ for some $N\in\mathbb N$, then \begin{align*} (\operatorname{ad}^{\overline{\mathfrak g}}_{\overline{x}})^N(\overline{y}) = \overline{(\operatorname{ad}_x)^N(y)} = \overline{0} \end{align*} for every $\overline{y}\in\overline{\mathfrak g}$. Hence every adjoint endomorphism of $\overline{\mathfrak g}$ is nilpotent. By the induction hypothesis, $\overline{\mathfrak g}$ is nilpotent. Therefore there exists $r\in\mathbb N$ such that \begin{align*} \gamma_r(\overline{\mathfrak g})=0. \end{align*} [/step] [step:Lift nilpotence of the quotient to nilpotence of $\mathfrak g$] Let \begin{align*} \pi:\mathfrak g &\to \overline{\mathfrak g} \\ x &\mapsto x+Z(\mathfrak g) \end{align*} be the quotient homomorphism. We claim that \begin{align*} \pi(\gamma_k(\mathfrak g))=\gamma_k(\overline{\mathfrak g}) \end{align*} for every $k\in\mathbb N$. For $k=1$, this is the surjectivity of $\pi$. If the equality holds for some $k$, then using that $\pi$ is a Lie algebra homomorphism, \begin{align*} \pi(\gamma_{k+1}(\mathfrak g)) &= \pi([\mathfrak g,\gamma_k(\mathfrak g)]) \\ &= [\pi(\mathfrak g),\pi(\gamma_k(\mathfrak g))] \\ &= [\overline{\mathfrak g},\gamma_k(\overline{\mathfrak g})] \\ &= \gamma_{k+1}(\overline{\mathfrak g}). \end{align*} Thus the claim follows by induction on $k$. Since $\gamma_r(\overline{\mathfrak g})=0$, the claim gives \begin{align*} \pi(\gamma_r(\mathfrak g))=0. \end{align*} Hence \begin{align*} \gamma_r(\mathfrak g)\subseteq \ker \pi = Z(\mathfrak g). \end{align*} Therefore \begin{align*} \gamma_{r+1}(\mathfrak g) = [\mathfrak g,\gamma_r(\mathfrak g)] \subseteq [\mathfrak g,Z(\mathfrak g)] = 0. \end{align*} Thus the lower central series of $\mathfrak g$ terminates, so $\mathfrak g$ is nilpotent. This completes the proof of Engel's Theorem. [/step]