[proofplan] One direction follows directly from the lower central series: repeated adjoint action by a fixed element pushes vectors deeper into the lower central series, which eventually vanishes for a nilpotent Lie algebra. For the converse, we prove [Engel's theorem](/theorems/3798) in the special form needed here: if every adjoint map is nilpotent, then every nonzero quotient of $\mathfrak g$ has nonzero center. Applying this to successive quotients constructs the upper central series until it reaches all of $\mathfrak g$, which is equivalent to nilpotence. [/proofplan]

[step:Show nilpotence forces every adjoint map to be nilpotent]Assume that $\mathfrak g$ is nilpotent. Define the lower central series $(\gamma_m(\mathfrak g))_{m \geq 1}$ by \begin{align*} \gamma_1(\mathfrak g) &:= \mathfrak g, \\ \gamma_{m+1}(\mathfrak g) &:= [\mathfrak g,\gamma_m(\mathfrak g)] \qquad \text{for } m \geq 1. \end{align*} Since $\mathfrak g$ is nilpotent, there exists an integer $r \geq 1$ such that $\gamma_r(\mathfrak g)=0$. Fix $x \in \mathfrak g$. We prove by induction on $m \geq 0$ that \begin{align*} (\operatorname{ad}x)^m(\mathfrak g) \subseteq \gamma_{m+1}(\mathfrak g). \end{align*} For $m=0$, this says $\mathfrak g \subseteq \gamma_1(\mathfrak g)$, which holds by definition. If the inclusion holds for $m$, then \begin{align*} (\operatorname{ad}x)^{m+1}(\mathfrak g) &= [x,(\operatorname{ad}x)^m(\mathfrak g)] \\ &\subseteq [\mathfrak g,\gamma_{m+1}(\mathfrak g)] \\ &= \gamma_{m+2}(\mathfrak g). \end{align*} Thus the induction is complete. Taking $m=r-1$ gives \begin{align*} (\operatorname{ad}x)^{r-1}(\mathfrak g) \subseteq \gamma_r(\mathfrak g)=0. \end{align*} Hence $\operatorname{ad}x$ is nilpotent.[/step]

[guided]Assume first that $\mathfrak g$ is nilpotent. The lower central series records all iterated commutators with elements of $\mathfrak g$. Define \begin{align*} \gamma_1(\mathfrak g) &:= \mathfrak g, \\ \gamma_{m+1}(\mathfrak g) &:= [\mathfrak g,\gamma_m(\mathfrak g)] \qquad \text{for } m \geq 1. \end{align*} Nilpotence means precisely that this descending sequence eventually reaches zero: there is an integer $r \geq 1$ with $\gamma_r(\mathfrak g)=0$. Fix $x \in \mathfrak g$. The map $\operatorname{ad}x:\mathfrak g \to \mathfrak g$ is given by $y \mapsto [x,y]$. Applying it once to an element of $\mathfrak g$ produces an element of $[\mathfrak g,\mathfrak g]=\gamma_2(\mathfrak g)$. Applying it again produces a bracket of an element of $\mathfrak g$ with something already in $\gamma_2(\mathfrak g)$, hence lands in $\gamma_3(\mathfrak g)$. Formally, we prove by induction that \begin{align*} (\operatorname{ad}x)^m(\mathfrak g) \subseteq \gamma_{m+1}(\mathfrak g) \end{align*} for every integer $m \geq 0$. For $m=0$, this is the identity inclusion $\mathfrak g \subseteq \gamma_1(\mathfrak g)$. Suppose it holds for $m$. Then \begin{align*} (\operatorname{ad}x)^{m+1}(\mathfrak g) &= [x,(\operatorname{ad}x)^m(\mathfrak g)] \\ &\subseteq [\mathfrak g,\gamma_{m+1}(\mathfrak g)] \\ &= \gamma_{m+2}(\mathfrak g). \end{align*} The first equality is the definition of the adjoint action, the inclusion uses $x \in \mathfrak g$ together with the induction hypothesis, and the final equality is the recursive definition of the lower central series. Now choose $m=r-1$. Since $\gamma_r(\mathfrak g)=0$, the inclusion gives \begin{align*} (\operatorname{ad}x)^{r-1}(\mathfrak g) \subseteq 0. \end{align*} Thus $(\operatorname{ad}x)^{r-1}=0$ as an endomorphism of $\mathfrak g$, so $\operatorname{ad}x$ is nilpotent.[/guided]

[step:Prove the Engel fixed vector lemma for nilpotent linear transformations] [claim:Engel fixed vector lemma] Let $V$ be a nonzero finite-dimensional [vector space](/page/Vector%20Space) over $F$, and let $\mathfrak l \subseteq \mathfrak{gl}(V)$ be a Lie subalgebra such that every $T \in \mathfrak l$ is nilpotent as an endomorphism of $V$. Then there exists $0 \neq v \in V$ such that $T(v)=0$ for every $T \in \mathfrak l$. [/claim] [proof] We argue by induction on $d:=\dim_F \mathfrak l$. If $d=0$, then $\mathfrak l=0$, and every nonzero vector of $V$ has the required property. Assume $d>0$ and that the result holds for all Lie algebras of nilpotent endomorphisms of dimension smaller than $d$. Choose a maximal proper Lie subalgebra $\mathfrak m \subsetneq \mathfrak l$. Such a subalgebra exists because $0$ is a proper Lie subalgebra when $d>0$ and $\mathfrak l$ is finite-dimensional. By the induction hypothesis applied to $\mathfrak m$, the subspace \begin{align*} W := \{v \in V : S(v)=0 \text{ for every } S \in \mathfrak m\} \end{align*} is nonzero. We prove that $\mathfrak m$ is an ideal of codimension $1$ in $\mathfrak l$. Define its normalizer in $\mathfrak l$ by \begin{align*} N_{\mathfrak l}(\mathfrak m) := \{T \in \mathfrak l : [T,S] \in \mathfrak m \text{ for every } S \in \mathfrak m\}. \end{align*} It suffices to show $N_{\mathfrak l}(\mathfrak m) \neq \mathfrak m$, because maximality of $\mathfrak m$ then implies $N_{\mathfrak l}(\mathfrak m)=\mathfrak l$. Let $Q:=\mathfrak l/\mathfrak m$ be the quotient vector space. For each $S \in \mathfrak m$, define the [linear map](/page/Linear%20Map) \begin{align*} \rho(S): Q &\to Q \\ T+\mathfrak m &\mapsto [S,T]+\mathfrak m. \end{align*} This is well-defined because $\mathfrak m$ is a Lie subalgebra: if $T-T' \in \mathfrak m$, then $[S,T-T'] \in \mathfrak m$. The assignment $S \mapsto \rho(S)$ is a Lie algebra homomorphism from $\mathfrak m$ to $\mathfrak{gl}(Q)$. For each $S \in \mathfrak m$, the map $\rho(S)$ is nilpotent. Indeed, $S:V\to V$ is nilpotent by hypothesis. Let $L_S:\mathfrak{gl}(V)\to\mathfrak{gl}(V)$ and $R_S:\mathfrak{gl}(V)\to\mathfrak{gl}(V)$ be the left and right multiplication maps $L_S(A)=SA$ and $R_S(A)=AS$. If $S^q=0$, then $L_S^q=0$ and $R_S^q=0$, and $L_S$ commutes with $R_S$. Hence $L_S-R_S$ is nilpotent, since \begin{align*} (L_S-R_S)^{2q-1} = \sum_{j=0}^{2q-1} (-1)^j \binom{2q-1}{j} L_S^{2q-1-j}R_S^j =0; \end{align*} in every summand either $2q-1-j \geq q$ or $j \geq q$. The restriction of $L_S-R_S$ to $\mathfrak l$ is $\operatorname{ad}S:T\mapsto [S,T]$, and $\rho(S)$ is the induced map on the quotient $Q$. Therefore $\rho(S)$ is nilpotent. If $Q$ were zero, then $\mathfrak m=\mathfrak l$, contrary to the choice of $\mathfrak m$; hence $Q\neq 0$. Since $\dim_F\mathfrak m<d$, the induction hypothesis applied to the Lie algebra $\rho(\mathfrak m)\subseteq\mathfrak{gl}(Q)$ gives a nonzero coset $T+\mathfrak m\in Q$ such that \begin{align*} \rho(S)(T+\mathfrak m)=0 \end{align*} for every $S\in\mathfrak m$. This means $T\notin\mathfrak m$ and $[S,T]\in\mathfrak m$ for every $S\in\mathfrak m$. Since $[T,S]=-[S,T]$, we have $T\in N_{\mathfrak l}(\mathfrak m)\setminus\mathfrak m$. Thus $N_{\mathfrak l}(\mathfrak m)$ strictly contains $\mathfrak m$, so maximality gives $N_{\mathfrak l}(\mathfrak m)=\mathfrak l$. Therefore $\mathfrak m$ is an ideal of $\mathfrak l$. The quotient Lie algebra $\mathfrak l/\mathfrak m$ has no nonzero proper Lie subalgebras, because the inverse image of any such subalgebra would be a Lie subalgebra strictly between $\mathfrak m$ and $\mathfrak l$. Since every one-dimensional subspace of a Lie algebra is a Lie subalgebra, it follows that $\dim_F(\mathfrak l/\mathfrak m)=1$. Choose $A\in\mathfrak l$ whose image spans $\mathfrak l/\mathfrak m$; then \begin{align*} \mathfrak l = \mathfrak m \oplus F A \end{align*} as vector spaces. Because $\mathfrak m$ is an ideal, $W$ is stable under $A$. Indeed, if $v \in W$ and $S \in \mathfrak m$, then \begin{align*} S(A(v)) = A(S(v)) - [A,S](v) = 0 - 0 = 0, \end{align*} since $S(v)=0$ and $[A,S]\in\mathfrak m$. Thus $A(W)\subseteq W$. The restriction $A|_W:W\to W$ is nilpotent because $A$ is nilpotent on $V$. Since $W\neq 0$, the kernel of $A|_W$ is nonzero. Choose $0\neq v\in W$ with $A(v)=0$. Then every element of $\mathfrak m$ kills $v$ by the definition of $W$, and $A$ kills $v$ by construction. Since $\mathfrak l=\mathfrak m\oplus F A$, every element of $\mathfrak l$ kills $v$. [/proof] [/step]

[step:Find a nonzero central element from nilpotent adjoint maps]Assume now that $\operatorname{ad}x$ is nilpotent for every $x \in \mathfrak g$. If $\mathfrak g=0$, then $\mathfrak g$ is nilpotent. Suppose $\mathfrak g \neq 0$. Apply the Engel fixed vector lemma with $V:=\mathfrak g$ and \begin{align*} \mathfrak l := \operatorname{ad}(\mathfrak g) = \{\operatorname{ad}x : x \in \mathfrak g\} \subseteq \mathfrak{gl}(\mathfrak g). \end{align*} This is a Lie subalgebra 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 of $\mathfrak l$ is nilpotent. Hence 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 $x \in \mathfrak g$, so $z \in Z(\mathfrak g)$. Therefore $Z(\mathfrak g) \neq 0$.[/step]

[guided]Now assume that each adjoint map $\operatorname{ad}x$ is nilpotent. We want to prove that $\mathfrak g$ is nilpotent. The first structural consequence is that the center is nonzero. If $\mathfrak g=0$, the conclusion is immediate, so assume $\mathfrak g \neq 0$. Consider the adjoint image \begin{align*} \mathfrak l := \operatorname{ad}(\mathfrak g) = \{\operatorname{ad}x : x \in \mathfrak g\} \subseteq \mathfrak{gl}(\mathfrak g). \end{align*} This is a Lie subalgebra of $\mathfrak{gl}(\mathfrak g)$. Indeed, for $x,y,u \in \mathfrak g$, the Jacobi identity gives \begin{align*} [\operatorname{ad}x,\operatorname{ad}y](u) &= [x,[y,u]] - [y,[x,u]] \\ &= [[x,y],u] \\ &= (\operatorname{ad}[x,y])(u). \end{align*} Thus $[\operatorname{ad}x,\operatorname{ad}y]=\operatorname{ad}[x,y]$, so $\mathfrak l$ is closed under the commutator bracket. The hypothesis says exactly that every element of $\mathfrak l$ is nilpotent as an endomorphism of the vector space $\mathfrak g$. The Engel fixed vector lemma therefore applies with $V=\mathfrak g$. It gives a nonzero element $z \in \mathfrak g$ such that \begin{align*} (\operatorname{ad}x)(z)=0 \end{align*} for every $x \in \mathfrak g$. Since $(\operatorname{ad}x)(z)=[x,z]$, this means $[x,z]=0$ for every $x \in \mathfrak g$. Hence $z$ lies in the center \begin{align*} Z(\mathfrak g):=\{u \in \mathfrak g : [x,u]=0 \text{ for every } x \in \mathfrak g\}. \end{align*} Thus $Z(\mathfrak g)$ contains the nonzero element $z$.[/guided]

[step:Apply the central element argument to all nonzero quotients]Let $\mathfrak a \trianglelefteq \mathfrak g$ be an ideal. Define the quotient Lie algebra $\mathfrak g/\mathfrak a$ with bracket \begin{align*} [x+\mathfrak a,y+\mathfrak a] := [x,y]+\mathfrak a. \end{align*} If $x \in \mathfrak g$, then the adjoint endomorphism of $\mathfrak g/\mathfrak a$ induced by $x+\mathfrak a$ satisfies \begin{align*} \operatorname{ad}_{\mathfrak g/\mathfrak a}(x+\mathfrak a)^m(y+\mathfrak a) = (\operatorname{ad}_{\mathfrak g}x)^m(y)+\mathfrak a \end{align*} for every $m \geq 1$ and every $y \in \mathfrak g$. Since $\operatorname{ad}_{\mathfrak g}x$ is nilpotent, $\operatorname{ad}_{\mathfrak g/\mathfrak a}(x+\mathfrak a)$ is nilpotent. Therefore every nonzero quotient $\mathfrak g/\mathfrak a$ has nonzero center by the previous step.[/step]

[guided]The previous step showed that a Lie algebra whose adjoint maps are all nilpotent has nonzero center. To build nilpotence of $\mathfrak g$, we need the same conclusion not only for $\mathfrak g$ itself but also for its quotients. Let $\mathfrak a \trianglelefteq \mathfrak g$ be an ideal. The quotient vector space $\mathfrak g/\mathfrak a$ becomes a Lie algebra under \begin{align*} [x+\mathfrak a,y+\mathfrak a] := [x,y]+\mathfrak a. \end{align*} This bracket is well-defined because $\mathfrak a$ is an ideal. For $x \in \mathfrak g$, the adjoint map on the quotient is \begin{align*} \operatorname{ad}_{\mathfrak g/\mathfrak a}(x+\mathfrak a): \mathfrak g/\mathfrak a &\to \mathfrak g/\mathfrak a \\ y+\mathfrak a &\mapsto [x,y]+\mathfrak a. \end{align*} We compare this map with $\operatorname{ad}_{\mathfrak g}x$. A direct induction on $m \geq 1$ gives \begin{align*} \operatorname{ad}_{\mathfrak g/\mathfrak a}(x+\mathfrak a)^m(y+\mathfrak a) = (\operatorname{ad}_{\mathfrak g}x)^m(y)+\mathfrak a. \end{align*} Because $\operatorname{ad}_{\mathfrak g}x$ is nilpotent by hypothesis, there exists an integer $m \geq 1$ such that $(\operatorname{ad}_{\mathfrak g}x)^m=0$. The displayed formula then gives \begin{align*} \operatorname{ad}_{\mathfrak g/\mathfrak a}(x+\mathfrak a)^m=0. \end{align*} Thus every adjoint map in the quotient is nilpotent. Applying the previous step to the nonzero Lie algebra $\mathfrak g/\mathfrak a$ shows that its center is nonzero.[/guided]

[step:Construct the upper central series until it reaches the whole algebra]Define the upper central series $(Z_m(\mathfrak g))_{m \geq 0}$ by \begin{align*} Z_0(\mathfrak g) &:= 0, \\ Z_{m+1}(\mathfrak g)/Z_m(\mathfrak g) &:= Z(\mathfrak g/Z_m(\mathfrak g)). \end{align*} Each $Z_m(\mathfrak g)$ is an ideal of $\mathfrak g$. If $Z_m(\mathfrak g) \neq \mathfrak g$, then the quotient $\mathfrak g/Z_m(\mathfrak g)$ is nonzero, so its center is nonzero by the previous step. Hence \begin{align*} Z_m(\mathfrak g) \subsetneq Z_{m+1}(\mathfrak g). \end{align*} Because $\mathfrak g$ is finite-dimensional, this strictly increasing chain cannot continue for more than $\dim_F \mathfrak g$ steps. Therefore there exists $c \geq 0$ such that \begin{align*} Z_c(\mathfrak g)=\mathfrak g. \end{align*}[/step]

[guided]The upper central series grows by repeatedly adding elements that are central modulo what has already been added. Define \begin{align*} Z_0(\mathfrak g) &:= 0, \\ Z_{m+1}(\mathfrak g)/Z_m(\mathfrak g) &:= Z(\mathfrak g/Z_m(\mathfrak g)). \end{align*} Equivalently, $Z_{m+1}(\mathfrak g)$ is the inverse image of the center of $\mathfrak g/Z_m(\mathfrak g)$ under the quotient map $\mathfrak g \to \mathfrak g/Z_m(\mathfrak g)$. This description shows that each $Z_m(\mathfrak g)$ is an ideal. Suppose $Z_m(\mathfrak g) \neq \mathfrak g$. Then the quotient $\mathfrak g/Z_m(\mathfrak g)$ is a nonzero finite-dimensional Lie algebra. By the quotient argument from the previous step, all its adjoint maps are nilpotent. Applying the nonzero-center result to this quotient gives \begin{align*} Z(\mathfrak g/Z_m(\mathfrak g)) \neq 0. \end{align*} By the definition of the upper central series, this means \begin{align*} Z_{m+1}(\mathfrak g)/Z_m(\mathfrak g) \neq 0, \end{align*} and therefore $Z_m(\mathfrak g) \subsetneq Z_{m+1}(\mathfrak g)$. Since $\mathfrak g$ has finite dimension over $F$, a strictly increasing chain of subspaces of $\mathfrak g$ has length at most $\dim_F \mathfrak g$. Thus this process must stop, and the only possible stopping point is $\mathfrak g$ itself. Hence there exists an integer $c \geq 0$ such that \begin{align*} Z_c(\mathfrak g)=\mathfrak g. \end{align*}[/guided]

[step:Conclude that the Lie algebra is nilpotent] We show that $Z_c(\mathfrak g)=\mathfrak g$ implies nilpotence. For every $m \geq 0$, the defining property of the upper central series gives \begin{align*} [\mathfrak g,Z_{m+1}(\mathfrak g)] \subseteq Z_m(\mathfrak g). \end{align*} We prove by induction on $k \geq 0$ that \begin{align*} \gamma_{k+1}(\mathfrak g) \subseteq Z_{c-k}(\mathfrak g) \end{align*} for $0 \leq k \leq c$. For $k=0$, this is $\mathfrak g \subseteq Z_c(\mathfrak g)$, which holds because $Z_c(\mathfrak g)=\mathfrak g$. If the inclusion holds for $k$, then \begin{align*} \gamma_{k+2}(\mathfrak g) &= [\mathfrak g,\gamma_{k+1}(\mathfrak g)] \\ &\subseteq [\mathfrak g,Z_{c-k}(\mathfrak g)] \\ &\subseteq Z_{c-k-1}(\mathfrak g). \end{align*} Thus $\gamma_{c+1}(\mathfrak g) \subseteq Z_0(\mathfrak g)=0$. Therefore the lower central series reaches zero, so $\mathfrak g$ is nilpotent. Combining this with the first step proves the equivalence. [/step]