[proofplan] We compare the two central series directly. The key translation is that $x \in Z_{i+1}(\mathfrak g)$ exactly when every bracket $[y,x]$ lies in $Z_i(\mathfrak g)$. If the upper central series reaches $\mathfrak g$, this translation forces the lower central series downward through the upper central series until it reaches $0$. Conversely, if the lower central series reaches $0$, the same translation lifts each lower central layer into the corresponding upper central layer. [/proofplan] [step:Define the two central series used in the argument] We use the upper central series $(Z_i(\mathfrak g))_{i\geq 0}$ defined recursively by \begin{align*} Z_0(\mathfrak g)&=0, \\ Z_{i+1}(\mathfrak g)/Z_i(\mathfrak g)&=Z(\mathfrak g/Z_i(\mathfrak g)) \quad \text{for } i\geq 0. \end{align*} Equivalently, $Z_{i+1}(\mathfrak g)$ is the inverse image of the center $Z(\mathfrak g/Z_i(\mathfrak g))$ under the quotient map $\mathfrak g\to \mathfrak g/Z_i(\mathfrak g)$. In particular, each $Z_i(\mathfrak g)$ is an ideal of $\mathfrak g$ by induction on $i$: $Z_0(\mathfrak g)=0$ is an ideal, and the inverse image of an ideal under a Lie algebra homomorphism is an ideal. We use the lower central series $(\gamma_r(\mathfrak g))_{r\geq 1}$ defined recursively by \begin{align*} \gamma_1(\mathfrak g)&=\mathfrak g, \\ \gamma_{r+1}(\mathfrak g)&=[\mathfrak g,\gamma_r(\mathfrak g)] \quad \text{for } r\geq 1, \end{align*} where $[\mathfrak g,\gamma_r(\mathfrak g)]$ denotes the linear span of all brackets $[y,x]$ with $y\in\mathfrak g$ and $x\in\gamma_r(\mathfrak g)$. [/step] [step:Record the commutator criterion for membership in the upper central series] For each integer $i \geq 0$, the definition of $Z_{i+1}(\mathfrak g)$ gives the equivalence \begin{align*} x \in Z_{i+1}(\mathfrak g) \quad\Longleftrightarrow\quad [y,x] \in Z_i(\mathfrak g) \text{ for every } y \in \mathfrak g. \end{align*} Indeed, $x \in Z_{i+1}(\mathfrak g)$ exactly means that the coset $x+Z_i(\mathfrak g)$ lies in the center of the quotient Lie algebra $\mathfrak g/Z_i(\mathfrak g)$. This is equivalent to \begin{align*} [y+Z_i(\mathfrak g),x+Z_i(\mathfrak g)] = [y,x]+Z_i(\mathfrak g) = Z_i(\mathfrak g) \end{align*} for every $y \in \mathfrak g$, which is precisely $[y,x] \in Z_i(\mathfrak g)$. Consequently, \begin{align*} [\mathfrak g,Z_{i+1}(\mathfrak g)] \subset Z_i(\mathfrak g) \end{align*} for every $i \geq 0$. [guided] The upper central series is defined through centers of quotient Lie algebras, so we first translate that quotient statement into an ordinary commutator statement inside $\mathfrak g$. Fix an integer $i \geq 0$ and an element $x \in \mathfrak g$. By definition, \begin{align*} x \in Z_{i+1}(\mathfrak g) \end{align*} means that the coset $x+Z_i(\mathfrak g)$ belongs to the center of $\mathfrak g/Z_i(\mathfrak g)$. The center of a Lie algebra consists of the elements whose bracket with every element is zero. In the quotient Lie algebra, this says \begin{align*} [y+Z_i(\mathfrak g),x+Z_i(\mathfrak g)] = Z_i(\mathfrak g) \end{align*} for every $y \in \mathfrak g$. The quotient bracket is defined by \begin{align*} [y+Z_i(\mathfrak g),x+Z_i(\mathfrak g)] = [y,x]+Z_i(\mathfrak g). \end{align*} Therefore the previous equality holds exactly when \begin{align*} [y,x]+Z_i(\mathfrak g)=Z_i(\mathfrak g), \end{align*} which is equivalent to $[y,x]\in Z_i(\mathfrak g)$. Hence \begin{align*} x \in Z_{i+1}(\mathfrak g) \quad\Longleftrightarrow\quad [y,x] \in Z_i(\mathfrak g) \text{ for every } y \in \mathfrak g. \end{align*} Applying this to every $x \in Z_{i+1}(\mathfrak g)$ gives the inclusion \begin{align*} [\mathfrak g,Z_{i+1}(\mathfrak g)] \subset Z_i(\mathfrak g). \end{align*} [/guided] [/step] [step:Push the lower central series down when the upper central series reaches $\mathfrak g$] Assume that $Z_c(\mathfrak g)=\mathfrak g$ for some $c \in \mathbb N$. We prove by induction on $r \in \{1,\dots,c+1\}$ that \begin{align*} \gamma_r(\mathfrak g) \subset Z_{c+1-r}(\mathfrak g). \end{align*} For $r=1$, this is \begin{align*} \gamma_1(\mathfrak g)=\mathfrak g=Z_c(\mathfrak g)=Z_{c+1-1}(\mathfrak g). \end{align*} Assume now that $1 \leq r \leq c$ and that $\gamma_r(\mathfrak g)\subset Z_{c+1-r}(\mathfrak g)$. Using the definition of the lower central series and the commutator inclusion from the previous step with $i=c-r$, we get \begin{align*} \gamma_{r+1}(\mathfrak g) &= [\mathfrak g,\gamma_r(\mathfrak g)] \\ &\subset [\mathfrak g,Z_{c+1-r}(\mathfrak g)] \\ &\subset Z_{c-r}(\mathfrak g) \\ &= Z_{c+1-(r+1)}(\mathfrak g). \end{align*} Thus the induction holds. Taking $r=c+1$ gives \begin{align*} \gamma_{c+1}(\mathfrak g) \subset Z_0(\mathfrak g)=0, \end{align*} so $\gamma_{c+1}(\mathfrak g)=0$. Hence $\mathfrak g$ is nilpotent. [guided] Assume that the upper central series reaches the whole Lie algebra, so $Z_c(\mathfrak g)=\mathfrak g$ for some $c \in \mathbb N$. We want to show that sufficiently long iterated commutators vanish. The lower central series measures exactly these iterated commutators, so it is enough to prove that the terms $\gamma_r(\mathfrak g)$ are forced into smaller and smaller terms of the upper central series. We prove by induction on $r \in \{1,\dots,c+1\}$ that \begin{align*} \gamma_r(\mathfrak g) \subset Z_{c+1-r}(\mathfrak g). \end{align*} For $r=1$, the statement is \begin{align*} \gamma_1(\mathfrak g)=\mathfrak g=Z_c(\mathfrak g)=Z_{c+1-1}(\mathfrak g), \end{align*} which is exactly the assumption that the upper central series reaches $\mathfrak g$. Now assume that $1 \leq r \leq c$ and that \begin{align*} \gamma_r(\mathfrak g)\subset Z_{c+1-r}(\mathfrak g). \end{align*} By the definition of the lower central series, \begin{align*} \gamma_{r+1}(\mathfrak g)=[\mathfrak g,\gamma_r(\mathfrak g)]. \end{align*} Since $\gamma_r(\mathfrak g)\subset Z_{c+1-r}(\mathfrak g)$, monotonicity of the bracket in the second argument gives \begin{align*} [\mathfrak g,\gamma_r(\mathfrak g)] \subset [\mathfrak g,Z_{c+1-r}(\mathfrak g)]. \end{align*} The commutator criterion from the previous step, applied with $i=c-r$, gives \begin{align*} [\mathfrak g,Z_{c+1-r}(\mathfrak g)]\subset Z_{c-r}(\mathfrak g). \end{align*} Combining these inclusions, \begin{align*} \gamma_{r+1}(\mathfrak g) &= [\mathfrak g,\gamma_r(\mathfrak g)] \\ &\subset [\mathfrak g,Z_{c+1-r}(\mathfrak g)] \\ &\subset Z_{c-r}(\mathfrak g) \\ &= Z_{c+1-(r+1)}(\mathfrak g). \end{align*} This completes the induction. At the final index $r=c+1$, we obtain \begin{align*} \gamma_{c+1}(\mathfrak g)\subset Z_0(\mathfrak g)=0. \end{align*} Therefore $\gamma_{c+1}(\mathfrak g)=0$, which is precisely nilpotence. [/guided] [/step] [step:Lift lower central layers into upper central layers when $\mathfrak g$ is nilpotent] Conversely, assume that $\mathfrak g$ is nilpotent. Then there exists $m \in \mathbb N$ such that \begin{align*} \gamma_{m+1}(\mathfrak g)=0. \end{align*} We prove by induction on $i \in \{0,\dots,m\}$ that \begin{align*} \gamma_{m+1-i}(\mathfrak g)\subset Z_i(\mathfrak g). \end{align*} For $i=0$, this is \begin{align*} \gamma_{m+1}(\mathfrak g)=0=Z_0(\mathfrak g). \end{align*} Assume that $0 \leq i < m$ and that \begin{align*} \gamma_{m+1-i}(\mathfrak g)\subset Z_i(\mathfrak g). \end{align*} Let $x \in \gamma_{m-i}(\mathfrak g)$. For every $y \in \mathfrak g$, \begin{align*} [y,x]\in [\mathfrak g,\gamma_{m-i}(\mathfrak g)] = \gamma_{m+1-i}(\mathfrak g) \subset Z_i(\mathfrak g). \end{align*} By the commutator criterion for the upper central series, $x \in Z_{i+1}(\mathfrak g)$. Hence \begin{align*} \gamma_{m-i}(\mathfrak g)\subset Z_{i+1}(\mathfrak g), \end{align*} which is the induction statement at $i+1$. Taking $i=m$ gives \begin{align*} \mathfrak g=\gamma_1(\mathfrak g)\subset Z_m(\mathfrak g)\subset \mathfrak g, \end{align*} so $Z_m(\mathfrak g)=\mathfrak g$. [guided] Now assume that $\mathfrak g$ is nilpotent. By definition, there is an integer $m \in \mathbb N$ such that \begin{align*} \gamma_{m+1}(\mathfrak g)=0. \end{align*} The goal is to prove that the upper central series reaches $\mathfrak g$. Instead of using [Engel's theorem](/theorems/3798), we use the lower central series directly: if a lower central layer brackets into a layer already known to lie in $Z_i(\mathfrak g)$, then that lower central layer itself lies in $Z_{i+1}(\mathfrak g)$. We prove by induction on $i \in \{0,\dots,m\}$ that \begin{align*} \gamma_{m+1-i}(\mathfrak g)\subset Z_i(\mathfrak g). \end{align*} For $i=0$, this says \begin{align*} \gamma_{m+1}(\mathfrak g)\subset Z_0(\mathfrak g). \end{align*} Both sides are $0$, since $\gamma_{m+1}(\mathfrak g)=0$ by nilpotence and $Z_0(\mathfrak g)=0$ by definition. Assume now that $0 \leq i < m$ and that \begin{align*} \gamma_{m+1-i}(\mathfrak g)\subset Z_i(\mathfrak g). \end{align*} We must show that \begin{align*} \gamma_{m-i}(\mathfrak g)\subset Z_{i+1}(\mathfrak g). \end{align*} Take an arbitrary element $x \in \gamma_{m-i}(\mathfrak g)$. To prove $x \in Z_{i+1}(\mathfrak g)$, the commutator criterion from the first step says that it suffices to prove \begin{align*} [y,x]\in Z_i(\mathfrak g) \end{align*} for every $y \in \mathfrak g$. But this follows from the definition of the lower central series: \begin{align*} [y,x]\in [\mathfrak g,\gamma_{m-i}(\mathfrak g)] = \gamma_{m+1-i}(\mathfrak g). \end{align*} By the induction hypothesis, \begin{align*} \gamma_{m+1-i}(\mathfrak g)\subset Z_i(\mathfrak g), \end{align*} and therefore $[y,x]\in Z_i(\mathfrak g)$ for every $y \in \mathfrak g$. The commutator criterion now gives $x \in Z_{i+1}(\mathfrak g)$. Since $x$ was arbitrary, \begin{align*} \gamma_{m-i}(\mathfrak g)\subset Z_{i+1}(\mathfrak g). \end{align*} This completes the induction. At $i=m$, the induction statement becomes \begin{align*} \gamma_1(\mathfrak g)\subset Z_m(\mathfrak g). \end{align*} Since $\gamma_1(\mathfrak g)=\mathfrak g$ and always $Z_m(\mathfrak g)\subset \mathfrak g$, we get \begin{align*} \mathfrak g=Z_m(\mathfrak g). \end{align*} Thus the upper central series reaches all of $\mathfrak g$. [/guided] [/step] [step:Conclude the equivalence] The first implication shows that if $Z_c(\mathfrak g)=\mathfrak g$ for some $c \in \mathbb N$, then $\mathfrak g$ is nilpotent. The second implication shows that if $\mathfrak g$ is nilpotent, then $Z_m(\mathfrak g)=\mathfrak g$ for some $m \in \mathbb N$. Therefore $\mathfrak g$ is nilpotent if and only if the upper central series reaches $\mathfrak g$ in finitely many steps. [/step]