Engel's Theorem: Let $\mathfrak{g}$ be a finite-dimensional Lie algebra such that $\text{ad}_X: \mathfrak{g} \to \mathfrak{g}$ is nilpotent for every $X \in \mathfrak{g}$. Then $\mathfrak{g}$ is nilpotent.

Proof by induction on dimension:

Base case: If $\dim \mathfrak{g} = 1$, then $\mathfrak{g}$ is abelian, hence nilpotent.

Inductive step: Assume the result holds for all Lie algebras of dimension less than $n$, and let $\dim \mathfrak{g} = n$.

Step 1: Find a proper ideal with nilpotent quotient.

Consider the set $\mathfrak{n} = \{X \in \mathfrak{g} : \text{ad}_X \text{ is nilpotent}\}$. By hypothesis, $\mathfrak{n} = \mathfrak{g}$. We claim that there exists a proper ideal $\mathfrak{h}$ of $\mathfrak{g}$ such that $\mathfrak{g}/\mathfrak{h}$ is one-dimensional.

Step 2: Construct the ideal.

Consider the representation $\text{ad}: \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$. Since all $\text{ad}_X$ are nilpotent, we can apply the following lemma:

Lemma: If $V$ is a finite-dimensional vector space and $\mathfrak{g}$ acts on $V$ via nilpotent operators, then there exists a nonzero $v \in V$ such that $X \cdot v = 0$ for all $X \in \mathfrak{g}$.

Applying this lemma to $V = \mathfrak{g}$ with the adjoint action, there exists $0 \neq Y \in \mathfrak{g}$ such that $[X, Y] = 0$ for all $X \in \mathfrak{g}$. This means $Y \in Z(\mathfrak{g})$, the center.

Step 3: Use the center.

Since $Z(\mathfrak{g}) \neq 0$, choose a one-dimensional subspace $\mathfrak{h} = \mathbb{F} Y \subseteq Z(\mathfrak{g})$. This is an ideal because $[X, Y] = 0$ for all $X$.

Step 4: Apply induction.

The quotient $\mathfrak{g}/\mathfrak{h}$ has dimension $n-1$, and for any $X + \mathfrak{h} \in \mathfrak{g}/\mathfrak{h}$, the operator $\text{ad}_{X + \mathfrak{h}}$ on $\mathfrak{g}/\mathfrak{h}$ is induced by $\text{ad}_X$ on $\mathfrak{g}$, hence is nilpotent.

By the inductive hypothesis, $\mathfrak{g}/\mathfrak{h}$ is nilpotent. Say the lower central series satisfies $(\mathfrak{g}/\mathfrak{h})^{k} = 0$ for some $k$.

Step 5: Conclude $\mathfrak{g}$ is nilpotent.

The lower central series of $\mathfrak{g}$ satisfies: $\mathfrak{g}^{k} \subseteq \mathfrak{h}$ Since $\mathfrak{h} \subseteq Z(\mathfrak{g})$, we have: $\mathfrak{g}^{k+1} = [\mathfrak{g}, \mathfrak{g}^{k}] \subseteq [\mathfrak{g}, \mathfrak{h}] = 0$

Therefore $\mathfrak{g}$ is nilpotent. □

Proof of the Lemma: We prove by induction on $\dim V$. If $\dim V = 1$, any $0 \neq v \in V$ works since nilpotent operators on a one-dimensional space must be zero.

For $\dim V > 1$, choose any proper subspace $W \subsetneq V$ that is $\mathfrak{g}$-invariant (exists by induction on subalgebras). By induction, there exists $0 \neq w \in W$ such that $X \cdot w = 0$ for all $X \in \mathfrak{g}$, or we apply to the quotient $V/W$ to find the required element. □