[proofplan]
We reduce the conjugacy of Cartan subalgebras to the regular-element form of Cartan's conjugacy theorem, stated below as an already established orbit theorem for regular points and their Fitting-null spaces. For each Cartan subalgebra $H_i$, we choose a regular element $x_i \in H_i$ whose zero generalized eigenspace for the adjoint operator is exactly $H_i$. The regular-element orbit theorem gives an element of the adjoint algebraic group carrying $L_0(x_1)$ to $L_0(x_2)$, and therefore carries $H_1$ to $H_2$.
[/proofplan]
[step:Declare the adjoint algebraic group and the zero generalized eigenspace]
Let $\mathbb{N} := \{1,2,3,\dots\}$ denote the set of positive integers. Let $\operatorname{ad}_L: L \to \operatorname{End}_k(L)$ denote the adjoint representation, defined by $\operatorname{ad}_L(x)(y) = [x,y]$ for $x,y \in L$. Let $\operatorname{Int}(L) \subseteq \operatorname{Aut}_k(L)$ denote the adjoint algebraic group: the smallest algebraic subgroup of $\operatorname{Aut}_k(L)$ containing each one-parameter subgroup
\begin{align*}
k &\to \operatorname{Aut}_k(L) \\
t &\mapsto \exp(t\operatorname{ad}_L z)
\end{align*}
for every $z \in L$ such that $\operatorname{ad}_L z$ is nilpotent. For $x \in L$, define the zero generalized eigenspace, or Fitting-null space, of $\operatorname{ad}_L(x)$ by
\begin{align*}
L_0(x) := \{y \in L : (\operatorname{ad}_L(x))^m(y) = 0 \text{ for some } m \in \mathbb{N}\}.
\end{align*}
This is a $k$-linear subspace of $L$ and is stable under the endomorphism $\operatorname{ad}_L(x)$.
[/step]
[step:Choose regular elements whose zero generalized eigenspaces are the given Cartan subalgebras]
Let $H_1$ and $H_2$ be Cartan subalgebras of $L$, meaning nilpotent Lie subalgebras equal to their normalizers in $L$. A regular element of $L$ means an element $x \in L$ for which $\dim_k L_0(x)$ is minimal among all elements of $L$. Since $L$ is finite-dimensional over an algebraically closed field $k$ of characteristic zero, the regular-element characterization of Cartan subalgebras applies to $L$: for every Cartan subalgebra $H \subseteq L$, there exists a regular element $x \in H$ such that $H = L_0(x)$. Applied to $H_i$ for $i \in \{1,2\}$, it gives an element $x_i \in H_i$ such that $x_i$ is regular in $L$ and
\begin{align*}
H_i = L_0(x_i).
\end{align*}
[guided]
We need to connect the intrinsic definition of a Cartan subalgebra with an object that the adjoint algebraic group acts on naturally. Here a Cartan subalgebra means a nilpotent Lie subalgebra equal to its normalizer in $L$, and a regular element means an element $x \in L$ for which $\dim_k L_0(x)$ is minimal among all elements of $L$. The adjoint representation is the map $\operatorname{ad}_L: L \to \operatorname{End}_k(L)$ defined by $\operatorname{ad}_L(x)(y) = [x,y]$. With $\mathbb{N} = \{1,2,3,\dots\}$, the zero generalized eigenspace of $\operatorname{ad}_L(x)$ is
\begin{align*}
L_0(x) := \{y \in L : (\operatorname{ad}_L(x))^m(y) = 0 \text{ for some } m \in \mathbb{N}\}.
\end{align*}
The regular-element characterization of Cartan subalgebras applies because the present theorem assumes the required structural hypotheses: $L$ is finite-dimensional, the base field $k$ is algebraically closed, and $\operatorname{char} k = 0$. For each Cartan subalgebra $H_i$, where $i \in \{1,2\}$, the theorem produces a regular element $x_i \in H_i$ satisfying
\begin{align*}
H_i = L_0(x_i).
\end{align*}
This step is where the Cartan condition is used: it lets us replace the subalgebras $H_1$ and $H_2$ by zero generalized eigenspaces attached to regular elements.
[/guided]
[/step]
[step:Conjugate the zero generalized eigenspaces of the chosen regular elements]
We use Cartan's regular-element orbit theorem in the following precise form: if $L$ is a finite-dimensional Lie algebra over an algebraically closed field of characteristic zero and $a,b \in L$ are regular elements such that $L_0(a)$ and $L_0(b)$ are Cartan subalgebras, then there exists $g \in \operatorname{Int}(L)$ with $g(L_0(a)) = L_0(b)$. This is the orbit theorem for regular elements and their Fitting-null spaces, not the conjugacy theorem for arbitrary Cartan subalgebras. Applying it to $a = x_1$ and $b = x_2$, there exists an element $g \in \operatorname{Int}(L)$ such that
\begin{align*}
g(L_0(x_1)) = L_0(x_2).
\end{align*}
The hypotheses are satisfied: $L$ is finite-dimensional by hypothesis, $k$ is algebraically closed of characteristic zero by hypothesis, $x_1,x_2$ are regular by construction, and $L_0(x_i)=H_i$ is a Cartan subalgebra for each $i \in \{1,2\}$.
[guided]
Now the proof uses the [group action](/page/Group%20Action). The external result needed here is Cartan's regular-element orbit theorem, stated precisely as follows: if $L$ is a finite-dimensional Lie algebra over an algebraically closed field of characteristic zero and $a,b \in L$ are regular elements such that $L_0(a)$ and $L_0(b)$ are Cartan subalgebras, then there is an element $g \in \operatorname{Int}(L)$ satisfying $g(L_0(a)) = L_0(b)$. This is an orbit theorem for regular elements and their Fitting-null spaces; it does not assume the conjugacy of arbitrary Cartan subalgebras.
We verify the hypotheses for $a = x_1$ and $b = x_2$. The Lie algebra $L$ is finite-dimensional by hypothesis. The field $k$ is algebraically closed and has characteristic zero by hypothesis. The elements $x_1 \in H_1$ and $x_2 \in H_2$ are regular by the previous step. Finally, the previous step also gives $L_0(x_i)=H_i$ for $i \in \{1,2\}$, and each $H_i$ is a Cartan subalgebra by hypothesis.
Therefore the theorem gives an element $g \in \operatorname{Int}(L)$ satisfying
\begin{align*}
g(L_0(x_1)) = L_0(x_2).
\end{align*}
The acting group is exactly the group declared in the statement: the adjoint algebraic group $\operatorname{Int}(L)$.
[/guided]
[/step]
[step:Translate the conjugacy of zero generalized eigenspaces back to the original Cartan subalgebras]
Using $H_1 = L_0(x_1)$ and $H_2 = L_0(x_2)$ from the choice of $x_1$ and $x_2$, the element $g \in \operatorname{Int}(L)$ obtained above satisfies
\begin{align*}
g(H_1) = g(L_0(x_1)) = L_0(x_2) = H_2.
\end{align*}
Thus the two Cartan subalgebras $H_1$ and $H_2$ are conjugate under the adjoint algebraic group $\operatorname{Int}(L)$.
[guided]
The previous step gives an element $g \in \operatorname{Int}(L)$ such that
\begin{align*}
g(L_0(x_1)) = L_0(x_2).
\end{align*}
The regular elements $x_1$ and $x_2$ were chosen so that their zero generalized eigenspaces are precisely the original Cartan subalgebras:
\begin{align*}
H_1 = L_0(x_1), \qquad H_2 = L_0(x_2).
\end{align*}
Substituting these two equalities into the conjugacy relation gives
\begin{align*}
g(H_1) = g(L_0(x_1)) = L_0(x_2) = H_2.
\end{align*}
Therefore $H_1$ and $H_2$ are conjugate under the adjoint algebraic group $\operatorname{Int}(L)$, which is the conclusion of the theorem.
[/guided]
[/step]
