All cartan subalgebras of a real semisimple lie algebra are conjugate

Given: The real Lie algebra ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ with basis $h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, $e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, $f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$ and $k=e-f$, and the two lines $\mathbb{R}h$ and $\mathbb{R}k$.

${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ is semisimple: its Killing form has matrix $\left(\begin{array}{ccc}8& 0& 0\\ 0& 0& 4\\ 0& 4& 0\end{array}\right)$ in the basis $(h,e,f)$, which is nondegenerate, and a finite-dimensional Lie algebra over a characteristic-zero field is semisimple if and only if its Killing form is nondegenerate (Cartan's semisimplicity criterion, Real Cartan subalgebras need not be conjugate).

A Cartan subalgebra of a Lie algebra is a nilpotent subalgebra equal to its own normalizer, and a one-dimensional abelian subalgebra is nilpotent (Cartan subalgebra).

In ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ the lines $\mathbb{R}h$ and $\mathbb{R}k$ are Cartan subalgebras, $\mathbb{R}h\subseteq {\mathfrak{p}}_0$ is the split part and $\mathbb{R}k\subseteq {\mathfrak{k}}_0$ is the compact part of the Cartan decomposition attached to $\theta (X)=-X^T$, and there is no automorphism $\alpha$ of ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ with $\alpha (\mathbb{R}h)=\mathbb{R}k$: such an $\alpha$ would satisfy $B(h,h)=B(\alpha h,\alpha h)$ for the automorphism-invariant Killing form, that is $8=-8c^2$ for $\alpha (h)=ck$, $c\mathrm{\ne }0$, which is impossible (Real Cartan subalgebras need not be conjugate, The special linear Lie algebra sl_2).