$su(2) $ and $ sl(2;R)$ are not isomorphic?

Question

As real Lie algebras, both are three-dimensional. The basis of $su(2)$ is

$$ \left( \begin{matrix} i & 0 \\ 0 & -i \end{matrix} \right), \left( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right),\left( \begin{matrix} 0 & i \\ i & 0 \end{matrix} \right) . $$

The basis of $sl(2;R)$ is

$$ \left( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right), \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right),\left( \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right) . $$

But how to prove that there is no isomorphism mapping between the two algebras?

Answer 1

The first of these has no two-dimensional subalgebras, whereas the second does. See $\mathfrak{su}(2)$ not isomorphic to $\mathfrak{sl}(2,\mathbb R)$ for a fuller explanation.