The same Dynkin diagram forces isomorphic connected Lie groups

By [L1] the groups $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ are connected Lie groups with isomorphic Lie algebras, namely $\mathfrak{s}\mathfrak{u}(2)\cong \mathfrak{s}\mathfrak{o}(3)$, and they are not isomorphic.

Put $h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, $e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, and $f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$. The matrices $ih,e-f,i(e+f)$ form a real basis of $\mathfrak{s}\mathfrak{u}(2)$ and a complex basis of ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$, because $h=-i(ih)$, $e=((e-f)-i(i(e+f)))\mathrm{/}2$, and $f=(-(e-f)-i(i(e+f)))\mathrm{/}2$. Thus complexifying the inclusion gives $\mathfrak{s}\mathfrak{u}(2){\otimes }_{\mathbb{R}}\mathbb{C}\cong {\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$. The isomorphism in step 1.1 gives the same complexification for $\mathfrak{s}\mathfrak{o}(3)$, and [L2] identifies the Dynkin diagram of both as $A_1$.

Thus the two connected groups have the same Dynkin diagram $A_1$ but are not isomorphic, which refutes the claim; the missing global information is the lattice data of the simply connected form, here the central subgroup $\{\pm I\}$. The proof uses countable choice only through the covering-group supplier of [L1], and introduces no further choice.