The Lie algebra and root system do not determine the root datum

The Lie algebras agree and the abstract root systems agree: by [F2], $\mathrm{Lie}(G_1)\cong \mathrm{Lie}(G_2)\cong {\mathfrak{s}\mathfrak{l}}_2$, and both root systems are $\{\pm \alpha \}$, abstractly the root system $A_1$; the groups are not isomorphic because their centres differ, ${\mu }_2\mathrm{\ne }1$ by [F3].

The root data are nevertheless not isomorphic. Write $X_1=X(T_1)=\mathbb{Z}{\chi }_1$ and $X_2=X(T_2)=\mathbb{Z}{\chi }_2$, so a $\mathbb{Z}$-linear isomorphism $f:X_1\to X_2$ carries the generator ${\chi }_1$ to $\pm {\chi }_2$. An isomorphism of root data must carry ${\mathrm{\Phi }}_1$ onto ${\mathrm{\Phi }}_2$, in particular it must send the root ${\alpha }_1=2{\chi }_1$ to an element of ${\mathrm{\Phi }}_2=\{\pm {\chi }_2\}$. But $f(2{\chi }_1)=\pm 2{\chi }_2\mathrm{\notin }\{\pm {\chi }_2\}$ because ${\chi }_2$ generates a free $\mathbb{Z}$-lattice. Equivalently, ${\alpha }_1$ is divisible by $2$ in $X_1$ while ${\alpha }_2={\chi }_2$ is not divisible by $2$ in $X_2$: divisibility of a root in the character lattice is an invariant of root data. The same obstruction appears dually: $\mathbb{Z}{\alpha }_1^{\vee }=X_{\ast }(T_1)$ is the full cocharacter lattice while $\mathbb{Z}{\alpha }_2^{\vee }=2\mathbb{Z}{\chi }_2^{\vee }$ has index $2$ in $X_{\ast }(T_2)$. Hence there is no isomorphism of quadruples $(X,\mathrm{\Phi },X^{\vee },{\mathrm{\Phi }}^{\vee })$.

Therefore the Lie algebra ${\mathfrak{s}\mathfrak{l}}_2$ and the abstract root system $A_1$, which are the same for the two groups in characteristic $\mathrm{\ne }2$, do not determine the root datum or the group: the additional data are the position of the coroot lattice $\mathbb{Z}{\alpha }^{\vee }$ inside $X_{\ast }(T)$ and the divisibility of the root in $X(T)$, and these separate the simply connected group ${\mathrm{S}\mathrm{L}}_2$ from the adjoint group ${\mathrm{P}\mathrm{G}\mathrm{L}}_2$.