$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\diag{diag}$In [SGA3](https://en.wikipedia.org/wiki/S%C3%A9minaire_de_G%C3%A9om%C3%A9trie_Alg%C3%A9brique_du_Bois_Marie), Expose XXIV, Lemme 7.2.2 it says (let's say our base scheme $S$ is an algebraically closed field $k$): if $G$ is reductive algebraic group, $T$ a maximal torus, $G'$ the derived subgroup of $G$ and $T' = T \cap G$ a maximal torus of $G'$, then $G$ is a pushout of $G'$ and $T$ over $T'$ (even in the category of group sheaves).


Doesn't the following very basic example contradict this? Or what have I misunderstood?


Suppose $G = \GL\_2$, $G' = \SL\_2$, $T$ the diagonal torus and $T'$ the diagonal torus with determinant 1.


Define homomorphisms $\alpha : T = \mathbb G\_m^2 \to \GL\_2$ sending $\diag(x,y)$ to $\diag(x,x^{-1})$ and $\beta : G' = \SL\_2 \to \GL\_2$ the inclusion. Then $\alpha$ and $\beta$ agree on the intersection $T'$, but there is no homomorphism $G = \GL\_2 \to \GL\_2$ that restricts to $\alpha$ and $\beta$ because otherwise $\alpha(\diag(x,x))$ and $\beta(y)$ would commute for any $x$, $y$.


(See pages 39-40 of the [pdf of Expose XXIV](https://webusers.imj-prg.fr/%7Epatrick.polo/SGA3/Expo24-29mars.pdf) at the [SGA3 Réédition project page](https://webusers.imj-prg.fr/%7Epatrick.polo/SGA3/), or the [official published version](https://doi.org/10.1007/BFb0059033))



Pushout of group schemes (question on a lemma in SGA3)
The Lemma is clearly wrong. There is no way to recover $G$ from $G'$, $T$ and $T'$ alone (not even up to isomorphism) since that data do not determine the radical $R:={\rm rad}(G)\subseteq T$. Maybe the authors had in mind the amalgamated product of $T$ and $R\times G'$ over $R\times T'$.


