# Pushout of group schemes (question on a lemma in SGA3)

$$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\diag{diag}$$In SGA3, 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 at the SGA3 Réédition project page, or the official published version)

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 $$R\times T'$$.
• Thank you! That's a very good point: $GL_2$ and $SL_2 \times \mathbb G_m$ give rise to isomorphic pushout diagrams. – alpha101 Jun 21 at 20:24