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$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\diag{diag}$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)

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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 $R\times T'$.

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  • $\begingroup$ Thank you! That's a very good point: $GL_2$ and $SL_2 \times \mathbb G_m$ give rise to isomorphic pushout diagrams. $\endgroup$ – alpha101 Jun 21 at 20:24

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