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Let $S$ be a del Pezzo surface of degree six over $\mathbb{C}$. Then $S$ is the blow-up of $\mathbb{P}^2$ in three general points $p_1,p_2,p_3$.

Is it true that its automorphism group is $((\mathbb{C}^{*})^{2}\rtimes S_2)\times S_3$?

Here $(\mathbb{C}^{*})^{2}$ are the automorphisms of $\mathbb{P}^2$ fixing $p_1,p_2,p_3$, $S_2$ is the group generated by the standard Cremona centered at $p_1,p_2,p_3$, and $S_3$ are the permutations of $p_1,p_2,p_3$.

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Not exactly. The quadratic transformation commutes with the action of $S_3$, and they both act on $(\mathbb{C}^*)^2$; so the automorphism group is $(\mathbb{C}^{*})^{2}\rtimes (S_3\times S_2)$. You'll find a detailed study of the automorphisms of del Pezzo surfaces in Chapter 8 of Dolgachev's book "Classical Algebraic Geometry: a modern view".

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    $\begingroup$ It seems to me that the Cremona does not commute with $(\mathbb{C}^{*})^2$. If $\alpha\in (\mathbb{C}^{*})^2$ and $f$ is the Cremona we have the relation $f\circ\alpha = \alpha^{-1}\circ f$. Therefore $f\circ\alpha = \alpha\circ f$ if and only if $\alpha$ is the identity. $\endgroup$ – John Simis Jul 21 '15 at 17:03
  • $\begingroup$ Maybe a silly comment cause I know nothing about the subject, but can't $\alpha$ be an involution different from the identity? $\endgroup$ – Sylvain JULIEN Jul 21 '15 at 17:28
  • $\begingroup$ Yes, if and only if $\alpha^2 = Id$. Anyway for a general $\alpha\in (\mathbb{C}^{*})^2$ we have $f\circ\alpha \neq \alpha\circ f$. $\endgroup$ – John Simis Jul 21 '15 at 18:08
  • $\begingroup$ Sorry, I misplaced the parentheses, I meant $(\mathbb{C}^*)^2\rtimes (S_3\times S_2)$. $\endgroup$ – abx Jul 21 '15 at 18:35

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