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In Springer's 'Linear Algebraic Groups', he proves that if $G$ is a linear (=affine) algebraic group and $H$ is a closed normal subgroup of $G$, then $G/H$ has a linear algebraic group structure with the usual group structure. But I think there's some problem with the proof: enter image description hereenter image description here

I think there's a problem with the third line, since the map $(xH, yH)\mapsto xy^{-1}H$ doesn't give the usual group structure on $G/H$. I think it is okay with the latter part, so I want to know whether the first paragraph of the proof is right or wrong. Thanks in advance.

Seewoo Lee
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  • I could be wrong, but I think the morphism of varieties he defines is not the multiplication morphism, but you can get the inversion morphism out of it, and hence composing with that you can get the multiplication morphism. – Joppy Sep 27 '18 at 00:30
  • @Joppy So do you mean that the author showed that the map $(xH, yH)\mapsto xy^{-1}H$ is a morphism from $G/H\times G/H\to G/H$, and by composing with the map $(\mathrm{id}, \mathrm{inv})$, we may get the morphism $(xH, yH)\mapsto xyH$ which now gives the group structure on $G/H$, right? – Seewoo Lee Sep 27 '18 at 22:01
  • Yes, I think so. Note also that you can get the map inv out of that map by fixing the first argument to be the identity. – Joppy Sep 28 '18 at 00:05

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It doesn't matter because he's just trying to show that the variety is affine. The group structure is unimportant for this part of the argument.

Alex Youcis
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