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I was asking this on stack exchange but I didn't get the answer. Borel's book Linear Algebraic Groups contains the following result

10.9 Theorem. Let $G$ be a connected affine group of dimension one. Then $G$ is isomorphic to either $\mathbb{G}_a$ or $\mathbb{G}_1$.

Proof: $G$ is a dense open set in a unique complete smooth curve $\overline{G}$ (see AG.18.5(d)). It follows from (AG.18.5(f)) that the action of $G$ on itself by translation extends uniquely to an action of $G$ on $\overline{G}$...

Is the above extension of the action of $G$ as an action of abstract groups or also as algebraic groups. Why?

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    $\begingroup$ The action of $G$ on itself by translation is a morphism $L_g:G\rightarrow G$ (which is only a homomorphism when $g=e$ is the neutral element of $G$) and the claim that is used by Borel is that the action $L_g$ extends to the complete curve $\overline{G}$ as a morphism $L_g:\overline{G}\rightarrow G$, which again is almost never a homomorphism. $\endgroup$
    – F Zaldivar
    Commented Nov 22, 2022 at 4:20
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    $\begingroup$ Crossposted at: math.stackexchange.com/questions/4571989/… $\endgroup$
    – F Zaldivar
    Commented Nov 22, 2022 at 4:29
  • $\begingroup$ @FZaldivar thank you for your response $\endgroup$
    – Laurence PW
    Commented Nov 22, 2022 at 5:06
  • $\begingroup$ The map $g\mapsto L_g$ is a homomorphism. $\endgroup$
    – Wilberd van der Kallen
    Commented Nov 22, 2022 at 9:02

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