# Zariski closure of an algebraic linear group

Let $$G$$ and $$H$$ algebraic linear groups and $$\phi : G \to H$$ a regular group homomorphism. I wonder if $$\overline{\phi(G)}$$ (the Zariski closure of $$\phi(G)$$) is again a subgroup and how this could be proven. First of all, I know Zariski closure $$\overline{Z}$$ of a set $$Z$$ is defined as $$\overline{Z} = V(I(Z))$$. I imagine this definition is also valid for algebraic linear groups, since $$G$$ is an affine variety. However, I can’t seem to find a strategy for this proof: any ideas?

• Is $H$ another algebraic group? Is $\phi$ a group homomorphism? Which algebraic formalism are you using (schemes, or something more elementary?) Commented Jan 21, 2021 at 16:44
• Thanks, I just edited. I am not using schemes, I am studying "classic" algebraic geometry
– cip
Commented Jan 21, 2021 at 17:01
• The image is automatically closed. Are you not aware of this? Commented Jan 21, 2021 at 17:38
• @cip I mean the proof in Milne doesn't, at least not explicitly, but if you really just wanna know that you just need to note that since $m:H\times H\to H$ is continuous that since $m(\phi(G),\phi(G))\subseteq \overline{\phi(G)}$ that $m(\overline{\phi(G)},\overline{\phi(G)})\subseteq \overline{\phi(G)}$. The reason is that $\overline{\phi(G)\times \phi(G)}=\overline{\phi(G)}\times \overline{\phi(G)}$ and proofwiki.org/wiki/Continuity_Defined_by_Closure Commented Jan 21, 2021 at 22:19
• @cip that’s right. Commented Jan 21, 2021 at 22:47