Absolute Galois group is topologically generated by Frobenius map clarification

Question

I came across the following while reading a text on elliptic curves:

For the $q$th power Frobenius morphism on an elliptic curve $E$, $\phi: E \rightarrow E$ given by $\phi(x,y) = (x^q, y^q)$,the absolute Galois group $Gal(\bar{\mathbb{F}_q}/ \mathbb{F}_q)$ is topologically generated by this morphism. And we have

$$ P \in E(\mathbb{F}_q) \Leftrightarrow \phi(P) = P. $$

However, I haven't been able to find a proper definition of 'topologically generated' anywhere.

It'd be great if someone could help me on this.

Thank you.

Answer 1

That the Galois group is topologically generated by the Frobenius automorphism, means that the closure of the subgroup generated by this automorphism is the full Galois group.