could any one tell me why such expression for $g(z)$, specially I dont get what and why is $a_n(g)$? notations confusing me and also I am not understanding properly, and why $g$ is automorphism? where does $a_i$ belong? what is the role of $G$ in that expression of power series? Thank you, I am self reading the subject, and extremely thank you to all the users who are responding while learning and posting my trivial doubts. Thank you again.
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The element $g$ gives rise to a function $g: X \to X$. Since $g$ is a function, it has a Taylor expansion at the point $p$ in the coordinate $z$. We could write $$ g(z) = \sum_{n=0}^\infty a_nz^n. $$ for some complex numbers $a_n$. However, we are going to want to do this for all the elements of the group. If there's another element, $k$, we could write $$ k(z) = \sum_{n=0}^\infty b_nz^n. $$ It is terrible notation to pick different letters for each element of the group, and how do we even do it for an arbitrary group? What the author is saying is that, instead, we can pick one efficient notation by labeling the Taylor-series coefficient with the appropriate element of $G$. So what I called $a_n$ is what he calls $a_n(g)$; what I called $b_n$ is what he called $a_n(k)$. Do you now understand his arguments that $a_0(g) = 0$ (for any $g$) and $a_1(g)$ can't be?
But it's not just that his notation is better than mine - it's also that it makes it more obvious that we get a function $G \to \mathbb{C}^\times$ by taking the first Taylor coefficient (in his notation that's $g \mapsto a_1(g)$). It's this function which he is showing is a homomorphism.
Let me add something to countinghaus' fine answer.
You also asked "why g is an automorphism", something that has not been satisfactorily answered yet.
It is in fact true in much bigger generality. Recall that one axiom of a group action is that for all $x \in X$ and $g,h \in G$ we have $$ g \cdot_{\text{a}} (h\cdot_{\text{a}} x) = (g \cdot_{\text{g}} h) \cdot_{\text{a}} x $$ where i labeled the group multiplication with g and the action operation with a.
Applying this to $h = g^{-1}$ and using that the unit element of the group acts as the identity on $X$, we see that we have found an inverse for our homomorphic map given by $g$. Hence it is an automorphism.
Note that one needs to be precise. Of course $g^{-1}$ is the inverse of $g$ in the group, that we all know. However, we've just shown something else: that the holomorphic map given by $g^{-1}$ is an inverse of the holomorphic map given by $g$.
Lastly, it works in any category $\mathcal{C}$. If we say some group acts on some object in $\mathcal{C}$, we normally require that the elements of the group induce morphisms in $\mathcal{C}$, which have a two sided inverse by the reasoning above. Hence they are isomorphisms in $\mathcal{C}$.
Hope that helps.