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so far I understand about the statement: let $p_i,i=1\dots,n$ has non trivial stabilizers i.e $S_{p_1}=\{g:g.p=p, g\in G\}\neq\{e\}$, is non trivial subgroup of $G$ for $p_1$ and so forth upto $p_n$ we will get $S_{p_n}$,so we need to show $\{p_1,\dots,p_n\}$ is discrete.

could any one make me understand the 2nd line of the proof? and in 3rd line $g$ is continous, how come a point $g\in G$ is continuous?

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Second line: $G$ is finite, so there must be at least one $g\in G$ that fixes infinitely many of the $p_i$; otherwise there could only be finitely many $p_i$. These $p_i$ fixed by that $g$ form a subsequence all of whose terms are fixed by the same nontrivial element $g$.

Third line: This is a slight abuse of notation; $g$ here is not the element of $G$ but the map induced by it, which is continuous since $G$ acts holomorphically.

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THANK YOU SIR,the map you mentioned about g, is the map $G/K\times X\rightarrow X, (gK,x)\mapsto gx$ ? –  Une Femme Douce Jul 27 '12 at 18:53
    
@Patience: I don't know what $K$ is or what the significance of that map is; I was just thinking of the map $f_g:X\to X,x\to gx$. –  joriki Jul 27 '12 at 20:42
    
the kernel of the action of $G$ on $X$,$K=\{g\in G|g.p=p\forall p\in X\}$ –  Une Femme Douce Jul 27 '12 at 20:44

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