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Let $G$ be a group generated by $\{g_1,g_2,\ldots , g_n\}$. Let $X$ be a space with a $G$-action on it, i.e. $G$ is acting on $X$. Suppose for each $x\in X$, the set $\{g_i;g_i(x)=x\}$ is trivial. Does it imply that the stabilizer of $x$, i.e. $\{g\in G;g(x)=x\}$ is finite? In other words to calculate finite stabilizer subgruops is it enough to consider generators only?

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  • $\begingroup$ No. Assuming your group is finitely generated, the set $\{g_i : g_i(x) = x\}$ will always be finite regardless of the action, but the stabilizer may not be finite. $\endgroup$ – Prahlad Vaidyanathan Nov 25 '13 at 6:03
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Let $G = (\mathbb{Z}, +)$. It is generated by one element $g_1 = 1$.

Let $G$ act on the set $X = \{0, 1\}$ like this: if $g \in G$ is even, then $g$ acts trivially. Otherwise, it swaps $0$ and $1$.

Now you can see that $g_1$ doesn't fix any elements of $X$, but both elements in $X$ have infinite stabilizers.

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