Consider the following extract from this expository article: (This is a key step in proving the Banach-Tarski theorem.)
My question is: if the action of $G$ on $S^2 - C$ were not free, would the result not still follow? Or, more precisely: where do we use the fact that the action is free in the proof? Is it necessary? We still get a partition of $S^2 - C$ into $G$-orbits, and using the AC we can still choose representatives for each orbit.
I guess I am confused because in the following extract from this article, freeness of the action is stressed:
Freeness is used to conclude that $\Omega_1 \cap \Omega_2 = \emptyset$, for example. Arguing by contradiction, if that intersection is not empty, that means there exist $g_1 \in G_1$, $g_2 \in G_2$, and $x_1,x_2 \in X$, such that $g_1 x_1 = g_2 x_2$. Since $X$ has just one point from each $G$-orbit, it follows that $x_1=x_2$. Therefore $g_1^{-1} g_2$ fixes $x_1$. But $g_1^{-1} g_2$ is not the identity because $G_1 \cap G_2 = \emptyset$. That contradicts freeness.