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I proved that every group acts freely by left translation on its Cayley graph if and only if the generating set $S$ of the group $G$ has no elements of order $2$.

Is it true that every finitely generated nontrivial group has a finite generating set without elements of order 2? If not, please give me a counterexample.

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Am I missing something? What about $\mathbb{Z}_2$? – Derek Allums Aug 12 '12 at 16:15
I don't follow. Freeness has nothing to do with the edges; every group $G$ acts freely on itself, so every group $G$ acts freely on the vertices of the Cayley graph with respect to any generating set. – Qiaochu Yuan Aug 12 '12 at 16:16
@Qiaochu: he means, acts freely on the 1-skeleton. To avoid the issue, one solution is to consider the oriented Cayley graph. So a generator of order two gives rise to a pair of opposite edges and the action is always free. – YCor Aug 21 '12 at 9:13

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