# Different statements of the ping pong lemma

The typical statement that I have seen of the Ping Pong lemma is "Suppose a group G with generators $$\{g_1,...,g_n\}$$ acts on a set $$S$$ such that ... Then $$G$$ is the free group of rank $$n$$." However in https://link.springer.com/article/10.1007/s00039-012-0198-z the lemma is stated as "Suppose $$G$$ is a free group generated by $$\{g_1,...,g_n\}$$ of rank $$n$$, and that it acts on a set $$S$$ such that ... Then the action of $$G$$ on $$S$$ is faithful.

I don't see how these two statements are equivalent, since acting faithfully on some arbitrary set does not imply $$G$$ is free. Can someone please explain how one statement is equivalent to the other?

Suppose that $$G$$ is a group with generators $$\{g_1,...,g_n\}$$ that acts on a set $$S$$ satisfying BLAHBLAH.
Let $$F\langle g_1,...,g_n\rangle$$ denote the free group on the set $$\{g_1,...,g_n\}$$. By the universal property of the free group, the identity map on the set $$\{g_1,...,g_n\}$$ extends to a surjective homomorphism $$F\langle g_1,...,g_n\rangle \mapsto G$$. By composing this homomorphism with the action of $$G$$ on $$S$$, we obtain an action of the group $$F\langle g_1,....,g_n\rangle$$ on $$S$$.
Applying the properties BLAHBLAH for the action of $$G$$, we deduce the properties LAHDIDAH for the action of $$F\langle g_1,...,g_n\rangle$$ (this, of course, is something that needs to be checked!!). We may therefore conclude that the action of $$F\langle g_1,...,g_n\rangle$$ on $$S$$ is faithful.
But this implies that the surjective homomorphism $$F\langle g_1,...,g_n\rangle \mapsto G$$ is injective, and it is therefore an isomorphism, hence $$G$$ is free.