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?


1 Answer 1


Let me prove that your second quoted statement implies your first quoted statement, since that appears to be the serious issue.

I'll use BLAHBLAH to refer to the ... conditions in your first quoted statement, and LAHDIDAH to refer to the ... conditions in your second quoted statement.

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .