# Showing that two elements of a group have no relations between them using group actions

Suppose $$G$$ acts on a set $$X$$, and let $$X_a, X_b \subseteq X$$ be disjoint and non-empty. Suppose further that $$a, b \in G$$ and that for any $$k \neq 0$$, $$a^k \cdot X_b \subseteq X_a$$ and $$b^k \cdot X_a \subseteq X_b$$. (The $$\cdot$$ indicates group action). Then I want to show that $$\langle a, b \rangle \cong F(a, b)$$ (that is, the subgroup generated by $$a, b$$ in $$G$$ is isomorphic to the free group consisting of two letters $$a, b$$).

So the crux of the question is to show that words of the form $$a^{k_1}b^{s_1}\cdots a^{k_n}b^{s_n}$$ aren't identity in $$G$$.

This is easy enough for words like : $$a^{k_1}b^{s_1}\cdots a^{k_n}$$ or $$b^{s_1}\cdots b^{s_n}$$ (i.e, words that start and end with the same letter). In the former case, observe the action of the word on some $$x \in X_b$$ (the final result will be in $$X_a$$) and in the latter case, observe the action on some $$x \in X_a$$ (the final result will be in $$X_b$$).

However, I am having some confusion with words that start and end with different letters, i.e, $$a^{k_1}b^{s_1}\cdots a^{k_n}b^{s_n}$$ or $$b^{s_1}a^{k_1}\cdots b^{s_n}a^{k_n}$$. Since for these words, if you start with, say $$x \in X_a$$, you end up in $$X_a$$ again.

This feels like an easy five minute question, but I'm having a surprising amount of trouble with it. Would appreciate some hints.

It's the ping-pong lemma: https://en.wikipedia.org/wiki/Ping-pong_lemma contains a proof. The proof there notes that you should just conjugate your element by (say) $$a$$ to fix your problem. (Note that $$w=1$$ if and only if $$w^a=1$$.)