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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.

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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 $w^a=1$.)

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