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 $w^a=1$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.