The ping-pong-lemma for subroups is often stated with the following assumptions, e.g. taking this one from Wikipedia:
Let G be a group acting on a set $X$ and let $H_1, H_2, ..., H_k$ be subgroups of $G$ where $k ≥ 2$, such that at least one of these subgroups has order greater than $2$. Suppose there exist pairwise disjoint nonempty subsets $X_1, X_2, ...,X_k$ of $X$ such that the following holds:
For any $i ≠ s$ and for any $h$ in $H_i$, $h ≠ 1$ we have $h(Xs) ⊆ Xi$.
Note the requirement that one of the H_1 has at least three elements. It turns out that this requirement can be dropped if one instead requires $k \ge 3$.
Can the assumptions be made even weaker than $(k \ge 3) \vee(∃i, |H_i|\ge3) $?
Or, put differently, does the ping-pong-lemma hold for $k = 2$ and $|H_1| = |H_2| = 2$?
Without additional assumptions, the ping-pong-lemma does not work for $k=2$ and $|H_1| = |H_2| = 2$.
Counter-example (written additively): Consider $G = \mathbb{Z}_2 \times \mathbb{Z}_2$ with subgroups $H_1 = \langle (1,0) \rangle$, $H_2 = \langle (0,1) \rangle$ acting on $X = \mathbb Z_2$ via $(a,b) \bullet c = a + b + c \pmod 2$. The subsets $X_1 = \{0\}$ and $X_2 = \{1\}$ fulfill the assumptions of the lemma, but $G$ is not $H_1 \star H_2$.
As a possible way out, one can add the extra assumption of a additional point in X: Let $x \in X$ be so that for all $i$, $x \notin X_i$ and for all $h \in H_i, h \ne 1$, $h \bullet x \in X_i$. With this assumption, the ping-pong-lemma goes through even for $k=2$ and no further assumptions about the $H_i$. This is equivalent to considering a third, trivial subgroup ($H_3 = {1}$) and applying the lemma in the $k\ge 3$ variant.