$G$ is a permutation group of degree $n \geq 4$ which acts transitively and faithfully on a set $X$ with $|X| = n$.
Given indices $i < j < k \leq n$, elements $\alpha \neq \beta \neq \gamma \in X$, and permutations $\pi \neq \rho \neq \sigma \in G$, we say the permutations are syntactically transitive over $i, j, k$ if
- $\pi(i) = \alpha$ and $\pi(j) = \beta$
- $\rho(j) = \beta$ and $\rho(k) = \gamma$
- $\sigma(i) = \alpha $ and $\sigma(k) = \gamma$
and connected if there exists another permutation $\tau \in G$ such that
- $\tau(i) = \alpha$, $\tau(j) = \beta$ and $\tau(k) = \gamma$
For example, the first three permutations below are syntactically transitive over indices $1, 2, 3$ and connected by the fourth permutation
- $\mathbf{2~4}~1~3$
- $1~\mathbf{4~3}~2$
- $\mathbf{2}~1~\mathbf{3}~4$
- $\mathbf{2~4~3}~1$
For the group G, is it generally true that any set of syntactically transitive permutations is connected by another permutation in $G$?
Also, is there a more standard terminology for syntactically transitive and connected? I’ve been looking through papers on group theory, formal language theory, and combinatorics, but I haven’t been able to find anything relating to this concept.
No.
By conjugating $G$ by some element of $S_n$, we may assume that $i=1$, $j=2$, $k=3$, which makes notation easier.
Now set $\alpha=1$. Then $\pi$, $\sigma$ and any potential $\tau$ lie in the stabilizer $H=\mathrm{Stab}_G(1)$. Fix any two distinct, non-trivial $\pi,\sigma$ in $H$, and of course set $\beta=\pi(2)$ and $\gamma=\sigma(3)$.
Suppose that the $2$-point stabilizers are trivial. Then $\pi(2)\neq \sigma(2)$ because $\pi\neq\sigma$ and otherwise $\pi^{-1}\sigma$ fixes $1$ and $2$. Similarly, since $\pi(1)=\tau(1)$ and $\pi(2)=\tau(2)$, we must have $\pi=\tau$. But also we have $\sigma=\tau$ for the same reason. This is a clear contradiction.
So counterexamples are any groups where the two-point stabilizers are trivial, but $|H|>2$. Examples of these are sharply $2$-transitive groups.
(Edit: Note that the existence of $\rho$ is not required for this argument.)