Assume now that $G$ is imprimitive.
Let $B_1$ be a minimal non-trivial block of imprimitivity (of size $b$) and $B_1, . . . , B_k$ the corresponding system of blocks of imprimitivity.
Then $G$ permutes these blocks transitively, let $K$ denote the kernel of this action. We may assume that $K \neq 1$. Then the minimality of $B_1$ implies that $K$ acts transitively on each $B_i$. Up to this point I understand, but then the article (Source of the question, page 2, 3): says:
and the corresponding transitive groups are permutation equivalent.
Previously, in the same article it was written that:
$G$ will always be transitive and all point-stabilizers equivalent
as permutation groups
What does "equivalent as permutation groups" and "permutation equivalent" mean here?
When group act on domain of $n$ objects, that group is permutation group, i.e., a subgroup of symmetric group, why then it has to be told separately?
Best Answer
The standard definition is that $G \le {\rm Sym}(X)$ and $H \le {\rm Sym}(Y)$ are equivalent as permutation groups if there is a bijection $\tau:X \to Y$ and an isomorphism $f:G \to H$ such that $\tau(x^g) = \tau(x)^{f(g)}$ for all $g \in G$ and $x \in X$.